# LongBench v2 / 66ead7bd5a08c7b9b35dca7a

task_id: c6043162-b6b1-5efa-b17a-935cd67e477d
task_key: train--66ead7bd5a08c7b9b35dca7a
task_revision_id: 3

{"choice_A":"ξ(γ) = γ + 5","choice_B":"ξ(γ) = 5γ + 1","choice_C":"ξ(γ) = 5γ - 1","choice_D":"ξ(γ) = 4γ + 1","context":"Uncertainty Theory\nFifth Edition\n\nContents\nPreface\nxi\n1\nIntroduction\n1\n1.1\nUrn Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .\n1\n1.2\nHow to Choose Your Mathematical Tool . . . . . . . . . . . .\n4\n2\nUncertain Measure\n7\n2.1\nUncertain Measure . . . . . . . . . . . . . . . . . . . . . . . .\n7\n2.2\nUncertainty Space\n. . . . . . . . . . . . . . . . . . . . . . . .\n11\n2.3\nProduct Uncertain Measure . . . . . . . . . . . . . . . . . . .\n12\n2.4\nIndependence . . . . . . . . . . . . . . . . . . . . . . . . . . .\n20\n2.5\nConditional Uncertain Measure . . . . . . . . . . . . . . . . .\n22\n2.6\nBibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n26\n3\nUncertain Variable\n27\n3.1\nUncertain Variable . . . . . . . . . . . . . . . . . . . . . . . .\n27\n3.2\nUncertainty Distribution . . . . . . . . . . . . . . . . . . . . .\n30\n3.3\nInverse Uncertainty Distribution\n. . . . . . . . . . . . . . . .\n46\n3.4\nIndependence . . . . . . . . . . . . . . . . . . . . . . . . . . .\n50\n3.5\nOperational Law . . . . . . . . . . . . . . . . . . . . . . . . .\n53\n3.6\nExpected Value . . . . . . . . . . . . . . . . . . . . . . . . . .\n78\n3.7\nVariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n83\n3.8\nMoments\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n86\n3.9\nDistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n88\n3.10 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n89\n3.11 Uncertain Sequence . . . . . . . . . . . . . . . . . . . . . . . .\n95\n3.12 Uncertain Vector . . . . . . . . . . . . . . . . . . . . . . . . .\n100\n3.13 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n102\n4\nUncertain Statistics\n103\n4.1\nEmpirical Uncertainty Distribution . . . . . . . . . . . . . . .\n103\n4.2\nMethod of Moments\n. . . . . . . . . . . . . . . . . . . . . . .\n105\n4.3\nMethod of Least Squares . . . . . . . . . . . . . . . . . . . . .\n109\n\n\nvi\nContents\n4.4\nUncertain Hypothesis Test . . . . . . . . . . . . . . . . . . . .\n110\n4.5\nUncertain Regression Analysis . . . . . . . . . . . . . . . . . .\n116\n4.6\nUncertain Time Series Analysis . . . . . . . . . . . . . . . . .\n131\n4.7\nBibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n139\n5\nUncertain Programming\n141\n5.1\nUncertain Programming . . . . . . . . . . . . . . . . . . . . .\n141\n5.2\nNumerical Method . . . . . . . . . . . . . . . . . . . . . . . .\n144\n5.3\nMachine Scheduling Problem\n. . . . . . . . . . . . . . . . . .\n146\n5.4\nVehicle Routing Problem . . . . . . . . . . . . . . . . . . . . .\n149\n5.5\nProject Scheduling Problem . . . . . . . . . . . . . . . . . . .\n153\n5.6\nUncertain Multiobjective Programming\n. . . . . . . . . . . .\n156\n5.7\nUncertain Goal Programming . . . . . . . . . . . . . . . . . .\n158\n5.8\nUncertain Multilevel Programming . . . . . . . . . . . . . . .\n159\n5.9\nBibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n160\n6\nUncertain Risk Analysis\n161\n6.1\nLoss Function . . . . . . . . . . . . . . . . . . . . . . . . . . .\n161\n6.2\nRisk Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n162\n6.3\nSeries System . . . . . . . . . . . . . . . . . . . . . . . . . . .\n164\n6.4\nParallel System . . . . . . . . . . . . . . . . . . . . . . . . . .\n164\n6.5\nStandby System\n. . . . . . . . . . . . . . . . . . . . . . . . .\n165\n6.6\nStructural Risk Analysis . . . . . . . . . . . . . . . . . . . . .\n165\n6.7\nValue-at-Risk . . . . . . . . . . . . . . . . . . . . . . . . . . .\n169\n6.8\nExpected Loss\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n170\n6.9\nBibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n171\n7\nUncertain Reliability Analysis\n173\n7.1\nStructure Function . . . . . . . . . . . . . . . . . . . . . . . .\n173\n7.2\nReliability Index\n. . . . . . . . . . . . . . . . . . . . . . . . .\n174\n7.3\nSeries System . . . . . . . . . . . . . . . . . . . . . . . . . . .\n175\n7.4\nParallel System . . . . . . . . . . . . . . . . . . . . . . . . . .\n175\n7.5\nGeneral System . . . . . . . . . . . . . . . . . . . . . . . . . .\n176\n7.6\nBibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n176\n8\nUncertain Propositional Logic\n177\n8.1\nUncertain Proposition . . . . . . . . . . . . . . . . . . . . . .\n177\n8.2\nTruth Value . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n179\n8.3\nChen-Ralescu Theorem . . . . . . . . . . . . . . . . . . . . . .\n181\n8.4\nUncertain Entailment\n. . . . . . . . . . . . . . . . . . . . . .\n184\n8.5\nUncertain Modus Ponens\n. . . . . . . . . . . . . . . . . . . .\n187\n8.6\nUncertain Modus Tollens\n. . . . . . . . . . . . . . . . . . . .\n188\n8.7\nUncertain Hypothetical Syllogism . . . . . . . . . . . . . . . .\n189\n8.8\nBibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n191\n\n\nContents\nvii\n9\nUncertain Set\n193\n9.1\nUncertain Set . . . . . . . . . . . . . . . . . . . . . . . . . . .\n193\n9.2\nMembership Function\n. . . . . . . . . . . . . . . . . . . . . .\n201\n9.3\nInverse Membership Function . . . . . . . . . . . . . . . . . .\n215\n9.4\nIndependence . . . . . . . . . . . . . . . . . . . . . . . . . . .\n217\n9.5\nSet Operational Law . . . . . . . . . . . . . . . . . . . . . . .\n219\n9.6\nArithmetic Operational Law . . . . . . . . . . . . . . . . . . .\n225\n9.7\nInclusion Relation\n. . . . . . . . . . . . . . . . . . . . . . . .\n231\n9.8\nExpected Value . . . . . . . . . . . . . . . . . . . . . . . . . .\n234\n9.9\nDistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n242\n9.10 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n243\n9.11 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n245\n10 Uncertain Logic\n247\n10.1 Individual Feature Data . . . . . . . . . . . . . . . . . . . . .\n247\n10.2 Uncertain Quantiﬁer . . . . . . . . . . . . . . . . . . . . . . .\n248\n10.3 Uncertain Subject\n. . . . . . . . . . . . . . . . . . . . . . . .\n253\n10.4 Uncertain Predicate\n. . . . . . . . . . . . . . . . . . . . . . .\n254\n10.5 Uncertain Proposition . . . . . . . . . . . . . . . . . . . . . .\n257\n10.6 Truth Value . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n258\n10.7 Linguistic Summarizer . . . . . . . . . . . . . . . . . . . . . .\n263\n10.8 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n266\n11 Uncertain Inference Control\n267\n11.1 Uncertain Inference Rule . . . . . . . . . . . . . . . . . . . . .\n267\n11.2 Uncertain Inference Controller\n. . . . . . . . . . . . . . . . .\n268\n11.3 Inverted Pendulum . . . . . . . . . . . . . . . . . . . . . . . .\n272\n11.4 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n274\n12 Uncertain Process\n275\n12.1 Uncertain Process\n. . . . . . . . . . . . . . . . . . . . . . . .\n275\n12.2 Uncertainty Distribution . . . . . . . . . . . . . . . . . . . . .\n278\n12.3 Independent Increment Process . . . . . . . . . . . . . . . . .\n281\n12.4 Extreme Value Theorem . . . . . . . . . . . . . . . . . . . . .\n283\n12.5 First Hitting Time . . . . . . . . . . . . . . . . . . . . . . . .\n286\n12.6 Time Integral . . . . . . . . . . . . . . . . . . . . . . . . . . .\n287\n12.7 Stationary Independent Increment Process . . . . . . . . . . .\n291\n12.8 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n296\n13 Uncertain Renewal Process\n297\n13.1 Uncertain Renewal Process\n. . . . . . . . . . . . . . . . . . .\n297\n13.2 Uncertain Renewal Reward Process . . . . . . . . . . . . . . .\n300\n13.3 Uncertain Insurance Model\n. . . . . . . . . . . . . . . . . . .\n304\n13.4 Uncertain Production Model\n. . . . . . . . . . . . . . . . . .\n309\n13.5 Uncertain Queueing Model\n. . . . . . . . . . . . . . . . . . .\n314\n\n\nviii\nContents\n13.6 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n321\n14 Uncertain Calculus\n323\n14.1 Liu Process . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n323\n14.2 Liu Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n325\n14.3 Diﬀerential\n. . . . . . . . . . . . . . . . . . . . . . . . . . . .\n329\n14.4 Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . .\n332\n14.5 Chain Rule\n. . . . . . . . . . . . . . . . . . . . . . . . . . . .\n333\n14.6 Change of Variables\n. . . . . . . . . . . . . . . . . . . . . . .\n334\n14.7 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . .\n335\n14.8 Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .\n336\n14.9 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n339\n15 Uncertain Diﬀerential Equation\n341\n15.1 Uncertain Diﬀerential Equation . . . . . . . . . . . . . . . . .\n341\n15.2 α-Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n353\n15.3 Yao-Chen Formula . . . . . . . . . . . . . . . . . . . . . . . .\n356\n15.4 Numerical Solution . . . . . . . . . . . . . . . . . . . . . . . .\n364\n15.5 Residual . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n364\n15.6 Uncertain Hypothesis Test . . . . . . . . . . . . . . . . . . . .\n367\n15.7 Parameter Estimation . . . . . . . . . . . . . . . . . . . . . .\n368\n15.8 Real-Life Examples . . . . . . . . . . . . . . . . . . . . . . . .\n369\n15.9 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n373\n16 Uncertain Finance\n377\n16.1 Uncertain Stock Model . . . . . . . . . . . . . . . . . . . . . .\n377\n16.2 European Options\n. . . . . . . . . . . . . . . . . . . . . . . .\n378\n16.3 American Options\n. . . . . . . . . . . . . . . . . . . . . . . .\n384\n16.4 Asian Options . . . . . . . . . . . . . . . . . . . . . . . . . . .\n390\n16.5 Uncertain Interest Rate Model\n. . . . . . . . . . . . . . . . .\n396\n16.6 Uncertain Currency Model . . . . . . . . . . . . . . . . . . . .\n402\n16.7 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n406\nA Chance Theory\n409\nA.1 Chance Measure\n. . . . . . . . . . . . . . . . . . . . . . . . .\n409\nA.2 Uncertain Random Variable . . . . . . . . . . . . . . . . . . .\n413\nA.3 Chance Distribution . . . . . . . . . . . . . . . . . . . . . . .\n414\nA.4 Operational Law . . . . . . . . . . . . . . . . . . . . . . . . .\n416\nA.5 Expected Value . . . . . . . . . . . . . . . . . . . . . . . . . .\n421\nA.6 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n425\nA.7 Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . .\n428\nA.8 Ellsberg Experiment . . . . . . . . . . . . . . . . . . . . . . .\n430\nA.9 Bibliographic Notes . . . . . . . . . . . . . . . . . . . . . . . .\n436\n\n\nContents\nix\nB Frequently Asked Questions\n439\nB.1\nWhat is belief degree? . . . . . . . . . . . . . . . . . . . . . .\n439\nB.2\nWhat is the diﬀerence between probability theory and uncer-\ntainty theory? . . . . . . . . . . . . . . . . . . . . . . . . . . .\n440\nB.3\nHow do we distinguish between randomness and uncertainty\nin practice? . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n441\nB.4\nWhy is stochastic diﬀerential equation not suitable for mod-\nelling physical systems?\n. . . . . . . . . . . . . . . . . . . . .\n441\nB.5\nWhy is stochastic diﬀerential equation not suitable for mod-\nelling ﬁnancial markets? . . . . . . . . . . . . . . . . . . . . .\n443\nB.6\nWhat is the diﬀerence between uncertainty theory and possi-\nbility theory? . . . . . . . . . . . . . . . . . . . . . . . . . . .\n449\nB.7\nWhy do I think fuzzy set theory is wrong? . . . . . . . . . . .\n449\nB.8\nWhy is fuzzy variable not suitable for modelling anything in\nthe real world? . . . . . . . . . . . . . . . . . . . . . . . . . .\n451\nB.9\nHow do we handle interval numbers by uncertainty theory? .\n453\nB.10 Why do I think none of interval analysis, rough set theory and\ngrey system is self-consistent in mathematics? . . . . . . . . .\n456\nB.11 How did “uncertainty” evolve over the past 100 years? . . . .\n457\nC Ye Lemma\n459\nBibliography\n463\nList of Frequently Used Symbols\n481\nIndex\n482\n\n\n\n\nPreface\nSomething is called random if its frequency of occurrence is known. Other-\nwise, it is called uncertain. The outcome of tossing a coin is an example of\nrandomness since the frequency that the coin will come up heads is known.\nThe outcome of a falling cake is an example of uncertainty since the frequency\nthat the cake will land butter-side down is unknown. In order to rationally\ndeal with those phenomena, there exist two mathematical systems, one is\nprobability theory and the other is uncertainty theory. Probability theory is\na branch of mathematics concerned with the analysis of random phenomena,\nwhile uncertainty theory is a branch of mathematics concerned with the anal-\nysis of uncertain phenomena. In order to use them to handle some quantity\n(e.g., stock price) in practice, the ﬁrst action is to produce a distribution\nfunction representing the possibility that the quantity falls into the left side\nof the current point. If you believe the distribution function is close enough\nto the future frequency, then you should use probability theory. Otherwise,\nyou have to use uncertainty theory. Numerous empirical studies show that\nthe real world is far from frequency stability. This fact makes the distribu-\ntion function obtained in practice usually deviate from the future frequency\neven when numerous observed data are available, and consequently provides\na motivation to learn and use uncertainty theory.\nUncertain Measure\nChapter 2 will provide normality, duality, subadditivity and product axioms\nof uncertainty theory. From those four axioms, Chapter 2 will also introduce\nthe tool of uncertain measure that is used to indicate the belief degree that\nsomething may happen. In addition, conditional uncertain measure will be\nexplored.\nUncertain Variable\nUncertain variable is a measurable function from an uncertainty space to\nthe set of real numbers. Chapter 3 is devoted to uncertain variable, uncer-\ntainty distribution, independence, operational law, expected value, variance,\nmoments, distance, entropy, uncertain sequence, and uncertain vector.\n\n\nxii\nPreface\nUncertain Statistics\nUncertain statistics is a set of mathematical techniques for collecting, analyz-\ning and interpreting data by uncertainty theory. Chapter 4 will be devoted\nto the method of moments, the method of least squares, uncertain hypothesis\ntest, uncertain regression analysis, and uncertain times series analysis.\nUncertain Programming\nUncertain programming is a type of mathematical programming involving un-\ncertain variables. Chapter 5 will provide the tool of uncertain programming\nwith applications to machine scheduling problem, vehicle routing problem,\nand project scheduling problem. In addition, uncertain multiobjective pro-\ngramming, uncertain goal programming and uncertain multilevel program-\nming are also documented.\nUncertain Risk Analysis\nThe term risk has been used in diﬀerent ways in literature. In this book\nthe risk is deﬁned as the accidental loss plus the belief degree of such loss.\nChapter 6 will introduce uncertain risk analysis that is a tool to quantify risk\nvia uncertainty theory. As applications of uncertain risk analysis, Chapter 6\nwill also discuss structural risk analysis.\nUncertain Reliability Analysis\nChapter 7 will introduce uncertain reliability analysis that is a tool to deal\nwith system reliability via uncertainty theory. A reliability index theorem\nwill be provided for calculating system reliability index.\nUncertain Propositional Logic\nUncertain propositional logic is a generalization of propositional logic in\nwhich every proposition is abstracted into a Boolean uncertain variable and\nthe truth value is deﬁned as the uncertain measure that the proposition is\ntrue. Chapter 8 will present uncertain propositional logic. In addition, un-\ncertain entailment is a methodology for determining the truth value of an\nuncertain proposition via the maximum uncertainty principle when the truth\nvalues of other uncertain propositions are given. Chapter 8 will also present\nan uncertain entailment model from which uncertain modus ponens, uncer-\ntain modus tollens and uncertain hypothetical syllogism are deduced.\nUncertain Set\nUncertain set is a set-valued function on an uncertainty space, and attempts\nto model unsharp concepts like “young”, “tall”, “warm”, and “most”. The\n\n\nPreface\nxiii\nmain diﬀerence between uncertain set and uncertain variable is that the for-\nmer takes values of set and the latter takes values of point. Uncertain set\ntheory will be introduced in Chapter 9.\nUncertain Logic\nSome knowledge in human brain is actually an uncertain set. This fact en-\ncourages us to design an uncertain logic that is a methodology for calculating\nthe truth values of uncertain propositions via uncertain set theory. Uncertain\nlogic may provide a ﬂexible means for extracting linguistic summary from a\ncollection of raw data. Chapter 10 will be devoted to uncertain logic and\nlinguistic summarizer.\nUncertain Inference Control\nUncertain inference controller is a function that maps the state variables of\na process under control to the action variables by using human knowledge\nand uncertain set theory. Chapter 11 will present uncertain inference rule,\nand uncertain inference controller with application to an inverted pendulum\nsystem.\nUncertain Process\nAn uncertain process is essentially a sequence of uncertain variables indexed\nby time. Thus an uncertain process is usually used to model uncertain phe-\nnomena that vary with time. Chapter 12 will be devoted to basic concepts\nof uncertain process and uncertainty distribution. In addition, extreme value\ntheorem, ﬁrst hitting time and time integral of uncertain processes will also\nbe introduced. Chapter 13 will provide uncertain renewal process with ap-\nplications to uncertain insurance model, uncertain production model, and\nuncertain queueing model.\nUncertain Calculus\nUncertain calculus is a branch of mathematics that deals with diﬀerentiation\nand integration of uncertain processes. Chapter 14 will introduce Liu pro-\ncess that is a stationary independent increment process whose increments are\nnormal uncertain variables, and discuss Liu integral that is a type of uncer-\ntain integral with respect to Liu process. Chapter 14 will also present the\nfundamental theorem of uncertain calculus, chain rule, change of variables,\nintegration by parts, and Fubini theorem.\nUncertain Diﬀerential Equation\nUncertain diﬀerential equation is a type of diﬀerential equation involving\nuncertain processes. Chapter 15 will discuss the existence, uniqueness and\n\n\nxiv\nPreface\nstability of solutions of uncertain diﬀerential equations, and introduce Yao-\nChen formula that represents the solution of an uncertain diﬀerential equation\nby a family of solutions of ordinary diﬀerential equations. On the basis of\nthis formula, some formulas to calculate extreme value, ﬁrst hitting time,\nand time integral of solution will be provided. Furthermore, some numerical\nmethods for solving uncertain diﬀerential equations will be designed.\nIn\naddition, uncertain hypothesis test will be employed to determine whether\nan uncertain diﬀerential equation ﬁts the observed data of some uncertain\nprocess. Finally, we will present some parameter estimation methods for an\nuncertain diﬀerential equation that ﬁts the observed data as much as possible.\nUncertain Finance\nAs applications of uncertain diﬀerential equation, Chapter 16 will introduce\nuncertain stock model, uncertain interest rate model, and uncertain currency\nmodel. Based on the fair price principle, Chapter 16 will also price European\noptions, American options, Asian options, zero-coupon bond, interest rate\nceiling, and interest rate ﬂoor.\nLaw of Truth Conservation\nThe law of excluded middle tells us that a proposition is either true or false,\nand the law of contradiction tells us that a proposition cannot be both true\nand false. In the state of uncertainty, some people said, the law of excluded\nmiddle and the law of contradiction are no longer valid because the truth\nvalue of a proposition is no longer 0 or 1. I cannot gainsay this viewpoint to\na certain extent. But it does not mean that you might “go as you please”.\nThe truth values of a proposition and its negation should sum to unity. This is\nthe law of truth conservation that is weaker than the law of excluded middle\nand the law of contradiction. Furthermore, the law of truth conservation\nagrees with the law of excluded middle and the law of contradiction when\nthe uncertainty vanishes.\nMaximum Uncertainty Principle\nAn event has no uncertainty if its uncertain measure is 1 because we may be-\nlieve that the event happens. An event has no uncertainty too if its uncertain\nmeasure is 0 because we may believe that the event does not happen. An\nevent is the most uncertain if its uncertain measure is 0.5 because the event\nand its complement may be regarded as “equally likely”. In practice, if there\nis no information about the uncertain measure of an event, we should assign\n0.5 to it. Sometimes, only partial information is available. In this case, the\nvalue of uncertain measure may be speciﬁed in some range. What value does\nthe uncertain measure take? For any event, if there are multiple reasonable\nvalues that an uncertain measure may take, then the value as close to 0.5 as\npossible is assigned to the event. This is the maximum uncertainty principle.\n\n\nPreface\nxv\nPurpose\nThe purpose of this textbook is to equip the readers with a branch of math-\nematics to deal with uncertainty. The textbook is suitable for researchers,\nengineers, and students in the ﬁeld of mathematics, information science, op-\nerations research, industrial engineering, computer science, artiﬁcial intelli-\ngence, automation, economics, and management science.\nBaoding Liu\nTsinghua University\nliu@tsinghua.edu.cn\nFebruary 23, 2024\n\n\nA rational man behaves as if he used uncertainty theory.\n\n\nChapter 1\nIntroduction\nSomething is called random if its frequency of occurrence is known. Other-\nwise, it is called uncertain. The outcome of tossing a coin is an example of\nrandomness since the frequency that the coin will come up heads is known.\nThe outcome of a falling cake is an example of uncertainty since the frequency\nthat the cake will land butter-side down is unknown. In order to rationally\ndeal with those phenomena, there exist two mathematical systems, one is\nprobability theory and the other is uncertainty theory. Probability theory\nis a branch of mathematics concerned with the analysis of random phenom-\nena, while uncertainty theory is a branch of mathematics concerned with the\nanalysis of uncertain phenomena.\n1.1\nUrn Problems\nAssume I ﬁlled 100 urns each with 100 balls that are either red or black.\nYou are only told that the numbers of red balls in each urns are independent\nand identically distributed (iid), but the distribution function is completely\nunknown to you. Consider the following three urn problems:\n(i) How many balls do you think are red in the ﬁrst urn?\n(ii) How many balls do you think are red in the 100 urns?\n(iii) How likely do you think the total number of red balls is 10,000?\nHow do you solve those urn problems by probability theory?\nSince you do not know the number of red balls completely, Laplace criterion\nmakes you assign equal probabilities to the possible numbers of red balls\n0, 1, 2, · · · , 100. Thus, for each i with 1 ≤i ≤100, the number of red balls in\n\n\n2\nChapter 1 - Introduction\nthe ith urn is a random variable,\nξ1 =\n\n\n\n\n\n\n\n0\nwith probability 1/101\n1\nwith probability 1/101\n.\n.\n.\n100 with probability 1/101.\nNote that ξ1, ξ2, · · · , ξ100 are iid random variables according to my promise.\nThe total number of red balls in the 100 urns is the sum\nξ = ξ1 + ξ2 + · · · + ξ100\nthat can take any integer between 0 and 10,000. Since the total number of\nred balls is 10,000 if and only if the 100 urns each contain 100 red balls, the\nprobability of the total number of red balls being 10,000 is\nPr{ξ = 10, 000} = Pr {ξi = 100, i = 1, 2, · · · , 100}\n=\n100\nY\ni=1\nPr{ξi = 100} =\n100\nY\ni=1\n1\n101\n≈3.6 × 10−201\nwhere Pr{·} represents the probability measure.\nHow do you solve those urn problems by uncertainty theory?\nSince you do not know the number of red balls completely, you have to assign\nequal belief degrees to the possible numbers of red balls 0, 1, 2, · · · , 100. Thus,\nfor each i with 1 ≤i ≤100, the number of red balls in the ith urn is an\nuncertain variable,\nη1 =\n\n\n\n\n\n\n\n0\nwith belief degree 1/101\n1\nwith belief degree 1/101\n.\n.\n.\n100 with belief degree 1/101.\nNote that η1, η2, · · · , η100 are iid uncertain variables according to my promise.\nThe total number of red balls in the 100 urns is the sum\nη = η1 + η2 + · · · + η100\nthat can also take any integer between 0 and 10,000. Since the total number\nof red balls is 10,000 if and only if the 100 urns each contain 100 red balls,\n\n\nSection 1.1 - Urn Problems\n3\nthe belief degree of the total number of red balls being 10,000 is\nM{η = 10, 000} = M {ηi = 100, i = 1, 2, · · · , 100}\n=\n100\n^\ni=1\nM{ηi = 100} =\n100\n^\ni=1\n1\n101\n=\n1\n101\nwhere M{·} represents the belief degree (i.e., uncertain measure).\nWhich result is more reasonable?\nProbability theory tells you that the probability of the total number of red\nballs being 10,000 is 3.6 × 10−201, while uncertainty theory tells you that the\nbelief degree is 1/101. Which result is more reasonable? In order to answer\nthis question, I have to introduce the fourth urn problem. Assume there exist\ntwo options:\nA: You lose $1,000,000 if the total number of red balls is 10,000, and receive\n$1 otherwise;\nB: Don’t bet.\nWhat is your choice between A and B? If probability theory is used, then the\nprobability of the total number of red balls being 10,000 is 3.6 × 10−201, and\nthe expected income of A is\nA = 1 × (1 −3.6 × 10−201) −1000000 × 3.6 × 10−201 ≈1.\nSince the income of B is always 0, we have\nA > B.\nThat is, probability theory makes you choose A. If uncertainty theory is used,\nthen the belief degree of the total number of red balls being 10,000 is 1/101,\nand the expected income of A is\nA = 1 ×\n\u0012\n1 −\n1\n101\n\u0013\n−1000000 ×\n1\n101 ≈−9900.\nSince the income of B is always 0, we have\nA < B.\nThat is, uncertainty theory makes you choose B. Probability theory and\nuncertainty theory give you two diametrically opposed choices. Which choice\ndo you think is better?\n\n\n4\nChapter 1 - Introduction\nHow did I ﬁll the 100 urns?\nIn order to compare the decisions made by probability theory and uncertainty\ntheory, I would like to show you how I ﬁlled the 100 urns. First I took a\ndistribution function,\nΥ(x) =\n(\n0,\nif x < 100\n1,\nif x ≥100\nthat is just the constant 100 (please recognize that I have the option to choose\nmy preferred distribution function). Next I generated a random number k\nfrom the distribution function Υ, and ﬁlled the ﬁrst urn with k red balls and\n100 −k black balls. Then I generated a new random number k from Υ, and\nﬁlled the second urn with k red balls and 100 −k black balls. Repeated this\nprocess until 100 urns were ﬁlled. Since the generated number k from the\ndistribution function Υ is always 100, each urn contains 100 red balls. Since\n100, 100, · · · , 100 are indeed iid, I kept my promise. Note also that the total\nnumber of red balls happens to be 10,000.\nYou would lose $1,000,000 if you used probability theory (i.e., you chose\nA). If this experiment is repeated, then you have to choose A again and\ncontinue to lose $1,000,000 as long as you use probability theory.\nWhy does probability theory fail?\nThe root cause is that your uniform distribution function (approximatively)\nof the number of red balls in each urn,\nΦ(x) =\n\n\n\n\n\n0,\nif x < 0\nx/100,\nif 0 ≤x ≤100\n1,\nif x > 100\nis not close to the real frequency,\nΥ(x) =\n(\n0,\nif x < 100\n1,\nif x ≥100.\nIn this case, probability theory led to wrong results. However, uncertainty\ntheory was proven successful to deal with those urn problems.\n1.2\nHow to Choose Your Mathematical Tool\nIn order to use probability theory or uncertainty theory to deal with some\nquantity (e.g., stock price) in practice, the ﬁrst action we take is to produce a\ndistribution function representing the possibility that the quantity falls into\nthe left side of the current point. See Figure 1.1. Such a function will always\n\n\nSection 1.2 - How to Choose Your Mathematical Tool\n5\nhave bigger values as the current point moves from the left to right. If the\ndistribution function takes value 0, then it is completely impossible that the\nquantity falls into the left side of the current point; if the distribution function\ntakes value 1, then it is completely impossible that the quantity falls into the\nright side; if the distribution function takes value 0.6, then we are 60% sure\nthat the quantity falls into the left side and 40% sure that the quantity falls\ninto the right side.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\n1\nx\nα\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................................................................\n....................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 1.1: A Distribution Function\nHow do we distinguish between randomness and uncertainty in practice?\nIf you believe your distribution function (no matter how you get it) is close\nenough to the future frequency, then you should treat the quantity as a\nrandom variable. Otherwise, you have to regard the quantity as an uncertain\nvariable.\nThus you may choose your mathematical tool by the following\ncriterion:\nIf your distribution function is close enough to the future fre-\nquency, then you should use probability theory. Otherwise, you\nhave to use uncertainty theory.\nShould you be satisﬁed with this answer? Numerous empirical studies show\nthat the real world is far from frequency stability. This fact makes the distri-\nbution function obtained in practice usually deviate from the future frequency\neven when numerous observed data are available, and consequently provides\na motivation to use uncertainty theory. Do you agree with me? If so, are you\nwilling to learn uncertainty theory?\n\n\n\n\nChapter 2\nUncertain Measure\nUncertainty theory was founded by Liu [113] in 2007 and subsequently studied\nby many researchers.\nNowadays uncertainty theory has become a branch\nof mathematics concerned with the analysis of uncertain phenomena. This\nchapter will provide normality, duality, subadditivity and product axioms of\nuncertainty theory. From those four axioms, this chapter will also introduce\nan uncertain measure that is a fundamental concept in uncertainty theory.\nIn addition, conditional uncertain measure will be explored at the end of this\nchapter.\n2.1\nUncertain Measure\nFrom the mathematical viewpoint, uncertainty theory is essentially an al-\nternative theory of measure.\nLet Γ be a nonempty set (sometimes called\nuniversal set), and let L be a σ-algebra over Γ. Recall that each element Λ\nin L is called a measurable set. The ﬁrst action we take is to rename mea-\nsurable set as event in uncertainty theory. The second action is to deﬁne an\nuncertain measure M on the σ-algebra L. That is, a number M{Λ} will be\nassigned to each event Λ to indicate the belief degree with which we believe Λ\nwill happen. There is no doubt that the assignment is not arbitrary, and the\nuncertain measure M must have certain mathematical properties. In order\nto rationally deal with belief degrees, Liu [113] suggested the following three\naxioms:\nAxiom 1. (Normality Axiom) M{Γ} = 1 for the universal set Γ.\nAxiom 2. (Duality Axiom) M{Λ} + M{Λc} = 1 for any event Λ.\nAxiom 3. (Subadditivity Axiom) For every countable sequence of events Λ1,\nΛ2, · · · , we have\nM\n( ∞\n[\ni=1\nΛi\n)\n≤\n∞\nX\ni=1\nM{Λi}.\n(2.1)\n\n\n8\nChapter 2 - Uncertain Measure\nDeﬁnition 2.1 (Liu [113]) The set function M is called an uncertain mea-\nsure if it satisﬁes the normality, duality, and subadditivity axioms.\nRemark 2.1: Uncertain measure is interpreted as the belief degree of an\nuncertain event that may happen1. Thus uncertain measure and belief degree\nare synonymous, and will be used interchangeably in this book.\nRemark 2.2: Uncertain measure (i.e., belief degree) depends on the personal\nknowledge concerning the event, and will change2 if the state of knowledge\nchanges.\nRemark 2.3: Since “1” means “complete belief ” and we cannot be in more\nbelief than “complete belief ”, the belief degree of any event cannot exceed 1.\nFurthermore, the belief degree of the universal set takes value 1 because it is\ncompletely believable. Thus the belief degree meets the normality axiom.\nRemark 2.4: Duality axiom is in fact an application of the law of truth\nconservation in uncertainty theory.\nThe property ensures that the uncer-\ntainty theory is consistent with the law of excluded middle and the law of\ncontradiction. In addition, the human thinking is always dominated by the\nduality. For example, if someone tells us that a proposition is true with belief\ndegree 0.6, then all of us will think that the proposition is false with belief\ndegree 0.4.\nRemark 2.5: Given two events with known belief degrees, it is frequently\nasked that how the belief degree for their union is generated from the in-\ndividuals. Personally, I do not think there exists any rule to make it. A\nlot of surveys showed that, generally speaking, the belief degree of a union\nof events is neither the sum of belief degrees of the individual events (e.g.\nprobability measure) nor the maximum (e.g. possibility measure). It seems\nthat there is no explicit relation between the union and individuals except\nfor the subadditivity axiom.\nRemark 2.6: Pathology occurs if subadditivity axiom is not assumed. For\nexample, suppose that a universal set contains 3 elements. We deﬁne a set\nfunction that takes value 0 for each singleton, and 1 for each event with at\nleast 2 elements. Then such a set function satisﬁes all axioms but subaddi-\ntivity. Do you think it is strange if such a set function serves as a measure?\nRemark 2.7: Although probability measure satisﬁes the above three axioms,\nprobability theory is not a special case of uncertainty theory because the\nproduct probability measure does not satisfy the fourth axiom, namely the\nproduct axiom on Page 12.\n1In contrast, probability measure is interpreted as the frequency of a random event that\nmay happen.\n2In contrast, probability measure does not change with the personal knowledge and\npreference.\n\n\nSection 2.1 - Uncertain Measure\n9\nExercise 2.1: Let Γ = {γ1, γ2}. It is clear that the power set of Γ (i.e., all\nsubsets of Γ) consists of 4 events,\n{γ1}, {γ2}, ∅, Γ.\n(2.2)\nAssume c is a real number with 0 < c < 1, and deﬁne\nM{γ1} = c,\nM{γ2} = 1 −c,\nM{∅} = 0,\nM{Γ} = 1.\nShow that M is an uncertain measure.\n(Hint: Verify M meets the three\naxioms.)\nExercise 2.2:\nLet Γ = {γ1, γ2, γ3}.\nIt is clear that the power set of Γ\nconsists of 8 events,\n{γ1}, {γ2}, {γ3}, {γ1, γ2}, {γ1, γ3}, {γ2, γ3}, ∅, Γ.\n(2.3)\nAssume c1, c2, c3 are nonnegative numbers satisfying the consistency condi-\ntion\nci + cj ≤1 ≤c1 + c2 + c3,\n∀i ̸= j.\n(2.4)\nDeﬁne\nM{γ1} = c1,\nM{γ2} = c2,\nM{γ3} = c3,\nM{γ1, γ2} = 1 −c3,\nM{γ1, γ3} = 1 −c2,\nM{γ2, γ3} = 1 −c1,\nM{∅} = 0,\nM{Γ} = 1.\nShow that M is an uncertain measure.\nExercise 2.3: Let Γ = {γ1, γ2, · · · }, and let c1, c2, · · · be nonnegative num-\nbers such that c1 + c2 + · · · = 1. For each subset Λ of Γ, we deﬁne\nM{Λ} =\nX\nγi∈Λ\nci.\n(2.5)\nShow that M is an uncertain measure.\nExercise 2.4: Lebesgue measure, named after French mathematician Henri\nLebesgue, is the standard way of assigning a length, area or volume to subsets\nof Euclidean space. For example, the Lebesgue measure of the interval [a, b] of\nreal numbers is the length b−a. (i) Let Γ = (0, 1), and let M be the Lebesgue\nmeasure. Show that M is an uncertain measure. (ii) Can we replace (0, 1)\nwith [0, 1]?\nExercise 2.5: Let Γ be the set of real numbers. For each subset Λ of Γ, we\ndeﬁne\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\n(2.6)\n\n\n10\nChapter 2 - Uncertain Measure\nShow that M is an uncertain measure.\nExercise 2.6: Let Γ be the set of real numbers, and let c be a real number\nwith 0 < c ≤0.5. For each subset Λ of Γ, we deﬁne\nM{Λ} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif Λ = ∅\nc,\nif Λ is upper bounded and Λ ̸= ∅\n0.5,\nif both Λ and Λc are upper unbounded\n1 −c,\nif Λc is upper bounded and Λ ̸= Γ\n1,\nif Λ = Γ.\n(2.7)\nShow that M is an uncertain measure.\nExercise 2.7: Suppose ρ(x) is a nonnegative and integrable function on ℜ\nsuch that\nZ\nℜ\nρ(x)dx ≥1.\n(2.8)\nDeﬁne a set function\nM{Λ} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nZ\nΛ\nρ(x)dx,\nif\nZ\nΛ\nρ(x)dx < 0.5\n1 −\nZ\nΛc ρ(x)dx,\nif\nZ\nΛc ρ(x)dx < 0.5\n0.5,\notherwise\n(2.9)\nfor each Borel set Λ of real numbers. Show that M is an uncertain measure.\nTheorem 2.1 (Monotonicity Theorem) The uncertain measure is a mono-\ntone increasing set function. That is, for any events Λ1 and Λ2 with Λ1 ⊂Λ2,\nwe have\nM{Λ1} ≤M{Λ2}.\n(2.10)\nProof: The normality axiom says M{Γ} = 1, and the duality axiom says\nM{Λc\n1} = 1 −M{Λ1}. Since Λ1 ⊂Λ2, we have Γ = Λc\n1 ∪Λ2. By using the\nsubadditivity axiom, we obtain\n1 = M{Γ} ≤M{Λc\n1} + M{Λ2} = 1 −M{Λ1} + M{Λ2}.\nThus M{Λ1} ≤M{Λ2}.\nTheorem 2.2 The empty set ∅always has an uncertain measure zero. That\nis,\nM{∅} = 0.\n(2.11)\nProof: Since ∅= Γc and M{Γ} = 1, it follows from the duality axiom that\nM{∅} = 1 −M{Γ} = 1 −1 = 0.\n\n\nSection 2.2 - Uncertainty Space\n11\nTheorem 2.3 The uncertain measure takes values between 0 and 1. That\nis, for any event Λ, we have\n0 ≤M{Λ} ≤1.\n(2.12)\nProof: It follows from the monotonicity theorem that 0 ≤M{Λ} ≤1 because\n∅⊂Λ ⊂Γ and M{∅} = 0, M{Γ} = 1.\nTheorem 2.4 An uncertain measure remains unchanged if the event is en-\nlarged or reduced by an event with uncertain measure zero. That is, for any\nevents Λ and ∆, if M{∆} = 0, then\nM{Λ ∪∆} = M{Λ\\∆} = M{Λ}.\n(2.13)\nProof: It follows from the monotonicity theorem and subadditivity axiom\nthat\nM{Λ} ≤M{Λ ∪∆} ≤M{Λ} + M{∆} = M{Λ}.\nThus M{Λ ∪∆} = M{Λ}. Since (Λ\\∆) ⊂Λ ⊂(Λ\\∆) ∪∆, we have\nM{Λ\\∆} ≤M{Λ} ≤M{Λ\\∆} + M{∆} = M{Λ\\∆}.\nHence M{Λ\\∆} = M{Λ}.\n2.2\nUncertainty Space\nDeﬁnition 2.2 (Liu [113]) Let Γ be a nonempty set, let L be a σ-algebra\nover Γ, and let M be an uncertain measure. Then the triplet (Γ, L, M) is\ncalled an uncertainty space.\nExample 2.1: Let Γ be a two-point set {γ1, γ2}, let L be the power set of\n{γ1, γ2}, and let M be an uncertain measure determined by M{γ1} = 0.6 and\nM{γ2} = 0.4. Then (Γ, L, M) is an uncertainty space.\nExample 2.2: Let Γ be a three-point set {γ1, γ2, γ3}, let L be the power set\nof {γ1, γ2, γ3}, and let M be an uncertain measure determined by M{γ1} =\n0.6, M{γ2} = 0.3 and M{γ3} = 0.2. Then (Γ, L, M) is an uncertainty space.\nExample 2.3: Let Γ be the interval (0, 1), let L be the Borel algebra over Γ,\nand let M be the Lebesgue measure. Then (Γ, L, M) is an uncertainty space.\nDeﬁnition 2.3 (Gao [51]) An uncertainty space (Γ, L, M) is called contin-\nuous if for any events Λ1, Λ2, · · · , we have\nM\nn\nlim\ni→∞Λi\no\n= lim\ni→∞M{Λi}\n(2.14)\nprovided that limi→∞Λi exists.\n\n\n12\nChapter 2 - Uncertain Measure\nExercise 2.8: Show that an uncertainty space (Γ, L, M) is always continuous\nif Γ consists of a ﬁnite number of points.\nExercise 2.9: Let Γ = (0, 1), let L be the Borel algebra over Γ, and let M\nbe the Lebesgue measure. Show that (Γ, L, M) is a continuous uncertainty\nspace.\nExercise 2.10: Let Γ be the set of real numbers, and let L be the power set\nover Γ. For each subset Λ, we deﬁne\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\n(2.15)\nShow that (Γ, L, M) is a discontinuous uncertainty space.\n2.3\nProduct Uncertain Measure\nProduct uncertain measure was deﬁned by Liu [116] in 2009, thus producing\nthe fourth axiom of uncertainty theory.\nLet (Γk, Lk, Mk) be uncertainty\nspaces for k = 1, 2, · · · Write\nΓ = Γ1 × Γ2 × · · ·\n(2.16)\nthat is the set of all ordered tuples of the form (γ1, γ2, · · · ), where γk ∈Γk\nfor k = 1, 2, · · · Denote the product σ-algebra on Γ by\nL = L1 × L2 × · · ·\n(2.17)\nIn order to deﬁne product uncertain measure on the product σ-algebra L, we\nﬁrst deﬁne it for every measurable rectangle\nΛ = Λ1 × Λ2 × · · ·\n(2.18)\nwhere Λk ∈Lk for k = 1, 2, · · · by the following product axiom (Liu [116]).\nAxiom 4. (Product Axiom) Let (Γk, Lk, Mk) be uncertainty spaces for k =\n1, 2, · · · The product uncertain measure M is an uncertain measure satisfying\nM\n( ∞\nY\nk=1\nΛk\n)\n=\n∞\n^\nk=1\nMk{Λk}\n(2.19)\nwhere Λk are arbitrarily chosen events from Lk for k = 1, 2, · · · , respectively.\nRemark 2.8: Note that (2.19) deﬁnes a product uncertain measure only for\nmeasurable rectangles like (2.18). How do we extend the uncertain measure\n\n\nSection 2.3 - Product Uncertain Measure\n13\nM from the class of rectangles to the product σ-algebra L? In fact, for each\nevent Λ ∈L, we may set\nM{Λ} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk},\nif\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk} > 0.5\n1 −\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk},\nif\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk} > 0.5\n0.5,\notherwise.\n(2.20)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ\nΓ1\nΓ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ1 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n................\n.................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 2.1: Extension from Rectangles to Product σ-Algebra. The uncertain\nmeasure of Λ (the disk) is essentially the acreage of its inscribed rectangle\nΛ1×Λ2 if it is greater than 0.5. Otherwise, we have to examine its complement\nΛc. If the inscribed rectangle of Λc is greater than 0.5, then M{Λc} is just\nits inscribed rectangle and M{Λ} = 1 −M{Λc}. If there does not exist an\ninscribed rectangle of Λ or Λc greater than 0.5, then we set M{Λ} = 0.5.\nRemark 2.9: The sum of the uncertain measures of the maximum rectangles\nin Λ and Λc is always less than or equal to 1, i.e.,\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk} +\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk} ≤1.\nThis means that at most one of\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk}\nand\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk}\n\n\n14\nChapter 2 - Uncertain Measure\nis greater than 0.5. Thus the expression (2.20) is reasonable.\nRemark 2.10: It is clear that for each Λ ∈L, the uncertain measure M{Λ}\ndeﬁned by (2.20) takes possible values on the interval\n\u0014\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk}, 1 −\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk}\n\u0015\n.\nThus (2.20) coincides with the maximum uncertainty principle (Liu [113]),\nthat is, M{Λ} takes the value as close to 0.5 as possible within the above\ninterval.\nRemark 2.11: If the sum of the uncertain measures of the maximum rect-\nangles in Λ and Λc is just 1, i.e.,\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk} +\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk} = 1,\nthen the product uncertain measure (2.20) is simpliﬁed as\nM{Λ} =\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk},\n(2.21)\nand\nM{Λc} =\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk}.\n(2.22)\nRemark 2.12: The product uncertain measure M deﬁned by (2.19) will be\ndenoted as\nM = M1 ∧M2 ∧· · ·\n(2.23)\nRemark 2.13: The diﬀerence between uncertainty theory and probability\ntheory does not lie in whether the measures are additive or not, but how the\nproduct measures are deﬁned. Uncertainty theory assumes product uncertain\nmeasure is the minimum of uncertain measures of individual events, i.e.,\nM\n( ∞\nY\nk=1\nΛk\n)\n=\n∞\n^\nk=1\nM{Λk},\n(2.24)\nwhile probability theory assumes product probability measure is the multi-\nplication of probability measures of individual events, i.e.,\nPr\n( ∞\nY\nk=1\nΛk\n)\n=\n∞\nY\nk=1\nPr{Λk}\n(2.25)\nwhere Λ1, Λ2, · · · are events from diﬀerent spaces.\nTheorem 2.5 (Peng-Iwamura [187]) The product uncertain measure deﬁned\nby (2.20) is an uncertain measure.\n\n\nSection 2.3 - Product Uncertain Measure\n15\nProof: In order to prove that the product uncertain measure (2.20) is indeed\nan uncertain measure, we should verify that the product uncertain measure\nmeets the normality, duality and subadditivity axioms.\nStep 1: Since Γ = Γ1×Γ2×· · · is a rectangle, it follows from the product\naxiom that\nM{Γ} = M1{Γ1} ∧M2{Γ2} ∧· · · = 1.\nThus the product uncertain measure meets the normality axiom.\nStep 2: Let us prove the product uncertain measure meets the duality\naxiom, i.e., M{Λ}+M{Λc} = 1. The argument breaks down into three cases.\nCase 1: Assume\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk} > 0.5.\nIt follows from (2.20) that\nM{Λ} =\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk},\nM{Λc} = 1 −\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk}.\nThus M{Λ} + M{Λc} = 1. Case 2: Assume\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk} > 0.5.\nIt follows from (2.20) that\nM{Λ} = 1 −\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk},\nM{Λc} =\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk}.\nThus M{Λ} + M{Λc} = 1. Case 3: Assume\nsup\nΛ1×Λ2×···⊂Λ\nmin\nk≥1 Mk{Λk} ≤0.5,\nsup\nΛ1×Λ2×···⊂Λc min\nk≥1 Mk{Λk} ≤0.5.\nIt follows from (2.20) that\nM{Λ} = 0.5,\nM{Λc} = 0.5.\nThus M{Λ} + M{Λc} = 1. Therefore, the product uncertain measure meets\nthe duality axiom.\nStep 3: Let us prove that the product uncertain measure is an increasing\nset function. Suppose Λ1 and Λ2 are two events in L with Λ1 ⊂Λ2. The\nargument breaks down into three cases. Case 1: Assume\nsup\nΛ11×Λ12×···⊂Λ1\nmin\nk≥1 Mk{Λ1k} > 0.5.\n\n\n16\nChapter 2 - Uncertain Measure\nThen\nsup\nΛ21×Λ22×···⊂Λ2\nmin\nk≥1 Mk{Λ2k} ≥\nsup\nΛ11×Λ12×···⊂Λ1\nmin\nk≥1 Mk{Λ1k} > 0.5.\nIt follows from (2.20) that\nM{Λ1} =\nsup\nΛ11×Λ12×···⊂Λ1\nmin\nk≥1 Mk{Λ1k},\nM{Λ2} =\nsup\nΛ21×Λ22×···⊂Λ2\nmin\nk≥1 Mk{Λ2k}.\nThus M{Λ1} ≤M{Λ2}. Case 2: Assume\nsup\nΛ21×Λ22×···⊂Λc\n2\nmin\nk≥1 Mk{Λ2k} > 0.5.\nThen\nsup\nΛ11×Λ12×···⊂Λc\n1\nmin\nk≥1 Mk{Λ1k} ≥\nsup\nΛ21×Λ22×···⊂Λc\n2\nmin\nk≥1 Mk{Λ2k} > 0.5.\nIt follows from (2.20) that\nM{Λ1} = 1 −\nsup\nΛ11×Λ12×···⊂Λc\n1\nmin\nk≥1 Mk{Λ1k},\nM{Λ2} = 1 −\nsup\nΛ21×Λ22×···⊂Λc\n2\nmin\nk≥1 Mk{Λ2k}.\nThus M{Λ1} ≤M{Λ2}. Case 3: Assume\nsup\nΛ11×Λ12×···⊂Λ1\nmin\nk≥1 Mk{Λ1k} ≤0.5,\nsup\nΛ21×Λ22×···⊂Λc\n2\nmin\nk≥1 Mk{Λ2k} ≤0.5.\nIt follows from (2.20) that\nM{Λ1} ≤0.5,\nM{Λ2} ≥0.5.\nThus M{Λ1} ≤M{Λ2}. Therefore, the product uncertain measure is an\nincreasing set function.\nStep 4: Finally, let us prove the product uncertain measure meets the\nsubadditivity axiom. The argument breaks down into four cases. Case 1: For\nany countable sequence {Λi} of events, assume M{Λi} < 0.5, i = 1, 2, · · · It\nfollows from (2.20) that\nM{Λi} = 1 −\nsup\nΛi1×Λi2×···⊂Λc\ni\nmin\nk≥1 Mk{Λik},\ni = 1, 2, · · ·\n\n\nSection 2.3 - Product Uncertain Measure\n17\nThus, for any given ε > 0, there exist rectangles\nΛi1 × Λi2 × · · · ⊂Λc\ni,\ni = 1, 2, · · ·\n(2.26)\nsuch that\n1 −min\nk≥1 Mk{Λik} ≤M{Λi} + ε\n2i ,\ni = 1, 2, · · ·\nBy using (2.26), we also have\n ∞\n\\\ni=1\nΛi1\n!\n×\n ∞\n\\\ni=1\nΛi2\n!\n× · · · ⊂\n∞\n\\\ni=1\nΛc\ni =\n ∞\n[\ni=1\nΛi\n!c\n.\nSince the product uncertain measure has been proved to be dual and increas-\ning, we obtain\nM\n( ∞\n[\ni=1\nΛi\n)\n= 1 −M\n( ∞\n[\ni=1\nΛi\n!c)\n≤1 −M\n( ∞\n\\\ni=1\nΛi1\n!\n×\n ∞\n\\\ni=1\nΛi2\n!\n× · · ·\n)\n= 1 −min\nk≥1 Mk\n( ∞\n\\\ni=1\nΛik\n)\n= max\nk≥1 Mk\n( ∞\n[\ni=1\nΛc\nik\n)\n≤max\nk≥1\n∞\nX\ni=1\nMk{Λc\nik} ≤\n∞\nX\ni=1\nmax\nk≥1 Mk{Λc\nik}\n=\n∞\nX\ni=1\n\u0012\n1 −min\nk≥1 Mk{Λik}\n\u0013\n≤\n∞\nX\ni=1\nM{Λi} + ε.\nLetting ε →0, we obtain\nM\n( ∞\n[\ni=1\nΛi\n)\n≤\n∞\nX\ni=1\nM{Λi}.\nCase 2: Suppose there is one term greater than or equal to 0.5, say\nM{Λ1} ≥0.5,\nM{Λi} < 0.5,\ni = 2, 3, · · ·\nand\nM\n( ∞\n[\ni=1\nΛi\n)\n≤0.5.\n\n\n18\nChapter 2 - Uncertain Measure\nIn this case, we immediately have\nM\n( ∞\n[\ni=1\nΛi\n)\n≤0.5 ≤\n∞\nX\ni=1\nM{Λi}.\nCase 3: Suppose there is one term greater than or equal to 0.5, say\nM{Λ1} ≥0.5,\nM{Λi} < 0.5,\ni = 2, 3, · · ·\nand\nM\n( ∞\n[\ni=1\nΛi\n)\n> 0.5.\nThen\nM\n( ∞\n\\\ni=1\nΛc\ni\n)\n= 1 −M\n( ∞\n[\ni=1\nΛi\n)\n< 0.5.\nBy using\nΛc\n1 ⊂\n ∞\n\\\ni=1\nΛc\ni\n!\n∪\n ∞\n[\ni=2\nΛi\n!\nand Case 1, we get\nM {Λc\n1} ≤M\n( ∞\n\\\ni=1\nΛc\ni\n)\n+\n∞\nX\ni=2\nM {Λi} .\nThat is,\n1 −M{Λ1} ≤1 −M\n( ∞\n[\ni=1\nΛi\n)\n+\n∞\nX\ni=2\nM {Λi} .\nHence\nM\n( ∞\n[\ni=1\nΛi\n)\n≤\n∞\nX\ni=1\nM{Λi}.\nCase 4: Suppose that there are at least two terms greater than or equal to\n0.5, say\nM{Λ1} ≥0.5,\nM{Λ2} ≥0.5.\nIn this case, we immediately have\nM\n( ∞\n[\ni=1\nΛi\n)\n≤1 ≤\n∞\nX\ni=1\nM{Λi}.\nThus the product uncertain measure meets the subadditivity axiom. The\ntheorem is proved.\n\n\nSection 2.4 - Independence\n19\nDeﬁnition 2.4 Assume (Γk, Lk, Mk) are uncertainty spaces for k = 1, 2, · · ·\nand\nΓ = Γ1 × Γ2 × · · ·\n(2.27)\nL = L1 × L2 × · · ·\n(2.28)\nM = M1 ∧M2 ∧· · ·\n(2.29)\nThen the triplet (Γ, L, M) is called a product uncertainty space.\nExercise 2.11: Let (Γ1, L1, M1) be the interval (0, 1) with Borel algebra\nand Lebesgue measure, and let (Γ2, L2, M2) be also the interval (0, 1) with\nBorel algebra and Lebesgue measure. Then\nΛ = {(γ1, γ2) ∈Γ1 × Γ2 | γ1 + γ2 ≤1}\n(2.30)\nis an event on the product uncertainty space (Γ1, L1, M1) × (Γ2, L2, M2).\nShow that\nM{Λ} = 1\n2.\n(2.31)\nExercise 2.12: Let (Γ1, L1, M1) be the interval (0, 1) with Borel algebra\nand Lebesgue measure, and let (Γ2, L2, M2) be also the interval (0, 1) with\nBorel algebra and Lebesgue measure. Then\nΛ =\n\b\n(γ1, γ2) ∈Γ1 × Γ2 | (γ1 −0.5)2 + (γ2 −0.5)2 < 0.52\t\n(2.32)\nis an event on the product uncertainty space (Γ1, L1, M1) × (Γ2, L2, M2). (i)\nShow that\nM{Λ} =\n1\n√\n2.\n(2.33)\n(ii) It follows from the duality that M{Λc} = 1−1/\n√\n2. Please ﬁnd a rectangle\nΛ1 × Λ2 in Λc such that M{Λ1 × Λ2} = 1 −1/\n√\n2.\nExercise 2.13: For each positive integer k, let (Γk, Lk, Mk) be the interval\n(0, 1) with Borel algebra and Lebesgue measure.\nDeﬁne an event on the\nproduct uncertainty space (Γ, L, M) as follows,\nΛ =\n∞\n[\ni=1\n\u0014\n1\ni + 1,\ni\ni + 1\n\u0015\n×\n\u0014\n1\ni + 1,\ni\ni + 1\n\u0015\n× · · ·\n(2.34)\n(i) Show that\n\u00121\n2, 1\n3, 1\n4, · · ·\n\u0013\n̸∈Λ.\n(2.35)\nThat is, Λ ̸= Γ. (ii) Show that for each (γ1, γ2, · · · ) ∈Λ, we have\n∞\n^\nk=1\nγk > 0\nand\n∞\n_\nk=1\nγk < 1.\n(2.36)\n(iii) Show that\nM{Λ} = 1.\n(2.37)\n\n\n20\nChapter 2 - Uncertain Measure\n2.4\nIndependence\nThe independence of two events means that knowing the occurrence of one\ndoes not change our estimation of the other. What events meet this condi-\ntion? A typical case is that they belong to diﬀerent uncertainty spaces. For\nexample, let Λ1 and Λ2 be events on the uncertainty spaces (Γ1, L1, M1) and\n(Γ2, L2, M2), respectively. Then Λ1 and Λ2 can be understood as Λ1 × Γ2\nand Γ1 × Λ2 on the product uncertainty space (Γ1, L1, M1) × (Γ2, L2, M2),\nrespectively. See Figure 2.2.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ1 × Λ2\nΓ1\nΓ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ1 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 2.2: (Λ1 × Γ2) ∩(Γ1 × Λ2) = Λ1 × Λ2 (abbreviated as Λ1 ∩Λ2)\nIt follows from the product axiom that the product uncertain measure of\nthe intersection is\nM{(Λ1 × Γ2) ∩(Γ1 × Λ2)} = M{Λ1 × Λ2} = M1{Λ1} ∧M2{Λ2}.\nBy using M{Λ1 × Γ2} = M1{Λ1} and M{Γ1 × Λ2} = M2{Λ2}, we obtain\nM{(Λ1 × Γ2) ∩(Γ1 × Λ2)} = M{Λ1 × Γ2} ∧M{Γ1 × Λ2}.\nSimilarly, we may prove other three equations as follows,\nM{(Λ1 × Γ2)c ∩(Γ1 × Λ2)} = M{(Λ1 × Γ2)c} ∧M{Γ1 × Λ2},\nM{(Λ1 × Γ2) ∩(Γ1 × Λ2)c} = M{Λ1 × Γ2} ∧M{(Γ1 × Λ2)c},\nM{(Λ1 × Γ2)c ∩(Γ1 × Λ2)c} = M{(Λ1 × Γ2)c} ∧M{(Γ1 × Λ2)c}.\nFor simplicity, we denote Λ1 × Γ2 and Γ1 × Λ2 by Λ1 and Λ2, respectively.\nThen the above four equations become\nM{Λ1 ∩Λ2} = M{Λ1} ∧M{Λ2},\nM{Λc\n1 ∩Λ2} = M{Λc\n1} ∧M{Λ2},\nM{Λ1 ∩Λc\n2} = M{Λ1} ∧M{Λc\n2},\nM{Λc\n1 ∩Λc\n2} = M{Λc\n1} ∧M{Λc\n2}.\n\n\nSection 2.4 - Independence\n21\nThus we say two events Λ1 and Λ2 are independent if and only if those\nfour equations hold. Generally, we may deﬁne independence of events in the\nfollowing form.\nDeﬁnition 2.5 (Liu [120]) The events Λ1, Λ2, · · · , Λn are said to be inde-\npendent if\nM\n( n\n\\\ni=1\nΛ∗\ni\n)\n=\nn\n^\ni=1\nM{Λ∗\ni }\n(2.38)\nwhere Λ∗\ni are arbitrarily chosen from {Λi, Λc\ni, Γ}, i = 1, 2, · · · , n, respectively,\nand Γ is the universal set.\nExample 2.4: The impossible event ∅is independent of any event Λ because\nthe following four equations hold:\nM{∅∩Λ} = M{∅} = M{∅} ∧M{Λ},\nM{∅c ∩Λ} = M{Λ} = M{∅c} ∧M{Λ},\nM{∅∩Λc} = M{∅} = M{∅} ∧M{Λc},\nM{∅c ∩Λc} = M{Λc} = M{∅c} ∧M{Λc}.\nExample 2.5: The sure event Γ is independent of any event Λ because the\nfollowing four equations hold:\nM{Γ ∩Λ} = M{Λ} = M{Γ} ∧M{Λ},\nM{Γc ∩Λ} = M{Γc} = M{Γc} ∧M{Λ},\nM{Γ ∩Λc} = M{Λc} = M{Γ} ∧M{Λc},\nM{Γc ∩Λc} = M{Γc} = M{Γc} ∧M{Λc}.\nExercise 2.14: Let Λ1, Λ2, · · · , Λn be independent events. Show that Λi\nand Λj are independent for any indexes i and j with 1 ≤i < j ≤n.\nExercise 2.15: Let Λ be an event. Are Λ and Λc independent? Please\njustify your answer.\nExercise 2.16: Construct n independent events. Hint: Deﬁne them on the\nproduct uncertainty space (Γ1, L1, M1) × (Γ2, L2, M2) × · · · × (Γn, Ln, Mn).\nRemark 2.14: The diﬀerence between uncertainty theory and probability\ntheory is that the former assumes the joint uncertain measure of independent\nevents is the minimum of uncertain measures of individual events, i.e.,\nM\n( ∞\n\\\nk=1\nΛk\n)\n=\n∞\n^\nk=1\nM{Λk},\n(2.39)\n\n\n22\nChapter 2 - Uncertain Measure\nwhile the latter assumes the joint probability measure of independent events\nis the multiplication of probability measures of individual events, i.e.,\nPr\n( ∞\n\\\nk=1\nΛk\n)\n=\n∞\nY\nk=1\nPr{Λk}\n(2.40)\nwhere Λ1, Λ2, · · · are independent events.\nTheorem 2.6 (Liu [120]) The events Λ1, Λ2, · · · , Λn are independent if and\nonly if\nM\n( n\n[\ni=1\nΛ∗\ni\n)\n=\nn\n_\ni=1\nM{Λ∗\ni }\n(2.41)\nwhere Λ∗\ni are arbitrarily chosen from {Λi, Λc\ni, ∅}, i = 1, 2, · · · , n, respectively,\nand ∅is the impossible event.\nProof: Assume Λ1, Λ2, · · · , Λn are independent events. It follows from the\nduality axiom of uncertain measure that\nM\n( n\n[\ni=1\nΛ∗\ni\n)\n= 1 −M\n( n\n\\\ni=1\nΛ∗c\ni\n)\n= 1 −\nn\n^\ni=1\nM{Λ∗c\ni } =\nn\n_\ni=1\nM{Λ∗\ni }\nwhere Λ∗\ni are arbitrarily chosen from {Λi, Λc\ni, ∅}, i = 1, 2, · · · , n, respectively.\nThe equation (2.41) is proved. Conversely, if the equation (2.41) holds, then\nM\n( n\n\\\ni=1\nΛ∗\ni\n)\n= 1 −M\n( n\n[\ni=1\nΛ∗c\ni\n)\n= 1 −\nn\n_\ni=1\nM{Λ∗c\ni } =\nn\n^\ni=1\nM{Λ∗\ni }.\nwhere Λ∗\ni are arbitrarily chosen from {Λi, Λc\ni, Γ}, i = 1, 2, · · · , n, respectively.\nThe equation (2.38) is true. The theorem is proved.\n2.5\nConditional Uncertain Measure\nWe consider the uncertain measure of an event Λ after it has been learned\nthat some other event A has occurred. This new uncertain measure of Λ is\ncalled the conditional uncertain measure of Λ given A.\nIn order to deﬁne a conditional uncertain measure M{Λ|A}, at ﬁrst we\nhave to enlarge M{Λ ∩A} because M{Λ ∩A} < 1 for all events whenever\nM{A} < 1. It seems that we have no alternative but to divide M{Λ ∩A} by\nM{A}. Unfortunately, M{Λ∩A}/M{A} is not always an uncertain measure.\nHowever, the value M{Λ|A} should not be greater than M{Λ ∩A}/M{A}\n(otherwise the normality will be lost), i.e.,\nM{Λ|A} ≤M{Λ ∩A}\nM{A}\n.\n(2.42)\n\n\nSection 2.5 - Conditional Uncertain Measure\n23\nOn the other hand, in order to preserve the duality, we should have\nM{Λ|A} = 1 −M{Λc|A} ≥1 −M{Λc ∩A}\nM{A}\n.\n(2.43)\nFurthermore, since (Λ ∩A) ∪(Λc ∩A) = A, we have M{A} ≤M{Λ ∩A} +\nM{Λc ∩A} by using the subadditivity axiom. Thus\n0 ≤1 −M{Λc ∩A}\nM{A}\n≤M{Λ ∩A}\nM{A}\n≤1.\n(2.44)\nHence any numbers between 1−M{Λc ∩A}/M{A} and M{Λ∩A}/M{A} are\nreasonable values that the conditional uncertain measure may take. Based\non the maximum uncertainty principle (Liu [113]), we have the following\nconditional uncertain measure.\nDeﬁnition 2.6 (Liu [113]) Let (Γ, L, M) be an uncertainty space, and Λ, A ∈\nL. Then the conditional uncertain measure of Λ given A is deﬁned by\nM{Λ|A} =\n\n\n\n\n\n\n\n\n\n\n\n\n\nM{Λ ∩A}\nM{A}\n,\nif M{Λ ∩A}\nM{A}\n< 0.5\n1 −M{Λc ∩A}\nM{A}\n,\nif M{Λc ∩A}\nM{A}\n< 0.5\n0.5,\notherwise\n(2.45)\nprovided that M{A} > 0.\nRemark 2.15:\nIt follows immediately from the deﬁnition of conditional\nuncertain measure that\n1 −M{Λc ∩A}\nM{A}\n≤M{Λ|A} ≤M{Λ ∩A}\nM{A}\n.\n(2.46)\nRemark 2.16: The conditional uncertain measure M{Λ|A} yields the pos-\nterior uncertain measure of Λ after the occurrence of event A.\nTheorem 2.7 (Liu [113]) Let (Γ, L, M) be an uncertainty space, and let A\nbe an event with M{A} > 0. Then the conditional uncertain measure M{·|A}\nis an uncertain measure, and (Γ, L, M{·|A}) is an uncertainty space.\nProof: It is suﬃcient to prove that the conditional uncertain measure satis-\nﬁes the normality, duality and subadditivity axioms. At ﬁrst, we prove that\nthe conditional uncertain measure satisﬁes the normality axiom. Since\nM{Γc ∩A}\nM{A}\n= M{∅}\nM{A} = 0 < 0.5,\n\n\n24\nChapter 2 - Uncertain Measure\nit follows from (2.45) that\nM{Γ|A} = 1 −M{Γc ∩A}\nM{A}\n= 1 −0 = 1.\nTherefore, the conditional uncertain measure satisﬁes the normality axiom.\nNext, we prove the conditional uncertain measure satisﬁes the duality axiom.\nThe argument breaks down into three cases. Case 1: Assume\nM{Λ ∩A}\nM{A}\n< 0.5.\nIt follows from (2.45) that\nM{Λ|A} = M{Λ ∩A}\nM{A}\n,\nM{Λc|A} = 1 −M{Λ ∩A}\nM{A}\n.\nThus M{Λ|A} + M{Λc|A} = 1. Case 2: Assume\nM{Λc ∩A}\nM{A}\n< 0.5.\nIt follows from (2.45) that\nM{Λ|A} = 1 −M{Λc ∩A}\nM{A}\n,\nM{Λc|A} = M{Λc ∩A}\nM{A}\n.\nThus M{Λ|A} + M{Λc|A} = 1. Case 3: Assume\nM{Λ ∩A}\nM{A}\n≥0.5,\nM{Λc ∩A}\nM{A}\n≥0.5.\nIt follows from (2.45) that\nM{Λ|A} = 0.5,\nM{Λc|A} = 0.5.\nThus M{Λ|A} + M{Λc|A} = 1. Therefore, the conditional uncertain mea-\nsure satisﬁes the duality axiom. Finally, we prove the conditional uncertain\nmeasure satisﬁes the subadditivity axiom. The argument breaks down into\nfour cases.\nCase 1: For any countable sequence {Λi} of events, suppose\nM{Λi|A} < 0.5, i = 1, 2, · · · It follows from (2.45) that\nM{Λi|A} = M{Λi ∩A}\nM{A}\n,\ni = 1, 2, · · ·\n\n\nSection 2.5 - Conditional Uncertain Measure\n25\nBy using (2.46) and the subadditivity axiom, we get\nM\n( ∞\n[\ni=1\nΛi | A\n)\n≤\nM\n( ∞\n[\ni=1\nΛi ∩A\n)\nM{A}\n≤\n∞\nX\ni=1\nM{Λi ∩A}\nM{A}\n=\n∞\nX\ni=1\nM{Λi|A}.\nCase 2: Suppose there is one term greater than or equal to 0.5, say\nM{Λ1|A} ≥0.5,\nM{Λi|A} < 0.5,\ni = 2, 3, · · ·\nand\nM\n( ∞\n[\ni=1\nΛi | A\n)\n≤0.5.\nIn this case, we immediately have\nM\n( ∞\n[\ni=1\nΛi | A\n)\n≤0.5 ≤\n∞\nX\ni=1\nM{Λi|A}.\nCase 3: Suppose there is one term greater than or equal to 0.5, say\nM{Λ1|A} ≥0.5,\nM{Λi|A} < 0.5,\ni = 2, 3, · · ·\nand\nM\n( ∞\n[\ni=1\nΛi | A\n)\n> 0.5.\nIt follows from (2.45) and (2.46) that\nM{Λ1|A} ≥1 −M{Λc\n1 ∩A}\nM{A}\n,\n(2.47)\nM{Λi|A} = M{Λi ∩A}\nM{A}\n,\ni = 2, 3, · · ·\n(2.48)\nM\n( ∞\n[\ni=1\nΛi | A\n)\n= 1 −\nM\n( ∞\n\\\ni=1\nΛc\ni ∩A\n)\nM{A}\n.\n(2.49)\nSince\nΛc\n1 ∩A ⊂\n ∞\n\\\ni=1\nΛc\ni ∩A\n!\n∪\n ∞\n[\ni=2\nΛi ∩A\n!\n,\nwe have\nM{Λc\n1 ∩A} ≤M\n( ∞\n\\\ni=1\nΛc\ni ∩A\n)\n+\n∞\nX\ni=2\nM{Λi ∩A}\n\n\n26\nChapter 2 - Uncertain Measure\ndue to the subadditivity axiom of uncertain measure. That is,\nM\n( ∞\n\\\ni=1\nΛc\ni ∩A\n)\n≥M{Λc\n1 ∩A} −\n∞\nX\ni=2\nM{Λi ∩A}.\n(2.50)\nIt follows from (2.47), (2.48), (2.49) and (2.50) that\nM\n( ∞\n[\ni=1\nΛi | A\n)\n= 1 −\nM\n( ∞\n\\\ni=1\nΛc\ni ∩A\n)\nM{A}\n≤1 −M{Λc\n1 ∩A}\nM{A}\n+\n∞\nX\ni=2\nM{Λi ∩A}\nM{A}\n≤\n∞\nX\ni=1\nM{Λi|A}.\nCase 4: Suppose that there are at least two terms greater than or equal to\n0.5, say\nM{Λ1|A} ≥0.5,\nM{Λ2|A} ≥0.5.\nIn this case, we immediately have\nM\n( ∞\n[\ni=1\nΛi | A\n)\n≤1 ≤\n∞\nX\ni=1\nM{Λi|A}.\nThus the conditional uncertain measure satisﬁes the subadditivity axiom.\nThe theorem is proved.\n2.6\nBibliographic Notes\nIn order to model uncertain phenomena, uncertainty theory was founded by\nLiu [113] in 2007 and perfected by Liu [116] in 2009. The core of uncertainty\ntheory is uncertain measure deﬁned by the normality axiom, duality axiom,\nsubadditivity axiom, and product axiom.\nIn practice, uncertain measure\nis interpreted as the personal belief degree of an uncertain event that may\nhappen.\nNowadays, uncertain measure was well developed and became a\nrigorous footstone of uncertainty theory.\n\n\nChapter 3\nUncertain Variable\nUncertain variable is a fundamental concept in uncertainty theory. It is used\nto represent quantities with uncertainty (e.g., stock price, market demand,\nand product lifetime). The emphasis in this chapter is mainly on uncertain\nvariable, uncertainty distribution, independence, operational law, expected\nvalue, variance, moments, distance, entropy, uncertain sequence, and uncer-\ntain vector.\n3.1\nUncertain Variable\nRoughly speaking, an uncertain variable is a measurable function on an un-\ncertainty space. A formal deﬁnition is given as follows.\nDeﬁnition 3.1 (Liu [113]) An uncertain variable is a function ξ from an\nuncertainty space (Γ, L, M) to the set of real numbers such that {ξ ∈B} is\nan event for any Borel set B of real numbers.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΓ\nℜ\nξ(γ)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.1: An Uncertain Variable\n\n\n28\nChapter 3 - Uncertain Variable\nRemark 3.1: Note that the event {ξ ∈B} is a subset of the universal set\nΓ, i.e.,\n{ξ ∈B} = {γ ∈Γ | ξ(γ) ∈B}.\n(3.1)\nExample 3.1: Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with power\nset and M{γ1} = 0.6, M{γ2} = 0.4. Then\nξ(γ) =\n(\n0,\nif γ = γ1\n1,\nif γ = γ2\n(3.2)\nis an uncertain variable. Furthermore, we have\nM{ξ = 0} = M{γ | ξ(γ) = 0} = M{γ1} = 0.6,\n(3.3)\nM{ξ = 1} = M{γ | ξ(γ) = 1} = M{γ2} = 0.4.\n(3.4)\nExample 3.2: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nξ(γ) = 3γ,\n∀γ ∈Γ\n(3.5)\nis an uncertain variable. Furthermore, we have\nM{ξ = 1} = M{γ | ξ(γ) = 1} = M{1/3} = 0,\n(3.6)\nM{ξ ∈[1, 2]} = M{γ | ξ(γ) ∈[1, 2]} = M{[1/3, 2/3]} = 1/3,\n(3.7)\nM{ξ > 1} = M{γ | ξ(γ) > 1} = M{(1/3, 1)} = 2/3.\n(3.8)\nExample 3.3:\nA real number c may be regarded as a special uncertain\nvariable. In fact, it is the constant function\nξ(γ) ≡c\n(3.9)\non the uncertainty space (Γ, L, M). Furthermore, for any Borel set B of real\nnumbers, we have\nM{ξ ∈B} = M{γ | ξ(γ) ∈B} = M{Γ} = 1,\nif c ∈B,\n(3.10)\nM{ξ ∈B} = M{γ | ξ(γ) ∈B} = M{∅} = 0,\nif c ̸∈B.\n(3.11)\nExample 3.4: Let ξ be an uncertain variable, and let ℜbe the set of real\nnumbers. Then\n{ξ ∈ℜ} = {γ | ξ(γ) ∈ℜ} = Γ.\nThus\nM{ξ ∈ℜ} ≡1.\n(3.12)\n\n\nSection 3.2 - Uncertainty Distribution\n29\nExample 3.5: Let ξ be an uncertain variable and let b be a real number.\nThen\n{ξ = b}c = {γ | ξ(γ) = b}c = {γ | ξ(γ) ̸= b} = {ξ ̸= b}.\nThus {ξ = b} and {ξ ̸= b} are opposite events. Furthermore, by the duality\naxiom, we obtain\nM{ξ = b} + M{ξ ̸= b} = 1.\n(3.13)\nExercise 3.1: Let ξ be an uncertain variable and let B be a Borel set of\nreal numbers. Show that {ξ ∈B} and {ξ ∈Bc} are opposite events, and\nM{ξ ∈B} + M{ξ ∈Bc} = 1.\n(3.14)\nExercise 3.2: Let ξ and η be two uncertain variables. Show that {ξ ≥η}\nand {ξ < η} are opposite events, and\nM{ξ ≥η} + M{ξ < η} = 1.\n(3.15)\nDeﬁnition 3.2 An uncertain variable ξ on the uncertainty space (Γ, L, M) is\nsaid to be (a) nonnegative if M{ξ < 0} = 0; and (b) positive if M{ξ ≤0} = 0.\nDeﬁnition 3.3 Let ξ and η be uncertain variables deﬁned on the uncertainty\nspace (Γ, L, M). We say ξ = η if ξ(γ) = η(γ) for almost all γ ∈Γ.\nDeﬁnition 3.4 Let ξ1, ξ2, · · · , ξn be uncertain variables, and let f be a real-\nvalued measurable function. Then ξ = f(ξ1, ξ2, · · · , ξn) is an uncertain vari-\nable deﬁned by\nξ(γ) = f(ξ1(γ), ξ2(γ), · · · , ξn(γ)),\n∀γ ∈Γ.\n(3.16)\nExample 3.6: Let ξ1 and ξ2 be two uncertain variables. Then the sum\nξ = ξ1 + ξ2 is an uncertain variable deﬁned by\nξ(γ) = ξ1(γ) + ξ2(γ),\n∀γ ∈Γ.\nThe multiplication ξ = ξ1ξ2 is also an uncertain variable deﬁned by\nξ(γ) = ξ1(γ) · ξ2(γ),\n∀γ ∈Γ.\nThe reader may wonder whether ξ(γ) deﬁned by (3.16) is an uncertain\nvariable. The following theorem answers this question.\nTheorem 3.1 Let ξ1, ξ2, · · · , ξn be uncertain variables, and let f be a real-\nvalued measurable function. Then f(ξ1, ξ2, · · · , ξn) is an uncertain variable.\nProof: Since ξ1, ξ2, · · · , ξn are uncertain variables, they are measurable func-\ntions from an uncertainty space (Γ, L, M) to the set of real numbers. Thus\nf(ξ1, ξ2, · · · , ξn) is also a measurable function from the uncertainty space\n(Γ, L, M) to the set of real numbers. Hence f(ξ1, ξ2, · · · , ξn) is an uncertain\nvariable.\n\n\n30\nChapter 3 - Uncertain Variable\n3.2\nUncertainty Distribution\nThis section introduces a concept of uncertainty distribution in order to de-\nscribe uncertain variables. Mention that uncertainty distribution is a carrier\nof incomplete information of uncertain variable. However, in many cases, it\nis suﬃcient to know the uncertainty distribution rather than the uncertain\nvariable itself.\nDeﬁnition 3.5 (Liu [113]) The uncertainty distribution Φ of an uncertain\nvariable ξ is deﬁned by\nΦ(x) = M {ξ ≤x}\n(3.17)\nfor any real number x.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................................................................\nFigure 3.2: An Uncertainty Distribution\nExercise 3.3: A real number c is a special uncertain variable ξ(γ) ≡c.\nShow that such an uncertain variable has an uncertainty distribution\nΦ(x) =\n(\n0,\nif x < c\n1,\nif x ≥c.\nExercise 3.4: Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with power\nset and M{γ1} = 0.7, M{γ2} = 0.3. Show that the uncertain variable\nξ(γ) =\n\u001a\n0,\nif γ = γ1\n1,\nif γ = γ2\nhas an uncertainty distribution\nΦ(x) =\n\n\n\n\n\n0,\nif x < 0\n0.7,\nif 0 ≤x < 1\n1,\nif x ≥1.\n\n\nSection 3.2 - Uncertainty Distribution\n31\nExercise 3.5: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3} with\npower set and M{γ1} = 0.6, M{γ2} = 0.3, M{γ3} = 0.2. Show that the\nuncertain variable\nξ(γ) =\n\n\n\n1,\nif γ = γ1\n2,\nif γ = γ2\n3,\nif γ = γ3\nhas an uncertainty distribution\nΦ(x) =\n\n\n\n\n\n\n\n\n\n0,\nif x < 1\n0.6,\nif 1 ≤x < 2\n0.8,\nif 2 ≤x < 3\n1,\nif x ≥3.\nExercise 3.6: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. (i) Show that the uncertain variable\nξ(γ) = γ,\n∀γ ∈(0, 1)\n(3.18)\nhas an uncertainty distribution\nΦ(x) =\n\n\n\n\n\n0,\nif x ≤0\nx,\nif 0 < x < 1\n1,\nif x ≥1.\n(3.19)\n(ii) What is the uncertainty distribution of ξ(γ) = 1−γ? (iii) What do those\ntwo uncertain variables make you think about? (iv) Design a third uncertain\nvariable whose uncertainty distribution is also (3.19).\nExercise 3.7: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Show that the uncertain variable ξ(γ) = γ2\nhas an uncertainty distribution\nΦ(x) =\n\n\n\n\n\n\n\n0,\nif x ≤0\n√x,\nif 0 < x < 1\n1,\nif x ≥1.\n(3.20)\nExercise 3.8: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure.\nWhat is the uncertainty distribution of\nξ(γ) = 1/γ?\nExercise 3.9: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure.\nWhat is the uncertainty distribution of\nξ(γ) = ln γ?\n\n\n32\nChapter 3 - Uncertain Variable\nExercise 3.10: Let ξ be an uncertain variable with uncertainty distribution\nΦ, and let a and b be real numbers with a > 0. Show that aξ + b has an\nuncertainty distribution\nΨ(x) = Φ\n\u0012x −b\na\n\u0013\n,\n∀x ∈ℜ.\n(3.21)\nExercise 3.11: Let ξ be an uncertain variable with continuous uncertainty\ndistribution Φ, and let a and b be real numbers with a < 0. Show that aξ +b\nhas an uncertainty distribution\nΨ(x) = 1 −Φ\n\u0012x −b\na\n\u0013\n,\n∀x ∈ℜ.\n(3.22)\nExercise 3.12: Let ξ be an uncertain variable with uncertainty distribution\nΦ. Show that exp(ξ) has an uncertainty distribution\nΨ(x) = Φ(ln(x)),\n∀x > 0.\n(3.23)\nExercise 3.13: Let ξ be a positive uncertain variable with continuous un-\ncertainty distribution Φ. Show that 1/ξ has an uncertainty distribution\nΨ(x) = 1 −Φ\n\u0012 1\nx\n\u0013\n,\n∀x > 0.\n(3.24)\nExercise 3.14: Let ξ be an uncertain variable with uncertainty distribution\nΦ, and let f be a continuous and strictly increasing function. Show that f(ξ)\nhas an uncertainty distribution\nΨ(x) = Φ(f −1(x)),\n∀x ∈ℜ.\n(3.25)\nExercise 3.15: Let ξ be an uncertain variable with continuous uncertainty\ndistribution Φ, and let f be a continuous and strictly decreasing function.\nShow that f(ξ) has an uncertainty distribution\nΨ(x) = 1 −Φ(f −1(x)),\n∀x ∈ℜ.\n(3.26)\nDeﬁnition 3.6 Uncertain variables are said to be identically distributed if\nthey have the same uncertainty distribution.\nIt is clear that uncertain variables ξ and η are identically distributed if\nξ = η. However, identical distribution does not imply ξ = η. For example,\nlet (Γ, L, M) be {γ1, γ2} with power set and M{γ1} = M{γ2} = 0.5. Deﬁne\nξ(γ) =\n(\n1,\nif γ = γ1\n−1,\nif γ = γ2,\nη(γ) =\n(\n−1,\nif γ = γ1\n1,\nif γ = γ2.\n\n\nSection 3.2 - Uncertainty Distribution\n33\nThen ξ and η have the same uncertainty distribution,\nΦ(x) =\n\n\n\n\n\n0,\nif x < −1\n0.5,\nif −1 ≤x < 1\n1,\nif x ≥1.\nThus the two uncertain variables ξ and η are identically distributed but ξ ̸= η.\nMeasure Inversion Theorem\nTheorem 3.2 (Liu [120], Measure Inversion Theorem) Let ξ be an uncertain\nvariable with uncertainty distribution Φ. Then for any real number x, we have\nM{ξ ≤x} = Φ(x),\nM{ξ > x} = 1 −Φ(x).\n(3.27)\nProof: The equation M{ξ ≤x} = Φ(x) follows from the deﬁnition of uncer-\ntainty distribution immediately. By using the duality of uncertain measure,\nwe get\nM{ξ > x} = 1 −M{ξ ≤x} = 1 −Φ(x).\nThe theorem is veriﬁed.\nTheorem 3.3 Let ξ be an uncertain variable with uncertainty distribution\nΦ. Then for any real number x, we have\nlim\ny↑x Φ(y) ≤M{ξ < x} ≤Φ(x),\n(3.28)\n1 −Φ(x) ≤M{ξ ≥x} ≤1 −lim\ny↑x Φ(y).\n(3.29)\nIf Φ is continuous at x, then\nM{ξ < x} = Φ(x),\nM{ξ ≥x} = 1 −Φ(x).\n(3.30)\nProof:\nOn the one hand, it follows from the monotonicity theorem and\nmeasure inversion theorem that\nM{ξ < x} ≤M{ξ ≤x} = Φ(x).\nOn the other hand, for any small number ε > 0, we have\nM{ξ < x} ≥M{ξ ≤x −ε} = Φ(x −ε).\nLetting ε →0, we obtain\nM{ξ < x} ≥lim\ny↑x Φ(y).\nThus (3.28) is proved. Similarly, on the one hand, we have\nM{ξ ≥x} ≥M{ξ > x} = 1 −Φ(x).\n\n\n34\nChapter 3 - Uncertain Variable\nOn the other hand, for any small number ε > 0, we have\nM{ξ ≥x} ≤M{ξ > x −ε} = 1 −Φ(x −ε).\nLetting ε →0, we obtain\nM{ξ ≥x} ≤1 −lim\ny↑x Φ(y).\nThus (3.29) is proved. When Φ is continuous at x, we immediately obtain\nlim\ny↑x Φ(y) = Φ(x),\nand (3.30) is thus proved.\nRemark 3.2: Generally speaking, it is an impossible to get the exact value\nof M{a < ξ ≤b} (except a = −∞or b = +∞) if only an uncertainty\ndistribution is available. However, the lower bound is given by the following\ntheorem.\nTheorem 3.4 Let ξ be an uncertain variable with uncertainty distribution\nΦ. Then for any real numbers a and b with a < b, we have\nM{a < ξ ≤b} ≥Φ(b) −Φ(a).\n(3.31)\nProof:\nSince {ξ ≤b} = {ξ ≤a} ∪{a < ξ ≤b}, it follows from the\nsubadditivity axiom that\nM{ξ ≤b} ≤M{ξ ≤a} + M{a < ξ ≤b}.\nThat is,\nΦ(b) ≤Φ(a) + M{a < ξ ≤b}.\nThe inequality is proved.\nRemark 3.3: It is inappropriate to regard the derivative Φ′(x) as an un-\ncertainty density function because uncertain measure is not additive, i.e.,\ngenerally speaking,\nM{a < ξ ≤b} ̸=\nZ b\na\nΦ′(x)dx.\n(3.32)\nSuﬃcient and Necessary Condition\nTheorem 3.5 (Peng-Iwamura [186] and Liu-Lio [140], Suﬃcient and Nec-\nessary Condition) A real-valued function Φ(x) on ℜis an uncertainty distri-\nbution if and only if it is a monotone increasing function satisfying\n0 ≤Φ(x) ≤1,\n(3.33)\nΦ(x) ̸≡0,\n(3.34)\nΦ(x) ̸≡1,\n(3.35)\nΦ(x0) = 1 if Φ(x) = 1 for any x > x0.\n(3.36)\n\n\nSection 3.2 - Uncertainty Distribution\n35\nProof: Suppose Φ is an uncertainty distribution of some uncertain variable\nξ. For any points x1 and x2 with x1 < x2, by using the monotonicity theorem,\nwe have\nΦ(x1) = M{ξ ≤x1} ≤M{ξ ≤x2} = Φ(x2).\nThus Φ is a monotone increasing function. For any point x, by using Theo-\nrem 2.3, we have\n0 ≤M{ξ ≤x} ≤1.\nThus 0 ≤Φ(x) ≤1 is veriﬁed. By using the measure inversion theorem and\nsubadditivity axiom, we have\n1 = M{ξ ∈ℜ} = M\n( ∞\n[\nn=1\n(ξ ≤n)\n)\n≤\n∞\nX\nn=1\nM{ξ ≤n} =\n∞\nX\nn=1\nΦ(n).\nThus Φ(x) ̸≡0. Similarly, we have\n1 = M{ξ ∈ℜ} = M\n( ∞\n[\nn=1\n(ξ > −n)\n)\n≤\n∞\nX\nn=1\nM{ξ > −n} =\n∞\nX\nn=1\n(1 −Φ(−n)).\nThus Φ(x) ̸≡1.\nFurthermore, let x0 be a given point.\nIf Φ(x) = 1 for\nany x > x0, then by using the measure inversion theorem and subadditivity\naxiom, we obtain\n1 −Φ(x0) = M{ξ > x0}\n= M\n( ∞\n[\ni=1\n\u0012\nξ > x0 + 1\ni\n\u0013)\n≤\n∞\nX\ni=1\nM\n\u001a\nξ > x0 + 1\ni\n\u001b\n=\n∞\nX\ni=1\n\u0012\n1 −Φ\n\u0012\nx0 + 1\ni\n\u0013\u0013\n= 0.\nThus Φ(x0) = 1 and (3.36) is veriﬁed.\nConversely, suppose that Φ is a monotone increasing function satisfying\n(3.33) to (3.36).\nWe will prove that there is an uncertain variable whose\nuncertainty distribution is just Φ. Let C be a collection of all intervals of the\nform (−∞, a], (b, ∞), ∅and ℜ. We deﬁne a set function on C as follows,\nM{(−∞, a]} = Φ(a),\nM{(b, +∞)} = 1 −Φ(b),\nM{∅} = 0,\nM{ℜ} = 1.\n\n\n36\nChapter 3 - Uncertain Variable\nFor any Borel set B of real numbers, there exists a sequence {Ai} in C that\ncovers B, i.e.,\nB ⊂\n∞\n[\ni=1\nAi.\nNote that such a sequence is not unique. At ﬁrst, let us prove an inequality,\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} +\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≥1\n(3.37)\nwhere Ai ∈C, i = 1, 2, · · · For any covers {A′\ni} and {A′′\ni } of B and Bc,\nrespectively, i.e.,\nB ⊂\n∞\n[\ni=1\nA′\ni,\nBc ⊂\n∞\n[\ni=1\nA′′\ni ,\nwhere A′\ni, A′′\ni ∈C, i = 1, 2, · · · , it is clear that\nℜ⊂\n ∞\n[\ni=1\nA′\ni\n!\n∪\n ∞\n[\ni=1\nA′′\ni\n!\n,\nand\ninf\nℜ⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≤\n∞\nX\ni=1\nM{A′\ni} +\n∞\nX\ni=1\nM{A′′\ni }.\nIt follows from the arbitrariness of covers {A′\ni} and {A′′\ni } that\ninf\nℜ⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≤\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} +\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}.\nHence, in order to prove the inequality (3.37), it is suﬃcient to prove that\nfor any cover {Ai} of ℜ, we have\n∞\nX\ni=1\nM{Ai} ≥1.\n(3.38)\nWithout loss of generality, we suppose\nA2j−1 = (−∞, aj],\nA2j = (bj, +∞)\nwhere aj and bj are allowed to take values +∞or −∞for j = 1, 2, · · · It is\nclear that\nℜ⊂\n∞\n[\ni=1\nAi =\n\n\n∞\n[\nj=1\n(−∞, aj]\n\n∪\n\n\n∞\n[\nj=1\n(bj, +∞)\n\n\n(3.39)\n\n\nSection 3.2 - Uncertainty Distribution\n37\nand\nmax\nj\naj ≥min\nj\nbj.\nThe argument may break down into ﬁve cases. Case 1: Assume there exists\nan index j such that aj = +∞or bj = −∞, i.e., A2j−1 = ℜor A2j = ℜ.\nThen\n∞\nX\ni=1\nM{Ai} ≥M{ℜ} = 1.\nCase 2: Assume aj < +∞, j = 1, 2, · · · , but\nmax\nj\naj = +∞.\nIt follows from (3.34) that\n∞\nX\ni=1\nM{Ai} ≥\n∞\nX\nj=1\nM{(−∞, aj]} =\n∞\nX\nj=1\nΦ(aj) = +∞> 1.\nCase 3: Assume bj > −∞, j = 1, 2, · · · , but\nmin\nj\nbj = −∞.\nIt follows from (3.35) that\n∞\nX\ni=1\nM{Ai} ≥\n∞\nX\nj=1\nM{(bj, +∞)} =\n∞\nX\nj=1\n(1 −Φ(bj)) = +∞> 1.\nCase 4: Assume there exists a number c such that\n+∞> max\nj\naj > c > min\nj\nbj > −∞.\nSince Φ(x) is a monotone increasing function, we have\n∞\nX\ni=1\nM{Ai} =\n∞\nX\nj=1\nM{(−∞, aj]} +\n∞\nX\nj=1\nM{(bj, +∞)}\n≥Φ(c) + (1 −Φ(c)) = 1.\nCase 5: Assume there exists a number c such that\n+∞> max\nj\naj = c = min\nj\nbj > −∞.\nIt follows from (3.39) that there exists an index j such that aj = c.\nIf\nΦ(c) = 1, then\n∞\nX\ni=1\nM{Ai} ≥M{(−∞, aj]} = Φ(c) = 1.\n\n\n38\nChapter 3 - Uncertain Variable\nIf there exists another index k such that bk = c, then\n∞\nX\ni=1\nM{Ai} ≥M{(−∞, aj]} + M{(bk, +∞)}\n= Φ(c) + (1 −Φ(c)) = 1.\nIf Φ(c) < 1 and bj > c, j = 1, 2, · · · , then by using (3.36), there exists a small\npositive number ε such that Φ(c + ε) < 1, and there exist inﬁnitely many\nindexes j’s such that\nbj < c + ε,\nΦ(bj) ≤Φ(c + ε) < 1.\nThus\n∞\nX\ni=1\nM{Ai} ≥\n∞\nX\nj=1\nM{(bj, +∞)} =\n∞\nX\nj=1\n(1 −Φ(bj)) = +∞> 1.\nTherefore, the inequality (3.37) holds. From this inequality, we immediately\nhave\n1 −\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≤\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}.\nFor any Borel set B of real numbers, since M{B} must lie on the interval\n\n\n1 −\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai},\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}\n\n\n,\nit follows from the maximum uncertainty principle that we may deﬁne\nM{B} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai},\nif\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5\n1 −\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai},\nif\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5\n0.5,\notherwise\n(3.40)\nwhere Ai ∈C, i = 1, 2, · · · Let us verify that M meets the normality, duality\nand subadditivity axioms.\nStep 1: Let us prove that M deﬁned by (3.40) meets the normality axiom.\nSince ℜc = ∅and\ninf\n∅⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} = 0 < 0.5,\n\n\nSection 3.2 - Uncertainty Distribution\n39\nit follows from (3.40) that\nM{ℜ} = 1 −\ninf\n∅⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} = 1 −0 = 1.\nTherefore, M meets the normality axiom.\nStep 2: Let us prove M meets the duality axiom, i.e., M{B}+M{Bc} =\n1. By using the inequality (3.37), the argument may break down into three\ncases. Case 1: Assume\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5.\nIt follows from (3.40) that\nM{B} =\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai},\nM{Bc} = 1 −\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}.\nThus M{B} + M{Bc} = 1. Case 2: Assume\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5.\nIt follows from (3.40) that\nM{B} = 1 −\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai},\nM{Bc} =\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}.\nThus M{B} + M{Bc} = 1. Case 3: Assume\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≥0.5,\ninf\nBc⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≥0.5.\nIt follows from (3.40) that\nM{B} = 0.5,\nM{Bc} = 0.5.\n\n\n40\nChapter 3 - Uncertain Variable\nThus M{B} + M{Bc} = 1. Therefore, M meets the duality axiom.\nStep 3: Let us prove that M is an increasing set function. Suppose B1\nand B2 are two Borel sets of real numbers with B1 ⊂B2. The argument\nbreaks down into three cases. Case 1: Assume M{B2} < 0.5. It follows from\n(3.40) that\nM{B2} =\ninf\nB2⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5.\nSince\ninf\nB1⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≤\ninf\nB2⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5,\nwe have\nM{B1} =\ninf\nB1⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}\nby using (3.40) again. Thus M{B1} ≤M{B2}. Case 2: Assume M{B1} >\n0.5. By using the duality, we have M{Bc\n1} < 0.5. It follows from (3.40) that\nM{Bc\n1} =\ninf\nBc\n1⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5.\nNoting that Bc\n2 ⊂Bc\n1, we get\ninf\nBc\n2⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} ≤\ninf\nBc\n1⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai} < 0.5.\nBy using (3.40) again, we obtain\nM{Bc\n2} =\ninf\nBc\n2⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}.\nThus M{Bc\n1} ≥M{Bc\n2}. The duality implies that M{B1} ≤M{B2}. Case 3:\nAssume M{B1} ≤0.5 and M{B2} ≥0.5. In this case, we immediately have\nM{B1} ≤M{B2}. Therefore, M is an increasing set function.\nStep 4: Let us prove M meets the subadditivity axiom. Suppose {Bj}\nis a sequence of Borel sets of real numbers, and write\nB =\n∞\n[\nj=1\nBj.\n(3.41)\n\n\nSection 3.2 - Uncertainty Distribution\n41\nThe argument breaks down into four cases. Case 1: Assume M{Bj} < 0.5,\nj = 1, 2, · · · On the one hand, it follows from (3.40) that\nM{Bj} =\ninf\nBj⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai},\nj = 1, 2, · · ·\nFor any given ε > 0 and each j, there exists a sequence {Aj\ni} in C such that\nBj ⊂\n∞\n[\ni=1\nAj\ni\nand\n∞\nX\ni=1\nM{Aj\ni} ≤M{Bj} + ε\n2j .\nOn the other hand, it follows from (3.40) and (3.37) that we always have\nM{B} ≤\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}.\nSince {Aj\ni | i, j = 1, 2, · · · } is a cover of B, we get\nM{B} ≤\ninf\nB⊂\n∞\nS\ni=1\nAi\n∞\nX\ni=1\nM{Ai}\n≤\n∞\nX\nj=1\n∞\nX\ni=1\nM{Aj\ni}\n≤\n∞\nX\nj=1\n\u0010\nM{Bj} + ε\n2j\n\u0011\n=\n∞\nX\nj=1\nM{Bj} + ε.\nLetting ε →0, we obtain\nM{B} ≤\n∞\nX\nj=1\nM{Bj}.\nCase 2: Assume there is only one term greater than or equal to 0.5, say\nM{B1} ≥0.5,\nM{Bj} < 0.5,\nj = 2, 3, · · ·\nand\nM{B} ≤0.5.\n\n\n42\nChapter 3 - Uncertain Variable\nIn this case, we immediately have\nM{B} ≤0.5 ≤\n∞\nX\nj=1\nM{Bj}.\nCase 3: Assume there is only one term greater than or equal to 0.5, say\nM{B1} ≥0.5,\nM{Bj} < 0.5,\nj = 2, 3, · · ·\nand\nM{B} > 0.5.\nIt follows from the duality of M that\nM{Bc} = 1 −M{B} < 0.5.\nSince\nBc\n1 ⊂\n\n\n∞\n\\\nj=1\nBc\nj\n\n∪\n\n\n∞\n[\nj=2\nBj\n\n= Bc ∪\n\n\n∞\n[\nj=2\nBj\n\n\nand M is an increasing set function, we have\nM{Bc\n1} ≤M\n\n\nBc ∪\n\n\n∞\n[\nj=2\nBj\n\n\n\n\n.\nIt follows from Case 1 that\nM\n\n\nBc ∪\n\n\n∞\n[\nj=2\nBj\n\n\n\n\n≤M{Bc} +\n∞\nX\nj=2\nM{Bj}.\nThus\nM{Bc\n1} ≤M{Bc} +\n∞\nX\nj=2\nM{Bj}.\nIt follows from the duality of M that\n1 −M{B1} ≤1 −M{B} +\n∞\nX\nj=2\nM{Bj}.\nThat is,\nM{B} ≤\n∞\nX\nj=1\nM{Bj}.\nCase 4: Assume there are at least two terms greater than or equal to 0.5, say\nM{B1} ≥0.5,\nM{B2} ≥0.5.\n\n\nSection 3.2 - Uncertainty Distribution\n43\nIn this case, we immediately have\nM{B} ≤1 ≤\n∞\nX\nj=1\nM{Bj}.\nThat is, M meets the subadditivity axiom. Therefore, M is an uncertain\nmeasure since it meets the normality, duality and subadditivity axioms.\nLet L be the Borel algebra over ℜ, and let M be the uncertain measure de-\nﬁned by (3.40). Then (ℜ, L, M) is an uncertainty space. Deﬁne an uncertain\nvariable as an identity function\nξ(γ) = γ.\n(3.42)\nThen for any real number x, we have\nM{ξ ≤x} = M{γ ∈ℜ| γ ≤x} = M{(−∞, x]} = Φ(x).\nThat is, the uncertain variable ξ has the uncertainty distribution Φ. The\ntheorem is proved.\nExample 3.7: A “completely unknown number” may be regarded as an\nuncertain variable whose uncertainty distribution is\nΦ(x) ≡0.5.\n(3.43)\nLet C be a collection of all intervals of the form (−∞, a], (b, ∞), ∅and ℜ.\nDeﬁne a set function on C as follows,\nM{(−∞, a]} = 0.5,\nM{(b, +∞)} = 0.5,\nM{∅} = 0,\nM{ℜ} = 1.\nThen the set function M can be extended to the Borel algebra L over ℜby\n(3.40) and has the form\nM{B} =\n\n\n\n\n\n0,\nif B = ∅\n1,\nif B = Γ\n0.5,\notherwise.\n(3.44)\nThen the uncertain variable ξ(γ) = γ on the uncertainty space (ℜ, L, M) has\nthe uncertainty distribution Φ(x) ≡0.5.\nExercise 3.16: (i) Design an uncertain variable whose uncertainty distribu-\ntion is\nΦ(x) ≡0.4.\n(3.45)\n(ii) Design an uncertain variable whose uncertainty distribution is\nΦ(x) ≡0.6.\n(3.46)\n\n\n44\nChapter 3 - Uncertain Variable\nSome Special Uncertainty Distributions\nDeﬁnition 3.7 (Liu [120]) An uncertain variable ξ is called linear if it has\na linear uncertainty distribution\nΦ(x) =\n\n\n\n\n\n\n\n0,\nif x ≤a\nx −a\nb −a ,\nif a < x ≤b\n1,\nif b < x\n(3.47)\ndenoted by L(a, b) where a and b are real numbers with a < b.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\na\nb\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n..........................................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.3: Linear Uncertainty Distribution L(a, b)\nExample 3.8: In practice, some quantities are sometimes only given by\nlower and upper bounds.\nFor example, someone thinks John is neither\nyounger than 24 nor older than 28. Then John’s age is a linear uncertain\nvariable L(24, 28) whose uncertainty distribution is\nΦ(x) =\n\n\n\n\n\n\n\n0,\nif x ≤24\n(x −24)/4,\nif 24 < x ≤28\n1,\nif 28 < x.\n(3.48)\nExample 3.9: Someone thinks James’ height is between 180 and 190 cen-\ntimeters. Then James’ height is a linear uncertain variable L(180, 190) whose\nuncertainty distribution is\nΦ(x) =\n\n\n\n\n\n\n\n0,\nif x ≤180\n(x −180)/10,\nif 180 < x ≤190\n1,\nif 190 < x.\n(3.49)\n\n\nSection 3.2 - Uncertainty Distribution\n45\nDeﬁnition 3.8 (Liu [120]) An uncertain variable ξ is called zigzag if it has\na zigzag uncertainty distribution\nΦ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤a\nx −a\n2(b −a),\nif a < x ≤b\nx + c −2b\n2(c −b) ,\nif b < x ≤c\n1,\nif c < x\n(3.50)\ndenoted by Z(a, b, c) where a, b, c are real numbers with a < b < c.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\na\nc\nb\n0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n..........................................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.............................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.4: Zigzag Uncertainty Distribution Z(a, b, c)\nExample 3.10: If a quantity is only given by median1, lower and upper\nbounds, then it is of zigzag form. For example, someone thinks James’ height\nis between 180 and 190 centimeters, and the median is 187 centimeters. Then\nJames’ height is a zigzag uncertain variable Z(180, 187, 190) whose uncer-\ntainty distribution is\nΦ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤180\n(x −180)/14,\nif 180 < x ≤187\n(x −184)/6,\nif 187 < x ≤190\n1,\nif 190 < x.\n(3.51)\nDeﬁnition 3.9 (Liu [120]) An uncertain variable ξ is called normal if it has\na normal uncertainty distribution\nΦ(x) =\n\u0012\n1 + exp\n\u0012π(e −x)\n√\n3σ\n\u0013\u0013−1\n,\nx ∈ℜ\n(3.52)\n1Let ξ be an uncertain variable with uncertainty distribution Φ(x). The median of ξ is\na point x at which Φ(x) = 0.5. Thus the median may be thought of as the “middle” point.\nThat is, we are 50% sure that the quantity falls into the left side and 50% sure that the\nquantity falls into the right side of the median.\n\n\n46\nChapter 3 - Uncertain Variable\ndenoted by N(e, σ) where e and σ are real numbers with σ > 0. A normal\nuncertainty distribution is called standard if e = 0 and σ = 1.\n.................................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\n0.5\ne\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................................................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................................................................\nFigure 3.5: Normal Uncertainty Distribution N(e, σ)\nRegular Uncertainty Distribution\nDeﬁnition 3.10 (Liu [120]) An uncertainty distribution Φ(x) is said to be\nregular if it is a continuous and strictly increasing function with respect to x\nat which 0 < Φ(x) < 1, and\nlim\nx→−∞Φ(x) = 0,\nlim\nx→+∞Φ(x) = 1.\n(3.53)\nFor example, linear uncertainty distribution, zigzag uncertainty distri-\nbution, and normal uncertainty distribution are all regular. However, the\nuncertainty distribution Φ(x) ≡0.5 is not regular.\n3.3\nInverse Uncertainty Distribution\nIt is clear that a regular uncertainty distribution Φ(x) has an inverse function\non the range of x with 0 < Φ(x) < 1, and the inverse function Φ−1(α) exists\non the open interval (0, 1).\nDeﬁnition 3.11 (Liu [120]) Let ξ be an uncertain variable with regular un-\ncertainty distribution Φ(x). Then the inverse function Φ−1(α) is called the\ninverse uncertainty distribution of ξ.\nExample 3.11: (Liu [120]) The inverse uncertainty distribution of linear\nuncertain variable L(a, b) is\nΦ−1(α) = (1 −α)a + αb.\n(3.54)\n\n\nSection 3.3 - Inverse Uncertainty Distribution\n47\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nα\nΦ−1(α)\n0\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\na\nb\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................................................\nFigure 3.6: Inverse Linear Uncertainty Distribution\nExample 3.12: (Liu [120]) The inverse uncertainty distribution of zigzag\nuncertain variable Z(a, b, c) is\nΦ−1(α) =\n(\n(1 −2α)a + 2αb,\nif α < 0.5\n(2 −2α)b + (2α −1)c,\nif α ≥0.5.\n(3.55)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nα\nΦ−1(α)\n0\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\na\nb\nc\n0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.7: Inverse Zigzag Uncertainty Distribution\nExample 3.13: (Liu [120]) The inverse uncertainty distribution of normal\nuncertain variable N(e, σ) is\nΦ−1(α) = e +\n√\n3σ\nπ\nln\nα\n1 −α.\n(3.56)\nTheorem 3.6 Let ξ be an uncertain variable with inverse uncertainty dis-\ntribution Φ−1(α). Then\nM{ξ ≤c} ≥α\n(3.57)\nif and only if\nΦ−1(α) ≤c\n(3.58)\n\n\n48\nChapter 3 - Uncertain Variable\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nα\nΦ−1(α)\n0\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ne\n0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.8: Inverse Normal Uncertainty Distribution\nwhere α and c are constants with 0 < α < 1.\nProof: It follows from M{ξ ≤c} = Φ(c) that M{ξ ≤c} ≥α if and only if\nΦ(c) ≥α, i.e., Φ−1(α) ≤c. The theorem is thus proved.\nExercise 3.17:\nLet ξ be an uncertain variable with inverse uncertainty\ndistribution Φ−1(α). Show that\nM{ξ ≥c} ≥α\n(3.59)\nif and only if\nΦ−1(1 −α) ≥c\n(3.60)\nwhere α and c are constants with 0 < α < 1.\nTheorem 3.7 A function Φ−1 : (0, 1) →ℜis the inverse uncertainty distri-\nbution of an uncertain variable ξ if and only if it is continuous and\nM{ξ ≤Φ−1(α)} = α\n(3.61)\nfor all α ∈(0, 1).\nProof: If Φ−1 is the inverse uncertainty distribution of ξ, then it is continu-\nous, and its inverse function Φ is just the uncertainty distribution of ξ. Thus\nfor any α ∈(0, 1), we have\nM{ξ ≤Φ−1(α)} = Φ(Φ−1(α)) = α.\nTherefore, the equation (3.61) is veriﬁed. Conversely, if Φ−1 is continuous\nand meets (3.61), then it is strictly increasing with respect to α ∈(0, 1), and\nhas an inverse function Φ. Write x = Φ−1(α). Then α = Φ(x) and\nM{ξ ≤x} = α = Φ(x).\n\n\nSection 3.3 - Inverse Uncertainty Distribution\n49\nThat is, Φ is the uncertainty distribution of ξ.\nHence Φ−1 is its inverse\nuncertainty distribution. The theorem is veriﬁed.\nExercise 3.18:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ, and let a and b be real numbers with a > 0. Show that aξ +b\nhas an inverse uncertainty distribution\nΨ−1(α) = aΦ−1(α) + b.\n(3.62)\nExercise 3.19:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ, and let a and b be real numbers with a < 0. Show that aξ +b\nhas an inverse uncertainty distribution\nΨ−1(α) = aΦ−1(1 −α) + b.\n(3.63)\nExercise 3.20:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ. Show that exp(ξ) has an inverse uncertainty distribution\nΨ−1(α) = exp(Φ−1(α)).\n(3.64)\nExercise 3.21: Let ξ be a positive uncertain variable with regular uncer-\ntainty distribution Φ. Show that the reciprocal 1/ξ has an inverse uncertainty\ndistribution\nΨ−1(α) =\n1\nΦ−1(1 −α).\n(3.65)\nExercise 3.22:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ, and let f be a continuous and strictly increasing function.\nShow that f(ξ) has an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1(α)).\n(3.66)\nExercise 3.23:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ, and let f be a continuous and strictly decreasing function.\nShow that f(ξ) has an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1(1 −α)).\n(3.67)\nExercise 3.24:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ. Show that Φ(ξ) is always a linear uncertain variable L(0, 1)\nwhose inverse uncertainty distribution is\nΨ−1(α) = α.\n(3.68)\n\n\n50\nChapter 3 - Uncertain Variable\nTheorem 3.8 (Liu [124], Suﬃcient and Necessary Condition) A function\nΦ−1 : (0, 1) →ℜis an inverse uncertainty distribution if and only if it is a\ncontinuous and strictly increasing function.\nProof: Suppose Φ−1 is an inverse uncertainty distribution. It follows from\nthe deﬁnition of inverse uncertainty distribution that Φ−1(α) is a continuous\nand strictly increasing function with respect to α ∈(0, 1). Conversely, sup-\npose Φ−1(α) is a continuous and strictly increasing function on (0, 1). Take\nan uncertainty space (Γ, L, M) to be (0, 1) with Borel algebra and Lebesgue\nmeasure. Deﬁne an uncertain variable,\nξ(γ) = Φ−1(γ).\n(3.69)\nThen for each α ∈(0, 1), we have\nM{ξ ≤Φ−1(α)} = M{γ ∈(0, 1) | Φ−1(γ) ≤Φ−1(α)}\n= M{γ ∈(0, 1) | γ ≤α}\n= M{(0, α]}\n= α.\nIt follows from Theorem 3.7 that ξ has the inverse uncertainty distribution\nΦ−1(α). The theorem is veriﬁed.\nExercise 3.25: Construct an uncertain variable with standard normal un-\ncertainty distribution N(0, 1). (Hint: Use (3.69) as a reference.)\n3.4\nIndependence\nThe independence of two uncertain variables means that knowing the value\nof one does not change our estimation of the value of the other2. What un-\ncertain variables meet this condition? A typical case is that they are deﬁned\non diﬀerent uncertainty spaces. Let ξ1(γ1) and ξ2(γ2) be uncertain variables\non the uncertainty spaces (Γ1, L1, M1) and (Γ2, L2, M2), respectively. It is\nclear that they are also uncertain variables on the product uncertainty space\n(Γ1, L1, M1) × (Γ2, L2, M2). Then for any Borel sets B1 and B2 of real num-\n2For example, it is clear that f(γ1, γ2) = γ1 + 1 and g(γ1, γ2) = γ2 + 2 are always\nindependent on the product uncertainty space (Γ1, L1, M1) × (Γ2, L2, M2).\nHowever,\nf(γ1, γ2) = γ1 + 1 and g(γ1, γ2) = γ1 −γ2 are not.\n\n\nSection 3.4 - Independence\n51\nbers, we have\nM{(ξ1 ∈B1) ∩(ξ2 ∈B2)}\n= M {(γ1, γ2) | ξ1(γ1) ∈B1, ξ2(γ2) ∈B2}\n= M {(γ1 | ξ1(γ1) ∈B1) × (γ2 | ξ2(γ2) ∈B2)}\n= M1 {γ1 | ξ1(γ1) ∈B1} ∧M2 {γ2 | ξ2(γ2) ∈B2}\n(product axiom)\n= M1 {ξ1 ∈B1} ∧M2 {ξ2 ∈B2}\n= M {ξ1 ∈B1} ∧M {ξ2 ∈B2} .\nThat is,\nM{(ξ1 ∈B1) ∩(ξ2 ∈B2)} = M {ξ1 ∈B1} ∧M {ξ2 ∈B2} .\n(3.70)\nThus we say two uncertain variables are independent if the equation (3.70)\nholds. Generally, we may deﬁne independence in the following form.\nDeﬁnition 3.12 (Liu [116]) The uncertain variables ξ1, ξ2, · · · , ξn are said\nto be independent if\nM\n( n\n\\\ni=1\n(ξi ∈Bi)\n)\n=\nn\n^\ni=1\nM {ξi ∈Bi}\n(3.71)\nfor any Borel sets B1, B2, · · · , Bn of real numbers.\nExercise 3.26: Show that a constant (a special uncertain variable) is always\nindependent of any uncertain variable.\nExercise 3.27: John gives Tom 2 dollars. Thus John gets “−2 dollars” and\nTom “+2 dollars”. Are their incomes independent? Why?\nExercise 3.28: Let ξ1, ξ2, · · · , ξn be independent uncertain variables. Show\nthat ξi and ξj are independent for any indexes i and j with 1 ≤i < j ≤n.\nExercise 3.29: Construct 100 independent uncertain variables. (Hint: De-\nﬁne them on the product uncertainty space (Γ1, L1, M1)×(Γ2, L2, M2)×· · ·×\n(Γ100, L100, M100).)\nExercise 3.30: Let ξ be an uncertain variable. Are ξ and 1−ξ independent?\nPlease justify your answer.\nExercise 3.31: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3, γ4}\nwith power set and\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\n\n\n52\nChapter 3 - Uncertain Variable\nDeﬁne\nξ(γ) =\n\n\n\n\n\n0,\nif γ = γ1\n1,\nif γ = γ2 or γ3\n2,\nif γ = γ4,\nη(γ) =\n(\n0,\nif γ = γ1 or γ4\n1,\nif γ = γ2 or γ3.\n(i) Show that for any real numbers x and y, we have\nM{ξ ≤x, η ≤y} = M{ξ ≤x} ∧M{η ≤y}.\n(ii) Show that ξ and η are not independent.\nExercise 3.32: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3, γ4}\nwith power set and\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\nDeﬁne\nξ1(γ) =\n(\n0,\nif γ = γ1 or γ2\n1,\nif γ = γ3 or γ4,\nξ2(γ) =\n(\n0,\nif γ = γ1 or γ3\n1,\nif γ = γ2 or γ4,\nξ3(γ) =\n(\n0,\nif γ = γ1 or γ4\n1,\nif γ = γ2 or γ3.\n(i) Show that ξ1, ξ2, ξ3 are pairwise independent (i.e., any two of which are\nindependent). (ii) Show that ξ1, ξ2, ξ3 are not independent.\nTheorem 3.9 (Liu [116]) The uncertain variables ξ1, ξ2, · · · , ξn are inde-\npendent if and only if\nM\n( n\n[\ni=1\n(ξi ∈Bi)\n)\n=\nn\n_\ni=1\nM {ξi ∈Bi}\n(3.72)\nfor any Borel sets B1, B2, · · · , Bn of real numbers.\nProof: If ξ1, ξ2, · · · , ξn are independent, it follows from the duality of un-\ncertain measure that\nM\n( n\n[\ni=1\n(ξi ∈Bi)\n)\n= 1 −M\n( n\n\\\ni=1\n(ξi ∈Bc\ni )\n)\n= 1 −\nn\n^\ni=1\nM{ξi ∈Bc\ni } =\nn\n_\ni=1\nM {ξi ∈Bi} .\n\n\nSection 3.5 - Operational Law\n53\nThus (3.72) holds. Conversely, if (3.72) is assumed, then\nM\n( n\n\\\ni=1\n(ξi ∈Bi)\n)\n= 1 −M\n( n\n[\ni=1\n(ξi ∈Bc\ni )\n)\n= 1 −\nn\n_\ni=1\nM{ξi ∈Bc\ni } =\nn\n^\ni=1\nM {ξi ∈Bi} .\nThus ξ1, ξ2, · · · , ξn are independent.\nTheorem 3.10 Let ξ1, ξ2, · · · , ξn be independent uncertain variables, and\nlet f1, f2, · · · , fn be measurable functions.\nThen f1(ξ1), f2(ξ2), · · · , fn(ξn)\nare independent uncertain variables.\nProof: For any Borel sets B1, B2, · · · , Bn of real numbers, it follows from\nthe deﬁnition of independence that\nM\n( n\n\\\ni=1\n(fi(ξi) ∈Bi)\n)\n= M\n( n\n\\\ni=1\n(ξi ∈f −1\ni\n(Bi))\n)\n=\nn\n^\ni=1\nM{ξi ∈f −1\ni\n(Bi)} =\nn\n^\ni=1\nM{fi(ξi) ∈Bi}.\nThus f1(ξ1), f2(ξ2), · · · , fn(ξn) are independent uncertain variables.\n3.5\nOperational Law\nThis section provides some operational laws for calculating the uncertainty\ndistributions of strictly increasing function, strictly decreasing function, and\nstrictly monotone function of uncertain variables.\nStrictly Increasing Function of Uncertain Variables\nA real-valued function f(x1, x2, · · · , xn) is said to be strictly increasing if\nf(x1, x2, · · · , xn) ≤f(y1, y2, · · · , yn)\n(3.73)\nwhenever xi ≤yi for i = 1, 2, · · · , n, and\nf(x1, x2, · · · , xn) < f(y1, y2, · · · , yn)\n(3.74)\nwhenever xi < yi for i = 1, 2, · · · , n. The following are strictly increasing\nfunctions,\nf(x1, x2, · · · , xn) = x1 ∨x2 ∨· · · ∨xn,\nf(x1, x2, · · · , xn) = x1 ∧x2 ∧· · · ∧xn,\nf(x1, x2, · · · , xn) = x1 + x2 + · · · + xn,\nf(x1, x2, · · · , xn) = x1x2 · · · xn,\nx1, x2, · · · , xn ≥0.\n\n\n54\nChapter 3 - Uncertain Variable\nTheorem 3.11 (Liu [120]) Let ξ1, ξ2, · · · , ξn be independent uncertain vari-\nables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. If f\nis a continuous and strictly increasing function, then\nξ = f(ξ1, ξ2, · · · , ξn)\n(3.75)\nhas an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (α), · · · , Φ−1\nn (α)).\n(3.76)\nProof: For simplicity, we only prove the case of n = 2. It is clear that\nΨ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (α)) is a continuous function with respect to α. On\nthe one hand, let\nγ ∈{ξ1 ≤Φ−1\n1 (α)} ∩{ξ2 ≤Φ−1\n2 (α)}.\nThen\nξ1(γ) ≤Φ−1\n1 (α),\nξ2(γ) ≤Φ−1\n2 (α).\nSince f is a strictly increasing function, we have\nf(ξ1(γ), ξ2(γ)) ≤f(Φ−1\n1 (α), Φ−1\n2 (α)).\nThus\nξ(γ) ≤Ψ−1(α).\nThat is,\nγ ∈{ξ ≤Ψ−1(α)}.\nHence\n{ξ ≤Ψ−1(α)} ⊃{ξ1 ≤Φ−1\n1 (α)} ∩{ξ2 ≤Φ−1\n2 (α)}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ ≤Ψ−1(α)} ≥M{(ξ1 ≤Φ−1\n1 (α)) ∩(ξ2 ≤Φ−1\n2 (α))}\n= M{ξ1 ≤Φ−1\n1 (α)} ∧M{ξ2 ≤Φ−1\n2 (α)}\n= α ∧α = α,\ni.e.,\nM{ξ ≤Ψ−1(α)} ≥α.\n(3.77)\nOn the other hand, let\nγ ∈{ξ ≤Ψ−1(α)}.\nThen\nξ(γ) ≤Ψ−1(α).\nThat is,\nf(ξ1(γ), ξ2(γ)) ≤f(Φ−1\n1 (α), Φ−1\n2 (α)).\n\n\nSection 3.5 - Operational Law\n55\nSince f is a strictly increasing function, we have\nξ1(γ) ≤Φ−1\n1 (α)\nor\nξ2(γ) ≤Φ−1\n2 (α).\nThus\nγ ∈{ξ1 ≤Φ−1\n1 (α)}\nor\nγ ∈{ξ2 ≤Φ−1\n2 (α)}.\nThat is,\nγ ∈{ξ1 ≤Φ−1\n1 (α)} ∪{ξ2 ≤Φ−1\n2 (α)}.\nHence\n{ξ ≤Ψ−1(α)} ⊂{ξ1 ≤Φ−1\n1 (α)} ∪{ξ2 ≤Φ−1\n2 (α)}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ ≤Ψ−1(α)} ≤M{(ξ1 ≤Φ−1\n1 (α)) ∪(ξ2 ≤Φ−1\n2 (α))}\n= M{ξ1 ≤Φ−1\n1 (α)} ∨M{ξ2 ≤Φ−1\n2 (α)}\n= α ∨α = α,\ni.e.,\nM{ξ ≤Ψ−1(α)} ≤α.\n(3.78)\nIt follows from (3.77) and (3.78) that M{ξ ≤Ψ−1(α)} = α. Therefore, Ψ−1\nis just the inverse uncertainty distribution of ξ. The theorem is proved.\nExercise 3.33: Let ξ1, ξ2, · · · , ξn be independent uncertain variables with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. Show that the\nsum\nξ = ξ1 + ξ2 + · · · + ξn\n(3.79)\nhas an inverse uncertainty distribution\nΨ−1(α) = Φ−1\n1 (α) + Φ−1\n2 (α) + · · · + Φ−1\nn (α).\n(3.80)\nExercise 3.34: Let ξ1, ξ2, · · · , ξn be independent positive uncertain variables\nwith regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively.\nShow\nthat the multiplication\nξ = ξ1 × ξ2 × · · · × ξn\n(3.81)\nhas an inverse uncertainty distribution\nΨ−1(α) = Φ−1\n1 (α) × Φ−1\n2 (α) × · · · × Φ−1\nn (α).\n(3.82)\nExercise 3.35: Let ξ1, ξ2, · · · , ξn be independent uncertain variables with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. Show that the\nminimum\nξ = ξ1 ∧ξ2 ∧· · · ∧ξn\n(3.83)\n\n\n56\nChapter 3 - Uncertain Variable\nhas an inverse uncertainty distribution\nΨ−1(α) = Φ−1\n1 (α) ∧Φ−1\n2 (α) ∧· · · ∧Φ−1\nn (α).\n(3.84)\nExercise 3.36: Let ξ1, ξ2, · · · , ξn be independent uncertain variables with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. Show that the\nmaximum\nξ = ξ1 ∨ξ2 ∨· · · ∨ξn\n(3.85)\nhas an inverse uncertainty distribution\nΨ−1(α) = Φ−1\n1 (α) ∨Φ−1\n2 (α) ∨· · · ∨Φ−1\nn (α).\n(3.86)\nExample 3.14: The independence condition in Theorem 3.11 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be (0, 1) with\nBorel algebra and Lebesgue measure. Then ξ1(γ) = γ is a linear uncertain\nvariable with inverse uncertainty distribution\nΦ−1\n1 (α) = α,\n(3.87)\nand ξ2(γ) = 1 −γ is also a linear uncertain variable with inverse uncertainty\ndistribution\nΦ−1\n2 (α) = α.\n(3.88)\nNote that ξ1 and ξ2 are not independent, and ξ1 + ξ2 ≡1 whose inverse\nuncertainty distribution is Ψ−1(α) ≡1. Thus\nΨ−1(α) ̸= Φ−1\n1 (α) + Φ−1\n2 (α).\n(3.89)\nTherefore, the independence condition cannot be removed.\nTheorem 3.12 Assume that ξ1 and ξ2 are independent linear uncertain\nvariables L(a1, b1) and L(a2, b2), respectively. Then the sum ξ1 + ξ2 is also a\nlinear uncertain variable L(a1 + a2, b1 + b2), i.e.,\nL(a1, b1) + L(a2, b2) = L(a1 + a2, b1 + b2).\n(3.90)\nThe multiplication of a linear uncertain variable L(a, b) and a scalar number\nk > 0 is also a linear uncertain variable L(ka, kb), i.e.,\nk · L(a, b) = L(ka, kb).\n(3.91)\nProof:\nAssume that the uncertain variables ξ1 and ξ2 have uncertainty\ndistributions Φ1 and Φ2, respectively. Then\nΦ−1\n1 (α) = (1 −α)a1 + αb1,\nΦ−1\n2 (α) = (1 −α)a2 + αb2.\n\n\nSection 3.5 - Operational Law\n57\nIt follows from the operational law that the inverse uncertainty distribution\nof ξ1 + ξ2 is\nΨ−1(α) = Φ−1\n1 (α) + Φ−1\n2 (α) = (1 −α)(a1 + a2) + α(b1 + b2).\nHence the sum is also a linear uncertain variable L(a1 + a2, b1 + b2). The\nﬁrst part is veriﬁed. Next, suppose that the uncertainty distribution of the\nuncertain variable ξ ∼L(a, b) is Φ. It follows from the operational law that\nwhen k > 0, the inverse uncertainty distribution of kξ is\nΨ−1(α) = kΦ−1(α) = (1 −α)(ka) + α(kb).\nHence kξ is just a linear uncertain variable L(ka, kb).\nExercise 3.37: Show that the multiplication of linear uncertain variables is\nno longer a linear one even they are independent and positive. That is,\nL(a1, b1) × L(a2, b2) ̸= L(a1 × a2, b1 × b2).\n(3.92)\nTheorem 3.13 Assume that ξ1 and ξ2 are independent zigzag uncertain\nvariables Z(a1, b1, c1) and Z(a2, b2, c2), respectively. Then the sum ξ1 + ξ2 is\nalso a zigzag uncertain variable Z(a1 + a2, b1 + b2, c1 + c2), i.e.,\nZ(a1, b1, c1) + Z(a2, b2, c2) = Z(a1 + a2, b1 + b2, c1 + c2).\n(3.93)\nThe multiplication of a zigzag uncertain variable Z(a, b, c) and a scalar num-\nber k > 0 is also a zigzag uncertain variable Z(ka, kb, kc), i.e.,\nk · Z(a, b, c) = Z(ka, kb, kc).\n(3.94)\nProof:\nAssume that the uncertain variables ξ1 and ξ2 have uncertainty\ndistributions Φ1 and Φ2, respectively. Then\nΦ−1\n1 (α) =\n(\n(1 −2α)a1 + 2αb1,\nif α < 0.5\n(2 −2α)b1 + (2α −1)c1,\nif α ≥0.5,\nΦ−1\n2 (α) =\n(\n(1 −2α)a2 + 2αb2,\nif α < 0.5\n(2 −2α)b2 + (2α −1)c2,\nif α ≥0.5.\nIt follows from the operational law that the inverse uncertainty distribution\nof ξ1 + ξ2 is\nΨ−1(α) =\n(\n(1 −2α)(a1 + a2) + 2α(b1 + b2),\nif α < 0.5\n(2 −2α)(b1 + b2) + (2α −1)(c1 + c2),\nif α ≥0.5.\nHence the sum is also a zigzag uncertain variable Z(a1 + a2, b1 + b2, c1 + c2).\nThe ﬁrst part is veriﬁed. Next, suppose that the uncertainty distribution of\n\n\n58\nChapter 3 - Uncertain Variable\nthe uncertain variable ξ ∼Z(a, b, c) is Φ. It follows from the operational law\nthat when k > 0, the inverse uncertainty distribution of kξ is\nΨ−1(α) = kΦ−1(α) =\n(\n(1 −2α)(ka) + 2α(kb),\nif α < 0.5\n(2 −2α)(kb) + (2α −1)(kc),\nif α ≥0.5.\nHence kξ is just a zigzag uncertain variable Z(ka, kb, kc).\nTheorem 3.14 Let ξ1 and ξ2 be independent normal uncertain variables\nN(e1, σ1) and N(e2, σ2), respectively. Then the sum ξ1 + ξ2 is also a normal\nuncertain variable N(e1 + e2, σ1 + σ2), i.e.,\nN(e1, σ1) + N(e2, σ2) = N(e1 + e2, σ1 + σ2).\n(3.95)\nThe multiplication of a normal uncertain variable N(e, σ) and a scalar num-\nber k > 0 is also a normal uncertain variable N(ke, kσ), i.e.,\nk · N(e, σ) = N(ke, kσ).\n(3.96)\nProof:\nAssume that the uncertain variables ξ1 and ξ2 have uncertainty\ndistributions Φ1 and Φ2, respectively. Then\nΦ−1\n1 (α) = e1 +\n√\n3σ1\nπ\nln\nα\n1 −α,\nΦ−1\n2 (α) = e2 +\n√\n3σ2\nπ\nln\nα\n1 −α.\nIt follows from the operational law that the inverse uncertainty distribution\nof ξ1 + ξ2 is\nΨ−1(α) = Φ−1\n1 (α) + Φ−1\n2 (α) = (e1 + e2) +\n√\n3(σ1 + σ2)\nπ\nln\nα\n1 −α.\nHence the sum is also a normal uncertain variable N(e1 + e2, σ1 + σ2). The\nﬁrst part is veriﬁed. Next, suppose that the uncertainty distribution of the\nuncertain variable ξ ∼N(e, σ) is Φ. It follows from the operational law that,\nwhen k > 0, the inverse uncertainty distribution of kξ is\nΨ−1(α) = kΦ−1(α) = (ke) +\n√\n3(kσ)\nπ\nln\nα\n1 −α.\nHence kξ is just a normal uncertain variable N(ke, kσ).\nTheorem 3.15 (Liu [125], Extreme Value Theorem) Let ξ1, ξ2, · · · , ξn be\nindependent uncertain variables with regular uncertainty distributions Φ1, Φ2,\n· · · , Φn, respectively. Then\nSi = ξ1 + ξ2 + · · · + ξi\n(3.97)\n\n\nSection 3.5 - Operational Law\n59\nhave inverse uncertainty distributions\nΨ−1\ni (α) = Φ−1\n1 (α) + Φ−1\n2 (α) + · · · + Φ−1\ni (α),\n(3.98)\ni = 1, 2, · · · , n, respectively, and the maximum\nS = S1 ∨S2 ∨· · · ∨Sn\n(3.99)\nhas an inverse uncertainty distribution\nΥ−1(α) = Ψ−1\n1 (α) ∨Ψ−1\n2 (α) ∨· · · ∨Ψ−1\nn (α).\n(3.100)\nProof: Since f(x1, x2, · · · , xn) = x1 ∨(x1 + x2) ∨· · · ∨(x1 + x2 + · · · + xn)\nis a continuous and strictly increasing function, and\nS = f(ξ1, ξ2, · · · , ξn),\nit follows from Theorem 3.11 that S has an inverse uncertainty distribution\nΥ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (α), · · · , Φ−1\nn (α))\n= max\n1≤i≤n(Φ−1\n1 (α) + Φ−1\n2 (α) + · · · + Φ−1\ni (α))\n= max\n1≤i≤n Ψ−1\ni (α).\nThus (3.100) is veriﬁed.\nTheorem 3.16 (Liu [125], Extreme Value Theorem) Let ξ1, ξ2, · · · , ξn be\nindependent uncertain variables with regular uncertainty distributions Φ1, Φ2,\n· · · , Φn, respectively. Then\nSi = ξ1 + ξ2 + · · · + ξi\n(3.101)\nhave inverse uncertainty distributions\nΨ−1\ni (α) = Φ−1\n1 (α) + Φ−1\n2 (α) + · · · + Φ−1\ni (α),\n(3.102)\ni = 1, 2, · · · , n, respectively, and the minimum\nS = S1 ∧S2 ∧· · · ∧Sn\n(3.103)\nhas an inverse uncertainty distribution\nΥ−1(α) = Ψ−1\n1 (α) ∧Ψ−1\n2 (α) ∧· · · ∧Ψ−1\nn (α).\n(3.104)\nProof: Since f(x1, x2, · · · , xn) = x1 ∧(x1 + x2) ∧· · · ∧(x1 + x2 + · · · + xn)\nis a continuous and strictly increasing function, and\nS = f(ξ1, ξ2, · · · , ξn),\n\n\n60\nChapter 3 - Uncertain Variable\nit follows from Theorem 3.11 that S has an inverse uncertainty distribution\nΥ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (α), · · · , Φ−1\nn (α))\n= min\n1≤i≤n(Φ−1\n1 (α) + Φ−1\n2 (α) + · · · + Φ−1\ni (α))\n= min\n1≤i≤n Ψ−1\ni (α).\nThus (3.104) is veriﬁed.\nStrictly Decreasing Function of Uncertain Variables\nA real-valued function f(x1, x2, · · · , xn) is said to be strictly decreasing if\nf(x1, x2, · · · , xn) ≤f(y1, y2, · · · , yn)\n(3.105)\nwhenever xi ≥yi for i = 1, 2, · · · , n, and\nf(x1, x2, · · · , xn) < f(y1, y2, · · · , yn)\n(3.106)\nwhenever xi > yi for i = 1, 2, · · · , n. If f(x1, x2, · · · , xn) is a strictly increas-\ning function, then\n−f(x1, x2, · · · , xn)\nis a strictly decreasing function, and\n1\nf(x1, x2, · · · , xn)\nis also a strictly decreasing function provided that f is positive.\nTheorem 3.17 (Liu [120]) Let ξ1, ξ2, · · · , ξn be independent uncertain vari-\nables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. If f\nis a continuous and strictly decreasing function, then\nξ = f(ξ1, ξ2, · · · , ξn)\n(3.107)\nhas an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1\n1 (1 −α), Φ−1\n2 (1 −α), · · · , Φ−1\nn (1 −α)).\n(3.108)\nProof: For simplicity, we only prove the case of n = 2. It is clear that\nΨ−1(α) = f(Φ−1\n1 (1 −α), Φ−1\n2 (1 −α)) is a continuous function with respect\nto α. On the one hand, let\nγ ∈{ξ1 ≥Φ−1\n1 (1 −α)} ∩{ξ2 ≥Φ−1\n2 (1 −α)}.\nThen\nξ1(γ) ≥Φ−1\n1 (1 −α),\nξ2(γ) ≥Φ−1\n2 (1 −α).\n\n\nSection 3.5 - Operational Law\n61\nSince f is a strictly decreasing function, we have\nf(ξ1(γ), ξ2(γ)) ≤f(Φ−1\n1 (1 −α), Φ−1\n2 (1 −α)).\nThus\nξ(γ) ≤Ψ−1(α).\nThat is,\nγ ∈{ξ ≤Ψ−1(α)}.\nHence\n{ξ ≤Ψ−1(α)} ⊃{ξ1 ≥Φ−1\n1 (1 −α)} ∩{ξ2 ≥Φ−1\n2 (1 −α)}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ ≤Ψ−1(α)} ≥M{(ξ1 ≥Φ−1\n1 (1 −α)) ∩(ξ2 ≥Φ−1\n2 (1 −α))}\n= M{ξ1 ≥Φ−1\n1 (1 −α)} ∧M{ξ2 ≥Φ−1\n2 (1 −α)}\n= α ∧α = α,\ni.e.,\nM{ξ ≤Ψ−1(α)} ≥α.\n(3.109)\nOn the other hand, let\nγ ∈{ξ ≤Ψ−1(α)}.\nThen\nξ(γ) ≤Ψ−1(α).\nThat is,\nf(ξ1(γ), ξ2(γ)) ≤f(Φ−1\n1 (1 −α), Φ−1\n2 (1 −α)).\nSince f is a strictly decreasing function, we have\nξ1(γ) ≥Φ−1\n1 (1 −α)\nor\nξ2(γ) ≥Φ−1\n2 (1 −α).\nThus\nγ ∈{ξ1 ≥Φ−1\n1 (1 −α)}\nor\nγ ∈{ξ2 ≥Φ−1\n2 (1 −α)}.\nThat is,\nγ ∈{ξ1 ≥Φ−1\n1 (1 −α)} ∪{ξ2 ≥Φ−1\n2 (1 −α)}.\nHence\n{ξ ≤Ψ−1(α)} ⊂{ξ1 ≥Φ−1\n1 (1 −α)} ∪{ξ2 ≥Φ−1\n2 (1 −α)}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ ≤Ψ−1(α)} ≤M{(ξ1 ≥Φ−1\n1 (1 −α)) ∪(ξ2 ≥Φ−1\n2 (1 −α))}\n= M{ξ1 ≥Φ−1\n1 (1 −α)} ∨M{ξ2 ≥Φ−1\n2 (1 −α)}\n= α ∨α = α,\n\n\n62\nChapter 3 - Uncertain Variable\ni.e.,\nM{ξ ≤Ψ−1(α)} ≤α.\n(3.110)\nIt follows from (3.109) and (3.110) that M{ξ ≤Ψ−1(α)} = α. Therefore,\nΨ−1 is just the inverse uncertainty distribution of ξ. The theorem is proved.\nExercise 3.38: Let ξ1 and ξ2 be independent and positive uncertain vari-\nables with regular uncertainty distributions Φ1 and Φ2, respectively. Show\nthat\nξ =\n1\nξ1 + ξ2\n(3.111)\nhas an inverse uncertainty distribution\nΨ−1(α) =\n1\nΦ−1\n1 (1 −α) + Φ−1\n2 (1 −α).\n(3.112)\nExercise 3.39: Let ξ be a normal uncertain variable N(0, 1). Show that −ξ\nis also a normal uncertain variable N(0, 1).\nExercise 3.40:\nShow that the independence condition in Theorem 3.17\ncannot be removed.\nStrictly Monotone Function of Uncertain Variables\nA real-valued function f(x1, x2, · · · , xn) is said to be strictly monotone if it\nis strictly increasing with respect to x1, x2, · · · , xm and strictly decreasing\nwith respect to xm+1, xm+2, · · · , xn, that is,\nf(x1, · · · , xm, xm+1, · · · , xn) ≤f(y1, · · · , ym, ym+1, · · · , yn)\n(3.113)\nwhenever xi ≤yi for i = 1, 2, · · · , m and xi ≥yi for i = m + 1, m + 2, · · · , n,\nand\nf(x1, · · · , xm, xm+1, · · · , xn) < f(y1, · · · , ym, ym+1, · · · , yn)\n(3.114)\nwhenever xi < yi for i = 1, 2, · · · , m and xi > yi for i = m + 1, m + 2, · · · , n.\nThe following are strictly monotone functions,\nf(x1, x2) = x1 −x2,\nf(x1, x2) = x1/x2,\nx1, x2 > 0,\nf(x1, x2) = x1/(x1 + x2),\nx1, x2 > 0.\nNote that both strictly increasing function and strictly decreasing function\nare special cases of strictly monotone function.\nTheorem 3.18 (Liu [120]) Let ξ1, ξ2, · · · , ξn be independent uncertain vari-\nables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. If\n\n\nSection 3.5 - Operational Law\n63\nf(x1, x2, · · · , xn) is continuous, strictly increasing with respect to x1, x2, · · · ,\nxm and strictly decreasing with respect to xm+1, xm+2, · · · , xn, then\nξ = f(ξ1, ξ2, · · · , ξn)\n(3.115)\nhas an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α)).\n(3.116)\nProof: For simplicity, we only prove the case of m = 1 and n = 2. It is clear\nthat Ψ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (1 −α)) is a continuous function with respect\nto α. On the one hand, let\nγ ∈{ξ1 ≤Φ−1\n1 (α)} ∩{ξ2 ≥Φ−1\n2 (1 −α)}.\nThen\nξ1(γ) ≤Φ−1\n1 (α),\nξ2(γ) ≥Φ−1\n2 (1 −α).\nSince f is a strictly monotone function, we have\nf(ξ1(γ), ξ2(γ)) ≤f(Φ−1\n1 (α), Φ−1\n2 (1 −α)).\nThus\nξ(γ) ≤Ψ−1(α).\nThat is,\nγ ∈{ξ ≤Ψ−1(α)}.\nHence\n{ξ ≤Ψ−1(α)} ⊃{ξ1 ≤Φ−1\n1 (α)} ∩{ξ2 ≥Φ−1\n2 (1 −α)}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ ≤Ψ−1(α)} ≥M{(ξ1 ≤Φ−1\n1 (α)) ∩(ξ2 ≥Φ−1\n2 (1 −α))}\n= M{ξ1 ≤Φ−1\n1 (α)} ∧M{ξ2 ≥Φ−1\n2 (1 −α)}\n= α ∧α = α,\ni.e.,\nM{ξ ≤Ψ−1(α)} ≥α.\n(3.117)\nOn the other hand, let\nγ ∈{ξ ≤Ψ−1(α)}.\nThen\nξ(γ) ≤Ψ−1(α).\nThat is,\nf(ξ1(γ), ξ2(γ)) ≤f(Φ−1\n1 (α), Φ−1\n2 (1 −α)).\n\n\n64\nChapter 3 - Uncertain Variable\nSince f is a strictly monotone function, we have\nξ1(γ) ≤Φ−1\n1 (α)\nor\nξ2(γ) ≥Φ−1\n2 (1 −α).\nThus\nγ ∈{ξ1 ≤Φ−1\n1 (α)}\nor\nγ ∈{ξ2 ≥Φ−1\n2 (1 −α)}.\nThat is,\nγ ∈{ξ1 ≤Φ−1\n1 (α)} ∪{ξ2 ≥Φ−1\n2 (1 −α)}.\nHence\n{ξ ≤Ψ−1(α)} ⊂{ξ1 ≤Φ−1\n1 (α)} ∪{ξ2 ≥Φ−1\n2 (1 −α)}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ ≤Ψ−1(α)} ≤M{(ξ1 ≤Φ−1\n1 (α)) ∪(ξ2 ≥Φ−1\n2 (1 −α))}\n= M{ξ1 ≤Φ−1\n1 (α)} ∨M{ξ2 ≥Φ−1\n2 (1 −α)}\n= α ∨α = α,\ni.e.,\nM{ξ ≤Ψ−1(α)} ≤α.\n(3.118)\nIt follows from (3.117) and (3.118) that M{ξ ≤Ψ−1(α)} = α. Therefore,\nΨ−1 is just the inverse uncertainty distribution of ξ. The theorem is proved.\nExercise 3.41: Let ξ1 and ξ2 be independent uncertain variables with regu-\nlar uncertainty distributions Φ1 and Φ2, respectively. Show that the inverse\nuncertainty distribution of the diﬀerence ξ1 −ξ2 is\nΨ−1(α) = Φ−1\n1 (α) −Φ−1\n2 (1 −α).\n(3.119)\nExercise 3.42:\nLet ξ1 and ξ2 be independent linear uncertain variables\nL(a1, b1) and L(a2, b2), respectively. Show that the diﬀerence ξ1 −ξ2 is also\na linear uncertain variable L(a1 −b2, b1 −a2), i.e.,\nL(a1, b1) −L(a2, b2) = L(a1 −b2, b1 −a2).\n(3.120)\nExercise 3.43: Let ξ1 and ξ2 be independent zigzag uncertain variables\nZ(a1, b1, c1) and Z(a2, b2, c2), respectively. Show that the diﬀerence ξ1 −ξ2\nis also a zigzag uncertain variable Z(a1 −c2, b1 −b2, c1 −a2), i.e.,\nZ(a1, b1, c1) −Z(a2, b2, c2) = Z(a1 −c2, b1 −b2, c1 −a2).\n(3.121)\nExercise 3.44: Let ξ1 and ξ2 be independent normal uncertain variables\nN(e1, σ1) and N(e2, σ2), respectively. Show that the diﬀerence ξ1 −ξ2 is also\na normal uncertain variable N(e1 −e2, σ1 + σ2), i.e.,\nN(e1, σ1) −N(e2, σ2) = N(e1 −e2, σ1 + σ2).\n(3.122)\n\n\nSection 3.5 - Operational Law\n65\nExercise 3.45: Let ξ1 and ξ2 be independent positive uncertain variables\nwith regular uncertainty distributions Φ1 and Φ2, respectively. Show that\nthe inverse uncertainty distribution of the quotient ξ1/ξ2 is\nΨ−1(α) =\nΦ−1\n1 (α)\nΦ−1\n2 (1 −α).\n(3.123)\nExercise 3.46: Assume ξ1 and ξ2 are independent positive uncertain vari-\nables with regular uncertainty distributions Φ1 and Φ2, respectively. Show\nthat the inverse uncertainty distribution of ξ1/(ξ1 + ξ2) is\nΨ−1(α) =\nΦ−1\n1 (α)\nΦ−1\n1 (α) + Φ−1\n2 (1 −α).\n(3.124)\nExercise 3.47:\nShow that the independence condition in Theorem 3.18\ncannot be removed.\nOperational Law via Uncertainty Distributions\nTheorem 3.19 (Liu [120]) Let ξ1, ξ2, · · · , ξn be independent uncertain vari-\nables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. If\nf(x1, x2, · · · , xn) is continuous, strictly increasing with respect to x1, x2, · · · ,\nxm and strictly decreasing with respect to xm+1, xm+2, · · · , xn, then\nξ = f(ξ1, ξ2, · · · , ξn)\n(3.125)\nhas an uncertainty distribution\nΨ(x) =\nsup\nf(x1,x2,··· ,xn)=x\n\u0012\nmin\n1≤i≤m Φi(xi) ∧\nmin\nm+1≤i≤n(1 −Φi(xi))\n\u0013\n.\n(3.126)\nProof: For simplicity, we only prove the case of m = 1 and n = 2. It follows\nfrom Theorem 3.18 that ξ = f(ξ1, ξ2) has an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (1 −α)).\nWrite\nΨ−1(α) = f(Φ−1\n1 (α), Φ−1\n2 (1 −α)) = x.\n(3.127)\nThen\nΨ(x) = α.\n(3.128)\nOn the one hand, for any x1 and x2 with f(x1, x2) = x, since f is strictly\nincreasing with respect to x1 and strictly decreasing with respect to x2, we\nhave\nx1 ≤Φ−1\n1 (α)\nor\nx2 ≥Φ−1\n2 (1 −α).\n\n\n66\nChapter 3 - Uncertain Variable\nThat is,\nα ≥Φ1(x1)\nor\nα ≥1 −Φ2(x2).\nThus\nΨ(x) ≥Φ1(x1) ∧(1 −Φ2(x2)).\nBy the arbitrariness of x1 and x2 with f(x1, x2) = x, we obtain\nΨ(x) ≥\nsup\nf(x1,x2)=x\nΦ1(x1) ∧(1 −Φ2(x2)).\n(3.129)\nOn the other hand, take\nx′\n1 = Φ−1\n1 (α),\nx′\n2 = Φ−1\n2 (1 −α),\ni.e.,\nα = Φ1(x′\n1),\nα = 1 −Φ2(x′\n2).\nBy using (3.127) and (3.128), we obtain\nf(x′\n1, x′\n2) = x,\nΨ(x) = Φ1(x′\n1) ∧(1 −Φ2(x′\n2)).\nThus\nΨ(x) ≤\nsup\nf(x1,x2)=x\nΦ1(x1) ∧(1 −Φ2(x2)).\n(3.130)\nIt follows from (3.129) and (3.130) that\nΨ(x) =\nsup\nf(x1,x2)=x\nΦ1(x1) ∧(1 −Φ2(x2)).\nThe theorem is proved.\nRemark 3.4: It is possible that the equation f(x1, x2, · · · , xn) = x does not\nhave a root for some values of x. In this case, if\nf(x1, x2, · · · , xn) < x\n(3.131)\nfor any vector (x1, x2, · · · , xn), then we set Ψ(x) = 1; and if\nf(x1, x2, · · · , xn) > x\n(3.132)\nfor any vector (x1, x2, · · · , xn), then we set Ψ(x) = 0.\nExample 3.15: Let ξ1, ξ2, · · · , ξn be iid uncertain variables with a common\nregular uncertainty distribution Φ. It follows from the operational law that\nthe sum\nξ = ξ1 + ξ2 + · · · + ξn\n(3.133)\n\n\nSection 3.5 - Operational Law\n67\nhas an uncertainty distribution\nΨ(x) =\nsup\nx1+x2+···+xn=x Φ(x1) ∧Φ(x2) ∧· · · ∧Φ(xn).\n(3.134)\nOn the one hand, take\nxi = x\nn,\ni = 1, 2, · · · , n.\nThen x1 + x2 + · · · + xn = x and\nΦ(x1) ∧Φ(x2) ∧· · · ∧Φ(xn) = Φ\n\u0010x\nn\n\u0011\n.\nOn the other hand, for any x1, x2, · · · , xn with x1 + x2 + · · · + xn = x, there\nexists an index k such that\nxk ≤x\nn.\nThus\nΦ(x1) ∧Φ(x2) ∧· · · ∧Φ(xn) ≤Φ(xk) ≤Φ\n\u0010x\nn\n\u0011\n.\nTherefore,\nsup\nx1+x2+···+xn=x Φ(x1) ∧Φ(x2) ∧· · · ∧Φ(xn) = Φ\n\u0010x\nn\n\u0011\n.\n(3.135)\nIt follows from (3.134) and (3.135) that the sum ξ = ξ1 + ξ2 + · · · + ξn has\nan uncertainty distribution\nΨ(x) = Φ\n\u0010x\nn\n\u0011\n.\n(3.136)\nFurthermore, we have\nξ1 + ξ2 + · · · + ξn\nn\n∼Φ(x).\n(3.137)\nExercise 3.48: Let ξ1, ξ2, · · · , ξn be iid positive uncertain variables with a\ncommon regular uncertainty distribution Φ. Show that the multiplication\nξ = ξ1ξ2 · · · ξn\n(3.138)\nhas an uncertainty distribution\nΨ(x) = Φ\nn\n√x\n\u0001\n.\n(3.139)\nExercise 3.49: Let ξ1, ξ2, · · · , ξn be independent uncertain variables with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. Show that the\nminimum\nξ = ξ1 ∧ξ2 ∧· · · ∧ξn\n(3.140)\n\n\n68\nChapter 3 - Uncertain Variable\nhas an uncertainty distribution\nΨ(x) = Φ1(x) ∨Φ2(x) ∨· · · ∨Φn(x).\n(3.141)\nExercise 3.50: Let ξ1, ξ2, · · · , ξn be independent uncertain variables with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. Show that the\nmaximum\nξ = ξ1 ∨ξ2 ∨· · · ∨ξn\n(3.142)\nhas an uncertainty distribution\nΨ(x) = Φ1(x) ∧Φ2(x) ∧· · · ∧Φn(x).\n(3.143)\nExercise 3.51: Let ξ1 and ξ2 be independent uncertain variables with reg-\nular uncertainty distributions Φ1 and Φ2, respectively. Show that ξ1 −ξ2 has\nan uncertainty distribution\nΨ(x) = sup\ny∈ℜ\nΦ1(x + y) ∧(1 −Φ2(y)).\n(3.144)\nExercise 3.52: Let ξ1 and ξ2 be independent positive uncertain variables\nwith regular uncertainty distributions Φ1 and Φ2, respectively. Show that\nξ1/ξ2 has an uncertainty distribution\nΨ(x) = sup\ny>0\nΦ1(xy) ∧(1 −Φ2(y)).\n(3.145)\nExercise 3.53: Let ξ1 and ξ2 be independent positive uncertain variables\nwith regular uncertainty distributions Φ1 and Φ2, respectively. Show that\nξ1/(ξ1 + ξ2) has an uncertainty distribution\nΨ(x) = sup\ny>0\nΦ1(xy) ∧(1 −Φ2(y −xy)).\n(3.146)\nExercise 3.54: (Jia-Lv-Wang [81]) Let ξ1, ξ2, · · · , ξn be independent un-\ncertain variables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, re-\nspectively. Assume f(x1, x2, · · · , xn) is continuous, strictly increasing with\nrespect to x1, x2, · · · , xm and strictly decreasing with respect to xm+1, xm+2,\n· · · , xn. Show that\nξ = f(ξ1, ξ2, · · · , ξn)\n(3.147)\nhas an uncertainty distribution\nΨ(x) =\ninf\nf(x1,x2,··· ,xn)=x\n\u0012\nmax\n1≤i≤m Φi(xi) ∨\nmax\nm+1≤i≤n(1 −Φi(xi))\n\u0013\n.\n(3.148)\n\n\nSection 3.5 - Operational Law\n69\nTheorem 3.20 (Liu [125], Extreme Value Theorem) Let ξ1, ξ2, · · · , ξn be\nindependent uncertain variables with regular uncertainty distributions Φ1, Φ2,\n· · · , Φn, respectively. Then\nSi = ξ1 + ξ2 + · · · + ξi\n(3.149)\nhave uncertainty distributions\nΨi(x) =\nsup\nx1+x2+···+xi=x Φ1(x1) ∧Φ2(x2) ∧· · · ∧Φi(xi),\n(3.150)\ni = 1, 2, · · · , n, respectively, and the maximum\nS = S1 ∨S2 ∨· · · ∨Sn\n(3.151)\nhas an uncertainty distribution\nΥ(x) = Ψ1(x) ∧Ψ2(x) ∧· · · ∧Ψn(x).\n(3.152)\nProof: It follows from Theorem 3.15 that the maximum S has an inverse\nuncertainty distribution\nΥ−1(α) = Ψ−1\n1 (α) ∨Ψ−1\n2 (α) ∨· · · ∨Ψ−1\nn (α).\nWrite\nΥ−1(α) = Ψ−1\n1 (α) ∨Ψ−1\n2 (α) ∨· · · ∨Ψ−1\nn (α) = x.\nThen\nΥ(x) = α\nand\nΨ−1\n1 (α) ≤x, Ψ−1\n2 (α) ≤x, · · · , Ψ−1\nn (α) ≤x,\ni.e.,\nΨ1(x) ≥α, Ψ2(x) ≥α, · · · , Ψn(x) ≥α.\nSince at least one of the above inequalities holds as an equality, we get\nΥ(x) = α = Ψ1(x) ∧Ψ2(x) ∧· · · ∧Ψn(x).\nThus (3.152) is veriﬁed.\nTheorem 3.21 (Liu [125], Extreme Value Theorem) Let ξ1, ξ2, · · · , ξn be\nindependent uncertain variables with regular uncertainty distributions Φ1, Φ2,\n· · · , Φn, respectively. Then\nSi = ξ1 + ξ2 + · · · + ξi\n(3.153)\nhave uncertainty distributions\nΨi(x) =\nsup\nx1+x2+···+xi=x Φ1(x1) ∧Φ2(x2) ∧· · · ∧Φi(xi),\n(3.154)\n\n\n70\nChapter 3 - Uncertain Variable\ni = 1, 2, · · · , n, respectively, and the minimum\nS = S1 ∧S2 ∧· · · ∧Sn\n(3.155)\nhas an uncertainty distribution\nΥ(x) = Ψ1(x) ∨Ψ2(x) ∨· · · ∨Ψn(x).\n(3.156)\nProof: It follows from Theorem 3.16 that the minimum S has an inverse\nuncertainty distribution\nΥ−1(α) = Ψ−1\n1 (α) ∧Ψ−1\n2 (α) ∧· · · ∧Ψ−1\nn (α).\nWrite\nΥ−1(α) = Ψ−1\n1 (α) ∧Ψ−1\n2 (α) ∧· · · ∧Ψ−1\nn (α) = x.\nThen\nΥ(x) = α\nand\nΨ−1\n1 (α) ≥x, Ψ−1\n2 (α) ≥x, · · · , Ψ−1\nn (α) ≥x,\ni.e.,\nΨ1(x) ≤α, Ψ2(x) ≤α, · · · , Ψn(x) ≤α.\nSince at least one of the above inequalities holds as an equality, we get\nΥ(x) = α = Ψ1(x) ∨Ψ2(x) ∨· · · ∨Ψn(x).\nThus (3.156) is veriﬁed.\nOperational Law for Boolean System\nA function is said to be Boolean if it is a mapping from {0, 1}n to {0, 1}. For\nexample,\nf(x1, x2, x3) = x1 ∨x2 ∧x3\n(3.157)\nis a Boolean function.\nAn uncertain variable is said to be Boolean if it\ntakes values either 0 or 1. For example, the following is a Boolean uncertain\nvariable,\nξ =\n(\n1 with uncertain measure a\n0 with uncertain measure 1 −a\n(3.158)\nwhere a is a number between 0 and 1. This section introduces an operational\nlaw for Boolean system.\nTheorem 3.22 (Liu [120]) Assume ξ1, ξ2, · · · , ξn are independent Boolean\nuncertain variables, i.e.,\nξi =\n(\n1 with uncertain measure ai\n0 with uncertain measure 1 −ai\n(3.159)\n\n\nSection 3.5 - Operational Law\n71\nfor i = 1, 2, · · · , n. If f is a Boolean function (not necessarily monotone),\nthen ξ = f(ξ1, ξ2, · · · , ξn) is a Boolean uncertain variable such that\nM{ξ = 1} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) < 0.5\n1 −\nsup\nf(x1,x2,··· ,xn)=0\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) ≥0.5\n(3.160)\nwhere xi take values either 0 or 1, and νi are deﬁned by\nνi(xi) =\n(\nai,\nif xi = 1\n1 −ai,\nif xi = 0,\n(3.161)\ni = 1, 2, · · · , n, respectively.\nProof: At ﬁrst, please mention that νi(xi) = M{ξi = xi}, i = 1, 2, · · · , n.\nFor simplicity, we only prove the case of n = 2. If f(x1, x2) = 1 has no root,\nthen\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2) = 0 < 0.5\nand\nM{ξ = 1} = M{f(ξ1, ξ2) = 1} = M{∅} = 0.\nThus\nM{ξ = 1} =\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2).\nIf f(x1, x2) = 0 has no root, then\nsup\nf(x1,x2)=0\nν1(x1) ∧ν2(x2) = 0,\nf(x1, x2) ≡1.\nOn the one hand, for each i with 1 ≤i ≤2, we take\nx′\ni =\n(\n0,\nif νi(0) ≥0.5\n1,\notherwise.\nThen\nν1(x′\n1) ≥0.5, ν2(x′\n2) ≥0.5.\nThus\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2) ≥ν1(x′\n1) ∧ν2(x′\n2) ≥0.5.\n\n\n72\nChapter 3 - Uncertain Variable\nOn the other hand, it follows from f(x1, x2) ≡1 that\nM{ξ = 1} = M{f(ξ1, ξ2) = 1} = M{Γ} = 1.\nThus\nM{ξ = 1} = 1 −\nsup\nf(x1,x2)=0\nν1(x1) ∧ν2(x2).\nWhen both f(x1, x2) = 1 and f(x1, x2) = 0 have roots, the argument breaks\ndown into three cases. Case 1: Assume\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2) = c < 0.5.\n(3.162)\nThen there exists a vector (x∗\n1, x∗\n2) such that\nf(x∗\n1, x∗\n2) = 1,\nν1(x∗\n1) ≥c, ν2(x∗\n2) ≥c\nand at least one of the above two inequalities holds as an equality. On the\none hand, since f(x∗\n1, x∗\n2) = 1, we immediately have\n{ξ = 1} ≡{f(ξ1, ξ2) = 1} ⊃{ξ1 = x∗\n1} ∩{ξ2 = x∗\n2}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ = 1} ≥M{(ξ1 = x∗\n1) ∩(ξ2 = x∗\n2)}\n= M{ξ1 = x∗\n1} ∧M{ξ2 = x∗\n2}\n= ν1(x∗\n1) ∧ν2(x∗\n2)\n= c,\ni.e.,\nM{ξ = 1} ≥c.\n(3.163)\nOn the other hand, for each i with 1 ≤i ≤2, we take\nBi =\n\n\n\n\n\n\n\n{1 −x∗\ni },\nif νi(x∗\ni ) = c\n{0, 1},\nif c < νi(x∗\ni ) < 1 −c\n{x∗\ni },\nif νi(x∗\ni ) ≥1 −c.\nIt is clear that\nM{ξi ∈Bi} ≥1 −c,\ni = 1, 2,\nν1(x1) ∧ν2(x2) > c,\n∀(x1, x2) ∈B1 × B2.\nBy using (3.162), we get\nf(x1, x2) = 0,\n∀(x1, x2) ∈B1 × B2.\n\n\nSection 3.5 - Operational Law\n73\nTherefore,\n{ξ = 0} ≡{f(ξ1, ξ2) = 0} ⊃{ξ1 ∈B1} ∩{ξ2 ∈B2}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ = 0} ≥M{(ξ1 ∈B1) ∩(ξ2 ∈B2)}\n= M{ξ1 ∈B1} ∧M{ξ2 ∈B2}\n≥1 −c,\ni.e.,\nM{ξ = 0} ≥1 −c.\n(3.164)\nIt follows from the duality axiom, (3.163) and (3.164) that\nM{ξ = 1} = c =\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2).\nCase 2: Assume\nsup\nf(x1,x2)=0\nν1(x1) ∧ν2(x2) = c < 0.5.\n(3.165)\nThen there exists a vector (x∗\n1, x∗\n2) such that\nf(x∗\n1, x∗\n2) = 0,\nν1(x∗\n1) ≥c, ν2(x∗\n2) ≥c\nand at least one of the above two inequalities holds as an equality. For each\ni with 1 ≤i ≤2, we take\nx′\ni =\n(\n1 −x∗\ni ,\nif νi(x∗\ni ) < 0.5\nx∗\ni ,\nif νi(x∗\ni ) ≥0.5.\nThen\nν1(x′\n1) ≥0.5, ν2(x′\n2) ≥0.5,\nf(x′\n1, x′\n2) = 1.\nThus\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2) ≥ν1(x′\n1) ∧ν2(x′\n2) ≥0.5.\nOn the one hand, since f(x∗\n1, x∗\n2) = 0, we immediately have\n{ξ = 0} ≡{f(ξ1, ξ2) = 0} ⊃{ξ1 = x∗\n1} ∩{ξ2 = x∗\n2}.\n\n\n74\nChapter 3 - Uncertain Variable\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ = 0} ≥M{(ξ1 = x∗\n1) ∩(ξ2 = x∗\n2)}\n= M{ξ1 = x∗\n1} ∧M{ξ2 = x∗\n2}\n= ν1(x∗\n1) ∧ν2(x∗\n2)\n= c,\ni.e.,\nM{ξ = 0} ≥c.\n(3.166)\nOn the other hand, for each i with 1 ≤i ≤2, we take\nBi =\n\n\n\n\n\n\n\n{1 −x∗\ni },\nif νi(x∗\ni ) = c\n{0, 1},\nif c < νi(x∗\ni ) < 1 −c\n{x∗\ni },\nif νi(x∗\ni ) ≥1 −c.\nIt is clear that\nM{ξi ∈Bi} ≥1 −c,\ni = 1, 2,\nν1(x1) ∧ν2(x2) > c,\n∀(x1, x2) ∈B1 × B2.\nBy using (3.165), we get\nf(x1, x2) = 1,\n∀(x1, x2) ∈B1 × B2.\nTherefore,\n{ξ = 1} ≡{f(ξ1, ξ2) = 1} ⊃{ξ1 ∈B1} ∩{ξ2 ∈B2}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ = 1} ≥M{(ξ1 ∈B1) ∩(ξ2 ∈B2)}\n= M{ξ1 ∈B1} ∧M{ξ2 ∈B2}\n≥1 −c,\ni.e.,\nM{ξ = 1} ≥1 −c.\n(3.167)\nIt follows from the duality axiom, (3.166) and (3.167) that\nM{ξ = 1} = 1 −c = 1 −\nsup\nf(x1,x2)=0\nν1(x1) ∧ν2(x2).\nCase 3: Assume\nsup\nf(x1,x2)=1\nν1(x1) ∧ν2(x2) = c ≥0.5,\n(3.168)\n\n\nSection 3.5 - Operational Law\n75\nsup\nf(x1,x2)=0\nν1(x1) ∧ν2(x2) = d ≥0.5.\n(3.169)\nThen there exist two vectors (x′\n1, x′\n2) and (x′′\n1, x′′\n2) such that\nf(x′\n1, x′\n2) = 1,\nf(x′′\n1, x′′\n2) = 0,\nν1(x′\n1) ≥c, ν2(x′\n2) ≥c,\nν1(x′′\n1) ≥d, ν2(x′′\n2) ≥d.\nThus we have\n{ξ = 1} ≡{f(ξ1, ξ2) = 1} ⊃{ξ1 = x′\n1} ∩{ξ2 = x′\n2},\n{ξ = 0} ≡{f(ξ1, ξ2) = 0} ⊃{ξ1 = x′′\n1} ∩{ξ2 = x′′\n2}.\nBy using the monotonicity theorem and independence of ξ1 and ξ2, we get\nM{ξ = 1} ≥M{(ξ1 = x′\n1) ∩(ξ2 = x′\n2)}\n= M{ξ1 = x′\n1} ∧M{ξ2 = x′\n2}\n= ν1(x′\n1) ∧ν2(x′\n2)\n≥c,\ni.e.,\nM{ξ = 1} ≥c ≥0.5,\n(3.170)\nand\nM{ξ = 0} ≥M{(ξ1 = x′′\n1) ∩(ξ2 = x′′\n2)}\n= M{ξ1 = x′′\n1} ∧M{ξ2 = x′\n2}\n= ν1(x′′\n1) ∧ν2(x′′\n2)\n≥d,\ni.e.,\nM{ξ = 0} ≥d ≥0.5.\n(3.171)\nIt follows from the duality axiom, (3.170) and (3.171) that\nM{ξ = 1} = M{ξ = 0} = c = d = 0.5.\nThus\nM{ξ = 1} = 1 −\nsup\nf(x1,x2)=0\nν1(x1) ∧ν2(x2).\nAll above cases can be summarized as the equation (3.160). The theorem is\nproved.\nExample 3.16: Let ξ1, ξ2, · · · , ξn be independent Boolean uncertain vari-\nables deﬁned by (3.159), and\nξ = ξ1 ∧ξ2 ∧· · · ∧ξn.\n(3.172)\n\n\n76\nChapter 3 - Uncertain Variable\nCase 1: Assume\nsup\nx1∧x2∧···∧xn=1\nmin\n1≤i≤n νi(xi) < 0.5.\nSince\nsup\nx1∧x2∧···∧xn=1\nmin\n1≤i≤n νi(xi) = min\n1≤i≤n νi(1) = a1 ∧a2 ∧· · · ∧an,\nit follows from the operational law that\nM{ξ = 1} =\nsup\nx1∧x2∧···∧xn=1\nmin\n1≤i≤n νi(xi) = a1 ∧a2 ∧· · · ∧an.\nCase 2: Assume\nsup\nx1∧x2∧···∧xn=1\nmin\n1≤i≤n νi(xi) ≥0.5.\nThen\nν1(1) ∧ν2(1) ∧· · · ∧νn(1) ≥0.5,\ni.e.,\na1 ∧a2 ∧· · · ∧an ≥0.5.\nLet k be the index such that\nak = a1 ∧a2 ∧· · · ∧an.\nOn the one hand, take\nx∗\nk = 0,\nx∗\ni = 1 if i ̸= k.\nThen x∗\n1 ∧x∗\n2 ∧· · · ∧x∗\nn = 0 and\nνk(x∗\nk) = 1 −ak ≤0.5,\nνi(x∗\ni ) = ai ≥0.5 if i ̸= k.\nThus\nmin\n1≤i≤n νi(x∗\ni ) = 1 −ak.\nOn the other hand, for any x1, x2, · · · , xn with x1 ∧x2 ∧· · · ∧xn = 0, there\nexists an index j such that\nxj = 0.\nThus\nmin\n1≤i≤n νi(xi) ≤νj(0) = 1 −aj ≤1 −ak.\nTherefore,\nsup\nx1∧x2∧···∧xn=0\nmin\n1≤i≤n νi(xi) = 1 −ak.\n\n\nSection 3.6 - Expected Value\n77\nIt follows from the operational law that\nM{ξ = 1} = 1 −\nsup\nx1∧x2∧···∧xn=0\nmin\n1≤i≤n νi(xi)\n= 1 −(1 −ak) = ak\n= a1 ∧a2 ∧· · · ∧an.\nBoth above cases can be summarized as\nM{ξ = 1} = a1 ∧a2 ∧· · · ∧an.\n(3.173)\nExercise 3.55: Let ξ1, ξ2, · · · , ξn be independent Boolean uncertain vari-\nables deﬁned by (3.159), and\nξ = ξ1 ∨ξ2 ∨· · · ∨ξn.\n(3.174)\nShow that\nM{ξ = 1} = a1 ∨a2 ∨· · · ∨an.\n(3.175)\nExample 3.17: The independence condition in Theorem 3.22 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be {γ1, γ2}\nwith power set and M{γ1} = M{γ2} = 0.5. Then\nξ1(γ) =\n(\n0,\nif γ = γ1\n1,\nif γ = γ2\n(3.176)\nis a Boolean uncertain variable with\nM{ξ1 = 1} = 0.5,\n(3.177)\nand\nξ2(γ) =\n(\n1,\nif γ = γ1\n0,\nif γ = γ2\n(3.178)\nis also a Boolean uncertain variable with\nM{ξ2 = 1} = 0.5.\n(3.179)\nNote that ξ1 and ξ2 are not independent, and ξ1 ∧ξ2 ≡0 from which we\nobtain\nM{ξ1 ∧ξ2 = 1} = 0.\n(3.180)\nHowever, by using (3.160), we get\nM{ξ1 ∧ξ2 = 1} = 0.5.\n(3.181)\nThus the independence condition cannot be removed.\n\n\n78\nChapter 3 - Uncertain Variable\n3.6\nExpected Value\nExpected value is the average value of uncertain variable in the sense of\nuncertain measure. A formal deﬁnition is given below.\nDeﬁnition 3.13 (Liu [113]) Let ξ be an uncertain variable. Then the ex-\npected value of ξ is deﬁned by\nE[ξ] =\nZ +∞\n0\nM{ξ ≥x}dx −\nZ 0\n−∞\nM{ξ ≤x}dx\n(3.182)\nprovided that at least one of the two integrals is ﬁnite.\nTheorem 3.23 (Liu [113]) Let ξ be an uncertain variable with uncertainty\ndistribution Φ. Then\nE[ξ] =\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx.\n(3.183)\nProof: It follows from the measure inversion theorem that for almost all\nnumbers x, we have M{ξ ≥x} = 1 −Φ(x) and M{ξ ≤x} = Φ(x). By using\nthe deﬁnition of expected value operator, we obtain\nE[ξ] =\nZ +∞\n0\nM{ξ ≥x}dx −\nZ 0\n−∞\nM{ξ ≤x}dx\n=\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx.\nSee Figure 3.9. The theorem is proved.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.9: E[ξ] =\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx\n\n\nSection 3.6 - Expected Value\n79\nTheorem 3.24 (Liu [120]) Let ξ be an uncertain variable with regular un-\ncertainty distribution Φ. Then\nE[ξ] =\nZ 1\n0\nΦ−1(α)dα.\n(3.184)\nProof: Since α = Φ(x) and x = Φ−1(α) represent the same curve in the\nrectangular coordinate system (x, α), we have\nZ +∞\n0\n(1 −Φ(x))dx =\nZ 1\nΦ(0)\nΦ−1(α)dα\n(3.185)\nbecause the two integrals make an identical acreage. See Figure 3.10. Simi-\nlarly, we also have\nZ 0\n−∞\nΦ(x)dx = −\nZ Φ(0)\n0\nΦ−1(α)dα.\n(3.186)\nIt follows from Theorem 3.23, (3.185) and (3.186) that the expected value is\nE[ξ] =\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx\n=\nZ 1\nΦ(0)\nΦ−1(α)dα +\nZ Φ(0)\n0\nΦ−1(α)dα\n=\nZ 1\n0\nΦ−1(α)dα.\nThe theorem is proved.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.10: E[ξ] =\nZ 1\n0\nΦ−1(α)dα\nExercise 3.56: Show that the linear uncertain variable ξ ∼L(a, b) has an\nexpected value\nE[ξ] = a + b\n2\n.\n(3.187)\n\n\n80\nChapter 3 - Uncertain Variable\nExercise 3.57: Show that the zigzag uncertain variable ξ ∼Z(a, b, c) has\nan expected value\nE[ξ] = a + 2b + c\n4\n.\n(3.188)\nExercise 3.58: Show that the normal uncertain variable ξ ∼N(e, σ) has\nan expected value e, i.e.,\nE[ξ] = e.\n(3.189)\nExercise 3.59:\nLet ξ be an uncertain variable with regular uncertainty\ndistribution Φ, and let f(x) be a continuous and strictly monotone (increasing\nor decreasing) function. Show that\nE[f(ξ)] =\nZ 1\n0\nf(Φ−1(α))dα.\n(3.190)\nExercise 3.60: Let ξ and η be independent and positive uncertain variables\nwith regular uncertainty distributions Φ and Ψ, respectively. Show that\nE[ξη] =\nZ 1\n0\nΦ−1(α)Ψ−1(α)dα.\n(3.191)\nExercise 3.61: Let ξ and η be independent and positive uncertain variables\nwith regular uncertainty distributions Φ and Ψ, respectively. Show that\nE\n\u0014ξ\nη\n\u0015\n=\nZ 1\n0\nΦ−1(α)\nΨ−1(1 −α)dα.\n(3.192)\nExercise 3.62: Assume ξ and η are independent and positive uncertain\nvariables with regular uncertainty distributions Φ and Ψ, respectively. Show\nthat\nE\n\u0014\nξ\nξ + η\n\u0015\n=\nZ 1\n0\nΦ−1(α)\nΦ−1(α) + Ψ−1(1 −α)dα.\n(3.193)\nLinearity of Expected Value Operator\nTheorem 3.25 (Liu [120]) Let ξ and η be independent uncertain variables\nwith ﬁnite expected values. Then for any real numbers a and b, we have\nE[aξ + bη] = aE[ξ] + bE[η].\n(3.194)\nProof: Without loss of generality, suppose ξ and η have regular uncertainty\ndistributions Φ and Ψ, respectively. Otherwise, we may give the uncertainty\ndistributions a small perturbation such that they become regular.\n\n\nSection 3.6 - Expected Value\n81\nStep 1: We ﬁrst prove E[aξ] = aE[ξ]. If a = 0, then the equation holds\ntrivially. If a > 0, then the inverse uncertainty distribution of aξ is\nΥ−1(α) = aΦ−1(α).\nIt follows from Theorem 3.24 that\nE[aξ] =\nZ 1\n0\naΦ−1(α)dα = a\nZ 1\n0\nΦ−1(α)dα = aE[ξ].\nIf a < 0, then the inverse uncertainty distribution of aξ is\nΥ−1(α) = aΦ−1(1 −α).\nIt follows from Theorem 3.24 that\nE[aξ] =\nZ 1\n0\naΦ−1(1 −α)dα = a\nZ 1\n0\nΦ−1(α)dα = aE[ξ].\nThus we always have E[aξ] = aE[ξ].\nStep 2: We prove E[ξ + η] = E[ξ] + E[η].\nThe inverse uncertainty\ndistribution of the sum ξ + η is\nΥ−1(α) = Φ−1(α) + Ψ−1(α).\nIt follows from Theorem 3.24 that\nE[ξ + η] =\nZ 1\n0\nΥ−1(α)dα =\nZ 1\n0\nΦ−1(α)dα +\nZ 1\n0\nΨ−1(α)dα = E[ξ] + E[η].\nStep 3: Finally, for any real numbers a and b, it follows from Steps 1\nand 2 that\nE[aξ + bη] = E[aξ] + E[bη] = aE[ξ] + bE[η].\nThe theorem is proved.\nExample 3.18:\nGenerally speaking, the expected value operator is not\nnecessarily linear if the independence is not assumed. For example, take an\nuncertainty space (Γ, L, M) to be {γ1, γ2, γ3} with power set and M{γ1} =\n0.6, M{γ2} = 0.3 and M{γ3} = 0.2. Deﬁne two uncertain variables as follows,\nξ(γ) =\n\n\n\n\n\n1,\nif γ = γ1\n0,\nif γ = γ2\n2,\nif γ = γ3,\nη(γ) =\n\n\n\n\n\n0,\nif γ = γ1\n2,\nif γ = γ2\n3,\nif γ = γ3.\nNote that ξ and η are not independent, and their sum is\n(ξ + η)(γ) =\n\n\n\n\n\n1,\nif γ = γ1\n2,\nif γ = γ2\n5,\nif γ = γ3.\n\n\n82\nChapter 3 - Uncertain Variable\nIt is easy to verify that E[ξ] = 0.9, E[η] = 1 and E[ξ + η] = 2. Thus we have\nE[ξ + η] > E[ξ] + E[η].\nIf the uncertain variables are deﬁned by\nξ(γ) =\n\n\n\n\n\n0,\nif γ = γ1\n1,\nif γ = γ2\n2,\nif γ = γ3,\nη(γ) =\n\n\n\n\n\n0,\nif γ = γ1\n3,\nif γ = γ2\n1,\nif γ = γ3.\nThen\n(ξ + η)(γ) =\n\n\n\n\n\n0,\nif γ = γ1\n4,\nif γ = γ2\n3,\nif γ = γ3.\nIt is easy to verify that E[ξ] = 0.6, E[η] = 1 and E[ξ + η] = 1.5. Thus we\nhave\nE[ξ + η] < E[ξ] + E[η].\nTherefore, the independence condition cannot be removed.\nAbsolute Value of Uncertain Variable\nLet ξ be an uncertain variable with uncertainty distribution Φ. Then the\nexpected value of |ξ| is\nE[|ξ|] =\nZ +∞\n0\nM{|ξ| ≥x}dx\n=\nZ +∞\n0\nM{(ξ ≥x) ∪(ξ ≤−x)}dx\n≤\nZ +∞\n0\n(M{ξ ≥x} + M{ξ ≤−x})dx\n=\nZ +∞\n0\n(1 −Φ(x) + Φ(−x))dx.\nThus we have the following stipulation.\nStipulation 3.1 (Liu [130]) Let ξ be an uncertain variable with uncertainty\ndistribution Φ. Then the expected value of |ξ| is\nE[|ξ|] =\nZ +∞\n0\n(1 −Φ(x) + Φ(−x))dx.\n(3.195)\nTheorem 3.26 (Liu [130]) Let ξ be an uncertain variable with regular un-\ncertainty distribution Φ. Then the expected value of |ξ| is\nE[|ξ|] =\nZ 1\n0\n|Φ−1(α)|dα.\n(3.196)\n\n\nSection 3.7 - Variance\n83\nProof: Since α = Φ(x) on (0, +∞) and x = |Φ−1(α)| on (Φ(0), 1) represent\nthe same curve in the rectangular coordinate system (x, α), we have\nZ +∞\n0\n(1 −Φ(x))dx =\nZ 1\nΦ(0)\n|Φ−1(α)|dα\n(3.197)\nbecause the two integrals make an identical acreage. Since α = Φ(−x) on\n(0, +∞) and x = |Φ−1(α)| on (0, Φ(0)) represent the same curve, we have\nZ +∞\n0\nΦ(−x)dx =\nZ Φ(0)\n0\n|Φ−1(α)|dα.\n(3.198)\nIt follows from Stipulation 3.1, (3.197) and (3.198) that the expected value\nis\nE[|ξ|] =\nZ +∞\n0\n(1 −Φ(x) + Φ(−x))dx\n=\nZ 1\nΦ(0)\n|Φ−1(α)|dα +\nZ Φ(0)\n0\n|Φ−1(α)|dα\n=\nZ 1\n0\n|Φ−1(α)|dα.\nThe theorem is proved.\n3.7\nVariance\nThe variance of uncertain variable provides a degree of the spread of the\ndistribution around its expected value. A small value of variance indicates\nthat the uncertain variable is tightly concentrated around its expected value;\nand a large value of variance indicates that the uncertain variable has a wide\nspread around its expected value.\nDeﬁnition 3.14 (Liu [113]) Let ξ be an uncertain variable with ﬁnite ex-\npected value e. Then the variance of ξ is\nV [ξ] = E[(ξ −e)2].\n(3.199)\nThis deﬁnition tells us that the variance is just the expected value of\n(ξ −e)2. Since (ξ −e)2 is a nonnegative uncertain variable, we also have\nV [ξ] =\nZ +∞\n0\nM{(ξ −e)2 ≥x}dx.\n(3.200)\nTheorem 3.27 (Liu [113]) If ξ is an uncertain variable with ﬁnite expected\nvalue, a and b are real numbers, then\nV [aξ + b] = a2V [ξ].\n(3.201)\n\n\n84\nChapter 3 - Uncertain Variable\nProof: Let e be the expected value of ξ. Then aξ + b has an expected value\nae + b. It follows from the deﬁnition of variance that\nV [aξ + b] = E\n\u0002\n(aξ + b −(ae + b))2\u0003\n= a2E[(ξ −e)2] = a2V [ξ].\nThe theorem is thus veriﬁed.\nTheorem 3.28 (Liu [113]) Let ξ be an uncertain variable with expected value\ne. Then V [ξ] = 0 if and only if M{ξ = e} = 1. That is, the uncertain variable\nξ is essentially the constant e.\nProof: We ﬁrst assume V [ξ] = 0. It follows from the equation (3.200) that\nZ +∞\n0\nM{(ξ −e)2 ≥x}dx = 0\nwhich implies M{(ξ −e)2 ≥x} = 0 for any x > 0. Hence we have\nM{(ξ −e)2 = 0} = 1.\nThat is, M{ξ = e} = 1.\nConversely, assume M{ξ = e} = 1.\nThen we\nimmediately have M{(ξ −e)2 = 0} = 1 and M{(ξ −e)2 ≥x} = 0 for any\nx > 0. Thus\nV [ξ] =\nZ +∞\n0\nM{(ξ −e)2 ≥x}dx = 0.\nThe theorem is proved.\nHow to Obtain Variance from Uncertainty Distribution?\nLet ξ be an uncertain variable with expected value e. If we only know its\nuncertainty distribution Φ, then the variance\nV [ξ] =\nZ +∞\n0\nM{(ξ −e)2 ≥x}dx\n=\nZ +∞\n0\nM{(ξ ≥e + √x) ∪(ξ ≤e −√x)}dx\n≤\nZ +∞\n0\n(M{ξ ≥e + √x} + M{ξ ≤e −√x})dx\n=\nZ +∞\n0\n(1 −Φ(e + √x) + Φ(e −√x))dx.\nThus we have the following stipulation.\nStipulation 3.2 (Liu [120]) Let ξ be an uncertain variable with uncertainty\ndistribution Φ and ﬁnite expected value e. Then\nV [ξ] =\nZ +∞\n0\n(1 −Φ(e + √x) + Φ(e −√x))dx.\n(3.202)\n\n\nSection 3.7 - Variance\n85\nTheorem 3.29 (Yao [259]) Let ξ be an uncertain variable with regular un-\ncertainty distribution Φ and ﬁnite expected value e. Then\nV [ξ] =\nZ 1\n0\n(Φ−1(α) −e)2dα.\n(3.203)\nProof: Since α = Φ(e + √x) on (0, +∞) and x = (Φ−1(α) −e)2 on (Φ(e), 1)\nrepresent the same curve in the rectangular coordinate system (x, α), we have\nZ +∞\n0\n(1 −Φ(e + √x))dx =\nZ 1\nΦ(e)\n(Φ−1(α) −e)2dα\n(3.204)\nbecause the two integrals make an identical acreage. Since α = Φ(e −√x)\non (0, +∞) and x = (Φ−1(α) −e)2 on (0, Φ(e)) represent the same curve, we\nhave\nZ +∞\n0\nΦ(e −√x)dx =\nZ Φ(e)\n0\n(Φ−1(α) −e)2dα.\n(3.205)\nIt follows from Stipulation 3.2, (3.204) and (3.205) that the variance is\nV [ξ] =\nZ +∞\n0\n(1 −Φ(e + √x) + Φ(e −√x))dx\n=\nZ 1\nΦ(e)\n(Φ−1(α) −e)2dα +\nZ Φ(e)\n0\n(Φ−1(α) −e)2dα\n=\nZ 1\n0\n(Φ−1(α) −e)2dα.\nThe theorem is proved.\nExercise 3.63: Show that the linear uncertain variable ξ ∼L(a, b) has a\nvariance\nV [ξ] = (b −a)2\n12\n.\n(3.206)\nExercise 3.64: Show that the zigzag uncertain variable ξ ∼Z(a, b, c) has a\nvariance\nV [ξ] = 5a2 + 4b2 + 5c2 −4ab −6ac −4bc\n48\n.\n(3.207)\nExercise 3.65: Show that the normal uncertain variable ξ ∼N(e, σ) has a\nvariance\nV [ξ] = σ2.\n(3.208)\nHint: Use the formula\nZ 1\n0\n\u0012\nln\nα\n1 −α\n\u00132\ndα = π2\n3 .\n(3.209)\n\n\n86\nChapter 3 - Uncertain Variable\nExercise 3.66: Let ξ and η be independent linear uncertain variables. Show\nthat\np\nV [ξ + η] =\np\nV [ξ] +\np\nV [η].\n(3.210)\nExercise 3.67:\nLet ξ and η be independent normal uncertain variables.\nShow that\np\nV [ξ + η] =\np\nV [ξ] +\np\nV [η].\n(3.211)\n3.8\nMoments\nDeﬁnition 3.15 (Liu [113]) Let ξ be an uncertain variable and let k be a\npositive integer. Then E[ξk] is called the k-th moment of ξ.\nTheorem 3.30 (Liu [130]) Let ξ be an uncertain variable with uncertainty\ndistribution Φ, and let k be an odd number. Then the k-th moment of ξ is\nE[ξk] =\nZ +∞\n0\n(1 −Φ( k\n√x))dx −\nZ 0\n−∞\nΦ( k\n√x)dx.\n(3.212)\nProof: Since k is an odd number, it follows from the deﬁnition of expected\nvalue operator that\nE[ξk] =\nZ +∞\n0\nM{ξk ≥x}dx −\nZ 0\n−∞\nM{ξk ≤x}dx\n=\nZ +∞\n0\nM{ξ ≥\nk\n√x}dx −\nZ 0\n−∞\nM{ξ ≤\nk\n√x}dx\n=\nZ +∞\n0\n(1 −Φ( k\n√x))dx −\nZ 0\n−∞\nΦ( k\n√x)dx.\nThe theorem is proved.\nHowever, when k is an even number, the k-th moment of ξ cannot be\nuniquely determined by the uncertainty distribution Φ. In this case, we have\nE[ξk] =\nZ +∞\n0\nM{ξk ≥x}dx\n=\nZ +∞\n0\nM{(ξ ≥\nk\n√x) ∪(ξ ≤−k\n√x)}dx\n≤\nZ +∞\n0\n(M{ξ ≥\nk\n√x} + M{ξ ≤−k\n√x})dx\n=\nZ +∞\n0\n(1 −Φ( k\n√x) + Φ(−k\n√x))dx.\nThus for the even number k, we have the following stipulation.\n\n\nSection 3.8 - Moments\n87\nStipulation 3.3 (Liu [130]) Let ξ be an uncertain variable with uncertainty\ndistribution Φ, and let k be an even number. Then the k-th moment of ξ is\nE[ξk] =\nZ +∞\n0\n(1 −Φ( k\n√x) + Φ(−k\n√x))dx.\n(3.213)\nTheorem 3.31 (Sheng-Kar [203]) Let ξ be an uncertain variable with regu-\nlar uncertainty distribution Φ, and let k be a positive integer. Then the k-th\nmoment of ξ is\nE[ξk] =\nZ 1\n0\n(Φ−1(α))kdα.\n(3.214)\nProof: When k is an odd number, since α = Φ( k\n√x) and x = (Φ−1(α))k\nrepresent the same curve in the rectangular coordinate system (x, α), we have\nZ +∞\n0\n(1 −Φ( k\n√x))dx =\nZ 1\nΦ(0)\n(Φ−1(α))kdα\n(3.215)\nbecause the two integrals make an identical acreage. Similarly, we also have\nZ 0\n−∞\nΦ( k\n√x)dx = −\nZ Φ(0)\n0\n(Φ−1(α))kdα.\n(3.216)\nIt follows from Theorem 3.30, (3.215) and (3.216) that the k-th moment is\nE[ξk] =\nZ +∞\n0\n(1 −Φ( k\n√x))dx −\nZ 0\n−∞\nΦ( k\n√x)dx\n=\nZ 1\nΦ(0)\n(Φ−1(α))kdα +\nZ Φ(0)\n0\n(Φ−1(α))kdα\n=\nZ 1\n0\n(Φ−1(α))kdα.\nWhen k is an even number, since α = Φ( k\n√x) on (0, +∞) and x = (Φ−1(α))k\non (Φ(0), 1) represent the same curve in the rectangular coordinate system\n(x, α), we have\nZ +∞\n0\n(1 −Φ( k\n√x))dx =\nZ 1\nΦ(0)\n(Φ−1(α))kdα.\n(3.217)\nSince α = Φ(−k\n√x) on (0, +∞) and x = (Φ−1(α))k on (0, Φ(0)) represent the\nsame curve, we have\nZ +∞\n0\nΦ(−k\n√x)dx =\nZ Φ(0)\n0\n(Φ−1(α))kdα.\n(3.218)\n\n\n88\nChapter 3 - Uncertain Variable\nIt follows from Stipulation 3.2, (3.217) and (3.218) that the k-th moment is\nE[ξk] =\nZ +∞\n0\n(1 −Φ( k\n√x) + Φ(−k\n√x))dx\n=\nZ 1\nΦ(0)\n(Φ−1(α))kdα +\nZ Φ(0)\n0\n(Φ−1(α))kdα\n=\nZ 1\n0\n(Φ−1(α))kdα.\nThe theorem is proved.\nExercise 3.68:\nShow that the kth moment of linear uncertain variable\nξ ∼L(a, b) is\nE[ξk] =\nbk+1 −ak+1\n(k + 1)(b −a).\n(3.219)\nExercise 3.69:\nShow that the kth moment of zigzag uncertain variable\nξ ∼Z(a, b, c) is\nE[ξk] =\nbk+1 −ak+1\n2(k + 1)(b −a) +\nck+1 −bk+1\n2(k + 1)(c −b).\n(3.220)\nExercise 3.70: (Ma-Yang-Yao [171]) Show that (i) the k-th moments of\nstandard normal uncertain variable ξ ∼N(0, 1) are\nE[ξk] = (2k −2)3\nk\n2 |Bk|\n(3.221)\nwhere Bk are the k-th Bernoulli numbers, k = 1, 2, · · · , respectively, and (ii)\nthe ﬁrst four moments are 0, 1, 0 and 4.2. Hint: Use the formula,\nZ 1\n0\n\u0012\nln\nα\n1 −α\n\u0013k\ndα = (2k −2)πk|Bk|.\n(3.222)\n3.9\nDistance\nDeﬁnition 3.16 (Liu [113]) The distance between uncertain variables ξ and\nη is deﬁned as\nd(ξ, η) = E[|ξ −η|].\n(3.223)\nThat is, the distance between ξ and η is just the expected value of |ξ −η|.\nSince |ξ −η| is a nonnegative uncertain variable, we always have\nd(ξ, η) =\nZ +∞\n0\nM{|ξ −η| ≥x}dx.\n(3.224)\n\n\nSection 3.10 - Entropy\n89\nTheorem 3.32 (Liu [113]) Let ξ, η, τ be uncertain variables, and let d(·, ·)\nbe the distance. Then we have\n(a) (Nonnegativity) d(ξ, η) ≥0;\n(b) (Identiﬁcation) d(ξ, η) = 0 if and only if ξ = η;\n(c) (Symmetry) d(ξ, η) = d(η, ξ);\n(d) (Triangle Inequality) d(ξ, η) ≤2d(ξ, τ) + 2d(η, τ).\nProof: The parts (a), (b) and (c) follow immediately from the deﬁnition.\nNow we prove the part (d). It follows from the subadditivity axiom that\nd(ξ, η) =\nZ +∞\n0\nM {|ξ −η| ≥x} dx\n≤\nZ +∞\n0\nM {|ξ −τ| + |τ −η| ≥x} dx\n≤\nZ +∞\n0\nM {(|ξ −τ| ≥x/2) ∪(|τ −η| ≥x/2)} dx\n≤\nZ +∞\n0\n(M{|ξ −τ| ≥x/2} + M{|τ −η| ≥x/2}) dx\n= 2E[|ξ −τ|] + 2E[|τ −η|] = 2d(ξ, τ) + 2d(τ, η).\nIt has been proved by Jia-Lio [82] that the triangle inequality is tight.\nTheorem 3.33 (Liu [130]) Let ξ and η be independent uncertain variables\nwith regular uncertainty distributions Φ and Ψ, respectively. Then the dis-\ntance between ξ and η is\nd(ξ, η) =\nZ 1\n0\n|Φ−1(α) −Ψ−1(1 −α)|dα.\n(3.225)\nProof: Note that the diﬀerence ξ −η has an inverse uncertainty distribution\nΥ−1(α) = Φ−1(α) −Ψ−1(1 −α). The equation (3.225) follows from d(ξ, η) =\nE[|ξ −η|] and Theorem 3.26 immediately.\nExercise 3.71: What is the distance between independent linear uncertain\nvariables L(a1, b1) and L(a2, b2)?\n3.10\nEntropy\nThis section deﬁnes an entropy as the degree of diﬃculty of predicting the\nrealization of an uncertain variable.\nDeﬁnition 3.17 (Liu [116]) Suppose that ξ is an uncertain variable with\nuncertainty distribution Φ. Then its entropy is deﬁned by\nH[ξ] =\nZ +∞\n−∞\nS(Φ(x))dx\n(3.226)\n\n\n90\nChapter 3 - Uncertain Variable\nwhere S(t) = −t ln t −(1 −t) ln(1 −t).\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\n0.5\n1\nt\nS(t)\nln 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. . . . . . . . . . . . . . . . . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 3.11: Function S(t) = −t ln t −(1 −t) ln(1 −t). It is easy to verify\nthat S(t) is a symmetric function about t = 0.5, strictly increasing on the\ninterval [0, 0.5], strictly decreasing on the interval [0.5, 1], and reaches its\nunique maximum ln 2 at t = 0.5.\nExample 3.19: Let ξ be an uncertain variable with uncertainty distribution\nΦ(x) =\n(\n0,\nif x < a\n1,\nif x ≥a.\n(3.227)\nEssentially, ξ is a constant a. It follows from the deﬁnition of entropy that\nH[ξ] = −\nZ a\n−∞\n(0 ln 0 + 1 ln 1) dx −\nZ +∞\na\n(1 ln 1 + 0 ln 0) dx = 0.\nThis means a constant has entropy 0.\nExample 3.20: Let ξ be a linear uncertain variable L(a, b). Then its entropy\nis\nH[ξ] = −\nZ b\na\n\u0012x −a\nb −a ln x −a\nb −a + b −x\nb −a ln b −x\nb −a\n\u0013\ndx = b −a\n2\n.\n(3.228)\nExercise 3.72: Show that the zigzag uncertain variable ξ ∼Z(a, b, c) has\nan entropy\nH[ξ] = c −a\n2\n.\n(3.229)\nExercise 3.73: Show that the normal uncertain variable ξ ∼N(e, σ) has\nan entropy\nH[ξ] = πσ\n√\n3.\n(3.230)\n\n\nSection 3.10 - Entropy\n91\nTheorem 3.34 Let ξ be an uncertain variable. Then H[ξ] ≥0 and equality\nholds if ξ is essentially a constant.\nProof: The nonnegativity is clear. In addition, when an uncertain variable\ntends to a constant, its entropy tends to the minimum 0.\nTheorem 3.35 Let ξ be an uncertain variable taking values on the interval\n[a, b]. Then\nH[ξ] ≤(b −a) ln 2\n(3.231)\nand equality holds if ξ has an uncertainty distribution Φ(x) = 0.5 on [a, b].\nProof: The theorem follows from the fact that the function S(t) reaches its\nmaximum ln 2 at t = 0.5.\nTheorem 3.36 Let ξ be an uncertain variable, and let c be a real number.\nThen\nH[ξ + c] = H[ξ].\n(3.232)\nThat is, the entropy is invariant under arbitrary translations.\nProof: Write the uncertainty distribution of ξ by Φ. Then the uncertain\nvariable ξ + c has an uncertainty distribution Φ(x −c). It follows from the\ndeﬁnition of entropy that\nH[ξ + c] =\nZ +∞\n−∞\nS (Φ(x −c)) dx =\nZ +∞\n−∞\nS(Φ(x))dx = H[ξ].\nThe theorem is proved.\nTheorem 3.37 (Dai-Chen [25]) Let ξ be an uncertain variable with regular\nuncertainty distribution Φ. Then\nH[ξ] =\nZ 1\n0\nΦ−1(α) ln\nα\n1 −αdα.\n(3.233)\nProof: It is clear that S(α) is a derivable function whose derivative has the\nform\nS′(α) = −ln\nα\n1 −α.\nSince\nS(Φ(x)) =\nZ Φ(x)\n0\nS′(α)dα = −\nZ 1\nΦ(x)\nS′(α)dα,\nwe have\nH[ξ] =\nZ +∞\n−∞\nS(Φ(x))dx =\nZ 0\n−∞\nZ Φ(x)\n0\nS′(α)dαdx −\nZ +∞\n0\nZ 1\nΦ(x)\nS′(α)dαdx.\n\n\n92\nChapter 3 - Uncertain Variable\nIt follows from Fubini theorem that\nH[ξ] =\nZ Φ(0)\n0\nZ 0\nΦ−1(α)\nS′(α)dxdα −\nZ 1\nΦ(0)\nZ Φ−1(α)\n0\nS′(α)dxdα\n= −\nZ Φ(0)\n0\nΦ−1(α)S′(α)dα −\nZ 1\nΦ(0)\nΦ−1(α)S′(α)dα\n= −\nZ 1\n0\nΦ−1(α)S′(α)dα =\nZ 1\n0\nΦ−1(α) ln\nα\n1 −αdα.\nThe theorem is veriﬁed.\nExercise 3.74: Let ξ and η be independent and positive uncertain variables\nwith regular uncertainty distributions Φ and Ψ, respectively. Show that\nH[ξη] =\nZ 1\n0\nΦ−1(α)Ψ−1(α) ln\nα\n1 −αdα.\nExercise 3.75: Let ξ and η be independent and positive uncertain variables\nwith regular uncertainty distributions Φ and Ψ, respectively. Show that\nH\n\u0014ξ\nη\n\u0015\n=\nZ 1\n0\nΦ−1(α)\nΨ−1(1 −α) ln\nα\n1 −αdα.\nExercise 3.76: Let ξ and η be independent and positive uncertain variables\nwith regular uncertainty distributions Φ and Ψ, respectively. Show that\nH\n\u0014\nξ\nξ + η\n\u0015\n=\nZ 1\n0\nΦ−1(α)\nΦ−1(α) + Ψ−1(1 −α) ln\nα\n1 −αdα.\nTheorem 3.38 (Dai-Chen [25]) Let ξ and η be independent uncertain vari-\nables. Then for any real numbers a and b, we have\nH[aξ + bη] = |a|H[ξ] + |b|H[η].\n(3.234)\nProof: Without loss of generality, suppose ξ and η have regular uncertainty\ndistributions Φ and Ψ, respectively. Otherwise, we may give the uncertainty\ndistributions a small perturbation such that they become regular.\nStep 1: We prove H[aξ] = |a|H[ξ]. If a > 0, then the inverse uncertainty\ndistribution of aξ is\nΥ−1(α) = aΦ−1(α).\nIt follows from Theorem 3.37 that\nH[aξ] =\nZ 1\n0\naΦ−1(α) ln\nα\n1 −αdα = a\nZ 1\n0\nΦ−1(α) ln\nα\n1 −αdα = |a|H[ξ].\n\n\nSection 3.10 - Entropy\n93\nIf a = 0, then we immediately have H[aξ] = 0 = |a|H[ξ]. If a < 0, then the\ninverse uncertainty distribution of aξ is\nΥ−1(α) = aΦ−1(1 −α).\nIt follows from Theorem 3.37 that\nH[aξ] =\nZ 1\n0\naΦ−1(1 −α) ln\nα\n1 −αdα =(−a)\nZ 1\n0\nΦ−1(α) ln\nα\n1 −αdα = |a|H[ξ].\nThus we always have H[aξ] = |a|H[ξ].\nStep 2: We prove H[ξ + η] = H[ξ] + H[η]. Note that the inverse uncer-\ntainty distribution of ξ + η is\nΥ−1(α) = Φ−1(α) + Ψ−1(α).\nIt follows from Theorem 3.37 that\nH[ξ + η] =\nZ 1\n0\n(Φ−1(α) + Ψ−1(α)) ln\nα\n1 −αdα = H[ξ] + H[η].\nStep 3: Finally, for any real numbers a and b, it follows from Steps 1\nand 2 that\nH[aξ + bη] = H[aξ] + H[bη] = |a|H[ξ] + |b|H[η].\nThe theorem is proved.\nExample 3.21: The independence condition in Theorem 3.38 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be (0, 1) with\nBorel algebra and Lebesgue measure. Then ξ(γ) = γ has an inverse uncer-\ntainty distribution Φ(α) = α and entropy\nH[ξ] = 0.5,\n(3.235)\nand η(γ) = 1 −γ has an inverse uncertainty distribution Ψ(α) = α and\nentropy\nH[η] = 0.5.\n(3.236)\nNote that ξ and η are not independent, and ξ + η ≡1 whose entropy is\nH[ξ + η] = 0.\n(3.237)\nThus\nH[ξ + η] ̸= H[ξ] + H[η].\n(3.238)\nTherefore, the independence condition cannot be removed.\n\n\n94\nChapter 3 - Uncertain Variable\nMaximum Entropy Principle\nGiven some constraints, for example, expected value and variance, there are\nusually multiple compatible uncertainty distributions.\nWhich uncertainty\ndistribution shall we take?\nThe maximum entropy principle attempts to\nselect the uncertainty distribution that has maximum entropy and satisﬁes\nthe prescribed constraints.\nTheorem 3.39 (Chen-Dai [11]) Let ξ be an uncertain variable whose un-\ncertainty distribution is arbitrary but the expected value e and variance σ2.\nThen the entropy\nH[ξ] ≤πσ\n√\n3\n(3.239)\nand the equality holds if ξ is a normal uncertain variable N(e, σ).\nProof: The proof was presented by Ma [172]. Without loss of generality, as-\nsume ξ has a regular uncertainty distribution Φ. It follows from Theorem 3.37\nand Cauchy-Schwarz inequality that\nH[ξ] =\nZ 1\n0\n(Φ−1(α) −e) ln\nα\n1 −αdα\n≤\nsZ 1\n0\n(Φ−1(α) −e)2dα ·\nZ 1\n0\n\u0012\nln\nα\n1 −α\n\u00132\ndα\n=\nr\nσ2 · π2\n3 = πσ\n√\n3\nand the equality holds if and only if\nΦ−1(α) −e = c ln\nα\n1 −α\n(3.240)\nfor any α ∈(0, 1) and some positive constant c, i.e.,\n(Φ−1(α) −e)2 = c2\n\u0012\nln\nα\n1 −α\n\u00132\n.\nIntegrating both sides of above equation, we obtain\nZ 1\n0\n(Φ−1(α) −e)2dα = c2\nZ 1\n0\n\u0012\nln\nα\n1 −α\n\u00132\ndα.\nThat is,\nσ2 = c2 · π2\n3 .\nThus\nc = σ\n√\n3\nπ\n.\n\n\nSection 3.11 - Uncertain Sequence\n95\nSubstituting the above equation into (3.240), we obtain\nΦ−1(α) = e +\n√\n3σ\nπ\nln\nα\n1 −α\nfor any α ∈(0, 1). Therefore, the maximum entropy distribution is just the\nnormal uncertainty distribution N(e, σ).\n3.11\nUncertain Sequence\nUncertain sequence is a sequence of uncertain variables indexed by integers.\nThis section introduces four convergence concepts of uncertain sequence: con-\nvergence almost surely (a.s.), convergence in measure, convergence in mean,\nand convergence in distribution.\nTable 3.1: Relationship among Convergence Concepts\nConvergence\n⇒\nConvergence\n⇒\nConvergence\nin Mean\nin Measure\nin Distribution\nConvergence Almost Surely\nDeﬁnition 3.18 (Liu [113]) The uncertain sequence {ξi} is said to be con-\nvergent a.s. to ξ if there exists an event Λ with M{Λ} = 1 such that\nlim\ni→∞|ξi(γ) −ξ(γ)| = 0\n(3.241)\nfor every γ ∈Λ. In that case we write ξi →ξ, a.s.\nDeﬁnition 3.19 (Liu [113]) The uncertain sequence {ξi} is said to be con-\nvergent in measure to ξ if\nlim\ni→∞M {|ξi −ξ| ≥ε} = 0\n(3.242)\nfor every ε > 0.\nDeﬁnition 3.20 (Liu [113]) The uncertain sequence {ξi} is said to be con-\nvergent in mean to ξ if\nlim\ni→∞E[|ξi −ξ|] = 0.\n(3.243)\nDeﬁnition 3.21 (Liu [113]) Let Φ, Φ1, Φ2, · · · be the uncertainty distribu-\ntions of uncertain variables ξ, ξ1, ξ2, · · · , respectively. We say the uncertain\nsequence {ξi} converges in distribution to ξ if\nlim\ni→∞Φi(x) = Φ(x)\n(3.244)\nfor all x at which Φ(x) is continuous.\n\n\n96\nChapter 3 - Uncertain Variable\nConvergence in Mean vs. Convergence in Measure\nTheorem 3.40 (Liu [113]) If the uncertain sequence {ξi} converges in mean\nto ξ, then {ξi} converges in measure to ξ.\nProof: Since {ξi} converges in mean to ξ, we have E[|ξi −ξ|] →0 as i →∞.\nFor any given number ε > 0, it follows from the deﬁnition of expected value\nthat\nE[|ξi −ξ|] =\nZ +∞\n0\nM{|ξi −ξ| ≥x}dx\n≥\nZ ε\n0\nM{|ξi −ξ| ≥x}dx\n≥\nZ ε\n0\nM{|ξi −ξ| ≥ε}dx\n= ε · M{|ξi −ξ| ≥ε}.\nThus\nM{|ξi −ξ| ≥ε} ≤E[|ξi −ξ|]\nε\n→0\nas i →∞. Therefore, {ξi} converges in measure to ξ. The theorem is proved.\nExample 3.22:\nConvergence in measure does not imply convergence in\nmean. Take an uncertainty space (Γ, L, M) to be {γ1, γ2, · · · } with power set\nand\nM{Λ} =\nX\nγj∈Λ\n1\n2j .\nDeﬁne uncertain variables as\nξi(γ) =\n(\n2i,\nif γ = γi\n0,\notherwise\nfor i = 1, 2, · · · and ξ ≡0. For any small number ε > 0, we have\nM{|ξi −ξ| ≥ε} = 1\n2i →0\nas i →∞. That is, the sequence {ξi} converges in measure to ξ. However,\nfor each i, we have\nE[|ξi −ξ|] = 1.\nThat is, the sequence {ξi} does not converge in mean to ξ.\nConvergence in Measure vs. Convergence in Distribution\nTheorem 3.41 (Liu [113]) If the uncertain sequence {ξi} converges in mea-\nsure to ξ, then {ξi} converges in distribution to ξ.\n\n\nSection 3.11 - Uncertain Sequence\n97\nProof: Let x be a continuity point of the uncertainty distribution Φ. On\nthe one hand, for any y > x, we have\n{ξi ≤x} ⊂{ξ ≤y} ∪{|ξi −ξ| ≥y −x}.\nIt follows from the subadditivity axiom that\nΦi(x) ≤Φ(y) + M{|ξi −ξ| ≥y −x}.\nSince {ξi} converges in measure to ξ, we have M{|ξi −ξ| ≥y −x} →0 as\ni →∞. Thus we obtain lim supi→∞Φi(x) ≤Φ(y) for any y > x. Letting\ny →x, we get\nlim sup\ni→∞\nΦi(x) ≤Φ(x).\n(3.245)\nOn the other hand, for any z < x, we have\n{ξ ≤z} ⊂{ξi ≤x} ∪{|ξi −ξ| ≥x −z}\nwhich implies that\nΦ(z) ≤Φi(x) + M{|ξi −ξ| ≥x −z}.\nSince M{|ξi −ξ| ≥x −z} →0, we obtain Φ(z) ≤lim infi→∞Φi(x) for any\nz < x. Letting z →x, we get\nΦ(x) ≤lim inf\ni→∞Φi(x).\n(3.246)\nIt follows from (3.245) and (3.246) that Φi(x) →Φ(x) as i →∞.\nThe\ntheorem is proved.\nExample 3.23: Convergence in distribution does not imply convergence in\nmeasure. Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with power set\nand M{γ1} = M{γ2} = 1/2. Deﬁne uncertain variables as\nξ(γ) =\n(\n−1,\nif γ = γ1\n1,\nif γ = γ2,\nand ξi = −ξ for i = 1, 2, · · · Then ξi and ξ have the same uncertainty\ndistribution. Thus {ξi} converges in distribution to ξ. However, for some\nsmall number ε > 0, we have\nM{|ξi −ξ| ≥ε} = 1.\nThat is, the sequence {ξi} does not converge in measure to ξ.\n\n\n98\nChapter 3 - Uncertain Variable\nConvergence Almost Surely vs. Convergence in Measure\nExample 3.24: Convergence a.s. does not imply convergence in measure.\nTake an uncertainty space (Γ, L, M) to be {γ1, γ2, · · · } with power set and\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\nDeﬁne uncertain variables as\nξi(γ) =\n(\ni,\nif γ = γi\n0,\notherwise\nfor i = 1, 2, · · · and ξ ≡0.\nThen the sequence {ξi} converges a.s. to ξ.\nHowever, for some small number ε > 0, we have\nM{|ξi −ξ| ≥ε} = 0.5\nfor each i. That is, the sequence {ξi} does not converge in measure to ξ.\nExample 3.25: Convergence in measure does not imply convergence a.s.\nTake an uncertainty space (Γ, L, M) to be [0, 1] with Borel algebra and\nLebesgue measure.\nFor any positive integer i, there is an integer j such\nthat i = 2j + k, where k is an integer between 0 and 2j −1. Deﬁne uncertain\nvariables as\nξi(γ) =\n(\n1,\nif k/2j ≤γ ≤(k + 1)/2j\n0,\notherwise\nfor i = 1, 2, · · · and ξ ≡0. Then for any small number ε > 0, we have\nM{|ξi −ξ| ≥ε} = 1\n2j →0\nas i →∞. That is, the sequence {ξi} converges in measure to ξ. However, for\nany γ ∈[0, 1], there is an inﬁnite number of intervals of the form [k/2j, (k +\n1)/2j] containing γ. Thus ξi(γ) does not converge to 0. In other words, the\nsequence {ξi} does not converge a.s. to ξ.\nConvergence Almost Surely vs. Convergence in Mean\nExample 3.26: Convergence a.s. does not imply convergence in mean. Take\nan uncertainty space (Γ, L, M) to be {γ1, γ2, · · · } with power set and\nM{Λ} =\nX\nγj∈Λ\n1\n2j .\n\n\nSection 3.11 - Uncertain Sequence\n99\nDeﬁne uncertain variables as\nξi(γ) =\n(\n2i,\nif γ = γi\n0,\notherwise\nfor i = 1, 2, · · · and ξ ≡0. Then ξi converges a.s. to ξ. However, the sequence\n{ξi} does not converge in mean to ξ because E[|ξi −ξ|] ≡1 for each i.\nExample 3.27: Convergence in mean does not imply convergence a.s. Take\nan uncertainty space (Γ, L, M) to be [0, 1] with Borel algebra and Lebesgue\nmeasure. For any positive integer i, there is an integer j such that i = 2j +k,\nwhere k is an integer between 0 and 2j −1. Deﬁne uncertain variables as\nξi(γ) =\n(\n1,\nif k/2j ≤γ ≤(k + 1)/2j\n0,\notherwise\nfor i = 1, 2, · · · and ξ ≡0. Then\nE[|ξi −ξ|] = 1\n2j →0\nas i →∞. That is, the sequence {ξi} converges in mean to ξ. However, for\nany γ ∈[0, 1], there is an inﬁnite number of intervals of the form [k/2j, (k +\n1)/2j] containing γ. Thus ξi(γ) does not converge to 0. In other words, the\nsequence {ξi} does not converge a.s. to ξ.\nConvergence Almost Surely vs. Convergence in Distribution\nExample 3.28:\nConvergence in distribution does not imply convergence\na.s. Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with power set and\nM{γ1} = M{γ2} = 1/2. Deﬁne uncertain variables as\nξ(γ) =\n(\n−1,\nif γ = γ1\n1,\nif γ = γ2\nand ξi = −ξ for i = 1, 2, · · · Then ξi and ξ have the same uncertainty\ndistribution. Thus {ξi} converges in distribution to ξ. However, the sequence\n{ξi} does not converge a.s. to ξ.\nExample 3.29: Convergence a.s. does not imply convergence in distribution.\nTake an uncertainty space (Γ, L, M) to be {γ1, γ2, · · · } with power set and\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\n\n\n100\nChapter 3 - Uncertain Variable\nDeﬁne uncertain variables as\nξi(γ) =\n(\ni,\nif γ = γi\n0,\notherwise\nfor i = 1, 2, · · · and ξ ≡0.\nThen the sequence {ξi} converges a.s. to ξ.\nHowever, the uncertainty distributions of ξi are\nΦi(x) =\n\n\n\n\n\n0,\nif x < 0\n0.5,\nif 0 ≤x < i\n1,\nif x ≥i\nfor i = 1, 2, · · · , respectively, and the uncertainty distribution of ξ is\nΦ(x) =\n(\n0,\nif x < 0\n1,\nif x ≥0.\nIt is clear that Φi(x) does not converge to Φ(x) at x > 0.\nThat is, the\nsequence {ξi} does not converge in distribution to ξ.\n3.12\nUncertain Vector\nAs an extension of uncertain variable, this section introduces a concept of\nuncertain vector whose components are uncertain variables.\nDeﬁnition 3.22 (Liu [113]) A k-dimensional uncertain vector is a function\nξ from an uncertainty space (Γ, L, M) to the set of k-dimensional real vectors\nsuch that {ξ ∈B} is an event for any Borel set B of k-dimensional real\nvectors.\nTheorem 3.42 (Liu [113]) The vector (ξ1, ξ2, · · · , ξk) is an uncertain vector\nif and only if ξ1, ξ2, · · · , ξk are uncertain variables.\nProof: Write ξ = (ξ1, ξ2, · · · , ξk). Suppose that ξ is an uncertain vector on\nthe uncertainty space (Γ, L, M). For any Borel set B of real numbers, the set\nB × ℜk−1 is a Borel set of k-dimensional real vectors. Thus the set\n{ξ1 ∈B} = {ξ1 ∈B, ξ2 ∈ℜ, · · · , ξk ∈ℜ} = {ξ ∈B × ℜk−1}\nis an event. Hence ξ1 is an uncertain variable. A similar process may prove\nthat ξ2, ξ3, · · · , ξk are uncertain variables.\nConversely, suppose that all ξ1, ξ2, · · · , ξk are uncertain variables on the\nuncertainty space (Γ, L, M). We deﬁne\nB =\n\b\nB ⊂ℜk \f\n\f {ξ ∈B} is an event\n\t\n.\n\n\nSection 3.12 - Uncertain Vector\n101\nThe vector ξ = (ξ1, ξ2, · · · , ξk) is proved to be an uncertain vector if we can\nprove that B contains all Borel sets of k-dimensional real vectors. First, the\nclass B contains all open intervals of ℜk because\n(\nξ ∈\nk\nY\ni=1\n(ai, bi)\n)\n=\nk\n\\\ni=1\n{ξi ∈(ai, bi)}\nis an event. Next, the class B is a σ-algebra over ℜk because (i) we have\nℜk ∈B since {ξ ∈ℜk} = Γ; (ii) if B ∈B, then {ξ ∈B} is an event, and\n{ξ ∈Bc} = {ξ ∈B}c\nis an event. This means that Bc ∈B; (iii) if Bi ∈B for i = 1, 2, · · · , then\n{ξ ∈Bi} are events and\n(\nξ ∈\n∞\n[\ni=1\nBi\n)\n=\n∞\n[\ni=1\n{ξ ∈Bi}\nis an event. This means that ∪iBi ∈B. Since the smallest σ-algebra con-\ntaining all open intervals of ℜk is just the Borel algebra over ℜk, the class B\ncontains all Borel sets of k-dimensional real vectors. The theorem is proved.\nDeﬁnition 3.23 (Liu [127]) The k-dimensional uncertain vectors ξ1, ξ2, · · · ,\nξn are said to be independent if for any Borel sets B1, B2, · · · , Bn of k-\ndimensional real vectors, we have\nM\n( n\n\\\ni=1\n(ξi ∈Bi)\n)\n=\nn\n^\ni=1\nM{ξi ∈Bi}.\n(3.247)\nExercise 3.77:\nLet (ξ1, ξ2, ξ3) and (η1, η2, η3) be independent uncertain\nvectors. Show that ξ1 and η2 are independent uncertain variables.\nExercise 3.78:\nLet (ξ1, ξ2, ξ3) and (η1, η2, η3) be independent uncertain\nvectors. Show that (ξ1, ξ2) and (η2, η3) are independent uncertain vectors.\nTheorem 3.43 (Liu [127]) The k-dimensional uncertain vectors ξ1, ξ2, · · · ,\nξn are independent if and only if\nM\n( n\n[\ni=1\n(ξi ∈Bi)\n)\n=\nn\n_\ni=1\nM {ξi ∈Bi}\n(3.248)\nfor any Borel sets B1, B2, · · · , Bn of k-dimensional real vectors.\n\n\n102\nChapter 3 - Uncertain Variable\nProof: If ξ1, ξ2, · · · , ξn are independent, it follows from the duality of un-\ncertain measure that\nM\n( n\n[\ni=1\n(ξi ∈Bi)\n)\n= 1 −M\n( n\n\\\ni=1\n(ξi ∈Bc\ni )\n)\n= 1 −\nn\n^\ni=1\nM{ξi ∈Bc\ni } =\nn\n_\ni=1\nM {ξi ∈Bi} .\nThus (3.248) holds. Conversely, if (3.248) is assumed, then\nM\n( n\n\\\ni=1\n(ξi ∈Bi)\n)\n= 1 −M\n( n\n[\ni=1\n(ξi ∈Bc\ni )\n)\n= 1 −\nn\n_\ni=1\nM{ξi ∈Bc\ni } =\nn\n^\ni=1\nM {ξi ∈Bi} .\nThus ξ1, ξ2, · · · , ξn are independent.\nTheorem 3.44 Let ξ1, ξ2, · · · , ξn be independent uncertain vectors, and let\nf1, f2, · · · , fn be vector-valued measurable functions. Then f1(ξ1), f2(ξ2), · · · ,\nfn(ξn) are also independent uncertain vectors.\nProof: For any Borel sets B1, B2, · · · , Bn of k-dimensional real vectors, it\nfollows from the deﬁnition of independence that\nM\n( n\n\\\ni=1\n(fi(ξi) ∈Bi)\n)\n= M\n( n\n\\\ni=1\n(ξi ∈f −1\ni\n(Bi))\n)\n=\nn\n^\ni=1\nM{ξi ∈f −1\ni\n(Bi)} =\nn\n^\ni=1\nM{fi(ξi) ∈Bi}.\nThus f1(ξ1), f2(ξ2), · · · , fn(ξn) are independent uncertain vectors.\n3.13\nBibliographic Notes\nAs a fundamental concept in uncertainty theory, uncertain variable was pre-\nsented by Liu [113] in 2007. Meanwhile, Liu [113] also proposed the concepts\nof uncertainty distribution, expected value, variance, and moments. Follow-\ning the independence concept of uncertain variables proposed by Liu [116],\nthe operational law was proved by Liu [120] for calculating the uncertainty\ndistribution of function of independent uncertain variables. Nowadays, un-\ncertainty theory has become a branch of mathematics concerned with the\nanalysis of uncertain phenomena.\n\n\nChapter 4\nUncertain Statistics\nUncertain statistics is a set of mathematical techniques for collecting, analyz-\ning and interpreting data by uncertainty theory. This chapter will introduce\nthe method of moments, the method of least squares, uncertain hypothesis\ntest, uncertain regression analysis, and uncertain times series analysis.\n4.1\nEmpirical Uncertainty Distribution\nLet ξ be an uncertain variable with unknown uncertainty distribution Φ, and\nlet\nz1, z2, · · · , zn\n(4.1)\nbe a set of observed data of the uncertain variable ξ. Then the empirical\nuncertainty distribution of ξ is deﬁned as a step function that jumps 1/n in\nheight at each observation, i.e.,\nΦ(z) = 1\nn\nn\nX\ni=1\nI(zi ≤z)\n(4.2)\nwhere I represents the indicator function, i.e.,\nI(zi ≤z) =\n(\n1,\nif zi ≤z\n0,\notherwise.\n(4.3)\nIt is easy to verify that the population mean (i.e., the expected value of\nξ with uncertainty distribution Φ) is just the sample mean, i.e.,\nµ = 1\nn\nn\nX\ni=1\nzi,\n(4.4)\nand the population variance is just the sample variance, i.e.,\nσ2 = 1\nn\nn\nX\ni=1\n(zi −µ)2.\n(4.5)\n\n\n104\nChapter 4 - Uncertain Statistics\nExample 4.1:\nTable 4.1 shows the interarrival times (days) between 59\ncompanies to ﬁle for initial public oﬀering (IPO) from December 22, 2016 to\nJune 16, 2021 in Shanghai Stock Exchange, Shanghai, China.\nTable 4.1: Interarrival Times between 59 Companies to File for IPO from\nDecember 22, 2016 to June 16, 2021 in Shanghai Stock Exchange, China\n148\n81\n53\n86\n87\n75\n198\n7\n0\n180\n0\n3\n220\n59\n35\n22\n7\n6\n0\n0\n4\n7\n2\n0\n0\n0\n0\n4\n0\n0\n11\n78\n68\n0\n0\n8\n4\n3\n0\n5\n2\n4\n0\n9\n26\n8\n0\n71\n5\n4\n12\n2\n12\n0\n0\n7\n14\n0\nDenote the interarrival times in Table 4.1 by z1, z2, · · · , z58.\nSee Fig-\nure 4.1. It is clear that the interarrival times do not have frequency stabil-\nity. This fact makes the generated distribution function deviate from the\nfrequency in the future. This is the reason why the interarrival times are\nregarded as uncertain variables rather than random variables.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ni\nz\n•\n•\n•\n•••\n•\n••\n•\n••\n•\n•\n•\n•\n•••••••••••••••\n••\n•••••••••••\n•\n••\n•\n•••••••••\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n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4.1: Plot of Interarrival Times between 59 Companies to File for\nIPO. Since the frequency is far from being stable, the interarrival times are\nregarded as uncertain variables rather than random variables.\nIt follows from (4.2) that the empirical uncertainty distribution of the\ninterarrival time ξ corresponding to z1, z2, · · · , z58 is\nΦ(z) = 1\n58\n58\nX\ni=1\nI(zi ≤z).\n(4.6)\n\n\nSection 4.2 - Method of Moments\n105\n4.2\nMethod of Moments\nThe method of moments is to estimate the values of unknown parameters\nof an uncertain statistical model by equating the sample moments with the\ncorresponding population moments.\nLet ξ be an uncertain variable with\nuncertainty distribution Φθ where θ = (θ1, θ2, · · · , θp) is an unknown vector\nof parameters, and let\nz1, z2, · · · , zn\n(4.7)\nbe a set of observed data of the uncertain variable ξ. In order to estimate\nthe value of θ based on the observed data (4.7), the method of moments\nwas suggested by Lio-Liu [104]. For each positive integer k, the k-th sample\nmoment of the observed data z1, z2, · · · , zn is\n1\nn\nn\nX\ni=1\nzk\ni ,\nand the k-th population moment of the uncertainty distribution Φθ is\nZ 1\n0\nΦ−1\nθ (α)\n\u0001k dα.\nThe moment estimate θ is then obtained by equating the ﬁrst p sample mo-\nments with the corresponding ﬁrst p population moments, where p is the\nnumber of unknown parameters.\nIn other words, the moment estimate θ\nshould solve the system of equations,\n1\nn\nn\nX\ni=1\nzk\ni =\nZ 1\n0\nΦ−1\nθ (α)\n\u0001k dα,\nk = 1, 2, · · · , p.\n(4.8)\nRemark 4.1: Sometimes the system of equations (4.8) has no solution. In\nthis case, it is suggested to use other methods, e.g., the maximum likelihood\nestimation (Lio-Liu [105] and Liu-Liu [147]) and the method of least squares\n(Liu-Liu [148]).\nTheorem 4.1 Let ξ be an uncertain variable that follows a normal uncer-\ntainty distribution N(e, σ) with unknown expected value e and unknown vari-\nance σ2, and let\nz1, z2, · · · , zn\n(4.9)\nbe a set of observed data of the uncertain variable ξ.\nThen the moment\nestimate of (e, σ) is\nˆ\ne = 1\nn\nn\nX\ni=1\nzi,\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n(zi −ˆ\ne)2.\n(4.10)\n\n\n106\nChapter 4 - Uncertain Statistics\nProof: Since the ﬁrst two population moments of N(e, σ) are e and e2 + σ2,\nit follows from (4.8) that\n\n\n\n\n\n\n\n\n\n\n\n1\nn\nn\nX\ni=1\nzi = e\n1\nn\nn\nX\ni=1\nz2\ni = e2 + σ2\nwhose root is just (4.10). The theorem is thus proved.\nExample 4.2: (Liu-Liu [147]) Table 4.2 shows Henry Hub monthly average\nnatural gas spot prices (US dollars per million BTU) from 2012 to 2016\nreported by US Energy Information Administration.\nTable 4.2: Henry Hub Monthly Average Natural Gas Spot Prices from 2012\nto 2016 reported by US Energy Information Administration\n2.67\n2.51\n2.17\n1.95\n2.43\n2.46\n2.95\n2.84\n2.85\n3.32\n3.54\n3.34\n3.33\n3.33\n3.81\n4.17\n4.04\n3.83\n3.62\n3.43\n3.62\n3.68\n3.64\n4.24\n4.71\n6.00\n4.90\n4.66\n4.58\n4.59\n4.05\n3.91\n3.92\n3.78\n4.12\n3.48\n2.99\n2.87\n2.83\n2.61\n2.85\n2.78\n2.84\n2.77\n2.66\n2.34\n2.09\n1.93\n2.28\n1.99\n1.73\n1.92\n1.92\n2.59\n2.82\n2.82\n2.99\n2.98\n2.55\n3.59\nDenote the gas prices in Table 4.2 by z1, z2, · · · , z60. See Figure 4.2. It\nis clear that the gas prices do not have frequency stability. This fact makes\nthe generated distribution function deviate from the frequency in the future.\nThis is the reason why the gas prices are regarded as uncertain variables\nrather than random variables.\nAssume the gas price ξ follows a normal uncertainty distribution N(e, σ)\nwith unknown expected value e and unknown variance σ2. It follows from\nthe method of moments that the moment estimate of (e, σ) is\nˆ\ne = 1\n60\n60\nX\ni=1\nzi = 3.2035,\n(4.11)\nˆ\nσ2 = 1\n60\n60\nX\ni=1\n(zi −ˆ\ne)2 = 0.75892.\n(4.12)\nThus ξ follows the normal uncertainty distribution N(3.2035, 0.7589), i.e.,\nΦ(z) =\n\u0012\n1 + exp\n\u0012π(3.2035 −z)\n0.7589\n√\n3\n\u0013\u0013−1\n.\n(4.13)\n\n\nSection 4.2 - Method of Moments\n107\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ni\nz\n••\n••\n••\n•••\n•••••\n•\n••••••••\n•\n•\n•\n••••\n••••\n•\n•\n•••••••••\n•••\n•\n••••\n•••••\n•\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n2\n4\n6\n1\n20\n40\n60\nFigure 4.2: Plot of Henry Hub Monthly Average Natural Gas Spot Prices.\nSince the frequency is far from being stable, the gas prices are regarded as\nuncertain variables rather than random variables.\nTheorem 4.2 Let ξ be an uncertain variable that follows a linear uncertainty\ndistribution L(a, b) with unknown parameters a and b, and let\nz1, z2, · · · , zn\n(4.14)\nbe a set of observed data of the uncertain variable ξ.\nThen the moment\nestimate of (a, b) is\nˆ\na = 1\nn\nn\nX\ni=1\nzi −\nv\nu\nu\nt 3\nn\nn\nX\ni=1\nz2\ni −3\nn2\n n\nX\ni=1\nzi\n!\n2\n,\n(4.15)\nˆ\nb = 1\nn\nn\nX\ni=1\nzi +\nv\nu\nu\nt 3\nn\nn\nX\ni=1\nz2\ni −3\nn2\n n\nX\ni=1\nzi\n!\n2\n.\n(4.16)\nProof: Since the ﬁrst two population moments of L(a, b) are (a + b)/2 and\n(a2 + ab + b2)/3, it follows from (4.8) that\n\n\n\n\n\n\n\n\n\n\n\n1\nn\nn\nX\ni=1\nzi = a + b\n2\n1\nn\nn\nX\ni=1\nz2\ni = a2 + ab + b2\n3\nwhose root is just (4.15) and (4.16). The theorem is thus proved.\nExample 4.3: (Liu-Liu [148]) Table 4.3 shows the monthly average elec-\ntricity prices (US dollars per kilowatt-hour) in US from December 2003 to\nMarch 2009 reported by US Bureau of Labor Statistics.\n\n\n108\nChapter 4 - Uncertain Statistics\nTable 4.3: Monthly Average Electricity Prices in US from December 2003 to\nMarch 2009 reported by US Bureau of Labor Statistics\n0.090\n0.091\n0.091\n0.091\n0.091\n0.093\n0.099\n0.099\n0.100\n0.099\n0.094\n0.092\n0.092\n0.094\n0.094\n0.094\n0.095\n0.097\n0.104\n0.105\n0.105\n0.106\n0.102\n0.102\n0.102\n0.108\n0.108\n0.109\n0.109\n0.110\n0.118\n0.118\n0.118\n0.118\n0.112\n0.110\n0.110\n0.113\n0.113\n0.113\n0.113\n0.115\n0.122\n0.122\n0.121\n0.121\n0.117\n0.115\n0.115\n0.116\n0.116\n0.116\n0.118\n0.120\n0.128\n0.131\n0.132\n0.130\n0.126\n0.123\n0.124\n0.126\n0.126\n0.126\nDenote the electricity prices in Table 4.3 by z1, z2, · · · , z64. See Figure 4.3.\nIt is clear that the electricity prices do not have frequency stability. This fact\nmakes the generated distribution function deviate from the frequency in the\nfuture. This is the reason why the electricity prices are regarded as uncertain\nvariables rather than random variables.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ni\nz\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0.09\n0.10\n0.11\n0.12\n0.13\n1\n20\n40\n64\nFigure 4.3: Plot of Monthly Average Electricity Prices. Since the frequency is\nfar from being stable, the electricity prices are regarded as uncertain variables\nrather than random variables.\nAssume the electricity price ξ follows a linear uncertainty distribution\nL(a, b) with unknown parameters a and b. It follows from the method of\nmoments that the moment estimate of (a, b) is\nˆ\na = 1\n64\n64\nX\ni=1\nzi −\nv\nu\nu\nt 3\n64\n64\nX\ni=1\nz2\ni −\n3\n642\n 64\nX\ni=1\nzi\n!2\n= 0.0890,\n\n\nSection 4.3 - Method of Least Squares\n109\nˆ\nb = 1\n64\n64\nX\ni=1\nzi +\nv\nu\nu\nt 3\n64\n64\nX\ni=1\nz2\ni −\n3\n642\n 64\nX\ni=1\nzi\n!2\n= 0.1306.\nTherefore, the electricity price ξ follows the linear uncertainty distribution\nL(0.0890, 0.1306), i.e.,\nΦ(z) =\n\n\n\n\n\n\n\n\n\n0,\nif z ≤0.0890\nz −0.0890\n0.0416\n,\nif 0.0890 < z ≤0.1306\n1,\nif z > 0.1306.\n(4.17)\n4.3\nMethod of Least Squares\nThe method of least squares is to estimate the values of unknown parameters\nof an uncertain statistical model by minimizing the sum of the squared devia-\ntions between the uncertainty distribution to be estimated and the empirical\nuncertainty distribution corresponding to the observed data. Let ξ be an\nuncertain variable with uncertainty distribution Φθ where θ is an unknown\nvector of parameters, and let\nz1, z2, · · · , zn\n(4.18)\nbe a set of observed data of the uncertain variable ξ. In order to estimate the\nvalue of θ based on the observed data (4.18), the method of least squares was\nsuggested by Liu-Liu [148]. Recall that the empirical uncertainty distribution\nof ξ corresponding to (4.18) is\nF(z) = 1\nn\nn\nX\ni=1\nI(zi ≤z)\n(4.19)\nwhere I represents the indicator function. In order to estimate the unknown\nvector θ of parameters, we have the following minimization problem,\nmin\nθ\nn\nX\ni=1\n(Φθ(zi) −F(zi))2 .\n(4.20)\nThe solution ˆ\nθ of the minimization problem (4.20) is called the least squares\nestimate of θ.\nExample 4.4: Let us reconsider Henry Hub monthly average natural gas\nspot prices (US dollars per million BTU) from 2012 to 2016 reported by US\nEnergy Information Administration. See Table 4.2 on Page 106. Assume the\ngas price ξ follows a normal uncertainty distribution N(e, σ) with unknown\n\n\n110\nChapter 4 - Uncertain Statistics\nexpected value e and unknown variance σ2.\nBy solving the minimization\nproblem (4.20), the least squares estimate of (e, σ) is\nˆ\ne = 3.1131,\nˆ\nσ = 0.9800.\nThus ξ follows the normal uncertainty distribution N(3.1131, 0.9800), i.e.,\nΦ(z) =\n\u0012\n1 + exp\n\u0012π(3.1131 −z)\n0.9800\n√\n3\n\u0013\u0013−1\n.\n(4.21)\nExample 4.5: Let us reconsider the monthly average electricity prices (US\ndollars per kilowatt-hour) in US from December 2003 to March 2009 reported\nby US Bureau of Labor Statistics.\nSee Table 4.3 on Page 108.\nAssume\nthe electricity price ξ follows a linear uncertainty distribution L(a, b) with\nunknown parameters a and b. By solving the minimization problem (4.20),\nthe least squares estimate of (a, b) is\nˆ\na = 0.0884,\nˆ\nb = 0.1297.\nThat is, the electricity price ξ follows the linear uncertainty distribution\nL(0.0884, 0.1297), i.e.,\nΦ(x) =\n\n\n\n\n\n\n\n\n\n0,\nif x ≤0.0884\nx −0.0884\n0.0413\n,\nif 0.0884 < x < 0.1297\n1,\nif x ≥0.1297.\n(4.22)\n4.4\nUncertain Hypothesis Test\nA statistical hypothesis is a formal statement relating to one or more uncer-\ntain variables. Uncertain hypothesis test was initialized by Ye-Liu [280] as\na statistical tool that uses uncertainty theory to determine whether or not\nthe hypothesis is correct on the basis of observed data of those uncertain\nvariables.\nLet ξ be an uncertain variable with uncertainty distribution Φθ where θ\nis an unknown vector of parameters. A hypothesis testing problem about θ\ncan be formulated as determining which of the following two statements is\ntrue:\nH0 : θ = θ0\nversus\nH1 : θ ̸= θ0.\n(4.23)\nThe statement H0 is called a null hypothesis, and H1 is called an alternative\nhypothesis. Assume\nz1, z2, · · · , zn\n(4.24)\n\n\nSection 4.4 - Uncertain Hypothesis Test\n111\nare a set of observed data of the uncertain variable ξ. A rejection region for\nthe null hypothesis H0 is a set W consisting of n-dimensional real vectors. If\nthe vector of observed data belongs to the rejection region W, i.e.,\n(z1, z2, · · · , zn) ∈W,\n(4.25)\nthen we reject H0.\nOtherwise, we accept H0.\nA core problem is how to\nchoose a suitable rejection region W for the null hypothesis H0.\nDeﬁnition 4.1 (Ye-Liu [280]) Let ξ be an uncertain variable with uncer-\ntainty distribution Φθ where θ is an unknown vector of parameters. A rejec-\ntion region W ⊂ℜn is said to be a test for the hypotheses\nH0 : θ = θ0\nversus\nH1 : θ ̸= θ0\n(4.26)\nat signiﬁcance level α (e.g., 0.05) if (i) for any (z1, z2, · · · , zn) ∈W, there\nare more than α of indexes i’s with 1 ≤i ≤n such that\nMθ0{ξ ≥zi} ∧Mθ0{ξ ≤zi} < α\n2 ;\n(4.27)\nand (ii) for some θ1 ̸= θ0 and some (z1, z2, · · · , zn) ∈W, there are at least\n1 −α of indexes i’s with 1 ≤i ≤n and more than α of indexes j’s with\n1 ≤j ≤n such that\nMθ1{ξ ≥zi} ∧Mθ1{ξ ≤zi} > Mθ0{ξ ≥zj} ∧Mθ0{ξ ≤zj}.\n(4.28)\nRemark 4.2: Condition (i) in Deﬁnition 4.1 means that θ0 is a bad estimate\nof the parameter θ. This condition is equivalent to\nmax\n|A|≥n(1−α) min\nz∈A Mθ0{ξ ≥z} ∧Mθ0{ξ ≤z} < α\n2\n(4.29)\nwhere A ⊂{z1, z2, · · · , zn}.\nRemark 4.3: Condition (ii) in Deﬁnition 4.1 means that there exists an es-\ntimate θ1 that is better than θ0 for some (z1, z2, · · · , zn) ∈W. This condition\nis equivalent to\nmax\n|A|≥n(1−α) min\nz∈A Mθ1{ξ ≥z} ∧Mθ1{ξ ≤z}\n>\nmax\n|A|≥n(1−α) min\nz∈A Mθ0{ξ ≥z} ∧Mθ0{ξ ≤z}\n(4.30)\nwhere A ⊂{z1, z2, · · · , zn}.\nTheorem 4.3 (Ye-Liu [280]) Let ξ be an uncertain variable that follows a\nnormal uncertainty distribution N(e, σ) with unknown expected value e and\nunknown variance σ2. Then the test for the hypotheses\nH0 : e = e0 and σ = σ0\nversus\nH1 : e ̸= e0 or σ ̸= σ0\n(4.31)\n\n\n112\nChapter 4 - Uncertain Statistics\nat signiﬁcance level α is\nW =\n\u001a\n(z1, z2, · · · , zn) : there are more than α of indexes i’s with\n1 ≤i ≤n such that zi < Φ−1\n0\n\u0010α\n2\n\u0011\nor zi > Φ−1\n0\n\u0010\n1 −α\n2\n\u0011 \u001b\nwhere Φ−1\n0\nis the inverse uncertainty distribution of N(e0, σ0), i.e.,\nΦ−1\n0 (α) = e0 + σ0\n√\n3\nπ\nln\nα\n1 −α.\n(4.32)\nProof: In order to prove that W is a test for the hypotheses (4.31), we need to\nverify that W simultaneously meets Conditions (i) and (ii) in Deﬁnition 4.1.\nFor any (z1, z2, · · · , zn) ∈W, it follows from the deﬁnition of W that there\nare more than α of indexes i’s with 1 ≤i ≤n such that\nzi < Φ−1\n0\n\u0010α\n2\n\u0011\nor\nzi > Φ−1\n0\n\u0010\n1 −α\n2\n\u0011\n.\nIf zi < Φ−1\n0 (α/2), then\nM0{ξ ≤zi} < Φ0\n\u0010\nΦ−1\n0\n\u0010α\n2\n\u0011\u0011\n= α\n2 .\nIf zi > Φ−1\n0 (1 −α/2), then\nM0{ξ ≥zi} < 1 −Φ0\n\u0010\nΦ−1\n0\n\u0010\n1 −α\n2\n\u0011\u0011\n= α\n2 .\nThus\nM0{ξ ≥zi} ∧M0{ξ ≤zi} < α\n2 .\nTherefore, W meets Condition (i) in Deﬁnition 4.1. In order to prove Con-\ndition (ii), we take\nzi = Φ−1\n0\n\u0010α\n2\n\u0011\n−1,\ni = 1, 2, · · · , n.\nIt is clear that (z1, z2, · · · , zn) ∈W. Let Φ1 denote the uncertainty distribu-\ntion of N(e1, σ1), where\ne1 = Φ−1\n0\n\u0010α\n2\n\u0011\n−1,\nσ1 = σ0.\nOn the one hand, we have\nM1{ξ ≥zi} = 1 −Φ1(zi) = 0.5,\nM1{ξ ≤zi} = Φ1(zi) = 0.5,\n\n\nSection 4.4 - Uncertain Hypothesis Test\n113\ni = 1, 2, · · · , n. Thus\nM1{ξ ≥zi} ∧M1{ξ ≤zi} = 0.5 ≥α\n2 ,\ni = 1, 2, · · · , n.\n(4.33)\nOn the other hand, we have\nM0{ξ ≤zj} = Φ0(zj) < α\n2 ,\nj = 1, 2, · · · , n.\nThus\nM0{ξ ≥zj} ∧M0{ξ ≤zj} < α\n2 ,\nj = 1, 2, · · · , n.\n(4.34)\nIt follows from (4.33) and (4.34) that\nM1{ξ ≥zi} ∧M1{ξ ≤zi} > M0{ξ ≥zj} ∧M0{ξ ≤zj},\ni, j = 1, 2, · · · , n. Therefore, W meets Condition (ii) in Deﬁnition 4.1. The\ntheorem is proved.\nExample 4.6: Let us reconsider Henry Hub monthly average natural gas\nspot prices (US dollars per million BTU) from 2012 to 2016 reported by US\nEnergy Information Administration. See Table 4.2 on Page 106. By using\nthe gas prices z1, z2, · · · , z60 in Table 4.2 and the method of moments, we\nhave inferred that the gas price follows the normal uncertainty distribution\nN(3.2035, 0.7589). Let us use uncertain hypothesis test to determine whether\nit ﬁts the observed gas prices. Note that the inverse uncertainty distribution\nof N(3.2035, 0.7589) is\nΦ−1\n0 (α) = 3.2035 + 0.7589\n√\n3\nπ\nln\nα\n1 −α.\nGiven a signiﬁcance level α = 0.05, we obtain\nΦ−1\n0\n\u0010α\n2\n\u0011\n= 1.6707,\nΦ−1\n0\n\u0010\n1 −α\n2\n\u0011\n= 4.7363.\nIt follows from α × 60 = 3 and Theorem 4.3 that the test is\nW =\nn\n(z1, z2, · · · , z60) : there are at least 4 of indexes i’s with\n1 ≤i ≤60 such that zi < 1.6707 or zi > 4.7363\no\n.\nSince only z26, z27 /\n∈[1.6707, 4.7363], we have (z1, z2, · · · , z60) /\n∈W. Thus\nthe normal uncertainty distribution N(3.2035, 0.7589) is a good ﬁt to the gas\nprices in Table 4.2.\nIf we use the method of least squares, then the estimated normal uncer-\ntainty distribution is N(3.1131, 0.9800). It follows from Theorem 4.3 that\nthe test is\nW =\nn\n(z1, z2, · · · , z60) : there are at least 4 of indexes i’s with\n1 ≤i ≤60 such that zi < 1.1337 or zi > 5.0925\no\n.\n\n\n114\nChapter 4 - Uncertain Statistics\nSince only z26 /\n∈[1.337, 5.0925], we have (z1, z2, · · · , z60) /\n∈W.Thus the nor-\nmal uncertainty distribution N(3.1131, 0.9800) is also a good ﬁt to the gas\nprices in Table 4.2.\nTheorem 4.4 (Ye-Liu [282]) Let ξ be an uncertain variable that follows\na linear uncertainty distribution L(a, b) with unknown parameters a and b.\nThen the test for the hypotheses\nH0 : a = a0 and b = b0\nversus\nH1 : a ̸= a0 or b ̸= b0\n(4.35)\nat signiﬁcance level α is\nW =\n\u001a\n(z1, z2, · · · , zn) : there are more than α of indexes i’s with\n1 ≤i ≤n such that zi < Φ−1\n0\n\u0010α\n2\n\u0011\nor zi > Φ−1\n0\n\u0010\n1 −α\n2\n\u0011 \u001b\nwhere Φ−1\n0\nis the inverse uncertainty distribution of L(a0, b0), i.e.,\nΦ−1\n0 (α) = (1 −α)a0 + αb0.\n(4.36)\nProof: In order to prove that W is a test for the hypotheses (4.35), we need to\nverify that W simultaneously meets Conditions (i) and (ii) in Deﬁnition 4.1.\nFor any (z1, z2, · · · , zn) ∈W, it follows from the deﬁnition of W that there\nare more than α of indexes i’s with 1 ≤i ≤n such that\nzi < Φ−1\n0\n\u0010α\n2\n\u0011\nor\nzi > Φ−1\n0\n\u0010\n1 −α\n2\n\u0011\n.\nIf zi < Φ−1\n0 (α/2), then\nM0{ξ ≤zi} < Φ0\n\u0010\nΦ−1\n0\n\u0010α\n2\n\u0011\u0011\n= α\n2 .\nIf zi > Φ−1\n0 (1 −α/2), then\nM0{ξ ≥zi} < 1 −Φ0\n\u0010\nΦ−1\n0\n\u0010\n1 −α\n2\n\u0011\u0011\n= α\n2 .\nThus\nM0{ξ ≥zi} ∧M0{ξ ≤zi} < α\n2 .\nTherefore, W meets Condition (i) in Deﬁnition 4.1. In order to prove Con-\ndition (ii), we take\nzi = a0,\ni = 1, 2, · · · , n.\nIt is clear that (z1, z2, · · · , zn) ∈W. Let Φ1 denote the uncertainty distribu-\ntion of L(a1, b1), where\na1 = 3a0 −b0\n2\n,\nb1 = a0 + b0\n2\n.\n\n\nSection 4.4 - Uncertain Hypothesis Test\n115\nOn the one hand, we have\nM1{ξ ≥zi} = 1 −Φ1(zi) = 0.5,\nM1{ξ ≤zi} = Φ1(zi) = 0.5,\ni = 1, 2, · · · , n. Thus\nM1{ξ ≥zi} ∧M1{ξ ≤zi} = 0.5 ≥α\n2 ,\ni = 1, 2, · · · , n.\n(4.37)\nOn the other hand, we have\nM0{ξ ≤zj} = Φ0(zj) < α\n2 ,\nj = 1, 2, · · · , n.\nThus\nM0{ξ ≥zj} ∧M0{ξ ≤zj} < α\n2 ,\nj = 1, 2, · · · , n.\n(4.38)\nIt follows from (4.37) and (4.38) that\nM1{ξ ≥zi} ∧M1{ξ ≤zi} > M0{ξ ≥zj} ∧M0{ξ ≤zj},\ni, j = 1, 2, · · · , n. Therefore, W meets Condition (ii) in Deﬁnition 4.1. The\ntheorem is proved.\nExample 4.7: Let us reconsider the monthly average electricity prices (US\ndollars per kilowatt-hour) in US from December 2003 to March 2009 reported\nby US Bureau of Labor Statistics. See Table 4.3 on Page 108. By using the\nelectricity prices z1, z2, · · · , z64 in Table 4.3 and the method of moments, we\nhave inferred that the electricity price follows the linear uncertainty distri-\nbution L(0.0890, 0.1306). Let us use uncertain hypothesis test to determine\nwhether it ﬁts the observed electricity prices. Note that the inverse uncer-\ntainty distribution of L(0.0890, 0.1306) is\nΦ−1\n0 (α) = 0.0890(1 −α) + 0.1306α.\nGiven a signiﬁcance level α = 0.05, we obtain\nΦ−1\n0\n\u0010α\n2\n\u0011\n= 0.0901,\nΦ−1\n0\n\u0010\n1 −α\n2\n\u0011\n= 0.1296.\nIt follows from α × 64 = 3.2 and Theorem 4.4 that the test is\nW =\nn\n(z1, z2, · · · , z64) : there are at least 4 of indexes i’s with\n1 ≤i ≤64 such that zi < 0.0901 or zi > 0.1296\no\n.\nSince z1, z56, z57, z58 /\n∈[0.0901, 0.1296], we have (z1, z2, · · · , z64) ∈W. Thus\nthe linear uncertainty distribution L(0.0890, 0.1306) is not a good ﬁt to the\nelectricity prices in Table 4.3.\n\n\n116\nChapter 4 - Uncertain Statistics\nHowever, if we use the method of least squares, then the estimated linear\nuncertainty distribution is L(0.0884, 0.1297). It follows from Theorem 4.4\nthat the test is\nW =\nn\n(z1, z2, · · · , z64) : there are at least 4 of indexes i’s with\n1 ≤i ≤64 such that zi < 0.0895 or zi > 0.1286\no\n.\nSince only z56, z57, z58 /\n∈[0.0895, 0.1286], we have (z1, z2, · · · , z64) /\n∈W. Thus\nthe linear uncertainty distribution L(0.0884, 0.1297) is a good ﬁt to the elec-\ntricity prices in Table 4.3.\n4.5\nUncertain Regression Analysis\nAs a branch of uncertain statistics, uncertain regression analysis is a set of\nstatistical techniques that use uncertainty theory to explore the relationship\nbetween explanatory variables and response variables.\nUncertain Regression Model\nLet (x1, x2, · · · , xp) be a vector of explanatory variables, and let y be a re-\nsponse variable.\nYao-Liu [270] suggested that the functional relationship\nbetween (x1, x2, · · · , xp) and y is expressed by an uncertain regression model\ny = f(x1, x2, · · · , xp|β) + ε\n(4.39)\nwhere β is a vector of parameters, and ε is an uncertain disturbance term\n(uncertain variable).\nExample 4.8: Linear regression model attempts to explain the relationship\nbetween response variable and explanatory variables by a linear function\ny = β0 + β1x1 + β2x2 + · · · + βpxp + ε.\n(4.40)\nExample 4.9: Exponential growth model attempts to describe situations in\nwhich growth begins slowly and then accelerates rapidly by an exponential\nfunction\ny = β1 exp(β2x) + ε,\nβ1 > 0, β2 > 0.\n(4.41)\nExample 4.10: Logarithmic growth model attempts to describe situations\nin which growth accelerates rapidly at ﬁrst and then slowly over time by a\nlogarithmic function\ny = β0 + β1 ln x + ε,\nβ1 > 0, x > 0.\n(4.42)\n\n\nSection 4.5 - Uncertain Regression Analysis\n117\nExample 4.11: Logistic growth model attempts to describe situations in\nwhich growth accelerates gradually at ﬁrst, more rapidly in the middle growth\nperiod and slowly at the end by a logistic function\ny = β0/(1 + β1 exp(−β2x)) + ε,\nβ0 > 0, β1 > 0, β2 > 0.\n(4.43)\nExample 4.12:\nLogistic decay model attempts to describe situations in\nwhich decay accelerates gradually at ﬁrst, more rapidly in the middle decay\nperiod and slowly at the end by a logistic function\ny = β0/(1 + β1 exp(β2x)) + ε,\nβ0 > 0, β1 > 0, β2 > 0.\n(4.44)\nParameter Estimation\nAssume we have a set of observed data (xi1, xi2, · · · , xip, yi), i = 1, 2, · · · , n.\nThe least squares estimate of β in the uncertain regression model\ny = f(x1, x2, · · · , xp|β) + ε\n(4.45)\nis the solution, ˆ\nβ, of the minimization problem,\nmin\nβ\nn\nX\ni=1\n(yi −f(xi1, xi2, · · · , xip|β))2 .\n(4.46)\nThus the ﬁtted regression model is determined by\ny = f(x1, x2, · · · , xp|ˆ\nβ).\n(4.47)\nExample 4.13: Let (xi1, xi2, · · · , xip, yi), i = 1, 2, · · · , n be a set of observed\ndata. The least squares estimate of (β0, β1, · · · , βp) in the linear regression\nmodel\ny = β0 +\np\nX\nj=1\nβjxj + ε\n(4.48)\nsolves the minimization problem,\nmin\nβ0,β1,··· ,βp\nn\nX\ni=1\n\nyi −β0 −\np\nX\nj=1\nβjxij\n\n\n2\n.\n(4.49)\nExample 4.14: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data. The\nleast squares estimate of (β1, β2) in the exponential growth model\ny = β1 exp(β2x) + ε,\nβ1 > 0, β2 > 0\n(4.50)\n\n\n118\nChapter 4 - Uncertain Statistics\nsolves the minimization problem,\nmin\nβ1>0,β2>0\nn\nX\ni=1\n(yi −β1 exp(β2xi))2 .\n(4.51)\nExample 4.15: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data. The\nleast squares estimate of (β0, β1) in the logarithmic growth model\ny = β0 + β1 ln x + ε,\nβ1 > 0, x > 0\n(4.52)\nsolves the minimization problem,\nmin\nβ0,β1>0\nn\nX\ni=1\n(yi −β0 −β1 ln xi)2 .\n(4.53)\nExample 4.16: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data. The\nleast squares estimate of (β0, β1, β2) in the logistic growth model\ny = β0/(1 + β1 exp(−β2x)) + ε,\nβ0 > 0, β1 > 0, β2 > 0\n(4.54)\nsolves the minimization problem,\nmin\nβ0>0,β1>0,β2>0\nn\nX\ni=1\n(yi −β0/(1 + β1 exp(−β2xi)))2 .\n(4.55)\nExample 4.17: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data. The\nleast squares estimate of (β0, β1, β2) in the logistic decay model\ny = β0/(1 + β1 exp(β2x)) + ε,\nβ0 > 0, β1 > 0, β2 > 0\n(4.56)\nsolves the minimization problem,\nmin\nβ0>0,β1>0,β2>0\nn\nX\ni=1\n(yi −β0/(1 + β1 exp(β2xi)))2 .\n(4.57)\nResidual Analysis\nA residual is the diﬀerence between an actual observed value and a value\npredicted by a ﬁtted regression model.\nDeﬁnition 4.2 Let (xi1, xi2, · · · , xip, yi), i = 1, 2, · · · , n be a set of observed\ndata, and let the ﬁtted regression model be\ny = f(x1, x2, · · · , xp|ˆ\nβ).\n(4.58)\nThen for each index i (1 ≤i ≤n), the term\nεi = yi −f(xi1, xi2, · · · , xip|ˆ\nβ)\n(4.59)\nis called the i-th residual.\n\n\nSection 4.5 - Uncertain Regression Analysis\n119\nThe residuals ε1, ε2, · · · , εn will be regarded as the samples of the uncer-\ntain disturbance term ε in the uncertain regression model\ny = f(x1, x2, · · · , xp|ˆ\nβ) + ε.\n(4.60)\nLet us further assume that the uncertain disturbance term ε follows a nor-\nmal uncertainty distribution N(e, σ). Lio-Liu [104] suggested the method of\nmoments that says the expected value of the uncertain disturbance term ε\ncan be estimated as\nˆ\ne = 1\nn\nn\nX\ni=1\nεi\n(4.61)\nand the variance can be estimated as\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n(εi −ˆ\ne)2.\n(4.62)\nTherefore, we obtain an uncertain regression model\ny = f(x1, x2, · · · , xp|ˆ\nβ) + N(ˆ\ne, ˆ\nσ).\n(4.63)\nExample 4.18: Let (xi1, xi2, · · · , xip, yi), i = 1, 2, · · · , n be a set of observed\ndata, and let the ﬁtted linear regression model be\ny = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβjxj.\n(4.64)\nBy using (4.63), we obtain an uncertain linear regression model\ny = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβjxj + N(ˆ\ne, ˆ\nσ)\n(4.65)\nwhere\nˆ\ne = 1\nn\nn\nX\ni=1\n\nyi −ˆ\nβ0 −\np\nX\nj=1\nˆ\nβjxij\n\n,\n(4.66)\nand\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n\nyi −ˆ\nβ0 −\np\nX\nj=1\nˆ\nβjxij −ˆ\ne\n\n\n2\n.\n(4.67)\nExample 4.19: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data, and\nlet the ﬁtted exponential growth model be\ny = ˆ\nβ1 exp(ˆ\nβ2x),\nˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.68)\n\n\n120\nChapter 4 - Uncertain Statistics\nBy using (4.63), we obtain an uncertain exponential growth model\ny = ˆ\nβ1 exp(ˆ\nβ2x) + N(ˆ\ne, ˆ\nσ)\n(4.69)\nwhere\nˆ\ne = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ1 exp(ˆ\nβ2xi)\n\u0011\n,\n(4.70)\nand\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ1 exp(ˆ\nβ2xi) −ˆ\ne\n\u00112\n.\n(4.71)\nExample 4.20: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data, and\nlet the ﬁtted logarithmic growth model be\ny = ˆ\nβ0 + ˆ\nβ1 ln x,\nˆ\nβ1 > 0, x > 0.\n(4.72)\nBy using (4.63), we obtain an uncertain logarithmic growth model\ny = ˆ\nβ0 + ˆ\nβ1 ln x + N(ˆ\ne, ˆ\nσ)\n(4.73)\nwhere\nˆ\ne = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ0 −ˆ\nβ1 ln xi\n\u0011\n,\n(4.74)\nand\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ0 −ˆ\nβ1 ln xi −ˆ\ne\n\u00112\n.\n(4.75)\nExample 4.21: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data, and\nlet the ﬁtted logistic growth model be\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2x)),\nˆ\nβ0 > 0, ˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.76)\nBy using (4.63), we obtain an uncertain logistic growth model\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2x)) + N(ˆ\ne, ˆ\nσ)\n(4.77)\nwhere\nˆ\ne = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2xi))\n\u0011\n,\n(4.78)\nand\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2xi)) −ˆ\ne\n\u00112\n.\n(4.79)\n\n\nSection 4.5 - Uncertain Regression Analysis\n121\nExample 4.22: Let (xi, yi), i = 1, 2, · · · , n be a set of observed data, and\nlet the ﬁtted logistic decay model be\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2x)),\nˆ\nβ0 > 0, ˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.80)\nBy using (4.63), we obtain an uncertain logistic decay model\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2x)) + N(ˆ\ne, ˆ\nσ)\n(4.81)\nwhere\nˆ\ne = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2xi))\n\u0011\n,\n(4.82)\nand\nˆ\nσ2 = 1\nn\nn\nX\ni=1\n\u0010\nyi −ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2xi)) −ˆ\ne\n\u00112\n.\n(4.83)\nUncertain Hypothesis Test\nBased on the observed data (xi1, xi2, · · · , xip, yi), i = 1, 2, · · · , n, we have\ninferred that the uncertain regression model is\ny = f(x1, x2, · · · , xp|ˆ\nβ) + N(ˆ\ne, ˆ\nσ).\n(4.84)\nIn order to test whether the uncertain regression model (4.84) ﬁts the ob-\nserved data, we should test whether the normal uncertainty distribution\nN(ˆ\ne, ˆ\nσ) ﬁts the residuals ε1, ε2, · · · , εn determined by (4.59), i.e.,\nε1, ε2, · · · , εn ∼N(ˆ\ne, ˆ\nσ).\n(4.85)\nIn order to do so, Ye-Liu [280] suggested using uncertain hypothesis test.\nGiven a signiﬁcance level α (e.g. 0.05), it follows from Theorem 4.3 that the\ntest is\nW =\n\u001a\n(z1, z2, · · · , zn) : there are more than α of indexes i’s with\n1 ≤i ≤n such that zi < Φ−1 \u0010α\n2\n\u0011\nor zi > Φ−1 \u0010\n1 −α\n2\n\u0011 \u001b\nwhere Φ−1 is the inverse uncertainty distribution of N(ˆ\ne, ˆ\nσ), i.e.,\nΦ−1(α) = ˆ\ne + ˆ\nσ\n√\n3\nπ\nln\nα\n1 −α.\nIf the vector of the n residuals ε1, ε2, · · · , εn belongs to W, i.e.,\n(ε1, ε2, · · · , εn) ∈W,\n(4.86)\nthen the uncertain regression model (4.84) is not a good ﬁt to the observed\ndata. In this case, we have to re-choose an uncertain regression model. If\n(ε1, ε2, · · · , εn) ̸∈W,\n(4.87)\nthen the uncertain regression model (4.84) is a good ﬁt to the observed data.\n\n\n122\nChapter 4 - Uncertain Statistics\nForecast Uncertain Variable\nBased on the estimated uncertain regression model (4.63), Lio-Liu [104] sug-\ngested that the forecast uncertain variable of response variable y with respect\nto a new explanatory vector (ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp) is\nˆ\ny = f(ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp|ˆ\nβ) + N(ˆ\ne, ˆ\nσ).\n(4.88)\nExample 4.23: Assume the estimated uncertain linear regression model is\ny = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβjxj + N(ˆ\ne, ˆ\nσ).\n(4.89)\nThen the forecast uncertain variable of response variable y with respect to a\nnew explanatory vector (ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp) is\nˆ\ny = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβj ˆ\nxj + N(ˆ\ne, ˆ\nσ).\n(4.90)\nExample 4.24: Assume the estimated uncertain exponential growth model\nis\ny = ˆ\nβ1 exp(ˆ\nβ2x) + N(ˆ\ne, ˆ\nσ),\nˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.91)\nThen the forecast uncertain variable of response variable y with respect to a\nnew explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ1 exp(ˆ\nβ2ˆ\nx) + N(ˆ\ne, ˆ\nσ).\n(4.92)\nExample 4.25: Assume the estimated uncertain logarithmic growth model\nis\ny = ˆ\nβ0 + ˆ\nβ1 ln x + N(ˆ\ne, ˆ\nσ),\nˆ\nβ1 > 0, x > 0.\n(4.93)\nThen the forecast uncertain variable of response variable y with respect to a\nnew explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ0 + ˆ\nβ1 ln ˆ\nx + N(ˆ\ne, ˆ\nσ).\n(4.94)\nExample 4.26: Assume the estimated uncertain logistic growth model is\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2x)) + N(ˆ\ne, ˆ\nσ),\nˆ\nβ0 > 0, ˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.95)\nThen the forecast uncertain variable of response variable y with respect to a\nnew explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2ˆ\nx)) + N(ˆ\ne, ˆ\nσ).\n(4.96)\n\n\nSection 4.5 - Uncertain Regression Analysis\n123\nExample 4.27: Assume the estimated uncertain logistic decay model is\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2x)) + N(ˆ\ne, ˆ\nσ),\nˆ\nβ0 > 0, ˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.97)\nThen the forecast uncertain variable of response variable y with respect to a\nnew explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2ˆ\nx)) + N(ˆ\ne, ˆ\nσ).\n(4.98)\nForecast Value\nAssume the forecast uncertain variable of response variable y with respect to\na new explanatory vector (ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp) is\nˆ\ny = f(ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp|ˆ\nβ) + N(ˆ\ne, ˆ\nσ).\n(4.99)\nLio-Liu [104] suggested that the forecast value is deﬁned as the expected value\nof the forecast uncertain variable ˆ\ny, i.e.,\nˆ\nµ = f(ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp|ˆ\nβ) + ˆ\ne.\n(4.100)\nExample 4.28: Assume the forecast uncertain variable of response variable\ny with respect to a new explanatory vector (ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp) is\nˆ\ny = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβj ˆ\nxj + N(ˆ\ne, ˆ\nσ).\n(4.101)\nThen the forecast value of response variable y is\nˆ\nµ = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβj ˆ\nxj + ˆ\ne.\n(4.102)\nExample 4.29: Assume the forecast uncertain variable of response variable\ny with respect to a new explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ1 exp(ˆ\nβ2ˆ\nx) + N(ˆ\ne, ˆ\nσ).\n(4.103)\nThen the forecast value of response variable y is\nˆ\nµ = ˆ\nβ1 exp(ˆ\nβ2ˆ\nx) + ˆ\ne.\n(4.104)\nExample 4.30: Assume the forecast uncertain variable of response variable\ny with respect to a new explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ0 + ˆ\nβ1 ln ˆ\nx + N(ˆ\ne, ˆ\nσ).\n(4.105)\n\n\n124\nChapter 4 - Uncertain Statistics\nThen the forecast value of response variable y is\nˆ\nµ = ˆ\nβ0 + ˆ\nβ1 ln ˆ\nx + ˆ\ne.\n(4.106)\nExample 4.31: Assume the forecast uncertain variable of response variable\ny with respect to a new explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2ˆ\nx)) + N(ˆ\ne, ˆ\nσ).\n(4.107)\nThen the forecast value of response variable y is\nˆ\nµ = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2ˆ\nx)) + ˆ\ne.\n(4.108)\nExample 4.32: Assume the forecast uncertain variable of response variable\ny with respect to a new explanatory variable ˆ\nx is\nˆ\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2ˆ\nx)) + N(ˆ\ne, ˆ\nσ).\n(4.109)\nThen the forecast value of response variable y is\nˆ\nµ = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2ˆ\nx)) + ˆ\ne.\n(4.110)\nConﬁdence Interval\nAssume the forecast uncertain variable of response variable y with respect to\na new explanatory vector (ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp) is\nˆ\ny = f(ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp|ˆ\nβ) + N(ˆ\ne, ˆ\nσ),\n(4.111)\nand the forecast value of response variable y is\nˆ\nµ = f(ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp|ˆ\nβ) + ˆ\ne.\n(4.112)\nIt follows from (4.111) and (4.112) that ˆ\ny has a normal uncertainty distribu-\ntion N(ˆ\nµ, ˆ\nσ), i.e.,\nˆ\nΨ(z) =\n\u0012\n1 + exp\n\u0012π(ˆ\nµ −z)\n√\n3ˆ\nσ\n\u0013\u0013−1\n.\n(4.113)\nTaking α (e.g., 95%) as the conﬁdence level, it follows from Theorem 3.4 that\nM\n\u001a\nˆ\nΨ−1\n\u00121 −α\n2\n\u0013\n≤ˆ\ny ≤ˆ\nΨ−1\n\u00121 + α\n2\n\u0013\u001b\n≥ˆ\nΨ\n\u0012\nˆ\nΨ−1\n\u00121 + α\n2\n\u0013\u0013\n−ˆ\nΨ\n\u0012\nˆ\nΨ−1\n\u00121 −α\n2\n\u0013\u0013\n= α.\n\n\nSection 4.5 - Uncertain Regression Analysis\n125\nThus Lio-Liu [104] suggested that the α conﬁdence interval of response vari-\nable y is\n\u0014\nˆ\nΨ−1\n\u00121 −α\n2\n\u0013\n, ˆ\nΨ−1\n\u00121 + α\n2\n\u0013\u0015\n.\n(4.114)\nSince ˆ\nΨ is a normal uncertainty distribution, the α conﬁdence interval is also\nwritten as\nˆ\nµ ± ˆ\nσ\n√\n3\nπ\nln 1 + α\n1 −α.\n(4.115)\nExercise 4.1: Let (ˆ\nx1, ˆ\nx2, · · · , ˆ\nxp) be a new explanatory vector. Assume\nthe forecast uncertain variable is\nˆ\ny = ˆ\nβ0 +\np\nX\nj=1\nˆ\nβj ˆ\nxj + N(ˆ\ne, ˆ\nσ).\n(4.116)\nWhat is the α conﬁdence interval of response variable y?\nExercise 4.2: Let ˆ\nx be a new explanatory variable. Assume the forecast\nuncertain variable is\nˆ\ny = ˆ\nβ1 exp(ˆ\nβ2ˆ\nx) + N(ˆ\ne, ˆ\nσ),\nˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.117)\nWhat is the α conﬁdence interval of response variable y?\nExercise 4.3: Let ˆ\nx be a new explanatory variable. Assume the forecast\nuncertain variable is\nˆ\ny = ˆ\nβ0 + ˆ\nβ1 ln ˆ\nx + N(ˆ\ne, ˆ\nσ),\nˆ\nβ1 > 0.\n(4.118)\nWhat is the α conﬁdence interval of response variable y?\nExercise 4.4: Let ˆ\nx be a new explanatory variable. Assume the forecast\nuncertain variable is\nˆ\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(−ˆ\nβ2ˆ\nx)) + N(ˆ\ne, ˆ\nσ),\nˆ\nβ0 > 0, ˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.119)\nWhat is the α conﬁdence interval of response variable y?\nExercise 4.5: Let ˆ\nx be a new explanatory variable. Assume the forecast\nuncertain variable is\nˆ\ny = ˆ\nβ0/(1 + ˆ\nβ1 exp(ˆ\nβ2ˆ\nx)) + N(ˆ\ne, ˆ\nσ),\nˆ\nβ0 > 0, ˆ\nβ1 > 0, ˆ\nβ2 > 0.\n(4.120)\nWhat is the α conﬁdence interval of response variable y?\n\n\n126\nChapter 4 - Uncertain Statistics\nSome Examples\nThis subsection will provide some real-world examples to illustrate the tool\nof uncertain regression analysis.\nExample 4.33: (Liu [165]) As a type of coronavirus, COVID-19 has quickly\nspread all over the world. Table 4.4 shows the cumulative numbers of COVID-\n19 infections in China from January 20 to March 15, 2020 reported by Na-\ntional Health Commission of China.\nTable 4.4: Cumulative Numbers of COVID-19 Infections in China (excluding\nimported cases) from January 20 to March 15, 2020 reported by National\nHealth Commission of China\n291\n440\n571\n830\n1287\n1975\n2744\n4515\n5974\n7711\n9692\n11791\n14380\n17205\n20438\n24324\n28018\n31161\n34546\n37198\n40171\n42638\n44653\n59804\n63851\n66492\n68500\n70548\n72436\n74185\n74576\n75465\n76288\n76936\n77150\n77658\n78064\n78497\n78824\n79251\n79824\n80026\n80151\n80270\n80389\n80516\n80591\n80632\n80668\n80685\n80699\n80708\n80725\n80729\n80733\n80737\nLet t = 1, 2, · · · , 56 represent the dates from January 20 to March 15,\n2020. For example, t = 1 and 25 represent January 20 and February 13,\nrespectively. In order to ﬁnd the functional relationship between t (the date)\nand y (the cumulative number of COVID-19 infections in China), we may\nuse the observed data\n(t, yt),\nt = 1, 2, · · · , 56\nwhere yt are the cumulative numbers shown in Table 4.4 on days t, t =\n1, 2, · · · , 56, respectively. For example,\ny1 = 291,\ny25 = 63851.\nHowever, since the cumulative numbers before February 13 (t = 25) are not\nreal-time data due to the limitation of testing ability, we have to use the last\n32 observed data, i.e.,\n(t, yt),\nt = 25, 26, · · · , 56.\n(4.121)\nIn order to ﬁt the above observed data, we employ the uncertain logistic\ngrowth model,\ny = β0/(1 + β1 exp(−β2t)) + ε,\nβ0 > 0, β1 > 0, β2 > 0\n(4.122)\n\n\nSection 4.5 - Uncertain Regression Analysis\n127\nwhere ε is an uncertain disturbance term. Using the observed data (4.121)\nand solving the minimization problem\nmin\nβ0>0,β1>0,β2>0\n56\nX\nt=25\n(yt −β0/(1 + β1 exp(−β2t)))2,\n(4.123)\nwe obtain a ﬁtted logistic growth model\ny = 80796/(1 + 25.366 exp(−0.1837t)).\n(4.124)\nSee Figure 4.4.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\ny\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n• • • • • • •\n•\n•\n•\n• •\n•\n• • • • • • • • • • • • • • • •\n20000\n40000\n60000\n80000\n0\n10\n20\n30\n40\n50\n60\nFigure 4.4:\nFitted Logistic Growth Model and Cumulative Numbers of\nCOVID-19 Infections in China from January 20 to March 15, 2020\nUsing εt = yt −80796/(1 + 25.366 exp(−0.1837t)), we obtain 32 residuals\nε25, ε26, · · · , ε56 shown in Figure 4.5. It follows from the method of moments\nthat the expected value of the uncertain disturbance term ε is\nˆ\ne = 1\n32\n56\nX\nt=25\nεt = 0.2311\n(4.125)\nand the variance is\nˆ\nσ2 = 1\n32\n56\nX\nt=25\n(εt −ˆ\ne)2 = 286.292.\n(4.126)\nTherefore, we obtain an uncertain logistic growth model for the cumulative\nnumber of COVID-19 infections in China as follows,\ny = 80796/(1 + 25.366 exp(−0.1837t)) + N(0.2311, 286.29).\n(4.127)\nLet us use uncertain hypothesis test to determine whether the uncertain\nlogistic growth model (4.127) ﬁts the observed data.\nThat is, we should\n\n\n128\nChapter 4 - Uncertain Statistics\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nε\n•\n••\n•\n•\n•\n••••\n•••••\n•\n••••••••••••••••\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n−500\n0\n500\n1000\n25\n30\n35\n40\n45\n50\n55\nFigure 4.5: Residual Plot of Uncertain Logistic Growth Model (4.127) Core-\nsponding to the Cumulative Numbers of COVID-19 Infections in China. Since\nthe frequency is far from being stable, the disturbance term is regarded as\nan uncertain variable rather than a random variable.\ntest whether the normal uncertainty distribution N(0.2311, 286.29) ﬁts the\nresiduals ε25, ε26, · · · , ε56. Given a signiﬁcance level α = 0.05, it follows from\nα × 32 = 1.6 and Theorem 4.3 that the test is\nW =\n\b\n(z25, z26, · · · , z56) : there are at least 2 of indexes t’s with\n25 ≤t ≤56 such that zt < −578.03 or zt > 578.49\n\t\n.\nSince only ε30 ̸∈[−578.03, 578.49], we have (ε25, ε26, · · · , ε56) ̸∈W. Thus the\nuncertain logistic growth model (4.127) is a good ﬁt to the observed data.\nBased on the estimated uncertain logistic growth model (4.127), the fore-\ncast uncertain variable of cumulative number of COVID-19 infections in\nChina on day 57 (March 16, 2020) is\nˆ\ny = 80796/(1 + 25.366 exp(−0.1837 × 57)) + N(0.2311, 286.29),\n(4.128)\ni.e., ˆ\ny ∼N(80738, 286.29). Furthermore, the forecast cumulative number ˆ\nµ\nis 80738, and the 95% conﬁdence interval is\n80738 ± 286.29\n√\n3\nπ\nln 1 + 0.95\n1 −0.95,\n(4.129)\ni.e., 80738 ± 578.\nExample 4.34: (Ye [286]) The labour income share, y, is deﬁned as the\nlabour income per unit of GDP. Some inﬂuence factors include (i) trade open-\nness, x1, deﬁned as the total value of exports of domestic source per unit of\nGDP, (ii) ﬁnancial development, x2, deﬁned as the loan balance of ﬁnancial\ninstitutions per unit of GDP, (iii) government intervention, x3, deﬁned as the\n\n\nSection 4.5 - Uncertain Regression Analysis\n129\nlocal government general budgetary expenditure per unit of GDP, and (iv)\nindustrial structure, x4, deﬁned as the value-added of the tertiary industry\nper unit of GDP. Table 4.5 shows the economic data of 22 provinces, 5 au-\ntonomous regions and 4 municipalities of Mainland China in 2014 reported\nby National Bureau of Statistics of China.\nTable 4.5: Economic Data of 22 Provinces, 5 Autonomous Regions and 4\nMunicipalities of Mainland China in 2014 reported by National Bureau of\nStatistics of China\nProvince\nx1\nx2\nx3\nx4\ny\nBeijing\n0.0848\n2.3402\n0.1974\n0.7997\n0.5094\nTianjin\n0.3003\n2.1825\n0.2711\n0.5513\n0.3933\nHebei\n0.1197\n1.1128\n0.1855\n0.4192\n0.4917\nShanxi\n0.0590\n1.3691\n0.2551\n0.4118\n0.5034\nInner Mongolia\n0.0323\n1.2392\n0.3191\n0.4446\n0.5085\nLiaoning\n0.1707\n1.6491\n0.2537\n0.4487\n0.5260\nJilin\n0.0385\n1.2738\n0.2923\n0.4908\n0.4415\nHeilongjiang\n0.0614\n1.1332\n0.2822\n0.3786\n0.5134\nShanghai\n0.4666\n1.8962\n0.1948\n0.6531\n0.4343\nJiangsu\n0.3322\n1.1098\n0.1307\n0.4654\n0.4498\nZhejiang\n0.4314\n1.7830\n0.1289\n0.4676\n0.4650\nAnhui\n0.0723\n1.0104\n0.2071\n0.4104\n0.5009\nFujian\n0.2403\n1.2048\n0.1326\n0.3978\n0.5246\nJiangxi\n0.1062\n1.0019\n0.2478\n0.3703\n0.4165\nShandong\n0.1875\n1.0569\n0.1414\n0.4436\n0.4190\nHenan\n0.0756\n0.7978\n0.1744\n0.3889\n0.5376\nHubei\n0.0522\n0.8912\n0.1747\n0.4331\n0.5322\nHunan\n0.0406\n0.8030\n0.1939\n0.4399\n0.5050\nGuangdong\n0.6716\n1.2457\n0.1343\n0.4871\n0.4791\nGuangxi\n0.0590\n1.1827\n0.2561\n0.4437\n0.5598\nHainan\n0.0745\n1.5632\n0.3189\n0.5304\n0.5137\nChongqing\n0.2180\n1.4108\n0.2260\n0.4690\n0.4820\nSichuan\n0.0780\n1.2028\n0.2352\n0.4252\n0.4705\nGuizhou\n0.0240\n1.3559\n0.3862\n0.4698\n0.5664\nYunnan\n0.0460\n1.3081\n0.3161\n0.4741\n0.5134\nTibet\n0.1338\n1.7226\n1.2616\n0.5760\n0.6176\nShaanxi\n0.0499\n1.1018\n0.2277\n0.3902\n0.4276\nGansu\n0.0196\n1.6992\n0.3899\n0.4601\n0.4892\nQinghai\n0.0105\n2.2578\n0.7292\n0.4963\n0.5388\nNingxia\n0.0662\n1.8628\n0.4044\n0.4622\n0.5291\nXinjiang\n0.1164\n1.3209\n0.3581\n0.4348\n0.5323\nLet i = 1, 2, · · · , 31 represent the provinces, autonomous regions and mu-\n\n\n130\nChapter 4 - Uncertain Statistics\nnicipalities in the ﬁrst column in Table 4.5. Denote the observed data by\nxi1, xi2, xi3, xi4, yi,\ni = 1, 2, · · · , 31.\n(4.130)\nIn order to ﬁnd the functional relationship between x1, x2, x3, x4 and y, we\nemploy the uncertain linear regression model,\ny = β0 + β1x1 + β2x2 + β3x3 + β4x4 + ε\n(4.131)\nwhere ε is an uncertain disturbance term. Using the observed data (4.130)\nand solving the minimization problem\nmin\nβ0,β1,β2,β3,β4\n31\nX\ni=1\n(yi −β0 −β1xi1 −β2xi2 −β3xi3 −β4xi4)2,\n(4.132)\nwe obtain a ﬁtted linear regression model\ny = 0.4757 −0.0672x1 −0.0304x2 + 0.1288x3 + 0.0750x4,\n(4.133)\nand 31 residuals ε1, ε2, · · · , ε31 shown in Figure 4.6.\nIt follows from the\nmethod of moments that the expected value of the uncertain disturbance\nterm ε is\nˆ\ne = 1\n31\n31\nX\ni=1\nεi = 0.0000\n(4.134)\nand the variance is\nˆ\nσ2 = 1\n31\n31\nX\ni=1\n(εi −ˆ\ne)2 = 0.03802.\n(4.135)\nTherefore, we obtain an uncertain linear regression model\ny = 0.4757−0.0672x1−0.0304x2+0.1288x3+0.0750x4+N(0, 0.0380). (4.136)\nFinally, let us use uncertain hypothesis test to determine whether the\nuncertain linear regression model (4.136) ﬁts the observed data. That is, we\nshould test whether the normal uncertainty distribution N(0.0000, 0.0380)\nﬁts the residuals ε1, ε2, · · · , ε31. Given a signiﬁcance level α = 0.05, it follows\nfrom α × 31 = 1.55 and Theorem 4.3 that the test is\nW =\n\b\n(z1, z2, · · · , z31) : there are at least 2 of indexes i’s with\n1 ≤i ≤31 such that zi < −0.0767 or zi > 0.0767\n\t\n.\nSince only ε14 /\n∈[−0.0767, 0.0767], we have (ε1, ε2, · · · , ε31) /\n∈W. Thus the\nuncertain linear regression model (4.136) is a good ﬁt to the observed data.\n\n\nSection 4.6 - Uncertain Time Series Analysis\n131\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. i\nε\n•\n•\n•\n•\n•\n• • •\n•\n•\n• • •\n•\n• •\n• • •\n•\n•\n•\n•\n•\n•\n•\n•\n• •\n•\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n−0.05\n0.00\n0.05\n1\n10\n20\n30\nFigure 4.6: Residual Plot with Descending Order by Labour Income Share (y)\nof Uncertain Linear Regression Model (4.136) Corresponding to the Economic\nData of Mainland China. Since the frequency is far from being stable, the\ndisturbance term is regarded as an uncertain variable rather than a random\nvariable.\n4.6\nUncertain Time Series Analysis\nAs a branch of uncertain statistics, uncertain time series analysis is a set\nof statistical techniques that use uncertainty theory to predict future values\nbased on the previously observed values. Assume Xt are observed values at\ntimes t, t = 1, 2, · · · , n, respectively. Then the sequence of observed values,\nX1, X2, · · · , Xn\n(4.137)\nis called a time series.\nA basic problem of uncertain time series analy-\nsis is to predict the value of Xn+1 based on previously observed values\nX1, X2, · · · , Xn.\nUncertain Time Series Model\nIn order to model the time series (4.137), Yang-Liu [243] suggested an uncer-\ntain time series model,\nXt = a0 +\nk\nX\ni=1\naiXt−i + ε\n(4.138)\nwhere k is called the order of the uncertain time series model, a0, a1, · · · , ak\nare unknown parameters, and ε is an uncertain disturbance term (uncertain\nvariable).\n\n\n132\nChapter 4 - Uncertain Statistics\nParameter Estimation\nBased on the observed values X1, X2, · · · , Xn, the least squares estimate of\n(a0, a1, · · · , ak) in the uncertain time series model (4.138) is the solution of\nthe minimization problem,\nmin\na0,a1,··· ,ak\nn\nX\nt=k+1\n \nXt −a0 −\nk\nX\ni=1\naiXt−i\n!\n2\n.\n(4.139)\nIf the minimization solution is (ˆ\na0, ˆ\na1, · · · , ˆ\nak), then the ﬁtted time series\nmodel is\nXt = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXt−i.\n(4.140)\nResidual Analysis\nA residual is the diﬀerence between an actual observed value and a value\npredicted by a ﬁtted time series model.\nDeﬁnition 4.3 Let X1, X2, · · · , Xn be a time series, and let the ﬁtted time\nseries model be\nXt = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXt−i.\n(4.141)\nThen for each index t (k + 1 ≤t ≤n), the diﬀerence between the actual\nobserved value and the value predicted by the model,\nεt = Xt −ˆ\na0 −\nk\nX\ni=1\nˆ\naiXt−i\n(4.142)\nis called the t-th residual.\nThe residuals εk+1, εk+2, · · · , εn will be regarded as the samples of the\nuncertain disturbance term ε in the uncertain time series model\nXt = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXt−i + ε.\n(4.143)\nLet us further assume that the uncertain disturbance term ε follows a normal\nuncertainty distribution N(e, σ). Yang-Liu [243] suggested the method of\nmoments that says the expected value of the uncertain disturbance term ε\ncan be estimated as\nˆ\ne =\n1\nn −k\nn\nX\nt=k+1\nεt\n(4.144)\n\n\nSection 4.6 - Uncertain Time Series Analysis\n133\nand the variance can be estimated as\nˆ\nσ2 =\n1\nn −k\nn\nX\nt=k+1\n(εt −ˆ\ne)2.\n(4.145)\nTherefore, we obtain an uncertain time series model\nXt = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXt−i + N(ˆ\ne, ˆ\nσ).\n(4.146)\nUncertain Hypothesis Test\nBased on the time series X1, X2, · · · , Xn, we have inferred that the uncertain\ntime series model is\nXt = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXt−i + N(ˆ\ne, ˆ\nσ).\n(4.147)\nIn order to test whether the uncertain time series model (4.147) ﬁts the\nobserved data, we should test whether the normal uncertainty distribution\nN(ˆ\ne, ˆ\nσ) ﬁts the n−k residuals εk+1, εk+2, · · · , εn determined by (4.142), i.e.,\nεk+1, εk+2, · · · , εn ∼N(ˆ\ne, ˆ\nσ).\n(4.148)\nIn order to do so, Ye-Liu [280] suggested using uncertain hypothesis test.\nGiven a signiﬁcance level α (e.g. 0.05), it follows from Theorem 4.3 that the\ntest is\nW =\n\u001a\n(zk+1, zk+2, · · · , zn) : there are more than α of indexes t’s with\nk + 1 ≤t ≤n such that zt < Φ−1 \u0010α\n2\n\u0011\nor zt > Φ−1 \u0010\n1 −α\n2\n\u0011 \u001b\nwhere\nΦ−1(α) = ˆ\ne + ˆ\nσ\n√\n3\nπ\nln\nα\n1 −α.\nIf the vector of the n −k residuals εk+1, εk+2, · · · , εn belongs to W, i.e.,\n(εk+1, εk+2, · · · , εn) ∈W,\n(4.149)\nthen the uncertain time series model (4.146) is not a good ﬁt to the observed\ndata. In this case, we have to re-choose an uncertain time series model. If\n(εk+1, εk+2, · · · , εn) ̸∈W,\n(4.150)\nthen the uncertain time series model (4.146) is a good ﬁt to the observed\ndata.\n\n\n134\nChapter 4 - Uncertain Statistics\nForecast Uncertain Variable\nBased on the estimated uncertain time series model (4.146), Yang-Liu [243]\nsuggested that the forecast uncertain variable of Xn+1 with respect to the\ntime series X1, X2, · · · , Xn is\nˆ\nXn+1 = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXn+1−i + N(ˆ\ne, ˆ\nσ).\n(4.151)\nForecast Value\nBased on the forecast uncertain variable (4.151), Yang-Liu [243] suggested\nthat the forecast value is deﬁned as the expected value of the forecast uncer-\ntain variable ˆ\nXn+1, i.e.,\nˆ\nµ = ˆ\na0 +\nk\nX\ni=1\nˆ\naiXn+1−i + ˆ\ne.\n(4.152)\nConﬁdence Interval\nIt follows from (4.151) and (4.152) that the forecast uncertain variable ˆ\nXn+1\nhas a normal uncertainty distribution N(ˆ\nµ, ˆ\nσ), i.e.,\nˆ\nΨ(z) =\n\u0012\n1 + exp\n\u0012π(ˆ\nµ −z)\n√\n3ˆ\nσ\n\u0013\u0013−1\n.\n(4.153)\nTaking α (e.g., 95%) as the conﬁdence level, it follows from Theorem 3.4 that\nM\n\u001a\nˆ\nΨ−1\n\u00121 −α\n2\n\u0013\n≤ˆ\nXn+1 ≤ˆ\nΨ−1\n\u00121 + α\n2\n\u0013\u001b\n≥ˆ\nΨ\n\u0012\nˆ\nΨ−1\n\u00121 + α\n2\n\u0013\u0013\n−ˆ\nΨ\n\u0012\nˆ\nΨ−1\n\u00121 −α\n2\n\u0013\u0013\n= α.\nThus Yang-Liu [243] suggested that the α conﬁdence interval of Xn+1 is\n\u0014\nˆ\nΨ−1\n\u00121 −α\n2\n\u0013\n, ˆ\nΨ−1\n\u00121 + α\n2\n\u0013\u0015\n.\n(4.154)\nSince ˆ\nΨ is a normal uncertainty distribution, the α conﬁdence interval is also\nwritten as\nˆ\nµ ± ˆ\nσ\n√\n3\nπ\nln 1 + α\n1 −α.\n(4.155)\n\n\nSection 4.6 - Uncertain Time Series Analysis\n135\nSome Examples\nThis subsection will provide some real-world examples to illustrate the tool\nof uncertain time series analysis.\nExample 4.35: (Ye-Zheng [283]) Table 4.6 shows the birth rates in Mainland\nChina from 1962 to 2020 reported by National Bureau of Statistics of China.\nTable 4.6: Birth Rates (‰) in Mainland China from 1962 to 2020 reported\nby National Bureau of Statistics of China\n37.22\n43.60\n39.34\n38.00\n35.21\n34.12\n35.75\n34.25\n33.59\n30.74\n29.92\n28.07\n24.95\n23.13\n20.01\n19.03\n18.25\n17.82\n18.21\n20.91\n22.28\n20.19\n19.90\n21.04\n22.43\n23.33\n22.37\n21.58\n21.06\n19.68\n18.24\n18.09\n17.70\n17.12\n16.98\n16.57\n15.64\n14.64\n14.03\n13.38\n12.86\n12.41\n12.29\n12.40\n12.09\n12.10\n12.14\n11.95\n11.90\n13.27\n14.57\n13.03\n13.83\n11.99\n13.57\n12.64\n10.86\n10.41\n8.52\nLet t = 1962, 1963, · · · , 2020 represent the years from 1962 to 2020. We\ndenote the observed data in Table 4.6 by\nXt,\nt = 1962, 1963, · · · , 2020\n(4.156)\nwhere Xt are the birth rates in years t, t = 1962, 1963, · · · , 2020, respectively.\nIn order to forecast the birth rate in 2021, we employ a 2-order uncertain time\nseries model,\nXt = a0 + a1Xt−1 + a2Xt−2 + ε\n(4.157)\nwhere ε is an uncertain disturbance term. Using the observed data (4.156)\nand solving the minimization problem\nmin\na0,a1,a2\n2020\nX\nt=1964\n(Xt −a0 −a1Xt−1 −a2Xt−2)\n2 ,\n(4.158)\nwe obtain a ﬁtted time series model\nXt = 0.7694 + 0.9198Xt−1 + 0.0111Xt−2,\n(4.159)\nand 57 residuals ε1964, ε1965, · · · , ε2020 shown in Figure 4.7. It follows from\nthe method of moments that the expected value of the uncertain disturbance\nterm ε is\nˆ\ne = 1\n57\n2020\nX\nt=1964\nεt = 0.0000\n(4.160)\n\n\n136\nChapter 4 - Uncertain Statistics\nand the variance is\nˆ\nσ2 = 1\n57\n2020\nX\nt=1964\n(εt −ˆ\ne)2 = 1.17512.\n(4.161)\nTherefore, we obtain an uncertain time series model\nXt = 0.7694 + 0.9198Xt−1 + 0.0111Xt−2 + N(0, 1.1751).\n(4.162)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nε\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\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\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•••\n•\n•\n•\n•\n•\n•\n•\n•\n•••\n••\n•••\n••\n••\n••••\n••\n•••••\n••\n•\n•\n•\n•\n•\n•\n•\n•\n−2\n0\n2\n1964\n1980\n2000\n2020\nFigure 4.7: Residual Plot of Uncertain Time Series Model (4.162) Corre-\nsponding to the Birth Rates in China. Since the frequency is far from being\nstable, the disturbance term is regarded as an uncertain variable rather than\na random variable.\nFinally, let us use uncertain hypothesis test to determine whether the un-\ncertain time series model (4.162) ﬁts the observed data. That is, we should\ntest whether the normal uncertainty distribution N(0, 1.1751) ﬁts the residu-\nals ε1964, ε1965, · · · , ε2020. Given a signiﬁcance level α = 0.05, it follows from\nα × 57 = 2.85 and Theorem 4.3 that the test is\nW = {(z1964, z1965, · · · , z2020) : there are at least 3 of indexes t’s with\n1964 ≤t ≤2020 such that zt < −2.3734 or zt > 2.3734}.\nSince only ε1968, ε1981 ̸∈[−2.3734, 2.3734], we have (ε1964, ε1965, · · · , ε2020) /\n∈\nW. Thus the uncertain time series model (4.162) is a good ﬁt to the birth\nrates in China.\nBased on the uncertain time series model (4.162), the forecast uncertain\nvariable of the birth rate in 2021 is\nˆ\nX2021 = 0.7694 + 0.9198 × 8.52 + 0.0111 × 10.41 + N(0, 1.1751),\n\n\nSection 4.6 - Uncertain Time Series Analysis\n137\ni.e., ˆ\nX2021 ∼N(8.7216, 1.1751). Furthermore, the forecast birth rate ˆ\nµ is\n8.7216, and the 95% conﬁdence interval is\n8.7216 ± 1.1751\n√\n3\nπ\nln 1 + 0.95\n1 −0.95,\ni.e., 8.7216 ± 2.3734.\nExample 4.36: (Liu [145]) Table 4.7 shows China’s populations from 1964\nto 2021 reported by National Bureau of Statistics of China.\nTable 4.7: China’s Populations (millions) from 1964 to 2021 reported by\nNational Bureau of Statistics of China\n704.99\n725.38\n745.42\n763.68\n785.34\n806.71\n829.92\n852.29\n871.77\n892.11\n908.59\n924.20\n937.17\n949.74\n962.59\n975.42\n987.05\n1000.72\n1016.54\n1030.08\n1043.57\n1058.51\n1075.07\n1093.00\n1110.26\n1127.04\n1143.33\n1158.23\n1171.71\n1185.17\n1198.50\n1211.21\n1223.89\n1236.26\n1247.61\n1257.86\n1267.43\n1276.27\n1284.53\n1292.27\n1299.88\n1307.56\n1314.48\n1321.29\n1328.02\n1334.50\n1340.91\n1349.16\n1359.22\n1367.26\n1376.46\n1383.26\n1392.32\n1400.11\n1405.41\n1410.08\n1412.12\n1412.60\nLet t = 1964, 1965, · · · , 2021 represent the years from 1964 to 2021. We\ndenote the observed data in Table 4.7 by\nXt,\nt = 1964, 1965, · · · , 2021\n(4.163)\nwhere Xt are China’s populations in years t, t = 1964, 1965, · · · , 2021, respec-\ntively. In order to forecast China’s population in 2022, we employ a 2-order\nuncertain time series model,\nXt = a0 + a1Xt−1 + a2Xt−2 + ε\n(4.164)\nwhere ε is an uncertain disturbance term. Using the observed data (4.163)\nand solving the minimization problem\nmin\na0,a1,a2\n2021\nX\nt=1966\n(Xt −a0 −a1Xt−1 −a2Xt−2)\n2 ,\n(4.165)\nwe obtain a ﬁtted time series model\nXt = 6.4232 + 1.8340Xt−1 −0.8382Xt−2,\n(4.166)\n\n\n138\nChapter 4 - Uncertain Statistics\nand 56 residuals ε1966, ε1967, · · · , ε2021 shown in Figure 4.8. It follows from\nthe method of moments that the expected value of the uncertain disturbance\nterm ε is\nˆ\ne = 1\n56\n2021\nX\nt=1966\nεt = 0.0000\n(4.167)\nand the variance is\nˆ\nσ2 = 1\n56\n2021\nX\nt=1966\n(εt −ˆ\ne)2 = 1.38642.\n(4.168)\nTherefore, we obtain an uncertain time series model\nXt = 6.4232 + 1.8340Xt−1 −0.8382Xt−2 + N(0, 1.3864).\n(4.169)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nε\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\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\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n••\n•\n•\n•\n•\n•\n•\n••\n•••\n••\n••\n•\n••\n••••••••\n•\n••••\n••\n•\n•\n•\n•\n•\n•\n•\n•\n•\n−3\n0\n3\n1966\n1984\n2003\n2021\nFigure 4.8: Residual Plot of Uncertain Time Series Model (4.169) Corre-\nsponding to China’s Populations. Since the frequency is far from being sta-\nble, the disturbance term is regarded as an uncertain variable rather than a\nrandom variable.\nFinally, let us use uncertain hypothesis test to determine whether the un-\ncertain time series model (4.169) ﬁts the observed data. That is, we should\ntest whether the normal uncertainty distribution N(0, 1.3864) ﬁts the residu-\nals ε1966, ε1967, · · · , ε2021. Given a signiﬁcance level α = 0.05, it follows from\nα × 56 = 2.8 and Theorem 4.3 that the test is\nW = {(z1966, z1967, · · · , z2021) : there are at least 3 of indexes t’s with\n1966 ≤t ≤2021 such that zt < −2.8003 or zt > 2.8003}.\nSince only ε1968, ε1974 ̸∈[−2.8003, 2.8003], we have (ε1966, ε1967, · · · , ε2021) /\n∈\nW. Thus the uncertain time series model (4.169) is a good ﬁt to China’s\npopulations.\n\n\nSection 4.7 - Bibliographic Notes\n139\nBased on the uncertain time series model (4.169), the forecast uncertain\nvariable of China’s population in 2022 is\nˆ\nX2022 = 6.4232 + 1.8340 × 1412.60 −0.8382 × 1412.12 + N(0, 1.3864),\ni.e., ˆ\nX2022 ∼N(1413.4926, 1.3864). Furthermore, the forecast population ˆ\nµ\nis 1413.4926, and the 95% conﬁdence interval is\n1413.4926 ± 1.3864\n√\n3\nπ\nln 1 + 0.95\n1 −0.95,\ni.e., 1413.4926 ± 2.8003.\n4.7\nBibliographic Notes\nSome people believe that probability distribution function can be generated\nas long as enough observed data are acquired. However, numerous empirical\nstudies show that the real world is far from frequency stability. This fact\nmakes the distribution function obtained in practice usually deviate from the\nfuture frequency. This is the reason why we use uncertainty theory rather\nthan probability theory even when numerous observed data are available.\nUncertain statistics is a set of mathematical techniques for collecting, an-\nalyzing and interpreting data by uncertainty theory. The study of uncertain\nstatistics was started by Liu [120] in 2010 and followed by many researchers.\nNowadays, uncertain statistics has achieved fruitful results in both theory\nand practice. Assume some realizations of uncertain variable or vector are\nobserved. In order to estimate the values of unknown parameters of uncer-\ntain statistical model based on those observed data, the method of moments\nwas suggested by Lio-Liu [104], the maximum likelihood estimation was pre-\nsented by Lio-Liu [105] and revised by Liu-Liu [147], and the method of least\nsquares was investigated by Liu-Liu [148]. In addition, Ye-Liu [280] initialized\nuncertain hypothesis test as a statistical tool that uses uncertainty theory to\ndetermine whether a statistical hypothesis is correct on the basis of observed\ndata.\nUncertain regression analysis is a set of statistical techniques that use\nuncertainty theory to explore the relationship between explanatory variables\nand response variables. The study of uncertain regression analysis was started\nby Yao-Liu [270] in 2018 by assuming that the disturbance term is an uncer-\ntain variable instead of a random variable. In order to estimate the uncertain\ndisturbance term, Lio-Liu [104] proposed the method of moments, Lio-Liu\n[105] and Liu-Liu [147] suggested the maximum likelihood estimation, and\nLiu-Liu [148] presented the method of least squares. In addition, Ye-Liu [280]\ndesigned an uncertain hypothesis test for evaluating the appropriateness of\nuncertain regression model, and Ye-Liu [284] presented an uncertain signiﬁ-\ncance test for regression coeﬃcients. Up to now, uncertain regression analysis\nhas been applied in many ﬁelds such as China’s birth rate (Ye-Zheng [283]),\n\n\n140\nChapter 4 - Uncertain Statistics\neconomics (Ye-Liu [284], Jiang-Ye [85], and Ye [286]), epidemic spread (Liu\n[165]), and grain yield (Liu [146]).\nUncertain time series analysis is a set of statistical techniques that use\nuncertainty theory to predict future values based on the previously observed\nvalues. The study of uncertain time series analysis was started by Yang-Liu\n[243] in 2019 by assuming that the disturbance term is an uncertain variable\ninstead of a random variable. This work was immediately followed by many\nresearchers. Up to now, uncertain time series analysis has been applied in\nmany ﬁelds such as China’s birth rate (Ye-Zheng [283]), China’s population\n(Liu [145]), crude oil price (Zhang-Gao [307]), epidemic spread (Ye-Yang\n[278]), grain yield (Ye-Kang [281]), motion analysis (Xie-Lio [229]), and water\ndemand (Li-Wang [100]).\n\n\nChapter 5\nUncertain Programming\nUncertain programming was founded by Liu [115] in 2009. This chapter will\nprovide the theory of uncertain programming, and present some uncertain\nprogramming models for machine scheduling problem, vehicle routing prob-\nlem, and project scheduling problem.\n5.1\nUncertain Programming\nUncertain programming is a type of mathematical programming involving\nuncertain variables. Assume that x is a decision vector, and ξ is an uncer-\ntain vector. Since an uncertain objective function f(x, ξ) cannot be directly\nminimized, we may minimize its expected value, i.e.,\nmin\nx E[f(x, ξ)].\n(5.1)\nIn addition, since the uncertain constraints gj(x, ξ) ≤0, j = 1, 2, · · · , p do not\ndeﬁne a crisp feasible set, it is naturally desired that the uncertain constraints\nhold with conﬁdence levels α1, α2, · · · , αp.\nThen we have a set of chance\nconstraints,\nM{gj(x, ξ) ≤0} ≥αj,\nj = 1, 2, · · · , p.\n(5.2)\nIn order to obtain a decision with minimum expected objective value subject\nto a set of chance constraints, Liu [115] proposed the following uncertain\nprogramming model,\n\n\n\n\n\n\n\nmin\nx E[f(x, ξ)]\nsubject to:\nM{gj(x, ξ) ≤0} ≥αj,\nj = 1, 2, · · · , p.\n(5.3)\n\n\n142\nChapter 5 - Uncertain Programming\nDeﬁnition 5.1 (Liu [115]) A vector x is called a feasible solution to the\nuncertain programming model (5.3) if\nM{gj(x, ξ) ≤0} ≥αj\n(5.4)\nfor j = 1, 2, · · · , p.\nDeﬁnition 5.2 (Liu [115]) A feasible solution x∗is called an optimal solu-\ntion to the uncertain programming model (5.3) if\nE[f(x∗, ξ)] ≤E[f(x, ξ)]\n(5.5)\nfor any feasible solution x.\nTheorem 5.1 Assume the objective function f(x, ξ1, ξ2, · · · , ξn) is continu-\nous, strictly increasing with respect to ξ1, ξ2, · · · , ξm and strictly decreasing\nwith respect to ξm+1, ξm+2, · · · , ξn. If ξ1, ξ2, · · · , ξn are independent uncertain\nvariables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively,\nthen the expected objective function E[f(x, ξ1, ξ2, · · · , ξn)] is equal to\nZ 1\n0\nf(x, Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α))dα.\n(5.6)\nProof: It follows from Theorem 3.18 that the inverse uncertainty distribution\nof f(x, ξ1, ξ2, · · · , ξn) is\nΨ−1(α) = f(x, Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α)).\nBy using Theorem 3.24, we obtain (5.6). The theorem is proved.\nExercise 5.1: Assume f(x, ξ) = h1(x)ξ1 + h2(x)ξ2 + · · · + hn(x)ξn + h0(x)\nwhere h1(x), h2(x), · · · , hn(x), h0(x) are real-valued functions and ξ1, ξ2, · · · ,\nξn are independent uncertain variables. Show that\nE[f(x, ξ)] = h1(x)E[ξ1] + h2(x)E[ξ2] + · · · + hn(x)E[ξn] + h0(x).\n(5.7)\nTheorem 5.2 Assume the constraint function g(x, ξ1, ξ2, · · · , ξn) is contin-\nuous, strictly increasing with respect to ξ1, ξ2, · · · , ξk and strictly decreasing\nwith respect to ξk+1, ξk+2, · · · , ξn. If ξ1, ξ2, · · · , ξn are independent uncertain\nvariables with regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively,\nthen the chance constraint\nM {g(x, ξ1, ξ2, · · · , ξn) ≤0} ≥α\n(5.8)\nholds if and only if\ng(x, Φ−1\n1 (α), · · · , Φ−1\nk (α), Φ−1\nk+1(1 −α), · · · , Φ−1\nn (1 −α)) ≤0.\n(5.9)\n\n\nSection 5.1 - Uncertain Programming\n143\nProof: It follows from the operational law of uncertain variables that the\ninverse uncertainty distribution of g(x, ξ1, ξ2, · · · , ξn) is\nΨ−1(α) = g(x, Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α)).\nThus (5.8) holds if and only if Ψ−1(α) ≤0. The theorem is thus veriﬁed.\nExercise 5.2: Assume x1, x2, · · · , xn are nonnegative decision variables, and\nξ1, ξ2, · · · , ξn, ξ are independent linear uncertain variables L(a1, b1), L(a2, b2),\n· · · , L(an, bn), L(a, b), respectively. Show that for any conﬁdence level α ∈\n(0, 1), the chance constraint\nM\n( n\nX\ni=1\nξixi ≤ξ\n)\n≥α\n(5.10)\nholds if and only if\nn\nX\ni=1\n((1 −α)ai + αbi)xi ≤αa + (1 −α)b.\n(5.11)\nExercise 5.3:\nAssume x1, x2, · · · , xn are nonnegative decision variables,\nand ξ1, ξ2, · · · , ξn, ξ are independent normal uncertain variables N(e1, σ1),\nN(e2, σ2), · · · , N(en, σn), N(e, σ), respectively. Show that for any conﬁdence\nlevel α ∈(0, 1), the chance constraint\nM\n( n\nX\ni=1\nξixi ≤ξ\n)\n≥α\n(5.12)\nholds if and only if\nn\nX\ni=1\n \nei + σi\n√\n3\nπ\nln\nα\n1 −α\n!\nxi ≤e −σ\n√\n3\nπ\nln\nα\n1 −α.\n(5.13)\nExercise 5.4:\nAssume ξ1, ξ2, · · · , ξn are independent uncertain variables\nwith regular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively, and h1(x),\nh2(x), · · · , hn(x), h0(x) are real-valued functions. Show that\nM\n( n\nX\ni=1\nhi(x)ξi ≤h0(x)\n)\n≥α\n(5.14)\nholds if and only if\nn\nX\ni=1\nh+\ni (x)Φ−1\ni (α) −\nn\nX\ni=1\nh−\ni (x)Φ−1\ni (1 −α) ≤h0(x)\n(5.15)\n\n\n144\nChapter 5 - Uncertain Programming\nwhere\nh+\ni (x) =\n(\nhi(x),\nif hi(x) > 0\n0,\nif hi(x) ≤0,\n(5.16)\nh−\ni (x) =\n(\n−hi(x),\nif hi(x) < 0\n0,\nif hi(x) ≥0\n(5.17)\nfor i = 1, 2, · · · , n.\nTheorem 5.3 Assume f(x, ξ1, ξ2, · · · , ξn) is continuous, strictly increasing\nwith respect to ξ1, ξ2, · · · , ξm and strictly decreasing with respect to ξm+1, ξm+2,\n· · · , ξn, and gj(x, ξ1, ξ2, · · · , ξn) are continuous, strictly increasing with re-\nspect to ξ1, ξ2, · · · , ξk and strictly decreasing with respect to ξk+1, ξk+2, · · · , ξn\nfor j = 1, 2, · · · , p. If ξ1, ξ2, · · · , ξn are independent uncertain variables with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively, then the uncer-\ntain programming\n\n\n\n\n\nmin\nx E[f(x, ξ1, ξ2, · · · , ξn)]\nsubject to:\nM{gj(x, ξ1, ξ2, · · · , ξn) ≤0} ≥αj,\nj = 1, 2, · · · , p\n(5.18)\nis equivalent to the crisp mathematical programming\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx\nZ 1\n0\nf(x, Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α))dα\nsubject to:\ngj(x, Φ−1\n1 (αj), · · · , Φ−1\nk (αj), Φ−1\nk+1(1 −αj), · · · , Φ−1\nn (1 −αj)) ≤0\nj = 1, 2, · · · , p.\nProof: It follows from Theorems 5.1 and 5.2 immediately.\n5.2\nNumerical Method\nWhen the objective functions and constraint functions are monotone with\nrespect to the uncertain parameters, the uncertain programming model may\nbe converted to a crisp mathematical programming.\nIt is fortunate for us that almost all objective and constraint functions\nin practical problems are indeed monotone with respect to the uncertain\nparameters (not decision variables).\nFrom the mathematical viewpoint, there is no diﬀerence between crisp\nmathematical programming and classical mathematical programming except\nfor an integral. Thus we may solve it by simplex method, branch-and-bound\nmethod, cutting plane method, implicit enumeration method, interior point\n\n\nSection 5.2 - Numerical Method\n145\nmethod, gradient method, genetic algorithm, particle swarm optimization,\nneural networks, tabu search, and so on.\nExample 5.1: Assume that x1, x2, x3 are nonnegative decision variables,\nξ1, ξ2, ξ3 are independent linear uncertain variables L(1, 2), L(2, 3), L(3, 4),\nand η1, η2, η3 are independent zigzag uncertain variables Z(1, 2, 3), Z(2, 3, 4),\nZ(3, 4, 5), respectively. Consider the uncertain programming,\n\n\n\n\n\n\n\n\n\n\n\nmax\nx1,x2,x3 E\n\u0002√x1 + ξ1 + √x2 + ξ2 + √x3 + ξ3\n\u0003\nsubject to:\nM{(x1 + η1)2 + (x2 + η2)2 + (x3 + η3)2 ≤100} ≥0.9\nx1, x2, x3 ≥0.\nNote that √x1 + ξ1 + √x2 + ξ2 + √x3 + ξ3 is a strictly increasing function\nwith respect to ξ1, ξ2, ξ3, and (x1 + η1)2 + (x2 + η2)2 + (x3 + η3)2 is a strictly\nincreasing function with respect to η1, η2, η3. It is easy to verify that the\nuncertain programming model can be converted to the crisp model,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmax\nx1,x2,x3\nZ 1\n0\n\u0012q\nx1 + Φ−1\n1 (α) +\nq\nx2 + Φ−1\n2 (α) +\nq\nx3 + Φ−1\n3 (α)\n\u0013\ndα\nsubject to:\n(x1 + Ψ−1\n1 (0.9))2 + (x2 + Ψ−1\n2 (0.9))2 + (x3 + Ψ−1\n3 (0.9))2 ≤100\nx1, x2, x3 ≥0\nwhere Φ−1\n1 , Φ−1\n2 , Φ−1\n3 , Ψ−1\n1 , Ψ−1\n2 , Ψ−1\n3\nare inverse uncertainty distributions of\nuncertain variables ξ1, ξ2, ξ3, η1, η2, η3, respectively. The optimal solution is\n(x∗\n1, x∗\n2, x∗\n3) = (2.9735, 1.9735, 0.9735)\nwhose objective value is 6.3419.\nExample 5.2: Assume that x1 and x2 are decision variables, ξ1 and ξ2 are iid\nlinear uncertain variables L(0, π/2). Consider the uncertain programming,\n\n\n\n\n\n\n\nmin\nx1,x2 E [x1 sin(x1 −ξ1) −x2 cos(x2 + ξ2)]\nsubject to:\n0 ≤x1 ≤π\n2 ,\n0 ≤x2 ≤π\n2 .\nIt is clear that x1 sin(x1 −ξ1) −x2 cos(x2 + ξ2) is strictly decreasing with\nrespect to ξ1 and strictly increasing with respect to ξ2. Thus the uncertain\nprogramming is equivalent to the crisp model,\n\n\n\n\n\n\n\n\n\n\n\nmin\nx1,x2\nZ 1\n0\nx1 sin(x1 −Φ−1\n1 (1 −α)) −x2 cos(x2 + Φ−1\n2 (α))\n\u0001\ndα\nsubject to:\n0 ≤x1 ≤π\n2 ,\n0 ≤x2 ≤π\n2\n\n\n146\nChapter 5 - Uncertain Programming\nwhere Φ−1\n1 , Φ−1\n2\nare inverse uncertainty distributions of ξ1, ξ2, respectively.\nThe optimal solution is\n(x∗\n1, x∗\n2) = (0.4026, 0.4026)\nwhose objective value is −0.2708.\n5.3\nMachine Scheduling Problem\nMachine scheduling problem is concerned with ﬁnding an eﬃcient schedule\nduring an uninterrupted period of time for a set of machines to process a set\nof jobs. A lot of research work has been done on this type of problem. The\nstudy of machine scheduling problem with uncertain processing times was\nstarted by Liu [120] in 2010.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nMachine\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nM1\nM2\nM3\nJ1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nJ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nJ3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nJ4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nJ5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nJ6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nJ7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nTime\nMakespan\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 5.1: A Machine Schedule with 3 Machines and 7 Jobs\nIn a machine scheduling problem, we assume that (a) each job can be\nprocessed on any machine without interruption; (b) each machine can process\nonly one job at a time; and (c) the processing times are uncertain variables\nwith known uncertainty distributions. We also use the following indices and\nparameters:\ni = 1, 2, · · · , n: jobs;\nk = 1, 2, · · · , m: machines;\nξik: uncertain processing time of job i on machine k;\nΦik: regular uncertainty distribution of ξik.\nHow to Represent a Schedule?\nLiu [111] suggested that a schedule should be represented by two decision\nvectors x and y, where\nx = (x1, x2, · · · , xn): integer decision vector representing n jobs with\n1 ≤xi ≤n and xi ̸= xj for all i ̸= j, i, j = 1, 2, · · · , n. That is, the sequence\n{x1, x2, · · · , xn} is a rearrangement of {1, 2, · · · , n};\n\n\nSection 5.3 - Machine Scheduling Problem\n147\ny = (y1, y2, · · · , ym−1): integer decision vector with y0 ≡0 ≤y1 ≤y2 ≤\n· · · ≤ym−1 ≤n ≡ym.\nWe note that the schedule is fully determined by the decision vectors x\nand y in the following way. For each k (1 ≤k ≤m), if yk = yk−1, then the\nmachine k is not used; if yk > yk−1, then the machine k is used and processes\njobs xyk−1+1, xyk−1+2, · · · , xyk in turn. Thus the schedule of all machines is\nas follows,\nMachine 1: xy0+1 →xy0+2 →· · · →xy1;\nMachine 2: xy1+1 →xy1+2 →· · · →xy2;\n· · ·\nMachine m: xym−1+1 →xym−1+2 →· · · →xym.\n(5.19)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nM-1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nM-2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nM-3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ny0\ny1\ny2\ny3\nFigure 5.2: Formulation of Schedule in which Machine 1 processes Jobs x1, x2,\nMachine 2 processes Jobs x3, x4 and Machine 3 processes Jobs x5, x6, x7.\nCompletion Times\nLet Ci(x, y, ξ) be the completion times of jobs i, i = 1, 2, · · · , n, respectively.\nFor each k with 1 ≤k ≤m, if the machine k is used (i.e., yk > yk−1), then\nwe have\nCxyk−1+1(x, y, ξ) = ξxyk−1+1k\n(5.20)\nand\nCxyk−1+j(x, y, ξ) = Cxyk−1+j−1(x, y, ξ) + ξxyk−1+jk\n(5.21)\nfor 2 ≤j ≤yk −yk−1.\nIf the machine k is used, then the completion time Cxyk−1+1(x, y, ξ) of\njob xyk−1+1 is an uncertain variable whose inverse uncertainty distribution is\nΨ−1\nxyk−1+1(x, y, α) = Φ−1\nxyk−1+1k(α).\n(5.22)\nGenerally, suppose the completion time Cxyk−1+j−1(x, y, ξ) has an in-\nverse uncertainty distribution Ψ−1\nxyk−1+j−1(x, y, α). Then the completion time\nCxyk−1+j(x, y, ξ) has an inverse uncertainty distribution\nΨ−1\nxyk−1+j(x, y, α) = Ψ−1\nxyk−1+j−1(x, y, α) + Φ−1\nxyk−1+jk(α).\n(5.23)\nThis recursive process may produce all inverse uncertainty distributions of\ncompletion times of jobs.\n\n\n148\nChapter 5 - Uncertain Programming\nMakespan\nNote that, for each k (1 ≤k ≤m), the value Cxyk (x, y, ξ) is just the time\nthat the machine k ﬁnishes all jobs assigned to it. Thus the makespan of the\nschedule (x, y) is determined by\nf(x, y, ξ) = max\n1≤k≤m Cxyk (x, y, ξ)\n(5.24)\nwhose inverse uncertainty distribution is\nΥ−1(x, y, α) = max\n1≤k≤m Ψ−1\nxyk (x, y, α).\n(5.25)\nMachine Scheduling Model\nIn order to minimize the expected makespan E[f(x, y, ξ)], we have the fol-\nlowing machine scheduling model,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx,y E[f(x, y, ξ)]\nsubject to:\n1 ≤xi ≤n,\ni = 1, 2, · · · , n\nxi ̸= xj,\ni ̸= j, i, j = 1, 2, · · · , n\n0 ≤y1 ≤y2 ≤· · · ≤ym−1 ≤n\nxi, yj,\ni = 1, 2, · · · , n,\nj = 1, 2, · · · , m −1,\nintegers.\n(5.26)\nSince Υ−1(x, y, α) is the inverse uncertainty distribution of f(x, y, ξ), the\nmachine scheduling model is simpliﬁed as follows,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx,y\nZ 1\n0\nΥ−1(x, y, α)dα\nsubject to:\n1 ≤xi ≤n,\ni = 1, 2, · · · , n\nxi ̸= xj,\ni ̸= j, i, j = 1, 2, · · · , n\n0 ≤y1 ≤y2 ≤· · · ≤ym−1 ≤n\nxi, yj,\ni = 1, 2, · · · , n,\nj = 1, 2, · · · , m −1,\nintegers.\n(5.27)\nNumerical Experiment\nAssume that there are 3 machines and 7 jobs with the following linear un-\ncertain processing times\nξik ∼L(i, i + k),\ni = 1, 2, · · · , 7, k = 1, 2, 3\nwhere i is the index of jobs and k is the index of machines. The optimal\nsolution is\nx∗= (1, 4, 5, 3, 7, 2, 6),\ny∗= (3, 5).\n(5.28)\n\n\nSection 5.4 - Vehicle Routing Problem\n149\nIn other words, the optimal machine schedule is\nMachine 1: 1 →4 →5\nMachine 2: 3 →7\nMachine 3: 2 →6\nwhose expected makespan is 12.\n5.4\nVehicle Routing Problem\nVehicle routing problem (VRP) is concerned with ﬁnding eﬃcient routes,\nbeginning and ending at a central depot, for a ﬂeet of vehicles to serve a\nnumber of customers.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 5.3: A Vehicle Routing Plan with Single Depot and 7 Customers\nDue to its wide applicability and economic importance, vehicle routing\nproblem has been extensively studied. Liu [120] ﬁrst introduced uncertainty\ntheory into the research area of vehicle routing problem in 2010.\nIn this\nsection, vehicle routing problem will be modelled by uncertain programming\nin which the travel times are assumed to be uncertain variables with known\nuncertainty distributions.\nWe assume that (a) a vehicle will be assigned for only one route on which\nthere may be more than one customer; (b) a customer will be visited by one\nand only one vehicle; (c) each route begins and ends at the depot; and (d) each\ncustomer speciﬁes its time window within which the delivery is permitted or\npreferred to start.\nLet us ﬁrst introduce the following indices and model parameters:\ni = 0: depot;\ni = 1, 2, · · · , n: customers;\nk = 1, 2, · · · , m: vehicles;\nDij: travel distance from customers i to j, i, j = 0, 1, 2, · · · , n;\nTij: uncertain travel time from customers i to j, i, j = 0, 1, 2, · · · , n;\nΦij: regular uncertainty distribution of Tij, i, j = 0, 1, 2, · · · , n;\n\n\n150\nChapter 5 - Uncertain Programming\n[ai, bi]: time window of customer i, i = 1, 2, · · · , n.\nOperational Plan\nLiu [111] suggested that an operational plan should be represented by three\ndecision vectors x, y and t, where\nx = (x1, x2, · · · , xn): integer decision vector representing n customers\nwith 1 ≤xi ≤n and xi ̸= xj for all i ̸= j, i, j = 1, 2, · · · , n. That is, the\nsequence {x1, x2, · · · , xn} is a rearrangement of {1, 2, · · · , n};\ny = (y1, y2, · · · , ym−1): integer decision vector with y0 ≡0 ≤y1 ≤y2 ≤\n· · · ≤ym−1 ≤n ≡ym;\nt = (t1, t2, · · · , tm): each tk represents the starting time of vehicle k at\nthe depot, k = 1, 2, · · · , m.\nWe note that the operational plan is fully determined by the decision\nvectors x, y and t in the following way. For each k (1 ≤k ≤m), if yk = yk−1,\nthen vehicle k is not used; if yk > yk−1, then vehicle k is used and starts from\nthe depot at time tk, and the tour of vehicle k is 0 →xyk−1+1 →xyk−1+2 →\n· · · →xyk →0. Thus the tours of all vehicles are as follows:\nVehicle 1: 0 →xy0+1 →xy0+2 →· · · →xy1 →0;\nVehicle 2: 0 →xy1+1 →xy1+2 →· · · →xy2 →0;\n· · ·\nVehicle m: 0 →xym−1+1 →xym−1+2 →· · · →xym →0.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. x7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nV-1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nV-2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nV-3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ny0\ny1\ny2\ny3\nFigure 5.4: Formulation of Operational Plan in which Vehicle 1 visits Cus-\ntomers x1, x2, Vehicle 2 visits Customers x3, x4 and Vehicle 3 visits Customers\nx5, x6, x7.\nIt is clear that this type of representation is intuitive, and the total number\nof decision variables is n + 2m −1. We also note that the above decision\nvariables x, y and t ensure that: (a) each vehicle will be used at most one\ntime; (b) all tours begin and end at the depot; (c) each customer will be\nvisited by one and only one vehicle; and (d) there is no subtour.\nArrival Times\nLet fi(x, y, t) be the arrival time function of some vehicles at customers i\nfor i = 1, 2, · · · , n. We remind readers that fi(x, y, t) are determined by the\ndecision variables x, y and t, i = 1, 2, · · · , n. Since unloading can start either\n\n\nSection 5.4 - Vehicle Routing Problem\n151\nimmediately, or later, when a vehicle arrives at a customer, the calculation of\nfi(x, y, t) is heavily dependent on the operational strategy. Here we assume\nthat the customer does not permit a delivery earlier than the time window.\nThat is, the vehicle will wait to unload until the beginning of the time window\nif it arrives before the time window. If a vehicle arrives at a customer after\nthe beginning of the time window, unloading will start immediately. For each\nk with 1 ≤k ≤m, if vehicle k is used (i.e., yk > yk−1), then we have\nfxyk−1+1(x, y, t) = tk + T0xyk−1+1\nand\nfxyk−1+j(x, y, t)=fxyk−1+j−1(x, y, t) ∨axyk−1+j−1 + Txyk−1+j−1xyk−1+j\nfor 2 ≤j ≤yk −yk−1. If the vehicle k is used, i.e., yk > yk−1, then the arrival\ntime fxyk−1+1(x, y, t) at the customer xyk−1+1 is an uncertain variable whose\ninverse uncertainty distribution is\nΨ−1\nxyk−1+1(x, y, t, α) = tk + Φ−1\n0xyk−1+1(α).\nGenerally, suppose the arrival time fxyk−1+j−1(x, y, t) has an inverse uncer-\ntainty distribution Ψ−1\nxyk−1+j−1(x, y, t, α). Then fxyk−1+j(x, y, t) has an in-\nverse uncertainty distribution\nΨ−1\nxyk−1+j(x, y, t, α)=Ψ−1\nxyk−1+j−1(x, y, t, α)∨axyk−1+j−1+Φ−1\nxyk−1+j−1xyk−1+j(α)\nfor 2 ≤j ≤yk −yk−1.\nThis recursive process may produce all inverse\nuncertainty distributions of arrival times at customers.\nTravel Distance\nLet g(x, y) be the total travel distance of all vehicles. Then we have\ng(x, y) =\nm\nX\nk=1\ngk(x, y)\n(5.29)\nwhere\ngk(x, y) =\n\n\n\nD0xyk−1+1 +\nyk−1\nP\nj=yk−1+1\nDxjxj+1 + Dxyk 0,\nif yk > yk−1\n0,\nif yk = yk−1\nfor k = 1, 2, · · · , m.\n\n\n152\nChapter 5 - Uncertain Programming\nVehicle Routing Model\nIf we hope that each customer i (1 ≤i ≤n) is visited within its time window\n[ai, bi] with conﬁdence level αi (i.e., the vehicle arrives at customer i before\ntime bi), then we have the following chance constraint,\nM{fi(x, y, t) ≤bi} ≥αi.\n(5.30)\nIf we want to minimize the total travel distance of all vehicles subject to the\ntime window constraint, then we have the following vehicle routing model,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx,y,t g(x, y)\nsubject to:\nM{fi(x, y, t) ≤bi} ≥αi,\ni = 1, 2, · · · , n\n1 ≤xi ≤n,\ni = 1, 2, · · · , n\nxi ̸= xj,\ni ̸= j, i, j = 1, 2, · · · , n\n0 ≤y1 ≤y2 ≤· · · ≤ym−1 ≤n\nxi, yj,\ni = 1, 2, · · · , n,\nj = 1, 2, · · · , m −1,\nintegers\n(5.31)\nwhich is equivalent to\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx,y,t g(x, y)\nsubject to:\nΨ−1\ni (x, y, t, αi) ≤bi,\ni = 1, 2, · · · , n\n1 ≤xi ≤n,\ni = 1, 2, · · · , n\nxi ̸= xj,\ni ̸= j, i, j = 1, 2, · · · , n\n0 ≤y1 ≤y2 ≤· · · ≤ym−1 ≤n\nxi, yj,\ni = 1, 2, · · · , n,\nj = 1, 2, · · · , m −1,\nintegers\n(5.32)\nwhere Ψ−1\ni (x, y, t, α) are the inverse uncertainty distributions of fi(x, y, t)\nfor i = 1, 2, · · · , n, respectively.\nNumerical Experiment\nAssume that there are 3 vehicles and 7 customers with time windows shown in\nTable 5.1, and each customer is visited within time windows with conﬁdence\nlevel 0.90.\nWe also assume that the distances are Dij = |i−j| for i, j = 0, 1, 2, · · · , 7,\nand the travel times are normal uncertain variables\nTij ∼N(2|i −j|, 1),\ni, j = 0, 1, 2, · · · , 7.\nThe optimal solution is\nx∗= (1, 3, 2, 5, 7, 4, 6),\ny∗= (2, 5),\nt∗= (6 : 18, 4 : 18, 8 : 18).\n(5.33)\n\n\nSection 5.5 - Project Scheduling Problem\n153\nTable 5.1: Time Windows of Customers\nNode\nWindow\nNode\nWindow\n1\n[7 : 00, 9 : 00]\n5\n[15 : 00, 17 : 00]\n2\n[7 : 00, 9 : 00]\n6\n[19 : 00, 21 : 00]\n3\n[15 : 00, 17 : 00]\n7\n[19 : 00, 21 : 00]\n4\n[15 : 00, 17 : 00]\nIn other words, the optimal operational plan is\nVehicle 1: depot →1 →3 →depot (the latest starting time is 6:18)\nVehicle 2: depot →2 →5 →7 →depot (the latest starting time is 4:18)\nVehicle 3: depot →4 →6 →depot (the latest starting time is 8:18)\nwhose total travel distance is 32.\n5.5\nProject Scheduling Problem\nProject scheduling problem is to determine the schedule of allocating re-\nsources so as to balance the total cost and the completion time. The study\nof project scheduling problem with uncertain factors was started by Liu [120]\nin 2010. This section presents an uncertain programming model for project\nscheduling problem in which the duration times are assumed to be uncertain\nvariables with known uncertainty distributions.\nProject scheduling is usually represented by a directed acyclic network\nwhere nodes correspond to milestones, and arcs to activities which are basi-\ncally characterized by the times and costs consumed.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 5 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 8\nFigure 5.5: A Project with 8 Milestones and 11 Activities\nLet (V, A) be a directed acyclic graph, where V = {1, 2, · · · , n, n + 1} is\nthe set of nodes, A is the set of arcs, (i, j) ∈A is the arc of the graph (V, A)\nfrom nodes i to j. It is well-known that we can rearrange the indexes of the\nnodes in V such that i < j for all (i, j) ∈A.\n\n\n154\nChapter 5 - Uncertain Programming\nBefore we begin to study project scheduling problem with uncertain ac-\ntivity duration times, we ﬁrst make some assumptions: (a) all of the costs\nneeded are obtained via loans with some given interest rate; and (b) each\nactivity can be processed only if the loan needed is allocated and all the\nforegoing activities are ﬁnished.\nIn order to model the project scheduling problem, we introduce the fol-\nlowing indices and parameters:\nξij: uncertain duration time of activity (i, j) in A;\nΦij: regular uncertainty distribution of ξij;\ncij: cost of activity (i, j) in A;\nr: interest rate;\nxi: integer decision variable representing the allocating time of all loans\nneeded for all activities (i, j) in A.\nStarting Times\nFor simplicity, we write ξ = {ξij : (i, j) ∈A} and x = (x1, x2, · · · , xn). Let\nTi(x, ξ) denote the starting time of all activities (i, j) in A. According to the\nassumptions, the starting time of the total project (i.e., the starting time of\nof all activities (1, j) in A) should be\nT1(x, ξ) = x1\n(5.34)\nwhose inverse uncertainty distribution may be written as\nΨ−1\n1 (x, α) = x1.\n(5.35)\nFrom the starting time T1(x, ξ), we deduce that the starting time of activity\n(2, 5) is\nT2(x, ξ) = x2 ∨(x1 + ξ12)\n(5.36)\nwhose inverse uncertainty distribution may be written as\nΨ−1\n2 (x, α) = x2 ∨(x1 + Φ−1\n12 (α)).\n(5.37)\nGenerally, suppose that the starting time Tk(x, ξ) of all activities (k, i) in A\nhas an inverse uncertainty distribution Ψ−1\nk (x, α). Then the starting time\nTi(x, ξ) of all activities (i, j) in A should be\nTi(x, ξ) = xi ∨max\n(k,i)∈A(Tk(x, ξ) + ξki)\n(5.38)\nwhose inverse uncertainty distribution is\nΨ−1\ni (x, α) = xi ∨max\n(k,i)∈A\nΨ−1\nk (x, α) + Φ−1\nki (α)\n\u0001\n.\n(5.39)\nThis recursive process may produce all inverse uncertainty distributions of\nstarting times of activities.\n\n\nSection 5.5 - Project Scheduling Problem\n155\nCompletion Time\nThe completion time T(x, ξ) of the total project (i.e, the ﬁnish time of all\nactivities (k, n + 1) in A) is\nT(x, ξ) =\nmax\n(k,n+1)∈A (Tk(x, ξ) + ξk,n+1)\n(5.40)\nwhose inverse uncertainty distribution is\nΨ−1(x, α) =\nmax\n(k,n+1)∈A\n\u0010\nΨ−1\nk (x, α) + Φ−1\nk,n+1(α)\n\u0011\n.\n(5.41)\nTotal Cost\nBased on the completion time T(x, ξ), the total cost of the project can be\nwritten as\nC(x, ξ) =\nX\n(i,j)∈A\ncij (1 + r)⌈T (x,ξ)−xi⌉\n(5.42)\nwhere ⌈a⌉represents the minimal integer greater than or equal to a. Note that\nC(x, ξ) is a discrete uncertain variable whose inverse uncertainty distribution\nis\nΥ−1(x, α) =\nX\n(i,j)∈A\ncij (1 + r)⌈Ψ−1(x;α)−xi⌉\n(5.43)\nfor 0 < α < 1.\nProject Scheduling Model\nIn order to minimize the expected cost of the project under the completion\ntime constraint, we may construct the following project scheduling model,\n\n\n\n\n\n\n\n\n\n\n\nmin\nx E[C(x, ξ)]\nsubject to:\nM{T(x, ξ) ≤T0} ≥α0\nx ≥0, integer vector\n(5.44)\nwhere T0 is a due date of the project, α0 is a predetermined conﬁdence level,\nT(x, ξ) is the completion time deﬁned by (5.40), and C(x, ξ) is the total cost\ndeﬁned by (5.42). This model is equivalent to\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx\nZ 1\n0\nΥ−1(x, α)dα\nsubject to:\nΨ−1(x, α0) ≤T0\nx ≥0, integer vector\n(5.45)\n\n\n156\nChapter 5 - Uncertain Programming\nwhere Ψ−1(x, α) is the inverse uncertainty distribution of T(x, ξ) determined\nby (5.41) and Υ−1(x, α) is the inverse uncertainty distribution of C(x, ξ)\ndetermined by (5.43).\nNumerical Experiment\nConsider a project scheduling problem shown by Figure 5.5 in which there are\n8 milestones and 11 activities. Assume that all duration times of activities\nare linear uncertain variables,\nξij ∼L(3i, 3j),\n∀(i, j) ∈A\nand the costs of activities are\ncij = i + j,\n∀(i, j) ∈A.\nIn addition, we also suppose that the interest rate is r = 0.02, the due date\nis T0 = 60, and the conﬁdence level is α0 = 0.85. The optimal solution is\nx∗= (7, 24, 17, 16, 35, 33, 30).\n(5.46)\nIn other words, the optimal allocating times of all loans needed for all activ-\nities are shown in Table 5.2 whose expected total cost is 190.6, and\nM{T(x∗, ξ) ≤60} = 0.88.\nTable 5.2: Optimal Allocating Times of Loans\nDate\n7\n16\n17\n24\n30\n33\n35\nNode\n1\n4\n3\n2\n7\n6\n5\nLoan\n12\n11\n27\n7\n15\n14\n13\n5.6\nUncertain Multiobjective Programming\nIt has been increasingly recognized that many real decision-making problems\ninvolve multiple, noncommensurable, and conﬂicting objectives which should\nbe considered simultaneously. In order to optimize multiple objectives, mul-\ntiobjective programming has been well developed and applied widely. For\nmodelling multiobjective decision-making problems with uncertain param-\neters, Liu-Chen [131] presented the following uncertain multiobjective pro-\ngramming,\n\n\n\n\n\nmin\nx (E[f1(x, ξ)], E[f2(x, ξ)], · · · , E[fm(x, ξ)])\nsubject to:\nM{gj(x, ξ) ≤0} ≥αj,\nj = 1, 2, · · · , p\n(5.47)\n\n\nSection 5.7 - Uncertain Goal Programming\n157\nwhere fi(x, ξ) are objective functions for i = 1, 2, · · · , m, gj(x, ξ) are con-\nstraint functions, and αj are conﬁdence levels for j = 1, 2, · · · , p.\nSince the objectives are usually in conﬂict, there is no optimal solution\nthat simultaneously minimizes all the objective functions. In this case, we\nhave to introduce the concept of Pareto solution, which means that it is\nimpossible to improve any one objective without sacriﬁcing on one or more\nof the other objectives.\nDeﬁnition 5.3 A feasible solution x∗is said to be Pareto to the uncertain\nmultiobjective programming (5.47) if there is no feasible solution x such that\nE[fi(x, ξ)] ≤E[fi(x∗, ξ)],\ni = 1, 2, · · · , m\n(5.48)\nand E[fj(x, ξ)] < E[fj(x∗, ξ)] for at least one index j.\nIf the decision maker has a real-valued preference function aggregating\nthe m objective functions, then we may minimize the aggregating preference\nfunction subject to the same set of chance constraints. This model is referred\nto as a compromise model whose solution is called a compromise solution.\nIt has been proved that the compromise solution is Pareto to the original\nmultiobjective model.\nThe ﬁrst well-known compromise model is set up by weighting the objec-\ntive functions, i.e.,\n\n\n\n\n\n\n\nmin\nx\nm\nP\ni=1\nλiE[fi(x, ξ)]\nsubject to:\nM{gj(x, ξ) ≤0} ≥αj,\nj = 1, 2, · · · , p\n(5.49)\nwhere the weights λ1, λ2, · · · , λm are nonnegative numbers with λ1 + λ2 +\n· · · + λm = 1, for example, λi ≡1/m for i = 1, 2, · · · , m.\nThe second way is related to minimizing the distance function from a\nsolution\n(E[f1(x, ξ)], E[f2(x, ξ)], · · · , E[fm(x, ξ)])\n(5.50)\nto an ideal vector (f ∗\n1 , f ∗\n2 , · · · , f ∗\nm), where f ∗\ni are the optimal values of the\nith objective functions without considering other objectives, i = 1, 2, · · · , m,\nrespectively. That is,\n\n\n\n\n\n\n\n\n\nmin\nx\nm\nP\ni=1\nλi(E[fi(x, ξ)] −f ∗\ni )2\nsubject to:\nM{gj(x, ξ) ≤0} ≥αj,\nj = 1, 2, · · · , p\n(5.51)\nwhere the weights λ1, λ2, · · · , λm are nonnegative numbers with λ1 + λ2 +\n· · · + λm = 1, for example, λi ≡1/m for i = 1, 2, · · · , m.\nBy the third way a compromise solution can be found via an interactive\napproach consisting of a sequence of decision phases and computation phases.\nVarious interactive approaches have been developed.\n\n\n158\nChapter 5 - Uncertain Programming\n5.7\nUncertain Goal Programming\nThe concept of goal programming was presented by Charnes-Cooper [4] in\n1961 and subsequently studied by many researchers. Goal programming can\nbe regarded as a special compromise model for multiobjective optimization\nand has been applied in a wide variety of real-world problems. In multiob-\njective decision-making problems, we assume that the decision-maker is able\nto assign a target level for each goal and the key idea is to minimize the de-\nviations (positive, negative, or both) from the target levels. In the real-world\nsituation, the goals are achievable only at the expense of other goals and\nthese goals are usually incompatible. In order to balance multiple conﬂicting\nobjectives, a decision-maker may establish a hierarchy of importance among\nthese incompatible goals so as to satisfy as many goals as possible in the\norder speciﬁed. For multiobjective decision-making problems with uncertain\nparameters, Liu-Chen [131] proposed an uncertain goal programming,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx\nl\nP\nj=1\nPj\nm\nP\ni=1\n(uijd+\ni + vijd−\ni )\nsubject to:\nE[fi(x, ξ)] + d−\ni −d+\ni = bi,\ni = 1, 2, · · · , m\nM{gj(x, ξ) ≤0} ≥αj,\nj = 1, 2, · · · , p\nd+\ni , d−\ni ≥0,\ni = 1, 2, · · · , m\n(5.52)\nwhere Pj are the preemptive priority factors, uij and vij are the weighting\nfactors, d+\ni are the positive deviations, d−\ni are the negative deviations, fi are\nthe functions in goal constraints, gj are the functions in real constraints, bi\nare the target values, αj are the conﬁdence levels, l is the number of priorities,\nm is the number of goal constraints, and p is the number of real constraints.\nNote that the positive and negative deviations are calculated by\nd+\ni =\n(\nE[fi(x, ξ)] −bi,\nif E[fi(x, ξ)] > bi\n0,\notherwise\n(5.53)\nand\nd−\ni =\n(\nbi −E[fi(x, ξ)],\nif E[fi(x, ξ)] < bi\n0,\notherwise\n(5.54)\nfor each i. Sometimes, the objective function in the goal programming model\nis written as follows,\nlexmin\n( m\nX\ni=1\n(ui1d+\ni + vi1d−\ni ),\nm\nX\ni=1\n(ui2d+\ni + vi2d−\ni ), · · · ,\nm\nX\ni=1\n(uild+\ni + vild−\ni )\n)\nwhere lexmin represents lexicographically minimizing the objective vector.\n\n\nSection 5.8 - Uncertain Multilevel Programming\n159\n5.8\nUncertain Multilevel Programming\nMultilevel programming oﬀers a means of studying decentralized decision\nsystems in which we assume that the leader and followers may have their\nown decision variables and objective functions, and the leader can only inﬂu-\nence the reactions of followers through his own decision variables, while the\nfollowers have full authority to decide how to optimize their own objective\nfunctions in view of the decisions of the leader and other followers.\nAssume that in a decentralized two-level decision system there is one\nleader and m followers. Let x and yi be the control vectors of the leader\nand the ith followers, i = 1, 2, · · · , m, respectively. We also assume that the\nobjective functions of the leader and ith followers are F(x, y1, · · · , ym, ξ) and\nfi(x, y1, · · · , ym, ξ), i = 1, 2, · · · , m, respectively, where ξ is an uncertain\nvector.\nLet the feasible set of control vector x of the leader be deﬁned by the\nchance constraint\nM{G(x, ξ) ≤0} ≥α\n(5.55)\nwhere G is a constraint function, and α is a predetermined conﬁdence level.\nThen for each decision x chosen by the leader, the feasibility of control vec-\ntors yi of the ith followers should be dependent on not only x but also\ny1, · · · , yi−1, yi+1, · · · , ym, and generally represented by the chance con-\nstraints,\nM{gi(x, y1, y2, · · · , ym, ξ) ≤0} ≥αi\n(5.56)\nwhere gi are constraint functions, and αi are predetermined conﬁdence levels,\ni = 1, 2, · · · , m, respectively.\nAssume that the leader ﬁrst chooses his control vector x, and the fol-\nlowers determine their control array (y1, y2, · · · , ym) after that. In order to\nminimize the expected objective of the leader, Liu-Yao [132] proposed the\nfollowing uncertain multilevel programming,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin\nx E[F(x, y∗\n1, y∗\n2, · · · , y∗\nm, ξ)]\nsubject to:\nM{G(x, ξ) ≤0} ≥α\n(y∗\n1, y∗\n2, · · · , y∗\nm) solves problems (i = 1, 2, · · · , m)\n\n\n\n\n\nmin\nyi E[fi(x, y1, y2, · · · , ym, ξ)]\nsubject to:\nM{gi(x, y1, y2, · · · , ym, ξ) ≤0} ≥αi.\n(5.57)\nDeﬁnition 5.4 Let x be a feasible control vector of the leader.\nA Nash\nequilibrium of followers is the feasible array (y∗\n1, y∗\n2, · · · , y∗\nm) with respect to\nx if\nE[fi(x, y∗\n1, · · · , y∗\ni−1, yi, y∗\ni+1, · · · , y∗\nm, ξ)]\n≥E[fi(x, y∗\n1, · · · , y∗\ni−1, y∗\ni , y∗\ni+1, · · · , y∗\nm, ξ)]\n(5.58)\n\n\n160\nChapter 5 - Uncertain Programming\nfor any feasible array (y∗\n1, · · · , y∗\ni−1, yi, y∗\ni+1, · · · , y∗\nm) and i = 1, 2, · · · , m.\nDeﬁnition 5.5 Suppose that x∗is a feasible control vector of the leader and\n(y∗\n1, y∗\n2, · · · , y∗\nm) is a Nash equilibrium of followers with respect to x∗. We call\nthe array (x∗, y∗\n1, y∗\n2, · · · , y∗\nm) a Stackelberg-Nash equilibrium to the uncertain\nmultilevel programming (5.57) if\nE[F(x, y1, y2, · · · , ym, ξ)] ≥E[F(x∗, y∗\n1, y∗\n2, · · · , y∗\nm, ξ)]\n(5.59)\nfor any feasible control vector x and the Nash equilibrium (y1, y2, · · · , ym)\nwith respect to x.\n5.9\nBibliographic Notes\nUncertain programming was ﬁrst proposed by Liu [115] in 2009 and was ap-\nplied to machine scheduling problem, vehicle routing problem and project\nscheduling problem by Liu [120] in 2010. As extensions of uncertain pro-\ngramming theory, Liu-Chen [131] developed uncertain multiobjective pro-\ngramming and uncertain goal programming. In addition, Liu-Yao [132] sug-\ngested uncertain multilevel programming for modeling decentralized decision\nsystems with uncertain factors. After that, uncertain programming has ob-\ntained fruitful results in both theory and practice.\n\n\nChapter 6\nUncertain Risk Analysis\nThe term risk has been used in diﬀerent ways in literature. Here the risk\nis deﬁned as the “accidental loss” plus “uncertain measure of such loss”.\nUncertain risk analysis is a tool to quantify risk via uncertainty theory. One\nmain feature of this topic is to model events that almost never occur. This\nchapter will introduce a deﬁnition of risk index and provide some useful\nformulas for calculating risk index. This chapter will also discuss structural\nrisk analysis in uncertain environments.\n6.1\nLoss Function\nA system usually contains some factors ξ1, ξ2, · · · , ξn that may be under-\nstood as lifetime, strength, demand, production rate, cost, proﬁt, and re-\nsource. Generally speaking, some speciﬁed loss is dependent on those factors.\nAlthough loss is a problem-dependent concept, usually such a loss may be\nrepresented by a loss function.\nDeﬁnition 6.1 Consider a system with factors ξ1, ξ2, · · · , ξn. A function f\nis called a loss function if some speciﬁed loss occurs if and only if\nf(ξ1, ξ2, · · · , ξn) > 0.\n(6.1)\nExample 6.1: Consider a series system in which there are n elements whose\nlifetimes are uncertain variables ξ1, ξ2, · · · , ξn. Such a system works whenever\nall elements work. Thus the system lifetime is\nξ = ξ1 ∧ξ2 ∧· · · ∧ξn.\n(6.2)\nIf the loss is understood as the case that the system fails before the time T,\nthen we have a loss function\nf(ξ1, ξ2, · · · , ξn) = T −ξ1 ∧ξ2 ∧· · · ∧ξn.\n(6.3)\n\n\n162\nChapter 6 - Uncertain Risk Analysis\nInput .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\nOutput\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 6.1: A Series System\nHence the system fails if and only if f(ξ1, ξ2, · · · , ξn) > 0.\nExample 6.2: Consider a parallel system in which there are n elements\nwhose lifetimes are uncertain variables ξ1, ξ2, · · · , ξn. Such a system works\nwhenever at least one element works. Thus the system lifetime is\nξ = ξ1 ∨ξ2 ∨· · · ∨ξn.\n(6.4)\nIf the loss is understood as the case that the system fails before the time T,\nthen the loss function is\nf(ξ1, ξ2, · · · , ξn) = T −ξ1 ∨ξ2 ∨· · · ∨ξn.\n(6.5)\nHence the system fails if and only if f(ξ1, ξ2, · · · , ξn) > 0.\nInput .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. Output\nFigure 6.2: A Parallel System\nExample 6.3: Consider a standby system in which there are n redundant\nelements whose lifetimes are ξ1, ξ2, · · · , ξn. For this system, only one element\nis active, and one of the redundant elements begins to work only when the\nactive element fails. Thus the system lifetime is\nξ = ξ1 + ξ2 + · · · + ξn.\n(6.6)\nIf the loss is understood as the case that the system fails before the time T,\nthen the loss function is\nf(ξ1, ξ2, · · · , ξn) = T −(ξ1 + ξ2 + · · · + ξn).\n(6.7)\nHence the system fails if and only if f(ξ1, ξ2, · · · , ξn) > 0.\n6.2\nRisk Index\nIn practice, the factors ξ1, ξ2, · · · , ξn of a system are usually uncertain vari-\nables rather than known constants. Thus the risk index is deﬁned as the\nuncertain measure that some speciﬁed loss occurs.\n\n\nSection 6.2 - Risk Index\n163\nInput .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. Output\nFigure 6.3: A Standby System\nDeﬁnition 6.2 (Liu [119]) Assume that a system contains uncertain factors\nξ1, ξ2, · · ·, ξn and has a loss function f. Then the risk index is the uncertain\nmeasure that the system is loss-positive, i.e.,\nRisk = M{f(ξ1, ξ2, · · · , ξn) > 0}.\n(6.8)\nTheorem 6.1 Assume that a system contains uncertain factors ξ1, ξ2, · · ·, ξn,\nand has a loss function f. If f(ξ1, ξ2, · · · , ξn) has an uncertainty distribution\nΦ, then the risk index is\nRisk = 1 −Φ(0).\n(6.9)\nProof: It follows from the deﬁnition of risk index and the duality axiom that\nRisk = M{f(ξ1, ξ2, · · · , ξn) > 0}\n= 1 −M{f(ξ1, ξ2, · · · , ξn) ≤0}\n= 1 −Φ(0).\nThe theorem is proved.\nTheorem 6.2 (Liu [119], Risk Index Theorem) Assume a system contains\nindependent uncertain variables ξ1, ξ2, · · · , ξn with regular uncertainty distri-\nbutions Φ1, Φ2, · · · , Φn, respectively. If the loss function f(ξ1, ξ2, · · · , ξn) is\ncontinuous, strictly increasing with respect to ξ1, ξ2, · · · , ξm and strictly de-\ncreasing with respect to ξm+1, ξm+2, · · · , ξn, then the risk index is just the\nroot α of the equation\nf(Φ−1\n1 (1 −α), · · · , Φ−1\nm (1 −α), Φ−1\nm+1(α), · · · , Φ−1\nn (α)) = 0.\n(6.10)\nProof: It follows from Theorem 3.18 that f(ξ1, ξ2, · · · , ξn) has an inverse\nuncertainty distribution\nΦ−1(α) = f(Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α)).\nSince Risk = 1 −Φ(0), it is the solution α of the equation Φ−1(1 −α) = 0.\nThe theorem is thus proved.\nRemark 6.1: Since f(Φ−1\n1 (1−α), · · · , Φ−1\nm (1−α), Φ−1\nm+1(α), · · · , Φ−1\nn (α)) is\na continuous and strictly decreasing function with respect to α, its root may\nbe estimated by the bisection method:\n\n\n164\nChapter 6 - Uncertain Risk Analysis\nStep 1. Set a = 0, b = 1 and c = (a + b)/2.\nStep 2. If f(Φ−1\n1 (1−c), · · · , Φ−1\nm (1−c), Φ−1\nm+1(c), · · · , Φ−1\nn (c)) > 0, then set\na = c. Otherwise, set b = c.\nStep 3. If |b −a| > ε (a predetermined precision), then set c = (b −a)/2\nand go to Step 2. Otherwise, output c as the root.\nRemark 6.2: Keep in mind that sometimes the equation (6.10) may not\nhave a root. In this case, if\nf(Φ−1\n1 (1 −α), · · · , Φ−1\nm (1 −α), Φ−1\nm+1(α), · · · , Φ−1\nn (α)) < 0\n(6.11)\nfor all α, then we set the root α = 0; and if\nf(Φ−1\n1 (1 −α), · · · , Φ−1\nm (1 −α), Φ−1\nm+1(α), · · · , Φ−1\nn (α)) > 0\n(6.12)\nfor all α, then we set the root α = 1.\n6.3\nSeries System\nConsider a series system in which there are n elements whose lifetimes are\nindependent uncertain variables ξ1, ξ2, · · · , ξn with regular uncertainty dis-\ntributions Φ1, Φ2, · · · , Φn, respectively. If the loss is understood as the case\nthat the system fails before the time T, then the loss function is\nf(ξ1, ξ2, · · · , ξn) = T −ξ1 ∧ξ2 ∧· · · ∧ξn\n(6.13)\nand the risk index is\nRisk = M{f(ξ1, ξ2, · · · , ξn) > 0}.\n(6.14)\nSince f is a strictly decreasing function with respect to ξ1, ξ2, · · · , ξn, the risk\nindex theorem says that the risk index is just the root α of the equation\nΦ−1\n1 (α) ∧Φ−1\n2 (α) ∧· · · ∧Φ−1\nn (α) = T.\n(6.15)\nIt is easy to verify that\nRisk = Φ1(T) ∨Φ2(T) ∨· · · ∨Φn(T).\n(6.16)\n6.4\nParallel System\nConsider a parallel system in which there are n elements whose lifetimes\nare independent uncertain variables ξ1, ξ2, · · · , ξn with regular uncertainty\ndistributions Φ1, Φ2, · · · , Φn, respectively. If the loss is understood as the\ncase that the system fails before the time T, then the loss function is\nf(ξ1, ξ2, · · · , ξn) = T −ξ1 ∨ξ2 ∨· · · ∨ξn\n(6.17)\n\n\nSection 6.6 - Structural Risk Analysis\n165\nand the risk index is\nRisk = M{f(ξ1, ξ2, · · · , ξn) > 0}.\n(6.18)\nSince f is a strictly decreasing function with respect to ξ1, ξ2, · · · , ξn, the risk\nindex theorem says that the risk index is just the root α of the equation\nΦ−1\n1 (α) ∨Φ−1\n2 (α) ∨· · · ∨Φ−1\nn (α) = T.\n(6.19)\nIt is easy to verify that\nRisk = Φ1(T) ∧Φ2(T) ∧· · · ∧Φn(T).\n(6.20)\n6.5\nStandby System\nConsider a standby system in which there are n elements whose lifetimes\nare independent uncertain variables ξ1, ξ2, · · · , ξn with regular uncertainty\ndistributions Φ1, Φ2, · · · , Φn, respectively. If the loss is understood as the\ncase that the system fails before the time T, then the loss function is\nf(ξ1, ξ2, · · · , ξn) = T −(ξ1 + ξ2 + · · · + ξn)\n(6.21)\nand the risk index is\nRisk = M{f(ξ1, ξ2, · · · , ξn) > 0}.\n(6.22)\nSince f is a strictly decreasing function with respect to ξ1, ξ2, · · · , ξn, the risk\nindex theorem says that the risk index is just the root α of the equation\nΦ−1\n1 (α) + Φ−1\n2 (α) + · · · + Φ−1\nn (α) = T.\n(6.23)\n6.6\nStructural Risk Analysis\nUncertain structural risk analysis was ﬁrst investigated by Liu [130]. Consider\na structural system in which the strengths and loads are assumed to be\nuncertain variables. We will suppose that a structural system fails whenever\nfor each rod, the load variable exceeds its strength variable. If the structural\nrisk index is deﬁned as the uncertain measure that the structural system fails,\nthen\nRisk = M\n( n\n[\ni=1\n(ξi < ηi)\n)\n(6.24)\nwhere ξ1, ξ2, · · · , ξn are strength variables, and η1, η2, · · · , ηn are load vari-\nables of the n rods.\nExample 6.4: (The Simplest Case) Assume there is only a single strength\nvariable ξ and a single load variable η with regular uncertainty distributions\nΦ and Ψ, respectively. In this case, the structural risk index is\nRisk = M{ξ < η}.\n\n\n166\nChapter 6 - Uncertain Risk Analysis\nIt follows from the risk index theorem that the risk index is just the root α\nof the equation\nΦ−1(α) = Ψ−1(1 −α).\n(6.25)\nEspecially, if the strength variable ξ has a normal uncertainty distribution\nN(es, σs) and the load variable η has a normal uncertainty distribution\nN(el, σl), then the structural risk index is\nRisk =\n\u0012\n1 + exp\n\u0012 π(es −el)\n√\n3(σs + σl)\n\u0013\u0013−1\n.\n(6.26)\nExample 6.5: (Constant Loads) Assume the uncertain strength variables\nξ1, ξ2, · · · , ξn are independent and have regular uncertainty distributions Φ1,\nΦ2, · · · , Φn, respectively. In many cases, the load variables η1, η2, · · · , ηn de-\ngenerate to crisp values c1, c2, · · · , cn (for example, weight limits allowed by\nthe legislation), respectively. In this case, it follows from (6.24) and indepen-\ndence that the structural risk index is\nRisk = M\n( n\n[\ni=1\n(ξi < ci)\n)\n=\nn\n_\ni=1\nM{ξi < ci}.\nThat is,\nRisk = Φ1(c1) ∨Φ2(c2) ∨· · · ∨Φn(cn).\n(6.27)\nExample 6.6: (Independent Load Variables) Assume the uncertain strength\nvariables ξ1, ξ2, · · · , ξn are independent and have regular uncertainty distri-\nbutions Φ1, Φ2, · · · , Φn, respectively. Also assume the uncertain load vari-\nables η1, η2, · · · , ηn are independent and have regular uncertainty distribu-\ntions Ψ1, Ψ2, · · · , Ψn, respectively. In this case, it follows from (6.24) and\nindependence that the structural risk index is\nRisk = M\n( n\n[\ni=1\n(ξi < ηi)\n)\n=\nn\n_\ni=1\nM{ξi < ηi}.\nThat is,\nRisk = α1 ∨α2 ∨· · · ∨αn\n(6.28)\nwhere αi are the roots of the equations\nΦ−1\ni (α) = Ψ−1\ni (1 −α)\n(6.29)\nfor i = 1, 2, · · · , n, respectively.\nHowever, generally speaking, the load variables η1, η2, · · · , ηn are neither\nconstants nor independent. For examples, the load variables η1, η2, · · · , ηn\nmay be functions of independent uncertain variables τ1, τ2, · · · , τm. In this\n\n\nSection 6.6 - Structural Risk Analysis\n167\ncase, the formula (6.28) is no longer valid. Thus we have to deal with those\nstructural systems case by case.\nExample 6.7: (Series System) Consider a structural system shown in Fig-\nure 6.4 that consists of n rods in series and an object.\nAssume that the\nstrength variables of the n rods are uncertain variables ξ1, ξ2, · · · , ξn with\nregular uncertainty distributions Φ1, Φ2, · · · , Φn, respectively. We also as-\nsume that the gravity of the object is an uncertain variable η with regular\nuncertainty distribution Ψ. For each i (1 ≤i ≤n), the load variable of the\nrod i is just the gravity η of the object. Thus the structural system fails\nwhenever the load variable η exceeds at least one of the strength variables\nξ1, ξ2, · · · , ξn. Hence the structural risk index is\nRisk = M\n( n\n[\ni=1\n(ξi < η)\n)\n= M{ξ1 ∧ξ2 ∧· · · ∧ξn < η}.\nDeﬁne the loss function as\nf(ξ1, ξ2, · · · , ξn, η) = η −ξ1 ∧ξ2 ∧· · · ∧ξn.\nThen\nRisk = M{f(ξ1, ξ2, · · · , ξn, η) > 0}.\nSince the loss function f is strictly increasing with respect to η and strictly\ndecreasing with respect to ξ1, ξ2, · · · , ξn, it follows from the risk index theo-\nrem that the risk index is just the root α of the equation\nΨ−1(1 −α) −Φ−1\n1 (α) ∧Φ−1\n2 (α) ∧· · · ∧Φ−1\nn (α) = 0.\n(6.30)\nOr equivalently, let αi be the roots of the equations\nΨ−1(1 −α) = Φ−1\ni (α)\n(6.31)\nfor i = 1, 2, · · · , n, respectively. Then the structural risk index is\nRisk = α1 ∨α2 ∨· · · ∨αn.\n(6.32)\nExample 6.8: Consider a structural system shown in Figure 6.5 that consists\nof 2 rods and an object.\nAssume that the strength variables of the left\nand right rods are uncertain variables ξ1 and ξ2 with regular uncertainty\ndistributions Φ1 and Φ2, respectively. We also assume that the gravity of the\nobject is an uncertain variable η with regular uncertainty distribution Ψ. In\nthis case, the load variables of left and right rods are respectively equal to\nη sin θ2\nsin(θ1 + θ2),\nη sin θ1\nsin(θ1 + θ2).\n\n\n168\nChapter 6 - Uncertain Risk Analysis\n/ / / / / / / / / / / / / / / /\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. · · · ·\n· · · ·\n· · · ·\n· · · ·\nFigure 6.4: A Structural System with n Rods and an Object\nThus the structural system fails whenever for any one rod, the load variable\nexceeds its strength variable. Hence the structural risk index is\nRisk = M\n\u001a\u0012\nξ1 <\nη sin θ2\nsin(θ1 + θ2)\n\u0013\n∪\n\u0012\nξ2 <\nη sin θ1\nsin(θ1 + θ2)\n\u0013\u001b\n= M\n\u001a\u0012\nξ1\nsin θ2\n<\nη\nsin(θ1 + θ2)\n\u0013\n∪\n\u0012\nξ2\nsin θ1\n<\nη\nsin(θ1 + θ2)\n\u0013\u001b\n= M\n\u001a\nξ1\nsin θ2\n∧\nξ2\nsin θ1\n<\nη\nsin(θ1 + θ2)\n\u001b\nDeﬁne the loss function as\nf(ξ1, ξ2, η) =\nη\nsin(θ1 + θ2) −\nξ1\nsin θ2\n∧\nξ2\nsin θ1\n.\nThen\nRisk = M{f(ξ1, ξ2, η) > 0}.\nSince the loss function f is strictly increasing with respect to η and strictly\ndecreasing with respect to ξ1, ξ2, it follows from the risk index theorem that\nthe risk index is just the root α of the equation\nΨ−1(1 −α)\nsin(θ1 + θ2) −Φ−1\n1 (α)\nsin θ2\n∧Φ−1\n2 (α)\nsin θ1\n= 0.\n(6.33)\nOr equivalently, let α1 be the root of the equation\nΨ−1(1 −α)\nsin(θ1 + θ2) = Φ−1\n1 (α)\nsin θ2\n(6.34)\n\n\nSection 6.7 - Value-at-Risk\n169\nand let α2 be the root of the equation\nΨ−1(1 −α)\nsin(θ1 + θ2) = Φ−1\n2 (α)\nsin θ1\n.\n(6.35)\nThen the structural risk index is\nRisk = α1 ∨α2.\n(6.36)\n/ / / / / / / / / / / / / / / /\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nθ1 θ2\n· · · ·\n· · · ·\n· · · ·\n· · · ·\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 6.5: A Structural System with 2 Rods and an Object\n6.7\nValue-at-Risk\nAs a substitute of risk index (6.8), a concept of value-at-risk is given by the\nfollowing deﬁnition.\nDeﬁnition 6.3 (Peng [184]) Assume that a system contains uncertain fac-\ntors ξ1, ξ2, · · ·, ξn and has a loss function f. Then the value-at-risk is deﬁned\nas\nVaR(α) = sup{x | M{f(ξ1, ξ2, · · · , ξn) ≥x} ≥α}.\n(6.37)\nNote that VaR(α) represents the maximum possible loss when α percent of\nthe right tail distribution is ignored. In other words, the loss f(ξ1, ξ2, · · · , ξn)\nwill exceed VaR(α) with uncertain measure α. See Figure 6.6. If the uncer-\ntainty distribution Φ(x) of f(ξ1, ξ2, · · · , ξn) is continuous, then\nVaR(α) = sup {x | Φ(x) ≤1 −α} .\n(6.38)\nIf its inverse uncertainty distribution Φ−1(α) exists, then\nVaR(α) = Φ−1(1 −α).\n(6.39)\nIt is also easy to show that VaR(α) is a monotone decreasing function with\nrespect to α.\n\n\n170\nChapter 6 - Uncertain Risk Analysis\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\nα\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nVaR(α)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n......................................................................\n..................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 6.6: Value-at-Risk\nTheorem 6.3 (Peng [184], Value-at-Risk Theorem) Assume a system con-\ntains independent uncertain variables ξ1, ξ2, · · · , ξn with regular uncertainty\ndistributions Φ1, Φ2, · · · , Φn, respectively. If the loss function f(ξ1, ξ2, · · · , ξn)\nis continuous, strictly increasing with respect to ξ1, ξ2, · · · , ξm and strictly de-\ncreasing with respect to ξm+1, ξm+2, · · · , ξn, then\nVaR(α) = f(Φ−1\n1 (1 −α), · · · , Φ−1\nm (1 −α), Φ−1\nm+1(α), · · · , Φ−1\nn (α)).\n(6.40)\nProof: It follows from the operational law of uncertain variables that the\nloss f(ξ1, ξ2, · · · , ξn) has an inverse uncertainty distribution\nΦ−1(α) = f(Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α)).\nThe theorem follows from (6.39) immediately.\n6.8\nExpected Loss\nLiu-Ralescu [158] proposed a concept of expected loss that is the expected\nvalue of the loss f(ξ1, ξ2, · · · , ξn) given f(ξ1, ξ2, · · · , ξn) > 0. A formal deﬁ-\nnition is given below.\nDeﬁnition 6.4 (Liu-Ralescu [158]) Assume that a system contains uncer-\ntain factors ξ1, ξ2, · · ·, ξn and has a loss function f. Then the expected loss is\ndeﬁned as\nL =\nZ +∞\n0\nM{f(ξ1, ξ2, · · · , ξn) ≥x}dx.\n(6.41)\nIf Φ(x) is the uncertainty distribution of the loss f(ξ1, ξ2, · · · , ξn), then\nwe immediately have\nL =\nZ +∞\n0\n(1 −Φ(x))dx.\n(6.42)\n\n\nSection 6.9 - Bibliographic Notes\n171\nIf its inverse uncertainty distribution Φ−1(α) exists, then the expected loss\nis\nL =\nZ 1\n0\nΦ−1(α)\n\u0001+ dα.\n(6.43)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nΦ(x)\n0\n1\nL\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 6.7: Expected Loss\nTheorem 6.4 (Liu-Ralescu [158], Expected Loss Theorem) Assume that a\nsystem contains independent uncertain variables ξ1, ξ2, · · · , ξn with regular\nuncertainty distributions Φ1, Φ2, · · · , Φn, respectively.\nIf the loss function\nf(ξ1, ξ2, · · · , ξn) is continuous, strictly increasing with respect to ξ1, ξ2, · · · , ξm\nand strictly decreasing with respect to ξm+1, ξm+2, · · · , ξn, then the expected\nloss is\nL =\nZ 1\n0\nf +(Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α))dα. (6.44)\nProof: It follows from the operational law of uncertain variables that the\nloss f(ξ1, ξ2, · · · , ξn) has an inverse uncertainty distribution\nΦ−1(α) = f(Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α)).\nThe theorem follows from (6.43) immediately.\n6.9\nBibliographic Notes\nUncertain risk analysis was proposed by Liu [119] in 2010 in which the risk\nindex was deﬁned as the uncertain measure that some speciﬁed loss occurs,\nand a risk index theorem was proved. This tool was also successfully applied\nby Liu [130] to structural risk analysis.\nAs a substitute of risk index, Peng [184] suggested the concept of value-\nat-risk that is the maximum possible loss when the right tail distribution is\nignored. In addition, Liu-Ralescu [158] investigated the concept of expected\nloss that takes into account not only the uncertain measure of the loss but\nalso its severity.\n\n\n\n\nChapter 7\nUncertain Reliability\nAnalysis\nUncertain reliability analysis is a tool to deal with system reliability via\nuncertainty theory.\nThis chapter will introduce a deﬁnition of reliability\nindex and provide some useful formulas for calculating the reliability index.\n7.1\nStructure Function\nMany real systems may be simpliﬁed to a Boolean system in which each\nelement (including the system itself) has two states: working and failure.\nWe denote the states of elements i by the Boolean variables\nxi =\n(\n1,\nif element i works\n0,\nif element i fails,\n(7.1)\ni = 1, 2, · · · , n, respectively. We also denote the state of the system by the\nBoolean variable\nX =\n(\n1,\nif the system works\n0,\nif the system fails.\n(7.2)\nUsually, the state of the system is completely determined by the states of its\nelements via the so-called structure function.\nDeﬁnition 7.1 Assume that X is a Boolean system containing elements\nx1, x2, · · · , xn.\nA Boolean function f is called a structure function of X\nif\nX = 1 if and only if f(x1, x2, · · · , xn) = 1.\n(7.3)\nIt is obvious that X = 0 if and only if f(x1, x2, · · · , xn) = 0 whenever f is\nindeed the structure function of the system.\n\n\n174\nChapter 7 - Uncertain Reliability Analysis\nExample 7.1: For a series system, the structure function is a mapping from\n{0, 1}n to {0, 1}, i.e.,\nf(x1, x2, · · · , xn) = x1 ∧x2 ∧· · · ∧xn.\n(7.4)\nInput .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\nOutput\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 7.1: A Series System\nExample 7.2: For a parallel system, the structure function is a mapping\nfrom {0, 1}n to {0, 1}, i.e.,\nf(x1, x2, · · · , xn) = x1 ∨x2 ∨· · · ∨xn.\n(7.5)\nInput .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. Output\nFigure 7.2: A Parallel System\n7.2\nReliability Index\nThe element in a Boolean system is usually represented by a Boolean uncer-\ntain variable, i.e.,\nξ =\n(\n1 with uncertain measure a\n0 with uncertain measure 1 −a.\n(7.6)\nIn this case, we will say ξ is an uncertain element with reliability a. Reliability\nindex is deﬁned as the uncertain measure that the system is working.\nDeﬁnition 7.2 (Liu [119]) Assume a Boolean system has uncertain ele-\nments ξ1, ξ2, · · · , ξn and a structure function f. Then the reliability index\nis the uncertain measure that the system is working, i.e.,\nReliability = M{f(ξ1, ξ2, · · · , ξn) = 1}.\n(7.7)\nTheorem 7.1 (Liu [119], Reliability Index Theorem) Assume that a system\ncontains uncertain elements ξ1, ξ2, · · ·, ξn, and has a structure function f. If\n\n\nSection 7.5 - General System\n175\nξ1, ξ2, · · · , ξn are independent uncertain elements with reliabilities a1, a2, · · · ,\nan, respectively, then the reliability index is\nReliability =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) < 0.5\n1 −\nsup\nf(x1,x2,··· ,xn)=0\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) ≥0.5\n(7.8)\nwhere xi take values either 0 or 1, and νi are deﬁned by\nνi(xi) =\n(\nai,\nif xi = 1\n1 −ai,\nif xi = 0\n(7.9)\nfor i = 1, 2, · · · , n, respectively.\nProof: Since ξ1, ξ2, · · · , ξn are independent Boolean uncertain variables and\nf is a Boolean function, the equation (7.8) follows from Deﬁnition 7.2 and\nTheorem 3.22 immediately.\n7.3\nSeries System\nConsider a series system having independent uncertain elements ξ1, ξ2, · · · , ξn\nwith reliabilities a1, a2, · · · , an, respectively. Note that the structure function\nis\nf(x1, x2, · · · , xn) = x1 ∧x2 ∧· · · ∧xn.\n(7.10)\nIt follows from the reliability index theorem that the reliability index is\nReliability = M{ξ1 ∧ξ2 ∧· · · ∧ξn = 1} = a1 ∧a2 ∧· · · ∧an.\n(7.11)\n7.4\nParallel System\nConsider a parallel system having independent uncertain elements ξ1, ξ2, · · · ,\nξn with reliabilities a1, a2, · · · , an, respectively. Note that the structure func-\ntion is\nf(x1, x2, · · · , xn) = x1 ∨x2 ∨· · · ∨xn.\n(7.12)\nIt follows from the reliability index theorem that the reliability index is\nReliability = M{ξ1 ∨ξ2 ∨· · · ∨ξn = 1} = a1 ∨a2 ∨· · · ∨an.\n(7.13)\n\n\n176\nChapter 7 - Uncertain Reliability Analysis\n7.5\nGeneral System\nIt is almost impossible to ﬁnd an analytic formula of reliability risk for general\nsystems. In this case, we have to employ a numerical method. Consider a\nbridge system shown in Figure 7.3 that consists of 5 independent uncertain\nelements whose states are denoted by ξ1, ξ2, ξ3, ξ4, ξ5.\nAssume each path\nworks if and only if all elements on which are working and the system works\nif and only if there is a path of working elements. Then the structure function\nof the bridge system is\nf(x1, x2, x3, x4, x5) = (x1 ∧x4) ∨(x2 ∧x5) ∨(x1 ∧x3 ∧x5) ∨(x2 ∧x3 ∧x4).\nAssume the 5 independent uncertain elements have reliabilities\n0.91, 0.92, 0.93, 0.94, 0.95\nin uncertain measure. The reliability index is\nReliability = M{f(ξ1, ξ2, · · · , ξ5) = 1} = 0.92\nin uncertain measure.\nInput .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\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.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. Output\nFigure 7.3: A Bridge System\n7.6\nBibliographic Notes\nUncertain reliability analysis was proposed by Liu [119] in 2010 in which\nthe reliability index was deﬁned as the uncertain measure that the system is\nworking, and a reliability index theorem was proved. After that, uncertain\nreliability analysis was signiﬁcantly developed by Zeng-Wen-Kang [293], Gao-\nYao [43], Zeng-Kang-Wen-Zio [294] and Gao-Yao-Zhou-Ke [40].\n\n\nChapter 8\nUncertain Propositional\nLogic\nPropositional logic, originated from the work of Aristotle (384-322 BC), is a\nbranch of logic that studies the properties of complex propositions composed\nof simpler propositions and logical connectives. Note that the propositions\nconsidered in propositional logic are not arbitrary statements but are the\nones that are either true or false and not both.\nUncertain propositional logic is a generalization of propositional logic in\nwhich every proposition is abstracted into a Boolean uncertain variable and\nthe truth value is deﬁned as the uncertain measure that the proposition is\ntrue. Uncertain entailment is a methodology for calculating the truth value\nof an uncertain formula via the maximum uncertainty principle when the\ntruth values of other uncertain formulas are given. In some sense, uncertain\npropositional logic and uncertain entailment are mutually inverse, the former\nattempts to compose a complex proposition from simpler ones, while the\nlatter attempts to decompose a complex proposition into simpler ones.\nThis chapter will deal with uncertain propositional logic, including un-\ncertain proposition, truth value deﬁnition, and truth value theorem. This\nchapter will also present an uncertain entailment model from which uncertain\nmodus ponens, uncertain modus tollens and uncertain hypothetical syllogism\nare deduced.\n8.1\nUncertain Proposition\nDeﬁnition 8.1 (Li-Liu [101]) An uncertain proposition is a statement whose\ntruth value is quantiﬁed by an uncertain measure.\nThat is, if we use X to express an uncertain proposition and use α to express\nits truth value in uncertain measure, then the uncertain proposition X is\n\n\n178\nChapter 8 - Uncertain Propositional Logic\nessentially a Boolean uncertain variable\nX =\n(\n1 with uncertain measure α\n0 with uncertain measure 1 −α\n(8.1)\nwhere X = 1 means X is true and X = 0 means X is false.\nExample 8.1: “Tom is tall with truth value 0.7” is an uncertain proposition,\nwhere “Tom is tall” is a statement, and its truth value is 0.7 in uncertain\nmeasure.\nExample 8.2: “John is young with truth value 0.8” is an uncertain propo-\nsition, where “John is young” is a statement, and its truth value is 0.8 in\nuncertain measure.\nExample 8.3: “Beijing is a big city with truth value 0.9” is an uncertain\nproposition, where “Beijing is a big city” is a statement, and its truth value\nis 0.9 in uncertain measure.\nConnective Symbols\nIn addition to the proposition symbols X and Y , we also need the negation\nsymbol ¬, conjunction symbol ∧, disjunction symbol ∨, conditional symbol\n→, and biconditional symbol ↔. Note that\n¬X means “not X”;\n(8.2)\nX ∧Y means “X and Y ”;\n(8.3)\nX ∨Y means “X or Y ”;\n(8.4)\nX →Y = (¬X) ∨Y means “if X then Y ”,\n(8.5)\nX ↔Y = (X →Y ) ∧(Y →X) means “X if and only if Y ”.\n(8.6)\nBoolean Function of Uncertain Propositions\nAssume X1, X2, · · · , Xn are uncertain propositions. Then their Boolean func-\ntion\nZ = f(X1, X2, · · · , Xn)\n(8.7)\nis a Boolean uncertain variable.\nThus Z is also an uncertain proposition\nprovided that it makes sense. Usually, such a Boolean function is a ﬁnite\nsequence of uncertain propositions and connective symbols. For example,\nZ = ¬X1,\nZ = X1 ∧(¬X2),\nZ = X1 →X2\n(8.8)\nare all uncertain propositions.\n\n\nSection 8.2 - Truth Value\n179\nIndependence of Uncertain Propositions\nUncertain propositions are called independent if they are independent uncer-\ntain variables. Assume X1, X2, · · · , Xn are independent uncertain proposi-\ntions. Then\nf1(X1), f2(X2) · · · , fn(Xn)\n(8.9)\nare also independent uncertain propositions for any Boolean functions f1, f2,\n· · · , fn. For example, if X1, X2, · · · , X5 are independent uncertain proposi-\ntions, then ¬X1, X2 ∨X3, X4 →X5 are also independent.\n8.2\nTruth Value\nTruth value is a key concept in uncertain propositional logic, and is deﬁned\nas the uncertain measure that the uncertain proposition is true.\nDeﬁnition 8.2 (Li-Liu [101]) Let X be an uncertain proposition. Then the\ntruth value of X is deﬁned as the uncertain measure that X is true, i.e.,\nT(X) = M{X = 1}.\n(8.10)\nTheorem 8.1 (Law of Excluded Middle) Let X be an uncertain proposition.\nThen X ∨¬X is a tautology, i.e.,\nT(X ∨¬X) = 1.\n(8.11)\nProof:\nIt follows from the deﬁnition of truth value and the property of\nuncertain measure that\nT(X ∨¬X) = M{X ∨¬X = 1} = M{(X = 1) ∪(X = 0)} = M{Γ} = 1.\nThe theorem is proved.\nTheorem 8.2 (Law of Contradiction) Let X be an uncertain proposition.\nThen X ∧¬X is a contradiction, i.e.,\nT(X ∧¬X) = 0.\n(8.12)\nProof:\nIt follows from the deﬁnition of truth value and the property of\nuncertain measure that\nT(X ∧¬X) = M{X ∧¬X = 1} = M{(X = 1) ∩(X = 0)} = M{∅} = 0.\nThe theorem is proved.\nTheorem 8.3 (Law of Truth Conservation) Let X be an uncertain proposi-\ntion. Then we have\nT(X) + T(¬X) = 1.\n(8.13)\n\n\n180\nChapter 8 - Uncertain Propositional Logic\nProof: It follows from the duality axiom of uncertain measure that\nT(¬X) = M{¬X = 1} = M{X = 0} = 1 −M{X = 1} = 1 −T(X).\nThe theorem is proved.\nTheorem 8.4 Let X be an uncertain proposition. Then X →X is a tau-\ntology, i.e.,\nT(X →X) = 1.\n(8.14)\nProof: It follows from the deﬁnition of conditional symbol and the law of\nexcluded middle that\nT(X →X) = T(¬X ∨X) = 1.\nThe theorem is proved.\nTheorem 8.5 Let X be an uncertain proposition. Then we have\nT(X →¬X) = 1 −T(X).\n(8.15)\nProof: It follows from the deﬁnition of conditional symbol and the law of\ntruth conservation that\nT(X →¬X) = T(¬X ∨¬X) = T(¬X) = 1 −T(X).\nThe theorem is proved.\nTheorem 8.6 Let X and Y be two independent uncertain propositions. Then\nT(X ∧Y ) = T(X) ∧T(Y ),\n(8.16)\nT(X ∨Y ) = T(X) ∨T(Y ),\n(8.17)\nT(X →Y ) = (1 −T(X)) ∨T(Y ).\n(8.18)\nProof:\nIt follows from the deﬁnition of truth value and the property of\nuncertain measure that\nT(X ∧Y ) = M{X ∧Y = 1} = M{(X = 1) ∩(Y = 1)}\n= M{X = 1} ∧M{Y = 1} = T(X) ∧T(Y ),\nT(X ∨Y ) = M{X ∨Y = 1} = M{(X = 1) ∪(Y = 1)}\n= M{X = 1} ∨M{Y = 1} = T(X) ∨T(Y ),\nT(X →Y ) = T(¬X ∨Y ) = T(¬X) ∨T(Y ) = (1 −T(X)) ∨T(Y ).\n\n\nSection 8.3 - Chen-Ralescu Theorem\n181\nTheorem 8.7 (De Morgan’s Law) For any uncertain propositions X and Y ,\nwe have\nT(¬(X ∧Y )) = T((¬X) ∨(¬Y )),\n(8.19)\nT(¬(X ∨Y )) = T((¬X) ∧(¬Y )).\n(8.20)\nProof:\nIt follows from the deﬁnition of truth value and the property of\nuncertain measure that\nT(¬(X ∧Y )) = M{X ∧Y = 0} = M{(X = 0) ∪(Y = 0)}\n= M{(¬X) ∨(¬Y ) = 1} = T((¬X) ∨(¬Y ))\nwhich proves the ﬁrst equality. A similar way may verify the second equality.\nTheorem 8.8 (Law of Contraposition) For any uncertain propositions X\nand Y , we have\nT(X →Y ) = T(¬Y →¬X).\n(8.21)\nProof: It follows from the deﬁnition of conditional symbol and basic prop-\nerties of uncertain measure that\nT(X →Y ) = M{(¬X) ∨Y = 1} = M{(X = 0) ∪(Y = 1)}\n= M{Y ∨(¬X) = 1} = T(¬Y →¬X).\nThe theorem is proved.\n8.3\nChen-Ralescu Theorem\nAn important contribution to uncertain propositional logic is the Chen-\nRalescu theorem that provides a numerical method for calculating the truth\nvalues of uncertain propositions.\nTheorem 8.9 (Chen-Ralescu Theorem [10]) Assume that X1, X2, · · · , Xn\nare independent uncertain propositions with truth values α1, α2, · · ·, αn, re-\nspectively. Then for a Boolean function f, the uncertain proposition\nZ = f(X1, X2, · · · , Xn).\n(8.22)\nhas a truth value\nT(Z) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) < 0.5\n1 −\nsup\nf(x1,x2,··· ,xn)=0\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) ≥0.5\n(8.23)\n\n\n182\nChapter 8 - Uncertain Propositional Logic\nwhere xi take values either 0 or 1, and νi are deﬁned by\nνi(xi) =\n(\nαi,\nif xi = 1\n1 −αi,\nif xi = 0\n(8.24)\nfor i = 1, 2, · · · , n, respectively.\nProof: Since Z = 1 if and only if f(X1, X2, · · · , Xn) = 1, we immediately\nhave\nT(Z) = M{f(X1, X2, · · · , Xn) = 1}.\nThus the equation (8.23) follows from Theorem 3.22 immediately.\nExample 8.4: Let X1 and X2 be independent uncertain propositions with\ntruth values α1 and α2, respectively. Then\nZ = X1 ↔X2\n(8.25)\nis an uncertain proposition. It is clear that Z = f(X1, X2) if we deﬁne\nf(1, 1) = 1,\nf(1, 0) = 0,\nf(0, 1) = 0,\nf(0, 0) = 1.\nAt ﬁrst, we have\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = max{α1 ∧α2, (1 −α1) ∧(1 −α2)},\nsup\nf(x1,x2)=0\nmin\n1≤i≤2 νi(xi) = max{(1 −α1) ∧α2, α1 ∧(1 −α2)}.\nWhen α1 ≥0.5 and α2 ≥0.5, we have\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = α1 ∧α2 ≥0.5.\nIt follows from Chen-Ralescu theorem that\nT(Z) = 1 −\nsup\nf(x1,x2)=0\nmin\n1≤i≤2 νi(xi) = 1 −(1 −α1) ∨(1 −α2) = α1 ∧α2.\nWhen α1 ≥0.5 and α2 < 0.5, we have\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = (1 −α1) ∨α2 ≤0.5.\nIt follows from Chen-Ralescu theorem that\nT(Z) =\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = (1 −α1) ∨α2.\nWhen α1 < 0.5 and α2 ≥0.5, we have\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = α1 ∨(1 −α2) ≤0.5.\n\n\nSection 8.3 - Chen-Ralescu Theorem\n183\nIt follows from Chen-Ralescu theorem that\nT(Z) =\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = α1 ∨(1 −α2).\nWhen α1 < 0.5 and α2 < 0.5, we have\nsup\nf(x1,x2)=1\nmin\n1≤i≤2 νi(xi) = (1 −α1) ∧(1 −α2) > 0.5.\nIt follows from Chen-Ralescu theorem that\nT(Z) = 1 −\nsup\nf(x1,x2)=0\nmin\n1≤i≤2 νi(xi) = 1 −α1 ∨α2 = (1 −α1) ∧(1 −α2).\nThus we have\nT(Z) =\n\n\n\n\n\n\n\n\n\nα1 ∧α2,\nif α1 ≥0.5 and α2 ≥0.5\n(1 −α1) ∨α2,\nif α1 ≥0.5 and α2 < 0.5\nα1 ∨(1 −α2),\nif α1 < 0.5 and α2 ≥0.5\n(1 −α1) ∧(1 −α2),\nif α1 < 0.5 and α2 < 0.5.\n(8.26)\nExample 8.5: The independence condition in Theorem 8.9 cannot be re-\nmoved. For example, take an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = M{γ2} = 0.5. Then\nX1(γ) =\n(\n0,\nif γ = γ1\n1,\nif γ = γ2\n(8.27)\nis an uncertain proposition with truth value\nT(X1) = 0.5,\n(8.28)\nand\nX2(γ) =\n(\n1,\nif γ = γ1\n0,\nif γ = γ2\n(8.29)\nis also an uncertain proposition with truth value\nT(X2) = 0.5.\n(8.30)\nNote that X1 and X2 are not independent, and X1 ∨X2 ≡1 from which we\nobtain\nT(X1 ∨X2) = 1.\n(8.31)\nHowever, by using (8.23), we get\nT(X1 ∨X2) = 0.5.\n(8.32)\n\n\n184\nChapter 8 - Uncertain Propositional Logic\nThus the independence condition cannot be removed.\nExercise 8.1: Let X1, X2, · · · , Xn be independent uncertain propositions\nwith truth values α1, α2, · · · , αn, respectively. Then\nZ = X1 ∧X2 ∧· · · ∧Xn\n(8.33)\nis an uncertain proposition. Show that the truth value of Z is\nT(Z) = α1 ∧α2 ∧· · · ∧αn.\n(8.34)\nExercise 8.2: Let X1, X2, · · · , Xn be independent uncertain propositions\nwith truth values α1, α2, · · · , αn, respectively. Then\nZ = X1 ∨X2 ∨· · · ∨Xn\n(8.35)\nis an uncertain proposition. Show that the truth value of Z is\nT(Z) = α1 ∨α2 ∨· · · ∨αn.\n(8.36)\nExercise 8.3: Let X1 and X2 be independent uncertain propositions with\ntruth values α1 and α2, respectively. (i) What is the truth value of (X1 ∧\nX2) →X2? (ii) What is the truth value of (X1 ∨X2) →X2? (iii) What\nis the truth value of X1 →(X1 ∧X2)?\n(iv) What is the truth value of\nX1 →(X1 ∨X2)?\nExercise 8.4: Let X1, X2, X3 be independent uncertain propositions with\ntruth values α1, α2, α3, respectively. What is the truth value of\nX1 ∧(X1 ∨X2) ∧(X1 ∨X2 ∨X3)?\n(8.37)\n8.4\nUncertain Entailment\nUncertain entailment is a methodology for calculating the truth value of an\nuncertain formula via the maximum uncertainty principle when the truth\nvalues of other uncertain formulas are given. Assume X1, X2, · · · , Xn are in-\ndependent uncertain propositions with unknown truth values α1, α2, · · · , αn,\nrespectively. Also assume that\nYj = fj(X1, X2, · · · , Xn)\n(8.38)\nare uncertain propositions with known truth values cj, j = 1, 2, · · · , m, re-\nspectively. Now let\nZ = f(X1, X2, · · · , Xn)\n(8.39)\n\n\nSection 8.4 - Uncertain Entailment\n185\nbe an additional uncertain proposition. What is the truth value of Z? This\nis just the uncertain entailment problem. In order to solve it, let us consider\nwhat values α1, α2, · · · , αn may take. The ﬁrst constraint is\n0 ≤αi ≤1,\ni = 1, 2, · · · , n.\n(8.40)\nThe second type of constraints is represented by\nT(Yj) = cj\n(8.41)\nwhere T(Yj) are determined by α1, α2, · · · , αn via\nT(Yj) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nfj(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi),\nif\nsup\nfj(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) < 0.5\n1 −\nsup\nfj(x1,x2,··· ,xn)=0\nmin\n1≤i≤n νi(xi),\nif\nsup\nfj(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) ≥0.5\n(8.42)\nfor j = 1, 2, · · · , m and\nνi(xi) =\n(\nαi,\nif xi = 1\n1 −αi,\nif xi = 0\n(8.43)\nfor i = 1, 2, · · · , n.\nPlease note that the additional uncertain proposition\nZ = f(X1, X2, · · · , Xn) has a truth value\nT(Z) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) < 0.5\n1 −\nsup\nf(x1,x2,··· ,xn)=0\nmin\n1≤i≤n νi(xi),\nif\nsup\nf(x1,x2,··· ,xn)=1\nmin\n1≤i≤n νi(xi) ≥0.5.\n(8.44)\nSince the truth values α1, α2, · · · , αn are not uniquely determined, the truth\nvalue T(Z) is not unique too. In this case, we have to use the maximum\nuncertainty principle to determine the truth value T(Z).\nThat is, T(Z)\nshould be assigned the value as close to 0.5 as possible.\nIn other words,\nwe should minimize the value |T(Z) −0.5| via choosing appreciate values of\nα1, α2, · · · , αn. The uncertain entailment model is thus written by Liu [117]\nas follows,\n\n\n\n\n\n\n\n\n\nmin |T(Z) −0.5|\nsubject to:\n0 ≤αi ≤1,\ni = 1, 2, · · · , n\nT(Yj) = cj,\nj = 1, 2, · · · , m\n(8.45)\n\n\n186\nChapter 8 - Uncertain Propositional Logic\nwhere T(Z), T(Yj), j = 1, 2, · · · , m are functions of unknown truth values\nα1, α2, · · · , αn.\nExample 8.6: Let A and B be independent uncertain propositions. It is\nknown that\nT(A ∨B) = a,\nT(A ∧B) = b.\n(8.46)\nWhat is the truth value of A →B? Denote the truth values of A and B by\nα1 and α2, respectively, and write\nY1 = A ∨B,\nY2 = A ∧B,\nZ = A →B.\nIt is clear that\nT(Y1) = α1 ∨α2 = a,\nT(Y2) = α1 ∧α2 = b,\nT(Z) = (1 −α1) ∨α2.\nIn this case, the uncertain entailment model (8.45) becomes\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin |(1 −α1) ∨α2 −0.5|\nsubject to:\n0 ≤α1 ≤1\n0 ≤α2 ≤1\nα1 ∨α2 = a\nα1 ∧α2 = b.\n(8.47)\nWhen a ≥b, there are only two feasible solutions (α1, α2) = (a, b) and\n(α1, α2) = (b, a). If a + b < 1, the optimal solution produces\nT(Z) = (1 −α∗\n1) ∨α∗\n2 = 1 −a;\nif a + b = 1, the optimal solution produces\nT(Z) = (1 −α∗\n1) ∨α∗\n2 = a or b;\nif a + b > 1, the optimal solution produces\nT(Z) = (1 −α∗\n1) ∨α∗\n2 = b.\nWhen a < b, there is no feasible solution and the truth values are ill-assigned.\nIn summary, from T(A ∨B) = a and T(A ∧B) = b we entail\nT(A →B) =\n\n\n\n\n\n\n\n\n\n1 −a,\nif a ≥b and a + b < 1\na or b,\nif a ≥b and a + b = 1\nb,\nif a ≥b and a + b > 1\nillness,\nif a < b.\n(8.48)\n\n\nSection 8.5 - Uncertain Modus Ponens\n187\nExercise 8.5:\nLet A, B, C be independent uncertain propositions.\nIt is\nknown that\nT(A →C) = a,\nT(B →C) = b,\nT(A ∨B) = c.\n(8.49)\nWhat is the truth value of C?\nExercise 8.6: Let A, B, C, D be independent uncertain propositions. It is\nknown that\nT(A →C) = a,\nT(B →D) = b,\nT(A ∨B) = c.\n(8.50)\nWhat is the truth value of C ∨D?\nExercise 8.7:\nLet A, B, C be independent uncertain propositions.\nIt is\nknown that\nT(A ∨B) = a,\nT(¬A ∨C) = b.\n(8.51)\nWhat is the truth value of B ∨C?\n8.5\nUncertain Modus Ponens\nUncertain modus ponens was presented by Liu [117]. Let A and B be inde-\npendent uncertain propositions. Assume A and A →B have truth values a\nand b, respectively. What is the truth value of B? Denote the truth values\nof A and B by α1 and α2, respectively, and write\nY1 = A,\nY2 = A →B,\nZ = B.\nIt is clear that\nT(Y1) = α1 = a,\nT(Y2) = (1 −α1) ∨α2 = b,\nT(Z) = α2.\nIn this case, the uncertain entailment model (8.45) becomes\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin |α2 −0.5|\nsubject to:\n0 ≤α1 ≤1\n0 ≤α2 ≤1\nα1 = a\n(1 −α1) ∨α2 = b.\n(8.52)\nWhen a+b > 1, there is a unique feasible solution (a, b) and then the optimal\nsolution is\nα∗\n1 = a,\nα∗\n2 = b.\n\n\n188\nChapter 8 - Uncertain Propositional Logic\nThus T(B) = α∗\n2 = b. When a + b = 1, the feasible set is {a} × [0, b] and the\noptimal solution is\nα∗\n1 = a,\nα∗\n2 = 0.5 ∧b.\nThus T(B) = α∗\n2 = 0.5 ∧b. When a + b < 1, there is no feasible solution and\nthe truth values are ill-assigned. In summary, from\nT(A) = a,\nT(A →B) = b\n(8.53)\nwe entail\nT(B) =\n\n\n\n\n\nb,\nif a + b > 1\n0.5 ∧b,\nif a + b = 1\nillness,\nif a + b < 1.\n(8.54)\nThis result coincides with the classical modus ponens that if both A and\nA →B are true, then B is true.\n8.6\nUncertain Modus Tollens\nUncertain modus tollens was presented by Liu [117]. Let A and B be inde-\npendent uncertain propositions. Assume A →B and B have truth values a\nand b, respectively. What is the truth value of A? Denote the truth values\nof A and B by α1 and α2, respectively, and write\nY1 = A →B,\nY2 = B,\nZ = A.\nIt is clear that\nT(Y1) = (1 −α1) ∨α2 = a,\nT(Y2) = α2 = b,\nT(Z) = α1.\nIn this case, the uncertain entailment model (8.45) becomes\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin |α1 −0.5|\nsubject to:\n0 ≤α1 ≤1\n0 ≤α2 ≤1\n(1 −α1) ∨α2 = a\nα2 = b.\n(8.55)\nWhen a > b, there is a unique feasible solution (1−a, b) and then the optimal\nsolution is\nα∗\n1 = 1 −a,\nα∗\n2 = b.\n\n\nSection 8.7 - Uncertain Hypothetical Syllogism\n189\nThus T(A) = α∗\n1 = 1 −a. When a = b, the feasible set is [1 −a, 1] × {b} and\nthe optimal solution is\nα∗\n1 = (1 −a) ∨0.5,\nα∗\n2 = b.\nThus T(A) = α∗\n1 = (1 −a) ∨0.5. When a < b, there is no feasible solution\nand the truth values are ill-assigned. In summary, from\nT(A →B) = a,\nT(B) = b\n(8.56)\nwe entail\nT(A) =\n\n\n\n\n\n1 −a,\nif a > b\n(1 −a) ∨0.5,\nif a = b\nillness,\nif a < b.\n(8.57)\nThis result coincides with the classical modus tollens that if A →B is true\nand B is false, then A is false.\n8.7\nUncertain Hypothetical Syllogism\nUncertain hypothetical syllogism was presented by Liu [117]. Let A, B, C be\nindependent uncertain propositions. Assume A →B and B →C have truth\nvalues a and b, respectively. What is the truth value of A →C? Denote the\ntruth values of A, B, C by α1, α2, α3, respectively, and write\nY1 = A →B,\nY2 = B →C,\nZ = A →C.\nIt is clear that\nT(Y1) = (1 −α1) ∨α2 = a,\nT(Y2) = (1 −α2) ∨α3 = b,\nT(Z) = (1 −α1) ∨α3.\nIn this case, the uncertain entailment model (8.45) becomes\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nmin |(1 −α1) ∨α3 −0.5|\nsubject to:\n0 ≤α1 ≤1\n0 ≤α2 ≤1\n0 ≤α3 ≤1\n(1 −α1) ∨α2 = a\n(1 −α2) ∨α3 = b.\n(8.58)\nIn order to solve this model, the argument may break down into three cases.\nCase 1: When a + b ≥1 but (a, b) ̸= (0.5, 0.5), the feasible set is the union\nof three sets,\n[1 −a, 1] × {a} × {b},\n\n\n190\nChapter 8 - Uncertain Propositional Logic\n{1 −a} × [1 −b, a] × {b},\n{1 −a} × {1 −b} × [0, b].\nIf a ≥b ≥0.5, then the optimal solution is\nα∗\n1 = 1 −b,\nα∗\n2 = a,\nα∗\n3 = b\nand\nT(A →C) = (1 −α∗\n1) ∨α∗\n3 = b.\nIf b > a ≥0.5, then the optimal solution is\nα∗\n1 = 1 −a,\nα∗\n2 = 1 −b,\nα∗\n3 = a\nand\nT(A →C) = (1 −α∗\n1) ∨α∗\n3 = a.\nIf a > 0.5 > b, then the optimal solution is\nα∗\n1 = 0.5,\nα∗\n2 = a,\nα∗\n3 = b\nand\nT(A →C) = (1 −α∗\n1) ∨α∗\n3 = 0.5.\nIf b > 0.5 > a, then the optimal solution is\nα∗\n1 = 1 −a,\nα∗\n2 = 1 −b,\nα∗\n3 = 0.5\nand\nT(A →C) = (1 −α∗\n1) ∨α∗\n3 = 0.5.\nCase 2: When (a, b) = (0.5, 0.5), the feasible set is [0.5, 1] × {0.5} × [0, 0.5].\nThus the optimal solution is\nα∗\n1 = 0.5,\nα∗\n2 = 0.5,\nα∗\n3 = 0.5\nand\nT(A →C) = (1 −α∗\n1) ∨α∗\n3 = 0.5.\nCase 3: When a + b < 1, there is no feasible solution and the truth values\nare ill-assigned. In summary, from\nT(A →B) = a,\nT(B →C) = b\n(8.59)\nwe entail\nT(A →C) =\n\n\n\n\n\n\n\na ∧b,\nif a > 0.5 and b > 0.5\n0.5,\nif a + b ≥1 and a ∧b ≤0.5\nillness,\nif a + b < 1.\n(8.60)\nThis result coincides with the classical hypothetical syllogism that if both\nA →B and B →C are true, then A →C is true.\n\n\nSection 8.8 - Bibliographic Notes\n191\n8.8\nBibliographic Notes\nUncertain propositional logic was designed by Li-Liu [101] in which every\nproposition is abstracted into a Boolean uncertain variable and the truth\nvalue is deﬁned as the uncertain measure that the proposition is true. An\nimportant contribution is Chen-Ralescu theorem [10] that provides a numeri-\ncal method for calculating the truth value of uncertain propositions. Another\ntopic is the uncertain predicate logic developed by Zhang-Li [304] in which an\nuncertain predicate proposition is deﬁned as a sequence of uncertain propo-\nsitions indexed by one or more parameters.\nUncertain entailment was proposed by Liu [117] for determining the truth\nvalue of an uncertain proposition via the maximum uncertainty principle\nwhen the truth values of other uncertain propositions are given. From the\nuncertain entailment model, Liu [117] deduced uncertain modus ponens, un-\ncertain modus tollens, and uncertain hypothetical syllogism.\nAfter that,\nYang-Gao-Ni [242] investigated the uncertain resolution principle.\n\n\n\n\nChapter 9\nUncertain Set\nUncertain set was ﬁrst proposed by Liu [118] in 2010 for modelling unsharp\nconcepts via uncertainty theory. This chapter will introduce the concepts of\nuncertain set, membership function, inverse membership function, indepen-\ndence, expected value, distance, and entropy. This chapter will also introduce\nthe operational law for uncertain sets via membership functions and inverse\nmembership functions.\n9.1\nUncertain Set\nRoughly speaking, an uncertain set is a set-valued function on an uncertainty\nspace, and attempts to model “unsharp concepts” that are essentially sets\nbut their boundaries are not sharply described (because of the ambiguity of\nhuman language). Some typical examples include “young”, “tall”, “warm”,\nand “most”. A formal deﬁnition is given as follows.\nDeﬁnition 9.1 (Liu [118]) An uncertain set is a function ξ from an uncer-\ntainty space (Γ, L, M) to a collection of sets of real numbers such that both\n{B ⊂ξ} and {ξ ⊂B} are events for any Borel set B of real numbers.\nRemark 9.1: Note that the events {B ⊂ξ} and {ξ ⊂B} are subsets of the\nuniversal set Γ, i.e.,\n{B ⊂ξ} = {γ ∈Γ | B ⊂ξ(γ)},\n(9.1)\n{ξ ⊂B} = {γ ∈Γ | ξ(γ) ⊂B}.\n(9.2)\nRemark 9.2: It is clear that uncertain set (Liu [118]) is very diﬀerent from\nrandom set (Robbins [193] and Matheron [173]) and fuzzy set (Zadeh [290]).\nThe essential diﬀerence among them is that diﬀerent measures are used, i.e.,\nrandom set uses probability measure, fuzzy set uses possibility measure, and\nuncertain set uses uncertain measure.\n\n\n194\nChapter 9 - Uncertain Set\nRemark 9.3: What is the diﬀerence between uncertain variable and un-\ncertain set? Both of them belong to the same broad category of uncertain\nconcepts. However, they are diﬀerentiated by their mathematical deﬁnitions:\nthe former refers to one value, while the latter to a collection of values. Es-\nsentially, the diﬀerence between uncertain variable and uncertain set focuses\non the property of exclusivity. If the concept has exclusivity, then it is an\nuncertain variable. Otherwise, it is an uncertain set. Consider the statement\n“John is a young man”. If we are interested in John’s real age, then “young”\nis an uncertain variable because it is an exclusive concept (John’s age can-\nnot be more than one value). For example, if John is 20 years old, then it\nis impossible that John is 25 years old. In other words, “John is 20 years\nold” does exclude the possibility that “John is 25 years old”. By contrast,\nif we are interested in what ages can be regarded “young”, then “young” is\nan uncertain set because the concept now has no exclusivity. For example,\nboth 20-year-old and 25-year-old men can be considered “young”. In other\nwords, “a 20-year-old man is young” does not exclude the possibility that “a\n25-year-old man is young”.\nExample 9.1: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3} with\npower set and M{γ1} = 0.6, M{γ2} = 0.3, M{γ3} = 0.2. Then\nξ(γ) =\n\n\n\n\n\n[1, 3],\nif γ = γ1\n[2, 4],\nif γ = γ2\n[3, 5],\nif γ = γ3\n(9.3)\nis an uncertain set. See Figure 9.1. Furthermore, we have\nM{2 ∈ξ} = M{γ | 2 ∈ξ(γ)} = M{γ1, γ2} = 0.8,\n(9.4)\nM{[3, 4] ⊂ξ} = M{γ | [3, 4] ⊂ξ(γ)} = M{γ2, γ3} = 0.4,\n(9.5)\nM{ξ ⊂[1, 5]} = M{γ | ξ(γ) ⊂[1, 5]} = M{γ1, γ2, γ3} = 1.\n(9.6)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. Γ\nℜ\nγ1\nγ2\nγ3\n1\n2\n3\n4\n5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n............\n...................................\n...................................\n.........................................................\n.........................................................\nFigure 9.1: An Uncertain Set\n\n\nSection 9.1 - Uncertain Set\n195\nExample 9.2: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nξ(γ) = [0, 3γ],\n∀γ ∈Γ\n(9.7)\nis an uncertain set. Furthermore, we have\nM{2 ∈ξ} = M{γ | 2 ∈ξ(γ)} = M{[2/3, 1)} = 1/3,\n(9.8)\nM{[0, 1] ⊂ξ} = M{γ | [0, 1] ⊂ξ(γ)} = M{[1/3, 1)} = 2/3,\n(9.9)\nM{ξ ⊂[0, 3)} = M{γ | ξ(γ) ⊂[0, 3)} = M{(0, 1])} = 1.\n(9.10)\nExample 9.3: A crisp set A of real numbers is a special uncertain set on\nan uncertainty space (Γ, L, M) deﬁned by\nξ(γ) ≡A,\n∀γ ∈Γ.\n(9.11)\nExample 9.4: Let ξ be an uncertain set and let x be a real number. Then\n{x ∈ξ}c = {γ | x ∈ξ(γ)}c = {γ | x ̸∈ξ(γ)} = {x ̸∈ξ}.\nThus {x ∈ξ} and {x ̸∈ξ} are opposite events.\nExercise 9.1: Let ξ be an uncertain set and let B be a Borel set of real\nnumbers. Show that {ξ ⊂B} and {ξ ̸⊂B} are opposite events.\nExercise 9.2: Let ξ and η be two uncertain sets. Show that {ξ ⊂η} and\n{ξ ̸⊂η} are opposite events.\nExercise 9.3: Let ∅be the empty set, and let ξ be an uncertain set. Show\nthat\nM{∅⊂ξ} = 1.\n(9.12)\nExercise 9.4: Let ξ be an uncertain set, and let ℜbe the set of real numbers.\nShow that\nM{ξ ⊂ℜ} = 1.\n(9.13)\nExercise 9.5: Let ξ be an uncertain set. Show that ξ is always included in\nitself, i.e.,\nM{ξ ⊂ξ} = 1.\n(9.14)\nTheorem 9.1 (Liu [133], Fundamental Relationship) Let ξ be an uncertain\nset, and let B be a crisp set of real numbers. Then\n{B ⊂ξ} =\n\\\nx∈B\n{x ∈ξ},\n(9.15)\n{ξ ⊂B} =\n\\\nx∈Bc\n{x ̸∈ξ}.\n(9.16)\n\n\n196\nChapter 9 - Uncertain Set\nProof: For any γ ∈{B ⊂ξ}, we have B ⊂ξ(γ). Thus for any x ∈B, we\nhave x ∈ξ(γ), i.e., γ ∈{x ∈ξ}. Therefore,\n{B ⊂ξ} ⊂{x ∈ξ},\n∀x ∈B,\ni.e.,\n{B ⊂ξ} ⊂\n\\\nx∈B\n{x ∈ξ}.\n(9.17)\nOn the other hand, for any\nγ ∈\n\\\nx∈B\n{x ∈ξ},\nwe have x ∈ξ(γ) whenever x ∈B. Thus B ⊂ξ(γ), i.e., γ ∈{B ⊂ξ}.\nTherefore,\n\\\nx∈B\n{x ∈ξ} ⊂{B ⊂ξ}.\n(9.18)\nIt follows from (9.17) and (9.18) that (9.15) holds.\nThe ﬁrst equation is\nproved.\nNext we verify the second equation.\nFor any γ ∈{ξ ⊂B}, we\nhave ξ(γ) ⊂B. Thus for any x ∈Bc, we have x ̸∈ξ(γ), i.e., γ ∈{x ̸∈ξ}.\nTherefore,\n{ξ ⊂B} ⊂{x ̸∈ξ},\n∀x ∈Bc,\ni.e.,\n{ξ ⊂B} ⊂\n\\\nx∈Bc\n{x ̸∈ξ}.\n(9.19)\nOn the other hand, for any\nγ ∈\n\\\nx∈Bc\n{x ̸∈ξ},\nif x ∈Bc, then x ̸∈ξ(γ). Thus Bc ⊂ξ(γ)c. This means ξ(γ) ⊂B, i.e.,\nγ ∈{ξ ⊂B}. Therefore,\n\\\nx∈Bc\n{x ̸∈ξ} ⊂{ξ ⊂B}.\n(9.20)\nIt follows from (9.19) and (9.20) that (9.16) holds. The theorem is proved.\nDeﬁnition 9.2 An uncertain set ξ on the uncertainty space (Γ, L, M) is said\nto be (i) nonempty if\nξ(γ) ̸= ∅\n(9.21)\nfor almost all γ ∈Γ, (ii) empty if\nξ(γ) = ∅\n(9.22)\nfor almost all γ ∈Γ, and (iii) half-empty if otherwise.\n\n\nSection 9.1 - Uncertain Set\n197\nExample 9.5: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nξ(γ) = [0, γ],\n∀γ ∈Γ\n(9.23)\nis a nonempty uncertain set,\nξ(γ) = ∅,\n∀γ ∈Γ\n(9.24)\nis an empty uncertain set, and\nξ(γ) =\n(\n∅,\nif γ > 0.8\n[0, γ],\nif γ ≤0.8\n(9.25)\nis a half-empty uncertain set.\nDeﬁnition 9.3 (Liu [133]) An uncertain set ξ deﬁned on the uncertainty\nspace (Γ, L, M) is called totally ordered if {ξ(γ) | γ ∈Γ} is a totally ordered\nset, i.e., for any given γ1 and γ2 ∈Γ, either ξ(γ1) ⊂ξ(γ2) or ξ(γ2) ⊂ξ(γ1)\nholds.\nExample 9.6: Let (Γ, L, M) be an uncertainty space, and let A be a crisp\nset of real numbers. The uncertain set ξ(γ) ≡A is of total order.\nExample 9.7: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3} with\npower set and M{γ1} = 0.6, M{γ2} = 0.3, M{γ3} = 0.2. The uncertain set\nξ(γ) =\n\n\n\n\n\n[2, 3],\nif γ = γ1\n[0, 5],\nif γ = γ2\n[1, 4],\nif γ = γ3\n(9.26)\nis of total order.\nExample 9.8: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. The uncertain set\nξ(γ) = [−γ, γ] ,\n∀γ ∈Γ\n(9.27)\nis of total order.\nExample 9.9: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. The uncertain set\nξ(γ) = [γ, γ + 1] ,\n∀γ ∈Γ\n(9.28)\nis not of total order.\nExercise 9.6:\nLet ξ be a totally ordered uncertain set.\nShow that its\ncomplement ξc is also of total order.\nExercise 9.7:\nLet ξ be a totally ordered uncertain set, and let f be a\nreal-valued function. Show that f(ξ) is also of total order.\n\n\n198\nChapter 9 - Uncertain Set\nTheorem 9.2 (Liu [133]) Let ξ be a totally ordered uncertain set. Then\n(i) the collection {x ∈ξ} indexed by x ∈ℜis of total order, and (ii) the\ncollection {x ̸∈ξ} indexed by x ∈ℜis also of total order.\nProof: If {x ∈ξ} indexed by x ∈ℜis not of total order, then there exist\ntwo numbers x1 and x2 such that neither {x1 ∈ξ} ⊂{x2 ∈ξ} nor {x2 ∈\nξ} ⊂{x1 ∈ξ} holds. This means there exist two points γ1 and γ2 in Γ such\nthat\nγ1 ∈{x1 ∈ξ},\nγ1 ̸∈{x2 ∈ξ},\nγ2 ∈{x2 ∈ξ},\nγ2 ̸∈{x1 ∈ξ}.\nThat is,\nx1 ∈ξ(γ1),\nx1 ̸∈ξ(γ2),\nx2 ∈ξ(γ2),\nx2 ̸∈ξ(γ1).\nThus neither ξ(γ1) ⊂ξ(γ2) nor ξ(γ2) ⊂ξ(γ1) holds. This result is in con-\ntradiction with that ξ is a totally ordered uncertain set. Therefore, {x ∈ξ}\nindexed by x ∈ℜis of total order. The ﬁrst part is proved. It follows from\n{x ̸∈ξ} = {x ∈ξ}c\nthat {x ̸∈ξ} indexed by x ∈ℜis also of total order. The second part is\nveriﬁed.\nUnion, Intersection and Complement\nDeﬁnition 9.4 Let ξ and η be two uncertain sets on the uncertainty space\n(Γ, L, M). Then (i) the union ξ ∪η of the uncertain sets ξ and η is\n(ξ ∪η)(γ) = ξ(γ) ∪η(γ),\n∀γ ∈Γ;\n(9.29)\n(ii) the intersection ξ ∩η of the uncertain sets ξ and η is\n(ξ ∩η)(γ) = ξ(γ) ∩η(γ),\n∀γ ∈Γ;\n(9.30)\n(iii) the complement ξc of the uncertain set ξ is\nξc(γ) = ξ(γ)c,\n∀γ ∈Γ.\n(9.31)\nExample 9.10: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3} with\npower set and M{γ1} = 0.6, M{γ2} = 0.3, M{γ3} = 0.2. Let ξ and η be two\nuncertain sets,\nξ(γ) =\n\n\n\n\n\n\n\n[1, 2],\nif γ = γ1\n[1, 3],\nif γ = γ2\n[1, 4],\nif γ = γ3,\nη(γ) =\n\n\n\n\n\n\n\n(2, 3),\nif γ = γ1\n(2, 4),\nif γ = γ2\n(2, 5),\nif γ = γ3.\n\n\nSection 9.1 - Uncertain Set\n199\nThen their union is\n(ξ ∪η)(γ) =\n\n\n\n\n\n\n\n[1, 3),\nif γ = γ1\n[1, 4),\nif γ = γ2\n[1, 5),\nif γ = γ3,\ntheir intersection is\n(ξ ∩η)(γ) =\n\n\n\n\n\n\n\n∅,\nif γ = γ1\n(2, 3],\nif γ = γ2\n(2, 4],\nif γ = γ3,\nand their complement sets are\nξc(γ) =\n\n\n\n\n\n\n\n(−∞, 1) ∪(2, +∞),\nif γ = γ1\n(−∞, 1) ∪(3, +∞),\nif γ = γ2\n(−∞, 1) ∪(4, +∞),\nif γ = γ3,\nηc(γ) =\n\n\n\n\n\n\n\n(−∞, 2] ∪[3, +∞),\nif γ = γ1\n(−∞, 2] ∪[4, +∞),\nif γ = γ2\n(−∞, 2] ∪[5, +∞),\nif γ = γ3.\nTheorem 9.3 (Law of Excluded Middle and Law of Contradiction) Let ξ be\nan uncertain set and let ξc be its complement. Then\nξ ∪ξc ≡ℜ,\nξ ∩ξc ≡∅.\n(9.32)\nProof: For each γ ∈Γ, it follows from the deﬁnition of uncertain set that\nthe union is\n(ξ ∪ξc)(γ) = ξ(γ) ∪ξc(γ) = ξ(γ) ∪ξ(γ)c ≡ℜ.\nThus we have ξ ∪ξc ≡ℜ. In addition, the intersection is\n(ξ ∩ξc)(γ) = ξ(γ) ∩ξc(γ) = ξ(γ) ∩ξ(γ)c ≡∅.\nThus we have ξ ∩ξc ≡∅.\nTheorem 9.4 (Double-Negation Law) Let ξ be an uncertain set. Then we\nhave\n(ξc)c = ξ.\n(9.33)\nProof: For each γ ∈Γ, it follows from the deﬁnition of complement that\n(ξc)c(γ) = (ξc(γ))c = (ξ(γ)c)c = ξ(γ).\nThus we have (ξc)c = ξ.\n\n\n200\nChapter 9 - Uncertain Set\nTheorem 9.5 (De Morgan’s Law) Let ξ and η be uncertain sets. Then we\nhave\n(ξ ∪η)c = ξc ∩ηc,\n(ξ ∩η)c = ξc ∪ηc.\n(9.34)\nProof: For each γ ∈Γ, it follows from the deﬁnition of complement that\n(ξ ∪η)c(γ) = ((ξ(γ) ∪η(γ))c = ξ(γ)c ∩η(γ)c = (ξc ∩ηc)(γ).\nThus we have (ξ ∪η)c = ξc ∩ηc. In addition, since\n(ξ ∩η)c(γ) = ((ξ(γ) ∩η(γ))c = ξ(γ)c ∪η(γ)c = (ξc ∪ηc)(γ),\nwe get (ξ ∩η)c = ξc ∪ηc.\nExercise 9.8: Let ξ be an uncertain set and let x be a real number. Show\nthat\n{x ∈ξc} = {x ̸∈ξ}.\n(9.35)\nExercise 9.9: Let ξ be an uncertain set and let x be a real number. Show\nthat {x ∈ξ} and {x ∈ξc} are opposite events.\nExercise 9.10: Let ξ and η be two uncertain sets. Show that {ξ ⊂η} and\n{ξ ⊂ηc} are not necessarily opposite events.\nFunction of Uncertain Sets\nDeﬁnition 9.5 Let ξ1, ξ2, · · · , ξn be uncertain sets on the uncertainty space\n(Γ, L, M), and let f be a measurable function. Then ξ = f(ξ1, ξ2, · · · , ξn) is\nan uncertain set deﬁned by\nξ(γ) = f(ξ1(γ), ξ2(γ), · · · , ξn(γ)),\n∀γ ∈Γ.\n(9.36)\nExample 9.11: Let ξ be an uncertain set on the uncertainty space (Γ, L, M)\nand let A be a crisp set of real numbers. Then ξ + A is also an uncertain set\ndetermined by\n(ξ + A)(γ) = ξ(γ) + A,\n∀γ ∈Γ.\n(9.37)\nExample 9.12: Note that the empty set ∅annihilates every other set. For\nexample, A + ∅= ∅and A × ∅= ∅. Take an uncertainty space (Γ, L, M) to\nbe {γ1, γ2, γ3} with power set and M{γ1} = 0.6, M{γ2} = 0.3, M{γ3} = 0.2.\nDeﬁne two uncertain sets,\nξ(γ) =\n\n\n\n\n\n∅,\nif γ = γ1\n[1, 3],\nif γ = γ2\n[1, 4],\nif γ = γ3,\nη(γ) =\n\n\n\n\n\n(2, 3),\nif γ = γ1\n(2, 4),\nif γ = γ2\n(2, 5),\nif γ = γ3.\n\n\nSection 9.2 - Membership Function\n201\nThen their sum is\n(ξ + η)(γ) =\n\n\n\n\n\n∅,\nif γ = γ1\n(3, 7),\nif γ = γ2\n(3, 9),\nif γ = γ3,\nand their multiplication is\n(ξ × η)(γ) =\n\n\n\n\n\n∅,\nif γ = γ1\n(2, 12),\nif γ = γ2\n(2, 20),\nif γ = γ3.\nExercise 9.11: Let ξ be an uncertain set. (i) Show that ξ + ξ ̸≡2ξ. (ii) Do\nyou think the same of crisp set?\n9.2\nMembership Function\nIt is well-known that a crisp set can be described by its indicator function.\nAs a generalization of indicator function, membership function will be used\nto describe an uncertain set.\nDeﬁnition 9.6 (Liu [123]) An uncertain set ξ is said to have a membership\nfunction µ if for any Borel set B of real numbers, we have\nM{B ⊂ξ} = inf\nx∈B µ(x),\n(9.38)\nM{ξ ⊂B} = 1 −sup\nx∈Bc µ(x).\n(9.39)\nThe above equations will be called measure inversion formulas.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\ninf\nx∈B µ(x)\n0\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nB\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n........\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\nsup\nx∈Bc µ(x)\n0\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nB\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n........\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.2: M{B ⊂ξ} = inf\nx∈B µ(x) and M{ξ ⊂B} = 1 −sup\nx∈Bc µ(x)\n\n\n202\nChapter 9 - Uncertain Set\nTheorem 9.6 Let ξ be an uncertain set whose membership function µ exists.\nThen\nµ(x) = M{x ∈ξ}\n(9.40)\nfor any number x.\nProof: For any number x, it follows from the ﬁrst measure inversion formula\nthat\nM{x ∈ξ} = M{{x} ⊂ξ} = inf\ny∈{x} µ(y) = µ(x).\nThe theorem is proved.\nRemark 9.4: The value of µ(x) is just the membership degree that x belongs\nto the uncertain set ξ. If µ(x) = 1, then x completely belongs to ξ; if µ(x) = 0,\nthen x does not belong to ξ at all. Thus the larger the value of µ(x) is, the\nmore true x belongs to ξ.\nTheorem 9.7 Let ξ be an uncertain set with membership function µ. Then\nM{x ̸∈ξ} = 1 −µ(x)\n(9.41)\nfor any number x.\nProof: Since {x ̸∈ξ} and {x ∈ξ} are opposite events, it follows from the\nduality axiom of uncertain measure that\nM{x ̸∈ξ} = 1 −M{x ∈ξ} = 1 −µ(x).\nThe theorem is proved.\nRemark 9.5: Theorem 9.7 states that if an element x belongs to an uncer-\ntain set with membership degree α, then x does not belong to the uncertain\nset with membership degree 1 −α.\nTheorem 9.8 Let ξ be an uncertain set with membership function µ. Then\nM{x ∈ξc} = 1 −µ(x)\n(9.42)\nfor any number x.\nProof: Since {x ∈ξc} and {x ∈ξ} are opposite events, it follows from the\nduality axiom of uncertain measure that\nM{x ∈ξc} = 1 −M{x ∈ξ} = 1 −µ(x).\nThe theorem is proved.\nRemark 9.6: Theorem 9.8 states that if an element x belongs to an un-\ncertain set with membership degree α, then x belongs to its complement set\nwith membership degree 1 −α.\n\n\nSection 9.2 - Membership Function\n203\nRemark 9.7: For any membership function µ, it is clear that 0 ≤µ(x) ≤1.\nWe will always take\ninf\nx∈∅µ(x) = 1,\nsup\nx∈∅\nµ(x) = 0.\n(9.43)\nThus\nM{∅⊂ξ} = 1 = inf\nx∈∅µ(x).\n(9.44)\nThat is, the ﬁrst measure inversion formula always holds for B = ∅. Further-\nmore, we have\nM{ξ ⊂ℜ} = 1 = 1 −sup\nx∈∅\nµ(x).\n(9.45)\nThat is, the second measure inversion formula always holds for B = ℜ.\nExample 9.13: The set ℜof real numbers is a special uncertain set ξ(γ) ≡ℜ.\nSuch an uncertain set has a membership function\nµ(x) ≡1\n(9.46)\nthat is just the indicator function of ℜ. In order to prove it, we must verify\nthat ℜand µ simultaneously satisfy the two measure inversion formulas (9.38)\nand (9.39). Let B be a Borel set of real numbers. Then\nM{B ⊂ξ} = M{Γ} = 1 = inf\nx∈B µ(x).\nThe ﬁrst measure inversion formula is veriﬁed. Next we prove the second\nmeasure inversion formula.\nWhen B = ℜ, the second measure inversion\nformula has been veriﬁed by (9.45). When B ̸= ℜ, we have\nM{ξ ⊂B} = M{∅} = 0 = 1 −sup\nx∈Bc µ(x).\nThus the second measure inversion formula holds for any Borel set B. There-\nfore, the uncertain set ξ(γ) ≡ℜhas the membership function µ(x) ≡1.\nExample 9.14: The empty set ∅is a special uncertain set ξ(γ) ≡∅. Such\nan uncertain set has a membership function\nµ(x) ≡0\n(9.47)\nthat is just the indicator function of ∅. In order to prove it, we must verify\nthat ∅and µ simultaneously satisfy the two measure inversion formulas (9.38)\nand (9.39). Let B be a Borel set of real numbers. When B = ∅, the ﬁrst\nmeasure inversion formula has been veriﬁed by (9.44). When B ̸= ∅, we have\nM{B ⊂ξ} = M{∅} = 0 = inf\nx∈B µ(x).\n\n\n204\nChapter 9 - Uncertain Set\nThus the ﬁrst measure inversion formula holds for any Borel set B. Next we\nprove the second measure inversion formula. For any Borel set B, we have\nM{ξ ⊂B} = M{Γ} = 1 = 1 −sup\nx∈Bc µ(x).\nThe second measure inversion formula is veriﬁed. Therefore, the uncertain\nset ξ(γ) ≡∅has the membership function µ(x) ≡0.\nExample 9.15: A crisp set A of real numbers is a special uncertain set\nξ(γ) ≡A. Such an uncertain set has a membership function\nµ(x) =\n(\n1,\nif x ∈A\n0,\nif x ̸∈A\n(9.48)\nthat is just the indicator function of A. In order to prove it, we must verify\nthat µ satisﬁes the two measure inversion formulas. Let B be a Borel set of\nreal numbers. When B ⊂A, we have\nM{B ⊂ξ} = M{Γ} = 1 = inf\nx∈B µ(x).\nWhen B ̸⊂A, we have\nM{B ⊂ξ} = M{∅} = 0 = inf\nx∈B µ(x).\nThus the ﬁrst measure inversion formula holds. When A ⊂B, we have\nM{ξ ⊂B} = M{Γ} = 1 = 1 −sup\nx∈Bc µ(x).\nWhen A ̸⊂B, we have\nM{ξ ⊂B} = M{∅} = 0 = 1 −sup\nx∈Bc µ(x).\nThe second measure inversion formula holds.\nTherefore, the membership\nfunction of a crisp set is identical to its indicator function.\nExample 9.16: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Consider an uncertain set\nξ(γ) = [−γ, γ] ,\n∀γ ∈(0, 1).\n(9.49)\nIf it does have a membership function, then µ(x) = M{x ∈ξ}.\nWhen\n−1 < x < 1, we have\nµ(x) = M {γ | x ∈[−γ, γ]} = M {[|x|, 1)} = 1 −|x|.\nWhen |x| ≥1, we have\nµ(x) = M {γ | x ∈[−γ, γ]} = M {∅} = 0.\n\n\nSection 9.2 - Membership Function\n205\nThus we get\nµ(x) =\n(\n1 −|x|,\nif −1 < x < 1\n0,\notherwise.\n(9.50)\nIn order to prove µ is a membership function of ξ, we should verify that ξ\nand µ satisfy the two measure inversion formulas. Let B be a Borel set of\nreal numbers. If\nsup\nx∈B\n|x| ≥1,\nthen\nM{B ⊂ξ} = M{γ ∈(0, 1) | B ⊂[−γ, γ]} = M {∅} = 0 = inf\nx∈B µ(x).\nIf\nsup\nx∈B\n|x| < 1,\nthen\nM{B ⊂ξ} = M{γ | B ⊂[−γ, γ]} ≤M\n\u001a\u0014\nsup\nx∈B\n|x|, 1\n\u0013\u001b\n= 1 −sup\nx∈B\n|x|,\nand\nM{B ⊂ξ} = M{γ | B ⊂[−γ, γ]} ≥M\n\u001a\u0012\nsup\nx∈B\n|x|, 1\n\u0013\u001b\n= 1 −sup\nx∈B\n|x|.\nThus\nM{B ⊂ξ} = 1 −sup\nx∈B\n|x| = inf\nx∈B(1 −|x|) = inf\nx∈B µ(x).\nThe ﬁrst measure inversion formula holds. If\ninf\nx∈Bc |x| ≥1,\nthen\nM{ξ ⊂B} = M{γ ∈(0, 1) | [−γ, γ] ⊂B} = M {Γ} = 1 = 1 −sup\nx∈Bc µ(x).\nIf\ninf\nx∈Bc |x| < 1,\nthen\nM{ξ ⊂B} = M{γ ∈(0, 1) | [−γ, γ] ⊂B} ≤M\n\u001a\u0012\n0, inf\nx∈Bc |x|\n\u0015\u001b\n= inf\nx∈Bc |x|,\nand\nM{ξ ⊂B} = M{γ ∈(0, 1) | [−γ, γ] ⊂B} ≥M\n\u001a\u0012\n0, inf\nx∈Bc |x|\n\u0013\u001b\n= inf\nx∈Bc |x|.\n\n\n206\nChapter 9 - Uncertain Set\nThus\nM{ξ ⊂B} = inf\nx∈Bc |x| = 1 −sup\nx∈Bc(1 −|x|) = 1 −sup\nx∈Bc µ(x).\nThe second measure inversion formula holds. Therefore, ξ has the member-\nship function (9.50).\nExercise 9.12:\nTake an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = 0.4, M{γ2} = 0.6. Show that the uncertain set\nξ(γ) =\n(\n∅,\nif γ = γ1\nA,\nif γ = γ2\nhas a membership function\nµ(x) =\n(\n0.6,\nif x ∈A\n0,\nif x ̸∈A\n(9.51)\nwhere A is a crisp set of real numbers.\nExercise 9.13:\nTake an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = 0.4, M{γ2} = 0.6. Deﬁne an uncertain set\nξ(γ) =\n(\n[2, 3],\nif γ = γ1\n[0, 5],\nif γ = γ2.\n(i) What is the membership function of ξ? (ii) Please justify your answer.\nExercise 9.14: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Deﬁne an uncertain set\nξ(γ) = [γ −1, 1 −γ] ,\n∀γ ∈(0, 1).\n(9.52)\n(i) What is the membership function of ξ? (ii) What do the two uncertain\nsets (9.49) and (9.52) make you think about?\nExercise 9.15: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Deﬁne an uncertain set\nξ(γ) =\nγ2, +∞\n\u0001\n.\n(9.53)\n(i) What is the membership function of ξ?\n(ii) What is the membership\nfunction of the complement set ξc? (iii) What do those two uncertain sets\nmake you think about?\nExercise 9.16: It is not true that every uncertain set has a membership\nfunction. Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with power set\n\n\nSection 9.2 - Membership Function\n207\nand M{γ1} = 0.4, M{γ2} = 0.6. Show that the uncertain set\nξ(γ) =\n(\n[1, 3],\nif γ = γ1\n[2, 4],\nif γ = γ2\n(9.54)\nhas no membership function. (Hint: If ξ does have a membership function,\nthen by using µ(x) = M{x ∈ξ}, we get\nµ(x) =\n\n\n\n\n\n\n\n\n\n0.4,\nif 1 ≤x < 2\n1,\nif 2 ≤x ≤3\n0.6,\nif 3 < x ≤4\n0,\notherwise.\n(9.55)\nVerify that ξ and µ cannot simultaneously satisfy the two measure inversion\nformulas (9.38) and (9.39).)\nExercise 9.17: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Show that the uncertain set\nξ(γ) = [γ, γ + 1] ,\n∀γ ∈Γ\n(9.56)\nhas no membership function.\nDeﬁnition 9.7 An uncertain set ξ is called triangular if it has a membership\nfunction\nµ(x) =\n\n\n\n\n\nx −a\nb −a ,\nif a ≤x ≤b\nx −c\nb −c ,\nif b < x ≤c\n(9.57)\ndenoted by (a, b, c) where a, b, c are real numbers with a < b < c.\nDeﬁnition 9.8 An uncertain set ξ is called trapezoidal if it has a member-\nship function\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\nx −a\nb −a ,\nif a ≤x ≤b\n1,\nif b < x ≤c\nx −d\nc −d ,\nif c < x ≤d\n(9.58)\ndenoted by (a, b, c, d) where a, b, c, d are real numbers with a < b < c < d.\n\n\n208\nChapter 9 - Uncertain Set\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\na\nb\nc\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\na\nb\nc\nd\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.3: Triangular and Trapezoidal Membership Functions\nWhat is “young”?\nSometimes we say “those students are young”. What ages can be considered\n“young”? In this case, “young” may be regarded as an uncertain set whose\nmembership function is\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/5,\nif 15 < x ≤20\n1,\nif 20 < x ≤35\n(45 −x)/10,\nif 35 < x ≤45\n0,\nif x > 45.\n(9.59)\nNote that we do not say “young” if the age is below 15.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n15yr 20yr\n35yr\n45yr\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.4: Membership Function of “young”\nWhat is “tall”?\nSometimes we say “those sportsmen are tall”. What heights (centimeters)\ncan be considered “tall”? In this case, “tall” may be regarded as an uncertain\n\n\nSection 9.2 - Membership Function\n209\nset whose membership function is\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤180\n(x −180)/5,\nif 180 < x ≤185\n1,\nif 185 < x ≤195\n(200 −x)/5,\nif 195 < x ≤200\n0,\nif x > 200.\n(9.60)\nNote that we do not say “tall” if the height is over 200cm.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n180cm 185cm\n195cm 200cm\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.5: Membership Function of “tall”\nWhat is “warm”?\nSometimes we say “those days are warm”. What temperatures can be con-\nsidered “warm”? In this case, “warm” may be regarded as an uncertain set\nwhose membership function is\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/3,\nif 15 < x ≤18\n1,\nif 18 < x ≤24\n(28 −x)/4,\nif 24 < x ≤28\n0,\nif 28 < x.\n(9.61)\nNote that we do not say “warm” if the temperature is above 28 degrees\nCelsius.\nWhat is “most”?\nSometimes we say “most students are boys”. What percentages can be con-\nsidered “most”? In this case, “most” may be regarded as an uncertain set\n\n\n210\nChapter 9 - Uncertain Set\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n15◦C 18◦C\n24◦C\n28◦C\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.6: Membership Function of “warm”\nwhose membership function is\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.7\n20(x −0.7),\nif 0.7 < x ≤0.75\n1,\nif 0.75 < x ≤0.85\n20(0.9 −x),\nif 0.85 < x ≤0.9\n0,\nif 0.9 < x ≤1.\n(9.62)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n70% 75%\n85% 90%\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.7: Membership Function of “most”\nRegular Membership Function\nDeﬁnition 9.9 (Liu [123]) A membership function µ is said to be regular\nif there exists a point x0 such that µ(x0) = 1 and µ(x) is unimodal about\nthe mode x0.\nThat is, µ(x) is increasing on (−∞, x0] and decreasing on\n[x0, +∞).\nFor example, both triangular and trapezoidal membership functions are\nregular.\nIn addition, the membership function µ(x) ≡1 is regular, but\nµ(x) ≡0 is not.\n\n\nSection 9.2 - Membership Function\n211\nExercise 9.18: Show that an uncertain set is nonempty if it has a regular\nmembership function.\nWhat uncertain sets have membership functions?\nIt is known that some uncertain sets do not have membership functions. This\nsubsection shows that totally ordered uncertain sets deﬁned on a continuous\nuncertainty space always have membership functions.\nTheorem 9.9 (Liu [133], Existence Theorem) Let ξ be a totally ordered un-\ncertain set on a continuous uncertainty space. Then its membership function\nalways exists, and\nµ(x) = M{x ∈ξ}.\n(9.63)\nProof: In order to prove that µ is the membership function of ξ, we must\nverify the two measure inversion formulas. Let B be any Borel set of real\nnumbers. Theorem 9.1 states that\n{B ⊂ξ} =\n\\\nx∈B\n{x ∈ξ}.\nSince the uncertain measure is assumed to be continuous, and {x ∈ξ} indexed\nby x ∈B is of total order, we obtain\nM{B ⊂ξ} = M\n( \\\nx∈B\n(x ∈ξ)\n)\n= inf\nx∈B M{x ∈ξ} = inf\nx∈B µ(x).\nThe ﬁrst measure inversion formula is veriﬁed. Next, Theorem 9.1 states that\n{ξ ⊂B} =\n\\\nx∈Bc\n{x ̸∈ξ}.\nSince the uncertain measure is assumed to be continuous, and {x ̸∈ξ} indexed\nby x ∈Bc is of total order, we obtain\nM{ξ ⊂B} = M\n( \\\nx∈Bc\n(x ̸∈ξ)\n)\n= inf\nx∈Bc M{x ̸∈ξ} = 1 −sup\nx∈Bc µ(x).\nThe second measure inversion formula is veriﬁed. Therefore, µ is the mem-\nbership function of ξ.\nExample 9.17: The continuity condition in Theorem 9.9 cannot be removed.\nFor example, take an uncertainty space (Γ, L, M) to be (0, 1) with power set\nand\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\n(9.64)\n\n\n212\nChapter 9 - Uncertain Set\nThen\nξ(γ) = (−γ, γ),\n∀γ ∈(0, 1)\n(9.65)\nis a totally ordered uncertain set on a discontinuous uncertainty space. If it\nindeed has a membership function, then by using µ(x) = M{x ∈ξ}, we get\nµ(x) =\n\n\n\n\n\n1,\nif x = 0\n0.5,\nif −1 < x < 0 or 0 < x < 1\n0,\notherwise.\n(9.66)\nHowever,\nM{(−1, 1) ⊂ξ} = M{∅} = 0 ̸= 0.5 =\ninf\nx∈(−1,1) µ(x).\n(9.67)\nThat is, the ﬁrst measure inversion formula is not valid and then ξ has\nno membership function. Therefore, the continuity condition cannot be re-\nmoved.\nExample 9.18: Some non-totally ordered uncertain sets may have mem-\nbership functions. For example, take an uncertainty space (Γ, L, M) to be\n{γ1, γ2, γ3, γ4} with power set and\nM{Λ} =\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.5,\notherwise.\n(9.68)\nThen\nξ(γ) =\n\n\n\n\n\n\n\n\n\n{1},\nif γ = γ1\n{1, 2},\nif γ = γ2\n{1, 3},\nif γ = γ3\n{1, 2, 3},\nif γ = γ4\n(9.69)\nis a non-totally ordered uncertain set. However, it has a membership function\nµ(x) =\n\n\n\n\n\n1,\nif x = 1\n0.5,\nif x = 2 or 3\n0,\notherwise\n(9.70)\nbecause ξ and µ can simultaneously satisfy the two measure inversion formu-\nlas (9.38) and (9.39).\nRemark 9.8: In practice, the unsharp concepts like “young”, “tall”, “warm”,\nand “most” can be regarded as totally ordered uncertain sets on a continuous\nuncertainty space.\n\n\nSection 9.2 - Membership Function\n213\nSuﬃcient and Necessary Condition\nTheorem 9.10 (Liu [121]) A real-valued function µ is a membership func-\ntion of uncertain set if and only if\n0 ≤µ(x) ≤1.\n(9.71)\nProof: If µ is a membership function of some uncertain set ξ, then µ(x) =\nM{x ∈ξ} and 0 ≤µ(x) ≤1. Conversely, suppose µ is a function such that\n0 ≤µ(x) ≤1. Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nξ(γ) = {x ∈ℜ| µ(x) ≥γ}\n(9.72)\nis a totally ordered uncertain set deﬁned on the continuous uncertainty space\n(Γ, L, M).\nSee Figure 9.8.\nBy using Theorem 9.9, it has a membership\nfunction\nb\nµ(x) = M{x ∈ξ} = M{γ ∈(0, 1) | µ(x) ≥γ} = M{(0, µ(x)]} = µ(x).\nThe theorem is proved.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nγ\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nξ(γ)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\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9.8: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel algebra\nand Lebesgue measure. Then ξ(γ) = {x ∈ℜ| µ(x) ≥γ} has the membership\nfunction µ.\nKeep in mind that ξ is not the unique uncertain set whose\nmembership function is µ.\nExample 9.19: Let c be a number between 0 and 1. It follows from the\nsuﬃcient and necessary condition that\nµ(x) ≡c\n(9.73)\nis a membership function. Take an uncertainty space (Γ, L, M) to be (0, 1)\nwith Borel algebra and Lebesgue measure. It follows from (9.72) that\nξ(γ) =\n(\nℜ,\nif 0 < γ ≤c\n∅,\nif c < γ < 1\n(9.74)\n\n\n214\nChapter 9 - Uncertain Set\nhas the membership function µ.\nExample 9.20: Let us design an uncertain set whose membership function\nis\nµ(x) = exp(−x2)\n(9.75)\nfor any real number x. Take an uncertainty space (Γ, L, M) to be (0, 1) with\nBorel algebra and Lebesgue measure. It follows from (9.72) that\nξ(γ) =\nh\n−\np\n−ln γ,\np\n−ln γ\ni\n,\n∀γ ∈(0, 1)\n(9.76)\nhas the membership function µ.\nExercise 9.19: Design an uncertain set whose membership function is just\nµ(x) = 1\n2 exp(−x2)\n(9.77)\nfor any real number x.\nExercise 9.20: Design an uncertain set whose membership function is just\nµ(x) = 1\n2 exp(−x2) + 1\n2\n(9.78)\nfor any real number x.\nTheorem 9.11 Let ξ be an uncertain set whose membership function µ ex-\nists. Then ξ is (i) nonempty if and only if\nsup\nx∈ℜ\nµ(x) = 1,\n(9.79)\n(ii) empty if and only if\nµ(x) ≡0,\n(9.80)\nand (iii) half-empty if and only if otherwise.\nProof: Since the membership function µ exists, it follows from the second\nmeasure inversion formula that\nM{ξ = ∅} = M{ξ ⊂∅} = 1 −sup\nx∈∅c µ(x) = 1 −sup\nx∈ℜ\nµ(x).\nThus ξ is (i) nonempty if and only if M{ξ = ∅} = 0, i.e., (9.79) holds, (ii)\nempty if and only if M{ξ = ∅} = 1, i.e., (9.80) holds, and (iii) half-empty if\nand only if otherwise.\nExercise 9.21:\nSome people prefer the uncertain set whose height (i.e.,\nthe supremum of the membership function) achieves 1. When the height is\nbelow 1, they divide all its membership values by the height and obtain a\n“normalized” membership function. Why is this idea wrong and harmful?\n\n\nSection 9.3 - Inverse Membership Function\n215\n9.3\nInverse Membership Function\nDeﬁnition 9.10 (Liu [123]) Let ξ be an uncertain set with membership func-\ntion µ. Then the set-valued function\nµ−1(α) =\n\b\nx ∈ℜ\n\f\n\f µ(x) ≥α\n\t\n,\n∀α ∈(0, 1]\n(9.81)\nis called the inverse membership function of ξ. For each given α, the set\nµ−1(α) is also called the α-cut of µ.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\nα\n0\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nµ−1(α)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n............\nFigure 9.9: Inverse Membership Function µ−1(α)\nRemark 9.9: Let ξ be an uncertain set with inverse membership function\nµ−1(α). Then the membership function of ξ is determined by\nµ(x) = sup\n\b\nα ∈(0, 1]\n\f\n\f x ∈µ−1(α)\n\t\n,\n∀x ∈ℜ.\n(9.82)\nExample 9.21: The triangular uncertain set ξ = (a, b, c) has an inverse\nmembership function\nµ−1(α) = [(1 −α)a + αb, αb + (1 −α)c].\n(9.83)\nExample 9.22: The trapezoidal uncertain set ξ = (a, b, c, d) has an inverse\nmembership function\nµ−1(α) = [(1 −α)a + αb, αc + (1 −α)d].\n(9.84)\nExample 9.23: Note that an inverse membership function may take value\nof the empty set ∅. Let ξ be an uncertain set with membership function\nµ(x) =\n(\n0.8,\nif 1 ≤x ≤2\n0,\notherwise.\n(9.85)\nThen its inverse membership function is\nµ−1(α) =\n(\n∅,\nif α > 0.8\n[1, 2],\notherwise.\n(9.86)\n\n\n216\nChapter 9 - Uncertain Set\nTheorem 9.12 (Liu [123]) An inverse membership function µ−1(α) is a\nmonotone decreasing set-valued function with respect to α ∈(0, 1]. That is,\nµ−1(α) ⊂µ−1(β),\nif α > β.\n(9.87)\nProof:\nFor any x ∈µ−1(α), we have µ(x) ≥α. Since α > β, we have\nµ(x) > β and then x ∈µ−1(β). Hence µ−1(α) ⊂µ−1(β). The theorem is\nproved.\nUncertain set does not necessarily take values of its α-cut!\nPlease keep in mind that uncertain set does not necessarily take values of its\nα-cuts. In fact, an α-cut is included in the uncertain set with uncertain mea-\nsure α. Conversely, the uncertain set is included in its α-cut with uncertain\nmeasure 1 −α. More precisely, we have the following theorem.\nTheorem 9.13 (Liu [123]) Let ξ be an uncertain set with inverse member-\nship function µ−1(α). Then for each α ∈(0, 1], we have\nM{µ−1(α) ⊂ξ} ≥α,\n(9.88)\nM{ξ ⊂µ−1(α)} ≥1 −α.\n(9.89)\nProof: For each x ∈µ−1(α), we have µ(x) ≥α. It follows from the ﬁrst\nmeasure inversion formula that\nM{µ−1(α) ⊂ξ} =\ninf\nx∈µ−1(α) µ(x) ≥α.\nFor each x ̸∈µ−1(α), we have µ(x) < α. It follows from the second measure\ninversion formula that\nM{ξ ⊂µ−1(α)} = 1 −\nsup\nx̸∈µ−1(α)\nµ(x) ≥1 −α.\nExample 9.24: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nξ(γ) = (−γ, γ),\n∀γ ∈(0, 1)\n(9.90)\nhas an inverse membership function\nµ−1(α) = [α −1, 1 −α],\n∀α ∈(0, 1].\n(9.91)\nIt is easy to verify that for each α ∈(0, 1], we have\nM{µ−1(α) ⊂ξ} = α,\n(9.92)\nM{ξ ⊂µ−1(α)} = 1 −α.\n(9.93)\n\n\nSection 9.4 - Independence\n217\nExample 9.25: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nξ(γ) ≡ℜ,\n∀γ ∈(0, 1)\n(9.94)\nhas an inverse membership function\nµ−1(α) ≡ℜ,\n∀α ∈(0, 1].\n(9.95)\nHowever, it is easy to verify that for each α ∈(0, 1), we have\nM{µ−1(α) ⊂ξ} = 1 > α,\n(9.96)\nM{ξ ⊂µ−1(α)} = 1 > 1 −α.\n(9.97)\n9.4\nIndependence\nThe independence of two uncertain sets means that knowing the value of\none does not change our estimation of the value of the other1. What un-\ncertain sets meet this condition?\nA typical case is that they are deﬁned\non diﬀerent uncertainty spaces.\nLet ξ1(γ1) and ξ2(γ2) be uncertain sets\non the uncertainty spaces (Γ1, L1, M1) and (Γ2, L2, M2), respectively.\nIt\nis clear that they are also uncertain sets on the product uncertainty space\n(Γ1, L1, M1) × (Γ2, L2, M2). Then for any Borel sets B1 and B2 of real num-\nbers, we have\nM{(ξ1 ⊂B1) ∩(ξ2 ⊂B2)}\n= M {(γ1, γ2) | ξ1(γ1) ⊂B1, ξ2(γ2) ⊂B2}\n= M {(γ1 | ξ1(γ1) ⊂B1) × (γ2 | ξ2(γ2) ⊂B2)}\n= M1 {γ1 | ξ1(γ1) ⊂B1} ∧M2 {γ2 | ξ2(γ2) ⊂B2}\n= M1 {ξ1 ⊂B1} ∧M2 {ξ2 ⊂B2}\n= M {ξ1 ⊂B1} ∧M {ξ2 ⊂B2} .\nThat is\nM{(ξ1 ⊂B1) ∩(ξ2 ⊂B2)} = M{ξ1 ⊂B1} ∧M{ξ2 ⊂B2}.\n(9.98)\nSimilarly, we may verify the following seven equations:\nM{(ξc\n1 ⊂B1) ∩(ξ2 ⊂B2)} = M{ξc\n1 ⊂B1} ∧M{ξ2 ⊂B2},\n(9.99)\nM{(ξ1 ⊂B1) ∩(ξc\n2 ⊂B2)} = M{ξ1 ⊂B1} ∧M{ξc\n2 ⊂B2},\n(9.100)\n1For example, it is clear that f(γ1, γ2) = [γ1, γ1 + 1] and g(γ1, γ2) = [γ2, γ2 + 2] are\nalways independent on the product uncertainty space (Γ1, L1, M1)×(Γ2, L2, M2). However,\nf(γ1, γ2) = [γ1, γ1 + 1] and g(γ1, γ2) = {γ1, γ2} are not.\n\n\n218\nChapter 9 - Uncertain Set\nM{(ξc\n1 ⊂B1) ∩(ξc\n2 ⊂B2)} = M{ξc\n1 ⊂B1} ∧M{ξc\n2 ⊂B2},\n(9.101)\nM{(ξ1 ⊂B1) ∪(ξ2 ⊂B2)} = M{ξ1 ⊂B1} ∨M{ξ2 ⊂B2},\n(9.102)\nM{(ξc\n1 ⊂B1) ∪(ξ2 ⊂B2)} = M{ξc\n1 ⊂B1} ∨M{ξ2 ⊂B2},\n(9.103)\nM{(ξ1 ⊂B1) ∪(ξc\n2 ⊂B2)} = M{ξ1 ⊂B1} ∨M{ξc\n2 ⊂B2},\n(9.104)\nM{(ξc\n1 ⊂B1) ∪(ξc\n2 ⊂B2)} = M{ξc\n1 ⊂B1} ∨M{ξc\n2 ⊂B2}.\n(9.105)\nThus we say two uncertain sets are independent if the above eight equations\nhold. Generally, we may deﬁne independence in the following form.\nDeﬁnition 9.11 (Liu [126]) The uncertain sets ξ1, ξ2, · · · , ξn are said to be\nindependent if for any Borel sets B1, B2, · · · , Bn of real numbers, we have\nM\n( n\n\\\ni=1\n(ξ∗\ni ⊂Bi)\n)\n=\nn\n^\ni=1\nM {ξ∗\ni ⊂Bi}\n(9.106)\nand\nM\n( n\n[\ni=1\n(ξ∗\ni ⊂Bi)\n)\n=\nn\n_\ni=1\nM {ξ∗\ni ⊂Bi}\n(9.107)\nwhere ξ∗\ni are arbitrarily chosen from {ξi, ξc\ni }, i = 1, 2, · · · , n, respectively.\nRemark 9.10: Note that (9.106) and (9.107) represent 2n+1 equations. For\nexample, when n = 2, they represent the 8 equations from (9.98) to (9.105).\nExercise 9.22: Show that a crisp set of real numbers (a special uncertain\nset) is always independent of any uncertain set.\nExercise 9.23: Let ξ1, ξ2, · · · , ξn be independent uncertain sets. Show that\nξi and ξj are independent for any indexes i and j with 1 ≤i < j ≤n.\nExercise 9.24: Let ξ be an uncertain set. Are ξ and ξc independent? Please\njustify your answer.\nExercise 9.25: Let ξ be an uncertain set, and let A be a crisp set. Are ξ\nand ξ + A independent? Please justify your answer.\nExercise 9.26:\nConstruct n independent uncertain sets.\n(Hint: Deﬁne\nthem on the product uncertainty space (Γ1, L1, M1) × (Γ2, L2, M2) × · · · ×\n(Γn, Ln, Mn).)\nExercise 9.27: Show that the following four statements are equivalent: (i)\nξ1 and ξ2 are independent; (ii) ξc\n1 and ξ2 are independent; (iii) ξ1 and ξc\n2 are\nindependent; and (iv) ξc\n1 and ξc\n2 are independent.\n\n\nSection 9.5 - Set Operational Law\n219\nTheorem 9.14 (Liu [126]) The uncertain sets ξ1, ξ2, · · · , ξn are independent\nif and only if for any Borel sets B1, B2, · · · , Bn of real numbers, we have\nM\n( n\n\\\ni=1\n(Bi ⊂ξ∗\ni )\n)\n=\nn\n^\ni=1\nM {Bi ⊂ξ∗\ni }\n(9.108)\nand\nM\n( n\n[\ni=1\n(Bi ⊂ξ∗\ni )\n)\n=\nn\n_\ni=1\nM {Bi ⊂ξ∗\ni }\n(9.109)\nwhere ξ∗\ni are arbitrarily chosen from {ξi, ξc\ni }, i = 1, 2, · · · , n, respectively.\nProof: Since {Bi ⊂ξ∗\ni } = {ξ∗c\ni\n⊂Bc\ni } for i = 1, 2, · · · , n, we immediately\nhave\nM\n( n\n\\\ni=1\n(Bi ⊂ξ∗\ni )\n)\n= M\n( n\n\\\ni=1\n(ξ∗c\ni\n⊂Bc\ni )\n)\n,\n(9.110)\nn\n^\ni=1\nM {Bi ⊂ξ∗\ni } =\nn\n^\ni=1\nM{ξ∗c\ni\n⊂Bc\ni },\n(9.111)\nM\n( n\n[\ni=1\n(Bi ⊂ξ∗\ni )\n)\n= M\n( n\n[\ni=1\n(ξ∗c\ni\n⊂Bc\ni )\n)\n,\n(9.112)\nn\n_\ni=1\nM {Bi ⊂ξ∗\ni } =\nn\n_\ni=1\nM{ξ∗c\ni\n⊂Bc\ni }.\n(9.113)\nIt follows from (9.110), (9.111), (9.112) and (9.113) that (9.108) and (9.109)\nare valid if and only if\nM\n( n\n\\\ni=1\n(ξ∗c\ni\n⊂Bc\ni )\n)\n=\nn\n^\ni=1\nM{ξ∗c\ni\n⊂Bc\ni },\n(9.114)\nM\n( n\n[\ni=1\n(ξ∗c\ni\n⊂Bc\ni )\n)\n=\nn\n_\ni=1\nM{ξ∗c\ni\n⊂Bc\ni }.\n(9.115)\nThe above two equations are also equivalent to the independence of the un-\ncertain sets ξ1, ξ2, · · · , ξn. The theorem is thus proved.\n9.5\nSet Operational Law\nThis section will discuss the union, intersection and complement of uncertain\nsets via membership functions.\n\n\n220\nChapter 9 - Uncertain Set\nUnion of Uncertain Sets\nTheorem 9.15 (Liu [123]) Let ξ and η be independent uncertain sets with\nmembership functions µ and ν, respectively. Then their union ξ ∪η has a\nmembership function\nλ(x) = µ(x) ∨ν(x).\n(9.116)\nProof: In order to prove µ ∨ν is the membership function of ξ ∪η, we must\nverify the two measure inversion formulas. Let B be any Borel set of real\nnumbers, and write\nβ = inf\nx∈B µ(x) ∨ν(x).\nIt is easy to verify that B ⊂µ−1(β) ∪ν−1(β). Thus\n{B ⊂(ξ ∪η)} ⊃{(µ−1(β) ∪ν−1(β)) ⊂(ξ ∪η)}.\nBy using the monotonicity theorem and independence of ξ and η, we have\nM{B ⊂(ξ ∪η)} ≥M{(µ−1(β) ∪ν−1(β)) ⊂(ξ ∪η)}\n≥M{(µ−1(β) ⊂ξ) ∩(ν−1(β) ⊂η)}\n= M{µ−1(β) ⊂ξ} ∧M{ν−1(β) ⊂η}\n≥β ∧β = β.\nHence\nM{B ⊂(ξ ∪η)} ≥β = inf\nx∈B µ(x) ∨ν(x).\n(9.117)\nOn the other hand, for any x ∈B, it holds that\n{B ⊂(ξ ∪η)} ⊂{x ∈(ξ ∪η)}.\nBy using the monotonicity theorem and independence of ξ and η, we have\nM{B ⊂(ξ ∪η)} ≤M{x ∈(ξ ∪η)} = M{(x ∈ξ) ∪(x ∈η)}\n= M{x ∈ξ} ∨M{x ∈η} = µ(x) ∨ν(x).\nThus\nM{B ⊂(ξ ∪η)} ≤inf\nx∈B µ(x) ∨ν(x).\n(9.118)\nIt follows from (9.117) and (9.118) that\nM{B ⊂(ξ ∪η)} = inf\nx∈B µ(x) ∨ν(x).\n(9.119)\nThe ﬁrst measure inversion formula is veriﬁed. Next we prove the second\nmeasure inversion formula. Since {(ξ ∪η) ⊂B} = {ξ ⊂B} ∩{η ⊂B}, by\n\n\nSection 9.5 - Set Operational Law\n221\nusing the independence of ξ and η, we have\nM{(ξ ∪η) ⊂B} = M{(ξ ⊂B) ∩(η ⊂B)} = M{ξ ⊂B} ∧M{η ⊂B}\n=\n\u0012\n1 −sup\nx∈Bc µ(x)\n\u0013\n∧\n\u0012\n1 −sup\nx∈Bc ν(x)\n\u0013\n= 1 −sup\nx∈Bc µ(x) ∨ν(x).\nThat is,\nM{(ξ ∪η) ⊂B} = 1 −sup\nx∈Bc µ(x) ∨ν(x).\n(9.120)\nThe second measure inversion formula is veriﬁed. Therefore, the union ξ ∪η\nis proved to have the membership function µ ∨ν.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nλ(x)\nµ(x)\nν(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n............\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.............\nFigure 9.10: Membership Function of Union of Uncertain Sets\nExample 9.26: The independence condition in Theorem 9.15 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be {γ1, γ2}\nwith power set and M{γ1} = M{γ2} = 0.5. Then\nξ(γ) =\n(\n[0, 1],\nif γ = γ1\n[0, 2],\nif γ = γ2\nis an uncertain set with membership function\nµ(x) =\n\n\n\n\n\n1,\nif 0 ≤x ≤1\n0.5,\nif 1 < x ≤2\n0,\notherwise,\nand\nη(γ) =\n(\n[0, 2],\nif γ = γ1\n[0, 1],\nif γ = γ2\n\n\n222\nChapter 9 - Uncertain Set\nis also an uncertain set with membership function\nν(x) =\n\n\n\n\n\n1,\nif 0 ≤x ≤1\n0.5,\nif 1 < x ≤2\n0,\notherwise.\nNote that ξ and η are not independent, and ξ ∪η ≡[0, 2] whose membership\nfunction is\nλ(x) =\n(\n1,\nif 0 ≤x ≤2\n0,\notherwise.\nThus\nλ(x) ̸= µ(x) ∨ν(x).\n(9.121)\nTherefore, the independence condition cannot be removed.\nExercise 9.28: Let ξ1, ξ2, · · · , ξn be independent uncertain sets with mem-\nbership functions µ1, µ2, · · · , µn, respectively. What is the membership func-\ntion of ξ1 ∪ξ2 ∪· · · ∪ξn?\nIntersection of Uncertain Sets\nTheorem 9.16 (Liu [123]) Let ξ and η be independent uncertain sets with\nmembership functions µ and ν, respectively. Then their intersection ξ ∩η has\na membership function\nλ(x) = µ(x) ∧ν(x).\n(9.122)\nProof: In order to prove µ ∧ν is the membership function of ξ ∩η, we\nmust verify the two measure inversion formulas. Let B be any Borel set of\nreal numbers. Since {B ⊂(ξ ∩η)} = {B ⊂ξ} ∩{B ⊂η}, by using the\nindependence of ξ and η, we have\nM{B ⊂(ξ ∩η)} = M{(B ⊂ξ) ∩(B ⊂η)} = M{B ⊂ξ} ∧M{B ⊂η}\n= inf\nx∈B µ(x) ∧inf\nx∈B ν(x) = inf\nx∈B µ(x) ∧ν(x).\nThat is,\nM{B ⊂(ξ ∩η)} = inf\nx∈B µ(x) ∧ν(x).\n(9.123)\nThe ﬁrst measure inversion formula is veriﬁed. In order to prove the second\nmeasure inversion formula, we write\nβ = sup\nx∈Bc µ(x) ∧ν(x).\nThen for any given number ε > 0, it holds that µ−1(β + ε) ∩ν−1(β + ε) ⊂B.\nThus\n{(ξ ∩η) ⊂B} ⊃{(ξ ∩η) ⊂(µ−1(β + ε) ∩ν−1(β + ε))}.\n\n\nSection 9.5 - Set Operational Law\n223\nBy using the monotonicity theorem and independence of ξ and η, we obtain\nM{(ξ ∩η) ⊂B} ≥M{(ξ ∩η) ⊂(µ−1(β + ε) ∩ν−1(β + ε))}\n≥M{(ξ ⊂µ−1(β + ε)) ∩(η ⊂ν−1(β + ε))}\n= M{ξ ⊂µ−1(β + ε)} ∧M{η ⊂ν−1(β + ε)}\n≥(1 −β −ε) ∧(1 −β −ε) = 1 −β −ε.\nLetting ε →0, we get\nM{(ξ ∩η) ⊂B} ≥1 −β = 1 −sup\nx∈Bc µ(x) ∧ν(x).\n(9.124)\nOn the other hand, for any γ ∈{(ξ ∩η) ⊂B}, we have (ξ ∩η)(γ) ⊂B. Thus\nfor any x ∈Bc, we have x ̸∈(ξ ∩η)(γ), i.e., γ ∈{x ̸∈(ξ ∩η)}. Hence\n{(ξ ∩η) ⊂B} ⊂{x ̸∈(ξ ∩η)}.\nBy using the monotonicity theorem and independence of ξ and η, we have\nM{(ξ ∩η) ⊂B} ≤M{x ̸∈(ξ ∩η)} = M{(x ̸∈ξ) ∪(x ̸∈η)}\n= M{x ̸∈ξ} ∨M{x ̸∈η} = (1 −µ(x)) ∨(1 −ν(x))\n= 1 −µ(x) ∧ν(x).\nThus\nM{(ξ ∩η) ⊂B} ≤1 −sup\nx∈Bc µ(x) ∧ν(x).\n(9.125)\nIt follows from (9.124) and (9.125) that\nM{(ξ ∩η) ⊂B} = 1 −sup\nx∈Bc µ(x) ∧(x).\n(9.126)\nThe second measure inversion formula is veriﬁed. Therefore, the intersection\nξ ∩η is proved to have the membership function µ ∧ν.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nλ(x)\nµ(x)\nν(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n............\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n............\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n...........\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.............\nFigure 9.11: Membership Function of Intersection of Uncertain Sets\n\n\n224\nChapter 9 - Uncertain Set\nExample 9.27: The independence condition in Theorem 9.16 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be {γ1, γ2}\nwith power set and M{γ1} = M{γ2} = 0.5. Then\nξ(γ) =\n(\n[0, 1],\nif γ = γ1\n[0, 2],\nif γ = γ2\nis an uncertain set with membership function\nµ(x) =\n\n\n\n\n\n1,\nif 0 ≤x ≤1\n0.5,\nif 1 < x ≤2\n0,\notherwise,\nand\nη(γ) =\n(\n[0, 2],\nif γ = γ1\n[0, 1],\nif γ = γ2\nis also an uncertain set with membership function\nν(x) =\n\n\n\n\n\n1,\nif 0 ≤x ≤1\n0.5,\nif 1 < x ≤2\n0,\notherwise.\nNote that ξ and η are not independent, and ξ ∩η ≡[0, 1] whose membership\nfunction is\nλ(x) =\n(\n1,\nif 0 ≤x ≤1\n0,\notherwise.\nThus\nλ(x) ̸= µ(x) ∧ν(x).\n(9.127)\nTherefore, the independence condition cannot be removed.\nExercise 9.29: Let ξ1, ξ2, · · · , ξn be independent uncertain sets with mem-\nbership functions µ1, µ2, · · · , µn, respectively. What is the membership func-\ntion of ξ1 ∩ξ2 ∩· · · ∩ξn?\nComplement of Uncertain Set\nTheorem 9.17 (Liu [123]) Let ξ be an uncertain set with membership func-\ntion µ. Then its complement ξc has a membership function\nλ(x) = 1 −µ(x).\n(9.128)\nProof: In order to prove 1 −µ is the membership function of ξc, we must\nverify the two measure inversion formulas.\nLet B be a Borel set of real\n\n\nSection 9.6 - Arithmetic Operational Law\n225\nnumbers. It follows from {B ⊂ξc} = {ξ ⊂Bc} and {ξc ⊂B} = {Bc ⊂ξ}\nthat\nM{B ⊂ξc} = M{ξ ⊂Bc} = 1 −\nsup\nx∈(Bc)c µ(x) = inf\nx∈B(1 −µ(x)),\nM{ξc ⊂B} = M{Bc ⊂ξ} = inf\nx∈Bc µ(x) = 1 −sup\nx∈Bc(1 −µ(x)).\nThus ξc has the membership function 1 −µ.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nλ(x)\nµ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.........................\n..\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n............................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n...\n..........................\nFigure 9.12: Membership Function of Complement of Uncertain Set\nExercise 9.30: Let ξ be an uncertain set with membership function µ(x).\nTheorem 9.17 tells us that ξc has a membership function 1 −µ(x). (i) It is\nknown that ξ ∪ξc ≡ℜwhose membership function is λ(x) ≡1, and\nλ(x) ̸= µ(x) ∨(1 −µ(x)).\n(9.129)\nWhy is Theorem 9.15 not applicable to the union of ξ and ξc? (ii) It is known\nthat ξ ∩ξc ≡∅whose membership function is λ(x) ≡0, and\nλ(x) ̸= µ(x) ∧(1 −µ(x)).\n(9.130)\nWhy is Theorem 9.16 not applicable to the intersection of ξ and ξc?\nExercise 9.31: Let ξ and η be independent uncertain sets with membership\nfunctions µ and ν, respectively. Then the set diﬀerence of ξ and η, denoted\nby ξ \\ η, is the set of all elements that are members of ξ but not members of\nη. That is,\nξ \\ η = ξ ∩ηc.\n(9.131)\nShow that ξ \\ η has a membership function\nλ(x) = µ(x) ∧(1 −ν(x)).\n(9.132)\n9.6\nArithmetic Operational Law\nThis section will present an arithmetic operational law of independent uncer-\ntain sets, including addition, subtraction, multiplication and division.\n\n\n226\nChapter 9 - Uncertain Set\nArithmetic Operational Law via Inverse Membership Functions\nTheorem 9.18 (Liu [123]) Let ξ1, ξ2, · · · , ξn be independent uncertain sets\nwith inverse membership functions µ−1\n1 , µ−1\n2 , · · · , µ−1\nn , respectively, and let f\nbe a measurable function. Then\nξ = f(ξ1, ξ2, · · · , ξn)\n(9.133)\nhas an inverse membership function,\nλ−1(α) = f(µ−1\n1 (α), µ−1\n2 (α), · · · , µ−1\nn (α)).\n(9.134)\nProof: For simplicity, we only prove the case n = 2. Let B be any Borel set\nof real numbers, and write\nβ = inf\nx∈B λ(x).\nIt holds that B ⊂λ−1(β). Thus {B ⊂ξ} ⊃{λ−1(β) ⊂ξ}. Since λ−1(β) =\nf(µ−1\n1 (β), µ−1\n2 (β)), by using the monotonicity theorem and independence of\nξ1 and ξ2, we have\nM{B ⊂ξ} ≥M{λ−1(β) ⊂ξ}\n= M{f(µ−1\n1 (β), µ−1\n2 (β)) ⊂f(ξ1, ξ2)}\n≥M{(µ−1\n1 (β) ⊂ξ1) ∩(µ−1\n2 (β) ⊂ξ2)}\n= M{µ−1\n1 (β) ⊂ξ1} ∧M{µ−1\n2 (β) ⊂ξ2}\n≥β ∧β = β.\nHence\nM{B ⊂ξ} ≥β = inf\nx∈B λ(x).\n(9.135)\nOn the other hand, for any given number ε > 0, we have B ̸⊂λ−1(β + ε).\nLet γ ∈{ξ ⊂λ−1(β + ε)}, i.e., ξ(γ) ⊂λ−1(β + ε). Then B ̸⊂ξ(γ), i.e.,\nγ ∈{B ̸⊂ξ}. Thus\n{B ̸⊂ξ} ⊃{ξ ⊂λ−1(β + ε)}.\nSince λ−1(β + ε) = f(µ−1\n1 (β + ε), µ−1\n2 (β + ε)), by using the monotonicity\ntheorem and independence of ξ1 and ξ2, we obtain\nM{B ̸⊂ξ} ≥M{ξ ⊂λ−1(β + ε)}\n= M{f(ξ1, ξ2) ⊂f(µ−1\n1 (β + ε), µ−1\n2 (β + ε))}\n≥M{(ξ1 ⊂µ−1\n1 (β + ε)) ∩(ξ2 ⊂µ−1\n2 (β + ε))}\n= M{ξ1 ⊂µ−1\n1 (β + ε)} ∧M{ξ2 ⊂µ−1\n2 (β + ε)}\n≥(1 −β −ε) ∧(1 −β −ε) = 1 −β −ε\nand then\nM{B ⊂ξ} = 1 −M{B ̸⊂ξ} ≤β + ε.\n\n\nSection 9.6 - Arithmetic Operational Law\n227\nLetting ε →0, we get\nM{B ⊂ξ} ≤β = inf\nx∈B λ(x).\n(9.136)\nIt follows from (9.135) and (9.136) that\nM{B ⊂ξ} = inf\nx∈B λ(x).\n(9.137)\nThe ﬁrst measure inversion formula is veriﬁed. In order to prove the second\nmeasure inversion formula, we write\nβ = sup\nx∈Bc λ(x).\nThen for any given number ε > 0, it holds that λ−1(β + ε) ⊂B. Thus\n{ξ ⊂B} ⊃{ξ ⊂λ−1(β + ε)}.\nSince λ−1(β + ε) = f(µ−1\n1 (β + ε), µ−1\n2 (β + ε)), by using the monotonicity\ntheorem and independence of ξ1 and ξ2, we obtain\nM{ξ ⊂B} ≥M{ξ ⊂λ−1(β + ε)}\n= M{f(ξ1, ξ2) ⊂f(µ−1\n1 (β + ε), µ−1\n2 (β + ε))}\n≥M{(ξ1 ⊂µ−1\n1 (β + ε)) ∩(ξ2 ⊂µ−1\n2 (β + ε))}\n= M{ξ1 ⊂µ−1\n1 (β + ε)} ∧M{ξ2 ⊂µ−1\n2 (β + ε)}\n≥(1 −β −ε) ∧(1 −β −ε) = 1 −β −ε.\nLetting ε →0, we get\nM{ξ ⊂B} ≥1 −β = 1 −sup\nx∈Bc λ(x).\n(9.138)\nOn the other hand, for any given number ε > 0, we have λ−1(β −ε) ̸⊂B.\nLet γ ∈{λ−1(β −ε) ⊂ξ}, i.e., λ−1(β −ε) ⊂ξ(γ). Then ξ(γ) ̸⊂B, i.e.,\nγ ∈{ξ ̸⊂B}. Thus\n{ξ ̸⊂B} ⊃{λ−1(β −ε) ⊂ξ}.\nSince λ−1(β −ε) = f(µ−1\n1 (β −ε), µ−1\n2 (β −ε)), by using the monotonicity\ntheorem and independence of ξ1 and ξ2, we obtain\nM{ξ ̸⊂B} ≥M{λ−1(β −ε) ⊂ξ}\n= M{f(µ−1\n1 (β −ε), µ−1\n2 (β −ε)) ⊂f(ξ1, ξ2)}\n≥M{(µ−1\n1 (β −ε) ⊂ξ1) ∩(µ−1\n2 (β −ε) ⊂ξ2)}\n= M{µ−1\n1 (β −ε) ⊂ξ1} ∧M{µ−1\n2 (β −ε) ⊂ξ2}\n≥(β −ε) ∧(β −ε) = β −ε\n\n\n228\nChapter 9 - Uncertain Set\nand then\nM{ξ ⊂B} = 1 −M{ξ ̸⊂B} ≤1 −β + ε.\nLetting ε →0, we get\nM{ξ ⊂B} ≤1 −β = 1 −sup\nx∈Bc λ(x).\n(9.139)\nIt follows from (9.138) and (9.139) that\nM{ξ ⊂B} = 1 −sup\nx∈Bc λ(x).\n(9.140)\nThe second measure inversion formula is veriﬁed. Therefore, ξ is proved to\nhave the membership function λ.\nExample 9.28: Let ξ = (a1, a2, a3) and η = (b1, b2, b3) be two independent\ntriangular uncertain sets. At ﬁrst, ξ has an inverse membership function,\nµ−1(α) = [(1 −α)a1 + αa2, αa2 + (1 −α)a3],\n(9.141)\nand η has an inverse membership function,\nν−1(α) = [(1 −α)b1 + αb2, αb2 + (1 −α)b3].\n(9.142)\nIt follows from the operational law that the sum ξ + η has an inverse mem-\nbership function,\nλ−1(α) = [(1−α)(a1 +b1)+α(a2 +b2), α(a2 +b2)+(1−α)(a3 +b3)]. (9.143)\nIn other words, the sum ξ + η is also a triangular uncertain set, and\nξ + η = (a1 + b1, a2 + b2, a3 + b3).\n(9.144)\nExample 9.29: Let ξ = (a1, a2, a3) and η = (b1, b2, b3) be two indepen-\ndent triangular uncertain sets. It follows from the operational law that the\ndiﬀerence ξ −η has an inverse membership function,\nλ−1(α) = [(1−α)(a1 −b3)+α(a2 −b2), α(a2 −b2)+(1−α)(a3 −b1)]. (9.145)\nIn other words, the diﬀerence ξ −η is also a triangular uncertain set, and\nξ −η = (a1 −b3, a2 −b2, a3 −b1).\n(9.146)\nExample 9.30: Let ξ = (a1, a2, a3) be a triangular uncertain set, and k a\nreal number. When k ≥0, the multiplication k ·ξ has an inverse membership\nfunction,\nλ−1(α) = [(1 −α)(ka1) + α(ka2), α(ka2) + (1 −α)(ka3)].\n(9.147)\n\n\nSection 9.6 - Arithmetic Operational Law\n229\nThat is, the multiplication k · ξ is a triangular uncertain set (ka1, ka2, ka3).\nWhen k < 0, the multiplication k · ξ has an inverse membership function,\nλ−1(α) = [(1 −α)(ka3) + α(ka2), α(ka2) + (1 −α)(ka1)].\n(9.148)\nThat is, the multiplication k · ξ is a triangular uncertain set (ka3, ka2, ka1).\nIn summary, we have\nk · ξ =\n(\n(ka1, ka2, ka3),\nif k ≥0\n(ka3, ka2, ka1),\nif k < 0.\n(9.149)\nExercise 9.32: Show that the multiplication of triangular uncertain sets is\nno longer a triangular one even they are independent and positive. That is,\n(a1, a2, a3) × (b1, b2, b3) ̸= (a1 × b1, a2 × b2, a3 × b3).\n(9.150)\nExercise 9.33: Let ξ = (a1, a2, a3, a4) and η = (b1, b2, b3, b4) be two inde-\npendent trapezoidal uncertain sets, and k a real number. Show that\nξ + η = (a1 + b1, a2 + b2, a3 + b3, a4 + b4),\n(9.151)\nξ −η = (a1 −b4, a2 −b3, a3 −b2, a4 −b1),\n(9.152)\nk · ξ =\n(\n(ka1, ka2, ka3, ka4),\nif k ≥0\n(ka4, ka3, ka2, ka1),\nif k < 0.\n(9.153)\nExample 9.31: The independence condition in Theorem 9.18 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be (0, 1) with\nBorel algebra and Lebesgue measure. Then\nξ1(γ) = [−γ, γ]\n(9.154)\nis a triangular uncertain set (−1, 0, 1) with inverse membership function\nµ−1\n1 (α) = [α −1, 1 −α],\n(9.155)\nand\nξ2(γ) = [γ −1, 1 −γ]\n(9.156)\nis also a triangular uncertain set (−1, 0, 1) with inverse membership function\nµ−1\n2 (α) = [α −1, 1 −α].\n(9.157)\nNote that ξ1 and ξ2 are not independent, and ξ1 + ξ2 ≡[−1, 1] whose inverse\nmembership function is\nλ−1(α) = [−1, 1].\n(9.158)\nThus\nλ−1(α) ̸= µ−1\n1 (α) + µ−1\n2 (α).\n(9.159)\nTherefore, the independence condition cannot be removed.\n\n\n230\nChapter 9 - Uncertain Set\nArithmetic Operational Law via Membership Functions\nTheorem 9.19 Let ξ1, ξ2, · · · , ξn be independent uncertain sets with mem-\nbership functions µ1(x), µ2(x), · · · , µn(x), respectively, and let f be a mea-\nsurable function. Then\nξ = f(ξ1, ξ2, · · · , ξn)\n(9.160)\nhas a membership function,\nλ(x) =\nsup\nf(x1,x2,··· ,xn)=x\nmin\n1≤i≤n µi(xi).\n(9.161)\nProof: Let λ be the membership function of ξ. For any given real number\nx, write β = λ(x). By using Theorem 9.18, we get\nλ−1(β) = f(µ−1\n1 (β), µ−1\n2 (β), · · · , µ−1\nn (β)).\nSince x ∈λ−1(β), there exist real numbers xi ∈µ−1\ni (β), i = 1, 2, · · · , n such\nthat f(x1, x2, · · · , xn) = x. Noting that µi(xi) ≥β for i = 1, 2, · · · , n, we\nhave\nλ(x) = β ≤min\n1≤i≤n µi(xi)\nand then\nλ(x) ≤\nsup\nf(x1,x2,··· ,xn)=x\nmin\n1≤i≤n µi(xi).\n(9.162)\nOn the other hand, assume x1, x2, · · · , xn are any given real numbers with\nf(x1, x2, · · · , xn) = x. Write\nβ = min\n1≤i≤n µi(xi).\nBy using Theorem 9.18, we get\nλ−1(β) = f(µ−1\n1 (β), µ−1\n2 (β), · · · , µ−1\nn (β)).\nNoting that xi ∈µ−1\ni (β) for i = 1, 2, · · · , n, we have\nx = f(x1, x2, · · · , xn) ∈f(µ−1\n1 (β), µ−1\n2 (β), · · · , µ−1\nn (β)) = λ−1(β).\nHence\nλ(x) ≥β = min\n1≤i≤n µi(xi)\nand then\nλ(x) ≥\nsup\nf(x1,x2,··· ,xn)=x\nmin\n1≤i≤n µi(xi).\n(9.163)\nIt follows from (9.162) and (9.163) that (9.161) holds.\nRemark 9.11: It is possible that the equation f(x1, x2, · · · , xn) = x does\nnot have a root for some values of x. In this case, we set λ(x) = 0.\n\n\nSection 9.7 - Inclusion Relation\n231\nExample 9.32: The independence condition in Theorem 9.19 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be (0, 1) with\nBorel algebra and Lebesgue measure. Then\nξ1(γ) = [−γ, γ]\n(9.164)\nis a triangular uncertain set (−1, 0, 1) with membership function\nµ1(x) =\n(\n1 −|x|,\nif −1 ≤x ≤1\n0,\notherwise,\n(9.165)\nand\nξ2(γ) = [γ −1, 1 −γ]\n(9.166)\nis also a triangular uncertain set (−1, 0, 1) with membership function\nµ2(x) =\n(\n1 −|x|,\nif −1 ≤x ≤1\n0,\notherwise.\n(9.167)\nNote that ξ1 and ξ2 are not independent, and ξ1 + ξ2 ≡[−1, 1] whose mem-\nbership function is\nλ(x) =\n(\n1,\nif −1 ≤x ≤1\n0,\notherwise.\n(9.168)\nThus\nλ(x) ̸=\nsup\nx1+x2=x µ1(x1) ∧µ2(x2).\n(9.169)\nTherefore, the independence condition cannot be removed.\nExercise 9.34: Let ξ and η be independent uncertain sets with membership\nfunctions µ(x) and ν(x), respectively. Show that ξ + η has a membership\nfunction,\nλ(x) = sup\ny∈ℜ\nµ(x −y) ∧ν(y).\n(9.170)\nExercise 9.35: Let ξ and η be independent uncertain sets with membership\nfunctions µ(x) and ν(x), respectively. Show that ξ −η has a membership\nfunction,\nλ(x) = sup\ny∈ℜ\nµ(x + y) ∧ν(y).\n(9.171)\n9.7\nInclusion Relation\nLet ξ be an uncertain set with membership function µ, and let B be a Borel\nset of real numbers. By using the deﬁnition of membership function, Liu\n\n\n232\nChapter 9 - Uncertain Set\n[123] presented two measure inversion formulas for calculating the uncertain\nmeasure of inclusion relation,\nM{B ⊂ξ} = inf\nx∈B µ(x),\n(9.172)\nM{ξ ⊂B} = 1 −sup\nx∈Bc µ(x).\n(9.173)\nEspecially, for any point x, Liu [123] also gave a formula for calculating the\nuncertain measure of containment relation,\nM{x ∈ξ} = µ(x).\n(9.174)\nA general formula was derived by Yao [261] for calculating the uncertain\nmeasure of inclusion relation between uncertain sets.\nTheorem 9.20 (Yao [261]) Let ξ and η be independent uncertain sets with\nmembership functions µ and ν, respectively. Then\nM{ξ ⊂η} = inf\nx∈ℜ(1 −µ(x)) ∨ν(x).\n(9.175)\nProof: Note that ξ ∩ηc has a membership function λ(x) = µ(x)∧(1−ν(x)).\nIt follows from {ξ ⊂η} ≡{ξ ∩ηc = ∅} and the second measure inversion\nformula that\nM{ξ ⊂η} = M{ξ ∩ηc = ∅}\n= M{ξ ∩ηc ⊂∅}\n= 1 −sup\nx∈∅c µ(x) ∧(1 −ν(x))\n= inf\nx∈ℜ(1 −µ(x)) ∨ν(x).\nThe theorem is proved.\nExample 9.33: Consider two special uncertain sets ξ = [1, 2] and η = [0, 3]\nthat are essentially crisp intervals whose membership functions are\nµ(x) =\n(\n1,\nif 1 ≤x ≤2\n0,\notherwise,\nν(x) =\n(\n1,\nif 0 ≤x ≤3\n0,\notherwise,\nrespectively. Mention that ξ ⊂η is a completely true relation since [1, 2] is\nindeed included in [0, 3]. By using (9.175), we also obtain\nM{ξ ⊂η} = inf\nx∈ℜ(1 −µ(x)) ∨ν(x) = 1.\n\n\nSection 9.7 - Inclusion Relation\n233\nExample 9.34: Consider two special uncertain sets ξ = [0, 2] and η = [1, 3]\nthat are essentially crisp intervals whose membership functions are\nµ(x) =\n(\n1,\nif 0 ≤x ≤2\n0,\notherwise,\nν(x) =\n(\n1,\nif 1 ≤x ≤3\n0,\notherwise,\nrespectively. Mention that ξ ⊂η is a completely false relation since [0, 2] is\nnot a subset of [1, 3]. By using (9.175), we also obtain\nM{ξ ⊂η} = inf\nx∈ℜ(1 −µ(x)) ∨ν(x) = 0.\nExample 9.35: Take an uncertainty space (Γ, L, M) to be {γ1, γ2, γ3, γ4}\nwith power set and\nM{Λ} =\n\n\n\n\n\n\n\n\n\n0,\nif Λ = ∅\n1,\nif Λ = Γ\n0.8,\nif γ1 ∈Λ ̸= Γ\n0.2,\nif γ1 ̸∈Λ ̸= ∅.\n(9.176)\nDeﬁne two uncertain sets,\nξ(γ) =\n(\n[0, 3],\nif γ = γ1 or γ2\n[1, 2],\nif γ = γ3 or γ4,\n(9.177)\nη(γ) =\n(\n[0, 3],\nif γ = γ1 or γ3\n[1, 2],\nif γ = γ2 or γ4.\n(9.178)\nWe may verify that ξ and η are independent, and share a common member-\nship function,\nµ(x) =\n\n\n\n\n\n1,\nif 1 ≤x ≤2\n0.8,\nif 0 ≤x < 1 or 2 < x ≤3\n0,\notherwise.\n(9.179)\nNote that\nM{ξ ⊂η} = M{γ1, γ3, γ4} = 0.8.\n(9.180)\nBy using (9.175), we also obtain\nM{ξ ⊂η} = inf\nx∈ℜ(1 −µ(x)) ∨µ(x) = 0.8.\n(9.181)\n\n\n234\nChapter 9 - Uncertain Set\nExercise 9.36: Let ξ and η be independent uncertain sets with membership\nfunctions µ and ν, respectively. Show that if µ ≤ν, then\nM{ξ ⊂η} ≥0.5.\n(9.182)\nExercise 9.37: Let ξ and η be independent uncertain sets with membership\nfunctions µ and ν, respectively, and let c be a number between 0.5 and 1. (i)\nConstruct ξ and η such that\nµ ≡ν\nand\nM{ξ ⊂η} = c.\n(9.183)\n(ii) Is it possible to construct ξ and η such that µ ≡ν and M{ξ ⊂η} = c\nwhen c is below 0.5? (iii) Is it stupid to think that ξ ⊂η if and only if\nµ(x) ≤ν(x) for all x? (iv) Is it stupid to think that ξ = η if and only if\nµ(x) = ν(x) for all x? (Hint: Use (9.176), (9.177) and (9.178) as a reference.)\nExample 9.36: The independence condition in Theorem 9.20 cannot be\nremoved. For example, take an uncertainty space (Γ, L, M) to be (0, 1) with\nBorel algebra and Lebesgue measure. Then\nξ(γ) = [−γ, γ]\n(9.184)\nis a triangular uncertain set (−1, 0, 1) with membership function\nµ(x) =\n(\n1 −|x|,\nif −1 ≤x ≤1\n0,\notherwise,\n(9.185)\nand\nη(γ) = [−γ, γ]\n(9.186)\nis also a triangular uncertain set (−1, 0, 1) with membership function\nν(x) =\n(\n1 −|x|,\nif −1 ≤x ≤1\n0,\notherwise.\n(9.187)\nNote that ξ and η are not independent (in fact, they are the same one), and\nM{ξ ⊂η} = 1. However, by using (9.175), we obtain\nM{ξ ⊂η} = inf\nx∈ℜ(1 −µ(x)) ∨ν(x) = 0.5 ̸= 1.\n(9.188)\nThus the independence condition cannot be removed.\n9.8\nExpected Value\nExpected value of an uncertain set is the center of gravity in the sense of\nuncertain measure (empty set and half-empty uncertain set have no expected\nvalue). A formal deﬁnition is given below.\n\n\nSection 9.8 - Expected Value\n235\nDeﬁnition 9.12 (Liu [118]) Let ξ be a nonempty uncertain set. Then the\nexpected value of ξ is deﬁned by\nE[ξ] =\nZ +∞\n0\nM{ξ ⪰x}dx −\nZ 0\n−∞\nM{ξ ⪯x}dx\n(9.189)\nprovided that at least one of the two integrals is ﬁnite.\nPlease note that ξ ⪰x represents “ξ is imaginarily included in [x, +∞)”,\nand ξ ⪯x represents “ξ is imaginarily included in (−∞, x]”. What are the\nappropriate values of M{ξ ⪰x} and M{ξ ⪯x}? Unfortunately, this problem\nis not as simple as you think.\nξ ̸< x\nξ ⪰x\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nξ ≥x\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 9.13: {ξ ≥x} ⊂{ξ ⪰x} ⊂{ξ ̸< x}\nIt is clear that the imaginary event {ξ ⪰x} is one between {ξ ≥x}\nand {ξ ̸< x}. See Figure 9.13. Intuitively, for the value of M{ξ ⪰x}, it is\ntoo conservative if we take M{ξ ≥x}, and it is too adventurous if we take\nM{ξ ̸< x}, i.e., 1 −M{ξ < x}. Thus we assign M{ξ ⪰x} the middle value\nbetween M{ξ ≥x} and 1 −M{ξ < x}. That is,\nM{ξ ⪰x} = 1\n2 (M{ξ ≥x} + 1 −M{ξ < x}) .\n(9.190)\nSimilarly, we also deﬁne\nM{ξ ⪯x} = 1\n2 (M{ξ ≤x} + 1 −M{ξ > x}) .\n(9.191)\nExample 9.37: Let (Γ, L, M) be an uncertainty space, and let [a, b] be a\ncrisp interval. Then\nξ(γ) ≡[a, b],\n∀γ ∈Γ\nis a special uncertain set.\nWhen a > 0, it follows from the deﬁnition of\nM{ξ ⪰x} and M{ξ ⪯x} that\nM{ξ ⪰x} =\n\n\n\n\n\n1,\nif 0 ≤x ≤a\n0.5,\nif a < x ≤b\n0,\nif x > b,\n\n\n236\nChapter 9 - Uncertain Set\nM{ξ ⪯x} ≡0,\n∀x ≤0.\nThus\nE[ξ] =\nZ a\n0\n1dx +\nZ b\na\n0.5dx = a + b\n2\n.\nWhen b < 0, we have\nM{ξ ⪰x} = 0,\n∀x ≥0,\nM{ξ ⪯x} =\n\n\n\n\n\n0,\nif x < a\n0.5,\nif a ≤x < b\n1,\nif b ≤x ≤0.\nThus\nE[ξ] = −\nZ b\na\n0.5dx −\nZ 0\nb\n1dx = a + b\n2\n.\nWhen a ≤0 ≤b, we have\nM{ξ ⪰x} =\n(\n0.5,\nif 0 < x ≤b\n0,\nif x > b,\nM{ξ ⪯x} =\n(\n0.5,\nif a ≤x < 0\n0,\nif x < a.\nThus\nE[ξ] =\nZ b\n0\n0.5dx −\nZ 0\na\n0.5dx = a + b\n2\n.\nIn summary, we always have\nE[ξ] = a + b\n2\n.\nExample 9.38: Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = 0.4, M{γ2} = 0.6. Deﬁne an uncertain set\nξ(γ) =\n(\n[2, 3],\nif γ = γ1\n[0, 5],\nif γ = γ2.\nIt follows from the deﬁnition of M{ξ ⪰x} and M{ξ ⪯x} that\nM{ξ ⪰x} =\n\n\n\n\n\n\n\n\n\n0.7,\nif 0 < x ≤2\n0.5,\nif 2 < x ≤3\n0.3,\nif 3 < x ≤5\n0,\nif x > 5,\n\n\nSection 9.8 - Expected Value\n237\nM{ξ ⪯x} ≡0,\n∀x < 0.\nThus\nE[ξ] =\nZ 2\n0\n0.7dx +\nZ 3\n2\n0.5dx +\nZ 5\n3\n0.3dx = 2.5.\nExample 9.39: Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = 0.4, M{γ2} = 0.6. Deﬁne an uncertain set\nξ(γ) =\n(\n[1, 3],\nif γ = γ1\n[2, 4],\nif γ = γ2.\nNote that this uncertain set has no membership function. It follows from the\ndeﬁnition of M{ξ ⪰x} and M{ξ ⪯x} that\nM{ξ ⪰x} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1,\nif 0 ≤x ≤1\n0.8,\nif 1 < x ≤2\n0.5,\nif 2 < x ≤3\n0.3,\nif 3 < x ≤4\n0,\nif x > 4,\nM{ξ ⪯x} ≡0,\n∀x ≤0.\nThus\nE[ξ] =\nZ 1\n0\n1dx +\nZ 2\n1\n0.8dx +\nZ 3\n2\n0.5dx +\nZ 4\n3\n0.3dx = 2.6.\nHow to Obtain Expected Value from Membership Function?\nLet ξ be an uncertain set with membership function µ. In order to calculate\nits expected value via (9.189), we must determine the values of M{ξ ⪰x}\nand M{ξ ⪯x} from the membership function µ.\nTheorem 9.21 (Liu [120]) Let ξ be a nonempty uncertain set with member-\nship function µ. Then for any real number x, we have\nM{ξ ⪰x} = 1\n2\n\u0012\nsup\ny≥x\nµ(y) + 1 −sup\ny<x µ(y)\n\u0013\n,\n(9.192)\nM{ξ ⪯x} = 1\n2\n\u0012\nsup\ny≤x\nµ(y) + 1 −sup\ny>x µ(y)\n\u0013\n.\n(9.193)\nProof: Since the uncertain set ξ has a membership function µ, the second\nmeasure inversion formula tells us that\nM{ξ ≥x} = 1 −sup\ny<x µ(y),\n\n\n238\nChapter 9 - Uncertain Set\nM{ξ < x} = 1 −sup\ny≥x\nµ(y).\nThus (9.192) follows from (9.190) immediately. We may also prove (9.193)\nsimilarly.\nTheorem 9.22 (Liu [120]) Let ξ be a nonempty uncertain set with member-\nship function µ. Then\nE[ξ] = x0 + 1\n2\nZ +∞\nx0\nsup\ny≥x\nµ(y)dx −1\n2\nZ x0\n−∞\nsup\ny≤x\nµ(y)dx\n(9.194)\nwhere x0 is a point such that µ(x0) = 1.\nProof: Since the membership function µ achieves 1 at the point x0, it follows\nfrom Theorem 9.21 that\nM{ξ ⪰x} =\n\n\n\n\n\n1 −sup\ny<x µ(y)/2,\nif x ≤x0\nsup\ny≥x\nµ(y)/2,\nif x > x0\n(9.195)\nand\nM{ξ ⪯x} =\n\n\n\n\n\nsup\ny≤x\nµ(y)/2,\nif x < x0\n1 −sup\ny>x µ(y)/2,\nif x ≥x0.\n(9.196)\nIf x0 ≥0, then\nE[ξ] =\nZ +∞\n0\nM{ξ ⪰x}dx −\nZ 0\n−∞\nM{ξ ⪯x}dx\n=\nZ x0\n0\n\u0012\n1 −sup\ny≤x\nµ(y)\n2\n\u0013\ndx +\nZ +∞\nx0\nsup\ny≥x\nµ(y)\n2\ndx −\nZ 0\n−∞\nsup\ny≤x\nµ(y)\n2\ndx\n= x0 + 1\n2\nZ +∞\nx0\nsup\ny≥x\nµ(y)dx −1\n2\nZ x0\n−∞\nsup\ny≤x\nµ(y)dx.\nIf x0 < 0, then\nE[ξ] =\nZ +∞\n0\nM{ξ ⪰x}dx −\nZ 0\n−∞\nM{ξ ⪯x}dx\n=\nZ +∞\n0\nsup\ny≥x\nµ(y)\n2\ndx −\nZ x0\n−∞\nsup\ny≤x\nµ(y)\n2\ndx −\nZ 0\nx0\n\u0012\n1 −sup\ny≥x\nµ(y)\n2\n\u0013\ndx\n= x0 + 1\n2\nZ +∞\nx0\nsup\ny≥x\nµ(y)dx −1\n2\nZ x0\n−∞\nsup\ny≤x\nµ(y)dx.\nThe theorem is thus proved.\n\n\nSection 9.8 - Expected Value\n239\nTheorem 9.23 (Liu [120]) Let ξ be an uncertain set with regular member-\nship function µ. Then\nE[ξ] = x0 + 1\n2\nZ +∞\nx0\nµ(x)dx −1\n2\nZ x0\n−∞\nµ(x)dx\n(9.197)\nwhere x0 is a point such that µ(x0) = 1.\nProof: Since µ is increasing on (−∞, x0] and decreasing on [x0, +∞), for\nalmost all x ≥x0, we have\nsup\ny≥x\nµ(y) = µ(x);\n(9.198)\nand for almost all x ≤x0, we have\nsup\ny≤x\nµ(y) = µ(x).\n(9.199)\nThus the theorem follows from (9.194) immediately.\nExercise 9.38: Show that the triangular uncertain set ξ = (a, b, c) has an\nexpected value\nE[ξ] = a + 2b + c\n4\n.\n(9.200)\nExercise 9.39: Show that the trapezoidal uncertain set ξ = (a, b, c, d) has\nan expected value\nE[ξ] = a + b + c + d\n4\n.\n(9.201)\nTheorem 9.24 (Liu [123]) Let ξ be a nonempty uncertain set with member-\nship function µ. If the expected value exists, then\nE[ξ] = 1\n2\nZ 1\n0\ninf µ−1(α) + sup µ−1(α)\n\u0001\ndα\n(9.202)\nwhere inf µ−1(α) and sup µ−1(α) are the inﬁmum and supremum of the α-cut,\nrespectively.\nProof: Since ξ is a nonempty uncertain set and has a ﬁnite expected value,\nwe may assume that there exists a point x0 such that µ(x0) = 1 (perhaps\nafter a small perturbation). It is clear that the two integrals\nZ +∞\nx0\nsup\ny≥x\nµ(y)dx\nand\nZ 1\n0\n(sup µ−1(α) −x0)dα\nmake an identical acreage. Thus\nZ +∞\nx0\nsup\ny≥x\nµ(y)dx =\nZ 1\n0\n(sup µ−1(α) −x0)dα =\nZ 1\n0\nsup µ−1(α)dα −x0.\n\n\n240\nChapter 9 - Uncertain Set\nSimilarly, we may prove\nZ x0\n−∞\nsup\ny≤x\nµ(y)dx =\nZ 1\n0\n(x0 −inf µ−1(α))dα = x0 −\nZ 1\n0\ninf µ−1(α)dα.\nIt follows from (9.194) that\nE[ξ] = x0 + 1\n2\nZ +∞\nx0\nsup\ny≥x\nµ(y)dx −1\n2\nZ x0\n−∞\nsup\ny≤x\nµ(y)dx\n= x0 + 1\n2\n\u0012Z 1\n0\nsup µ−1(α)dα −x0\n\u0013\n−1\n2\n\u0012\nx0 −\nZ 1\n0\ninf µ−1(α)dα\n\u0013\n= 1\n2\nZ 1\n0\n(inf µ−1(α) + sup µ−1(α))dα.\nThe theorem is thus veriﬁed.\nLinearity of Expected Value Operator\nTheorem 9.25 (Liu [123]) Let ξ and η be independent uncertain sets with\nﬁnite expected values. Then for any real numbers a and b, we have\nE[aξ + bη] = aE[ξ] + bE[η].\n(9.203)\nProof: Denote the membership functions of ξ and η by µ and ν, respectively.\nThen\nE[ξ] = 1\n2\nZ 1\n0\ninf µ−1(α) + sup µ−1(α)\n\u0001\ndα,\nE[η] = 1\n2\nZ 1\n0\ninf ν−1(α) + sup ν−1(α)\n\u0001\ndα.\nStep 1: We ﬁrst prove E[aξ] = aE[ξ]. The multiplication aξ has an\ninverse membership function,\nλ−1(α) = aµ−1(α).\nIt follows from Theorem 9.24 that\nE[aξ] = 1\n2\nZ 1\n0\ninf λ−1(α) + sup λ−1(α)\n\u0001\ndα\n= a\n2\nZ 1\n0\ninf µ−1(α) + sup µ−1(α)\n\u0001\ndα = aE[ξ].\nStep 2: We then prove E[ξ + η] = E[ξ] + E[η]. The sum ξ + η has an\ninverse membership function,\nλ−1(α) = µ−1(α) + ν−1(α).\n\n\nSection 9.8 - Expected Value\n241\nIt follows from Theorem 9.24 that\nE[ξ + η] = 1\n2\nZ 1\n0\ninf λ−1(α) + sup λ−1(α)\n\u0001\ndα\n= 1\n2\nZ 1\n0\ninf µ−1(α) + sup µ−1(α)\n\u0001\ndα\n+1\n2\nZ 1\n0\ninf ν−1(α) + sup ν−1(α)\n\u0001\ndα\n= E[ξ] + E[η].\nStep 3: Finally, for any real numbers a and b, it follows from Steps 1\nand 2 that\nE[aξ + bη] = E[aξ] + E[bη] = aE[ξ] + bE[η].\nThe theorem is proved.\nExample 9.40:\nGenerally speaking, the expected value operator is not\nnecessarily linear if the independence is not assumed. For example, take an\nuncertainty space (Γ, L, M) to be {γ1, γ2, γ3} with power set and M{γ1} =\n0.6, M{γ2} = 0.3, M{γ3} = 0.2. Deﬁne two uncertain sets as follows,\nξ(γ) =\n\n\n\n\n\n[1, 4],\nif γ = γ1\n[1, 3],\nif γ = γ2\n[1, 2],\nif γ = γ3,\nη(γ) =\n\n\n\n\n\n[1, 5],\nif γ = γ1\n[1, 2],\nif γ = γ2\n[1, 4],\nif γ = γ3.\nNote that ξ and η are not independent, and their sum is\n(ξ + η)(γ) =\n\n\n\n\n\n[2, 9],\nif γ = γ1\n[2, 5],\nif γ = γ2\n[2, 6],\nif γ = γ3.\nIt is easy to verify that E[ξ] = 2.2, E[η] = 2.5 and E[ξ + η] = 4.75. Thus we\nhave\nE[ξ + η] > E[ξ] + E[η].\nIf the uncertain sets are deﬁned by\nξ(γ) =\n\n\n\n\n\n[1, 4],\nif γ = γ1\n[1, 3],\nif γ = γ2\n[1, 2],\nif γ = γ3,\nη(γ) =\n\n\n\n\n\n[1, 4],\nif γ = γ1\n[1, 6],\nif γ = γ2\n[1, 2],\nif γ = γ3,\nthen\n(ξ + η)(γ) =\n\n\n\n\n\n[2, 8],\nif γ = γ1\n[2, 9],\nif γ = γ2\n[2, 4],\nif γ = γ3.\n\n\n242\nChapter 9 - Uncertain Set\nIt is easy to verify that E[ξ] = 2.2, E[η] = 2.6 and E[ξ + η] = 4.75. Thus we\nhave\nE[ξ + η] < E[ξ] + E[η].\nTherefore, the independence condition cannot be removed.\n9.9\nDistance\nDeﬁnition 9.13 (Liu [121]) The distance between nonempty uncertain sets\nξ and η is deﬁned as\nd(ξ, η) = E[|ξ −η|].\n(9.204)\nThat is, the distance between ξ and η is just the expected value of |ξ −η|.\nSince |ξ −η| is a nonnegative uncertain set, we have\nd(ξ, η) =\nZ +∞\n0\nM{|ξ −η| ⪰x}dx.\n(9.205)\nPlease note that |ξ −η| ⪰x represents “|ξ −η| is imaginarily included in\n[x, +∞)”. What is the appropriate value of M{|ξ −η| ⪰x}? Intuitively, it is\ntoo conservative if we take the value M{|ξ−η| ≥x}, and it is too adventurous\nif we take the value 1 −M{|ξ −η| < x}. Thus we assign M{|ξ −η| ⪰x} the\nmiddle value between them. That is,\nM{|ξ −η| ⪰x} = 1\n2 (M{|ξ −η| ≥x} + 1 −M{|ξ −η| < x}) .\n(9.206)\nTheorem 9.26 (Liu [130]) Let ξ and η be nonempty uncertain sets. Then\nfor any real number x, we have\nM{|ξ −η| ⪰x} = 1\n2\n \nsup\n|y|≥x\nλ(y) + 1 −sup\n|y|<x\nλ(y)\n!\n(9.207)\nwhere λ is the membership function of ξ −η.\nProof: Since ξ −η is an uncertain set with membership function λ, it follows\nfrom the measure inversion formula that for any real number x, we have\nM{|ξ −η| ≥x} = 1 −sup\n|y|<x\nµ(y),\nM{|ξ −η| < x} = 1 −sup\n|y|≥x\nµ(y).\nThe equation (9.207) is thus proved by (9.206).\n\n\nSection 9.10 - Entropy\n243\nTheorem 9.27 (Liu [130]) Let ξ and η be nonempty uncertain sets. Then\nthe distance between ξ and η is\nd(ξ, η) = 1\n2\nZ +∞\n0\n \nsup\n|y|≥x\nλ(y) + 1 −sup\n|y|<x\nλ(y)\n!\ndx\n(9.208)\nwhere λ is the membership function of ξ −η.\nProof: The theorem follows from (9.205) and (9.207) immediately.\nExercise 9.40: Let ξ be a nonempty uncertain set with membership function\nµ, and let b be a real number. Show that the distance between ξ and b is\nd(ξ, b) = 1\n2\nZ +∞\n0\n \nsup\n|y−b|≥x\nµ(y) + 1 −\nsup\n|y−b|<x\nµ(y)\n!\ndx.\n(9.209)\nExercise 9.41:\nLet ξ1 and ξ2 be independent triangular uncertain sets\n(a1, b1, c1) and (a2, b2, c2), respectively.\nWhat is the distance between ξ1\nand ξ2?\nExercise 9.42: Let ξ1 and ξ2 be independent trapezoidal uncertain sets\n(a1, b1, c1, d1) and (a2, b2, c2, d2), respectively. What is the distance between\nξ1 and ξ2?\n9.10\nEntropy\nThis section deﬁnes an entropy as the degree of diﬃculty of predicting the\nrealization of an uncertain set.\nDeﬁnition 9.14 (Liu [121]) Suppose that ξ is an uncertain set with mem-\nbership function µ. Then its entropy is deﬁned by\nH[ξ] =\nZ +∞\n−∞\nS(µ(x))dx\n(9.210)\nwhere S(t) = −t ln t −(1 −t) ln(1 −t).\nRemark 9.12: Note that the entropy (9.210) has the same form with de\nLuca and Termini’s entropy for fuzzy set [29].\nRemark 9.13: If ξ is a discrete uncertain set taking values in {x1, x2, · · · },\nthen the entropy becomes\nH[ξ] =\n∞\nX\ni=1\nS(µ(xi)).\n(9.211)\n\n\n244\nChapter 9 - Uncertain Set\nExample 9.41: A crisp set A of real numbers is a special uncertain set\nξ(γ) ≡A. Its membership function is\nµ(x) =\n(\n1,\nif x ∈A\n0,\nif x ̸∈A\nand entropy is\nH[ξ] =\nZ +∞\n−∞\nS(µ(x))dx =\nZ +∞\n−∞\n0dx = 0.\nThis means a crisp set has entropy 0.\nExercise 9.43: Let ξ = (a, b, c) be a triangular uncertain set. Show that its\nentropy is\nH[ξ] = c −a\n2\n.\n(9.212)\nExercise 9.44: Let ξ = (a, b, c, d) be a trapezoidal uncertain set. Show that\nits entropy is\nH[ξ] = b −a + d −c\n2\n.\n(9.213)\nTheorem 9.28 Let ξ be an uncertain set. Then H[ξ] ≥0 and equality holds\nif ξ is essentially a crisp set.\nProof: The nonnegativity is clear. In addition, when an uncertain set tends\nto a crisp set, its entropy tends to the minimum value 0.\nTheorem 9.29 Let ξ be an uncertain set on the interval [a, b]. Then\nH[ξ] ≤(b −a) ln 2\n(9.214)\nand equality holds if ξ has a membership function µ(x) = 0.5 on [a, b].\nProof: The theorem follows from the fact that the function S(t) reaches its\nmaximum value ln 2 at t = 0.5.\nTheorem 9.30 Let ξ be an uncertain set, and let ξc be its complement. Then\nH[ξc] = H[ξ].\n(9.215)\nProof: Write the membership function of ξ by µ. Then its complement ξc\nhas a membership function 1−µ(x). It follows from the deﬁnition of entropy\nthat\nH[ξc] =\nZ +∞\n−∞\nS (1 −µ(x)) dx =\nZ +∞\n−∞\nS(µ(x))dx = H[ξ].\nThe theorem is proved.\n\n\nSection 9.11 - Bibliographic Notes\n245\n9.11\nBibliographic Notes\nIn order to model unsharp concepts like “young”, “tall” and “most”, uncer-\ntain set was proposed by Liu [118] in 2010. Two years later, membership\nfunction was presented by Liu [123] to describe uncertain sets. Some uncer-\ntain sets have membership functions, and some uncertain sets do not. Liu\n[133] proved that totally ordered uncertain sets on a continuous uncertainty\nspace always have membership functions. In addition, Liu [126] deﬁned the\nindependence of uncertain sets, and provided the operational law through\nmembership functions. Yao [261] derived a formula for calculating the un-\ncertain measure of inclusion relation between uncertain sets.\nThe expected value of uncertain set was deﬁned by Liu [118]. Following\nthat, Liu [120][123] gave some formulas for calculating the expected value\nby membership function. Based on the expected value operator, Liu [121]\npresented the variance and distance between uncertain sets, and Yang-Gao\n[237] investigated the moments of uncertain set.\nEntropy was presented by Liu [121] as the degree of diﬃculty of predicting\nthe realization of an uncertain set. Some formulas were provided by Yao-Ke\n[256] for calculating the value of entropy.\n\n\n\n\nChapter 10\nUncertain Logic\nUncertain logic is a methodology for calculating the truth values of uncertain\npropositions via uncertain set theory. This chapter will introduce individual\nfeature data, uncertain quantiﬁer, uncertain subject, uncertain predicate,\nuncertain proposition, and truth value. Uncertain logic may provide a ﬂexible\nmeans for extracting linguistic summary from a collection of raw data.\n10.1\nIndividual Feature Data\nAt ﬁrst, we should have a universe A of individuals we are talking about.\nWithout loss of generality, we may assume that A consists of n individuals\nand is represented by\nA = {a1, a2, · · · , an}.\n(10.1)\nIn order to deal with the universe A, we should have feature data of all\nindividuals a1, a2, · · · , an. When we talk about “those days are warm”, we\nshould know the individual feature data of all days, for example,\nA = {22, 23, 25, 28, 30, 32, 36}\n(10.2)\nwhose elements are temperatures in centigrades. When we talk about “those\nstudents are young”, we should know the individual feature data of all stu-\ndents, for example,\nA = {21, 22, 22, 23, 24, 25, 26, 27, 28, 30, 32, 35, 36, 38, 40}\n(10.3)\nwhose elements are ages in years. When we talk about “those sportsmen\nare tall”, we should know the individual feature data of all sportsmen, for\nexample,\nA =\n\u001a 175, 178, 178, 180, 183, 184, 186, 186\n188, 190, 192, 192, 193, 194, 195, 196\n\u001b\n(10.4)\nwhose elements are heights in centimeters.\n\n\n248\nChapter 10 - Uncertain Logic\nSometimes the individual feature data are represented by vectors rather\na scalar number. When we talk about “those young students are tall”, we\nshould know the individual feature data of all students, for example,\nA =\n\n\n\n(24, 185), (25, 190), (26, 184), (26, 170), (27, 187), (27, 188)\n(28, 160), (30, 190), (32, 185), (33, 176), (35, 185), (36, 188)\n(38, 164), (38, 178), (39, 182), (40, 186), (42, 165), (44, 170)\n\n\n(10.5)\nwhose elements are ages and heights in years and centimeters, respectively.\n10.2\nUncertain Quantiﬁer\nIf we want to represent all individuals in the universe A, we use the universal\nquantiﬁer,\n∀= “for all”.\n(10.6)\nIf we want to represent some (at least one) individuals, we use the existential\nquantiﬁer,\n∃= “there exists at least one”.\n(10.7)\nIn addition to the two quantiﬁers, there are numerous imprecise quantiﬁers\nin human language, for example, many, several, some, most, a few, about\n10, and about 70%. This section will model them by the tool of uncertain\nquantiﬁer.\nDeﬁnition 10.1 (Liu [121]) Uncertain quantiﬁer is an uncertain set repre-\nsenting the number of individuals.\nExample 10.1: The universal quantiﬁer (∀) on the universe A of n individ-\nuals is a special uncertain quantiﬁer,\n∀≡{n}\n(10.8)\nwhose membership function is\nλ(x) =\n(\n1,\nif x = n\n0,\notherwise.\n(10.9)\nExample 10.2: The existential quantiﬁer (∃) on the universe A of n indi-\nviduals is a special uncertain quantiﬁer,\n∃≡{1, 2, · · · , n}\n(10.10)\nwhose membership function is\nλ(x) =\n(\n0,\nif x = 0\n1,\notherwise.\n(10.11)\n\n\nSection 10.2 - Uncertain Quantifier\n249\nExample 10.3: The quantiﬁer “there does not exist one” on the universe A\nis a special uncertain quantiﬁer\nQ ≡{0}\n(10.12)\nwhose membership function is\nλ(x) =\n(\n1,\nif x = 0\n0,\notherwise.\n(10.13)\nExample 10.4: The quantiﬁer “there exist exactly m” on the universe A is\na special uncertain quantiﬁer\nQ ≡{m}\n(10.14)\nwhose membership function is\nλ(x) =\n(\n1,\nif x = m\n0,\notherwise.\n(10.15)\nExample 10.5: The quantiﬁer “there exist at least m” on the universe A is\na special uncertain quantiﬁer\nQ ≡{m, m + 1, · · · , n}\n(10.16)\nwhose membership function is\nλ(x) =\n(\n1,\nif m ≤x ≤n\n0,\nif 0 ≤x < m.\n(10.17)\nExample 10.6: The quantiﬁer “there exist at most m” on the universe A is\na special uncertain quantiﬁer\nQ ≡{0, 1, 2, · · · , m}\n(10.18)\nwhose membership function is\nλ(x) =\n(\n1,\nif 0 ≤x ≤m\n0,\nif m < x ≤n.\n(10.19)\nExample 10.7: The uncertain quantiﬁer Q of “about 10 ” on the universe A\nmay have a membership function\nλ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤7\n(x −7)/2,\nif 7 ≤x ≤9\n1,\nif 9 ≤x ≤11\n(13 −x)/2,\nif 11 ≤x ≤13\n0,\nif 13 ≤x ≤n.\n(10.20)\n\n\n250\nChapter 10 - Uncertain Logic\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nλ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n7\n9\n10\n11\n13\n..............................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.1: Membership Function of Quantiﬁer “about 10 ”\nExample 10.8: In many cases, it is more convenient for us to use a per-\ncentage than an absolute quantity. For example, we may use the uncertain\nquantiﬁer Q of “about 70% ”. In this case, a possible membership function of\nQ is\nλ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.6\n20(x −0.6),\nif 0.6 ≤x ≤0.65\n1,\nif 0.65 ≤x ≤0.75\n20(0.8 −x),\nif 0.75 ≤x ≤0.8\n0,\nif 0.8 ≤x ≤1.\n(10.21)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nλ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n60% 65%\n75% 80%\n............................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.2: Membership Function of Quantiﬁer “about 70% ”\nNegated Quantiﬁer\nWhat is the negation of an uncertain quantiﬁer? The following deﬁnition\ngives a formal answer.\nDeﬁnition 10.2 (Liu [121]) Let Q be an uncertain quantiﬁer.\nThen the\nnegated quantiﬁer ¬Q is the complement of Q, i.e.,\n¬Q = Qc.\n(10.22)\n\n\nSection 10.2 - Uncertain Quantifier\n251\nExample 10.9: Let ∀= {n} be the universal quantiﬁer. Then its negated\nquantiﬁer\n¬∀≡{0, 1, 2, · · · , n −1}.\n(10.23)\nExample 10.10: Let ∃= {1, 2, · · · , n} be the existential quantiﬁer. Then\nits negated quantiﬁer is\n¬∃≡{0}.\n(10.24)\nTheorem 10.1 Let Q be an uncertain quantiﬁer whose membership function\nis λ. Then the negated quantiﬁer ¬Q has a membership function\n¬λ(x) = 1 −λ(x).\n(10.25)\nProof: This theorem follows from the operational law of uncertain set im-\nmediately.\nExample 10.11: Let Q be the uncertain quantiﬁer “about 70% ” deﬁned by\n(10.21). Then its negated quantiﬁer ¬Q has a membership function\n¬λ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1,\nif 0 ≤x ≤0.6\n20(0.65 −x),\nif 0.6 ≤x ≤0.65\n0,\nif 0.65 ≤x ≤0.75\n20(x −0.75),\nif 0.75 ≤x ≤0.8\n1,\nif 0.8 ≤x ≤1.\n(10.26)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\n¬λ(x)\n¬λ(x)\nλ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n60% 65%\n75% 80%\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.3: Membership Function of Negated Quantiﬁer of “about 70% ”\nTheorem 10.2 Let Q be an uncertain quantiﬁer. Then we have ¬¬Q = Q.\nProof: This theorem follows from ¬¬Q = ¬Qc = (Qc)c = Q.\n\n\n252\nChapter 10 - Uncertain Logic\nDual Quantiﬁer\nDeﬁnition 10.3 (Liu [121]) Let Q be an uncertain quantiﬁer. Then the dual\nquantiﬁer of Q is\nQ∗= ∀−Q.\n(10.27)\nRemark 10.1: Note that Q and Q∗are dependent uncertain sets such that\nQ + Q∗≡∀. If the cardinality of the universe A is n, then\nQ∗= {n} −Q.\n(10.28)\nExample 10.12: Since ∀≡{n}, we immediately have ∀∗= {0} = ¬∃. That\nis\n∀∗≡¬∃.\n(10.29)\nExample 10.13:\nSince ¬∀= {0, 1, 2, · · · , n −1}, we immediately have\n(¬∀)∗= {1, 2, · · · , n} = ∃. That is,\n(¬∀)∗≡∃.\n(10.30)\nExample 10.14: Since ∃≡{1, 2, · · · , n}, we have ∃∗= {0, 1, 2, · · · , n−1} =\n¬∀. That is,\n∃∗≡¬∀.\n(10.31)\nExample 10.15: Since ¬∃= {0}, we immediately have (¬∃)∗= {n} = ∀.\nThat is,\n(¬∃)∗= ∀.\n(10.32)\nTheorem 10.3 Let Q be an uncertain quantiﬁer whose membership function\nis λ. Then the dual quantiﬁer Q∗has a membership function\nλ∗(x) = λ(n −x)\n(10.33)\nwhere n is the cardinality of the universe A.\nProof: This theorem follows from the operational law of uncertain set im-\nmediately.\nExample 10.16: Let Q be the uncertain quantiﬁer “about 70% ” deﬁned by\n(10.21). Then its dual quantiﬁer Q∗has a membership function\nλ∗(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.2\n20(x −0.2),\nif 0.2 ≤x ≤0.25\n1,\nif 0.25 ≤x ≤0.35\n20(0.4 −x),\nif 0.35 ≤x ≤0.4\n0,\nif 0.4 ≤x ≤1.\n(10.34)\n\n\nSection 10.3 - Uncertain Subject\n253\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nλ∗(x)\nλ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n20%\n40%\n60%\n80%\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.4: Membership Function of Dual Quantiﬁer of “about 70% ”\nTheorem 10.4 Let Q be an uncertain quantiﬁer. Then we have Q∗∗= Q.\nProof: The theorem follows from Q∗∗= ∀−Q∗= ∀−(∀−Q) = Q.\n10.3\nUncertain Subject\nSometimes, we are interested in a subset of the universe of individuals, for\nexample, “warm days”, “young students” and “tall sportsmen”. This section\nwill model them by the concept of uncertain subject.\nDeﬁnition 10.4 (Liu [121]) Uncertain subject is an uncertain set containing\nsome speciﬁed individuals in the universe.\nExample 10.17: “Warm days are here again” is a statement in which “warm\ndays” is an uncertain subject that is an uncertain set on the universe of “all\ndays”, whose membership function may be deﬁned by\nν(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/3,\nif 15 ≤x ≤18\n1,\nif 18 ≤x ≤24\n(28 −x)/4,\nif 24 ≤x ≤28\n0,\nif 28 ≤x.\n(10.35)\nExample 10.18: “Young students are tall” is a statement in which “young\nstudents” is an uncertain subject that is an uncertain set on the universe of\n“all students”, whose membership function may be deﬁned by\nν(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/5,\nif 15 ≤x ≤20\n1,\nif 20 ≤x ≤35\n(45 −x)/10,\nif 35 ≤x ≤45\n0,\nif x ≥45.\n(10.36)\n\n\n254\nChapter 10 - Uncertain Logic\nExample 10.19: “Tall students are heavy” is a statement in which “tall\nstudents” is an uncertain subject that is an uncertain set on the universe of\n“all students”, whose membership function may be deﬁned by\nν(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤180\n(x −180)/5,\nif 180 ≤x ≤185\n1,\nif 185 ≤x ≤195\n(200 −x)/5,\nif 195 ≤x ≤200\n0,\nif x ≥200.\n(10.37)\nSubuniverse\nLet S be an uncertain subject with membership function ν on the universe\nA = {a1, a2, · · · , an} of individuals. In many cases, we are interested in some\nindividuals a’s with ν(a) ≥ω, where ω is a conﬁdence level. Thus we have a\nsubuniverse,\nSω = {a ∈A | ν(a) ≥ω}\n(10.38)\nthat will play a new universe of individuals we are talking about, and the\nindividuals out of Sω will be ignored at the conﬁdence level ω.\nTheorem 10.5 Let ω1 and ω2 be conﬁdence levels with ω1 > ω2, and let Sω1\nand Sω2 be subuniverses with conﬁdence levels ω1 an ω2, respectively. Then\nSω1 ⊂Sω2.\n(10.39)\nThat is, Sω is a decreasing sequence of sets with respect to ω.\nProof: If a ∈Sω1, then ν(a) ≥ω1 > ω2. Thus a ∈Sω2. It follows that\nSω1 ⊂Sω2. Note that Sω1 and Sω2 may be empty.\n10.4\nUncertain Predicate\nThere are numerous imprecise predicates in human language, for example,\nwarm, cold, hot, young, old, tall, small, and big. This section will model them\nby the concept of uncertain predicate.\nDeﬁnition 10.5 (Liu [121]) Uncertain predicate is an uncertain set repre-\nsenting a property that the individuals have in common.\nExample 10.20: “Today is warm” is a statement in which “warm” is an\nuncertain predicate that may be represented by a membership function\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/3,\nif 15 ≤x ≤18\n1,\nif 18 ≤x ≤24\n(28 −x)/4,\nif 24 ≤x ≤28\n0,\nif 28 ≤x.\n(10.40)\n\n\nSection 10.4 - Uncertain Predicate\n255\nExample 10.21: “John is young” is a statement in which “young” is an\nuncertain predicate that may be represented by a membership function\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/5,\nif 15 ≤x ≤20\n1,\nif 20 ≤x ≤35\n(45 −x)/10,\nif 35 ≤x ≤45\n0,\nif x ≥45.\n(10.41)\nExample 10.22: “Tom is tall” is a statement in which “tall” is an uncertain\npredicate that may be represented by a membership function\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤180\n(x −180)/5,\nif 180 ≤x ≤185\n1,\nif 185 ≤x ≤195\n(200 −x)/5,\nif 195 ≤x ≤200\n0,\nif x ≥200.\n(10.42)\nNegated Predicate\nDeﬁnition 10.6 (Liu [121]) Let P be an uncertain predicate.\nThen its\nnegated predicate ¬P is the complement of P, i.e.,\n¬P = P c.\n(10.43)\nTheorem 10.6 Let P be an uncertain predicate with membership function\nµ. Then its negated predicate ¬P has a membership function\n¬µ(x) = 1 −µ(x).\n(10.44)\nProof: The theorem follows from the deﬁnition of negated predicate and the\noperational law of uncertain set immediately.\nExample 10.23:\nLet P be the uncertain predicate “warm” deﬁned by\n(10.40). Then its negated predicate ¬P has a membership function\n¬µ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1,\nif x ≤15\n(18 −x)/3,\nif 15 ≤x ≤18\n0,\nif 18 ≤x ≤24\n(x −24)/4,\nif 24 ≤x ≤28\n1,\nif 28 ≤x.\n(10.45)\n\n\n256\nChapter 10 - Uncertain Logic\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n¬µ(x)\n¬µ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n15◦C 18◦C\n24◦C\n28◦C\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.5: Membership Function of Negated Predicate of “warm”\nExample 10.24:\nLet P be the uncertain predicate “young” deﬁned by\n(10.41). Then its negated predicate ¬P has a membership function\n¬µ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1,\nif x ≤15\n(20 −x)/5,\nif 15 ≤x ≤20\n0,\nif 20 ≤x ≤35\n(x −35)/10,\nif 35 ≤x ≤45\n1,\nif x ≥45.\n(10.46)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n¬µ(x)\n¬µ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n15yr 20yr\n35yr\n45yr\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.6: Membership Function of Negated Predicate of “young”\nExample 10.25: Let P be the uncertain predicate “tall ” deﬁned by (10.42).\nThen its negated predicate ¬P has a membership function\n¬µ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1,\nif x ≤180\n(185 −x)/5,\nif 180 ≤x ≤185\n0,\nif 185 ≤x ≤195\n(x −195)/5,\nif 195 ≤x ≤200\n1,\nif x ≥200.\n(10.47)\nTheorem 10.7 Let P be an uncertain predicate. Then we have ¬¬P = P.\nProof: The theorem follows from ¬¬P = ¬P c = (P c)c = P.\n\n\nSection 10.5 - Uncertain Proposition\n257\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx\nµ(x)\n¬µ(x)\n¬µ(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n180cm 185cm\n195cm 200cm\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 10.7: Membership Function of Negated Predicate of “tall ”\n10.5\nUncertain Proposition\nDeﬁnition 10.7 (Liu [121]) Assume that Q is an uncertain quantiﬁer, S is\nan uncertain subject, and P is an uncertain predicate. Then the triplet\n(Q, S, P) =“ Q of S are P”\n(10.48)\nis called an uncertain proposition.\nRemark 10.2: Let A be the universe of individuals. Then (Q, A, P) is a\nspecial uncertain proposition because A itself is a special uncertain subject.\nRemark 10.3:\nLet ∀be the universal quantiﬁer.\nThen (∀, A, P) is an\nuncertain proposition representing “all of A are P”.\nRemark 10.4:\nLet ∃be the existential quantiﬁer.\nThen (∃, A, P) is an\nuncertain proposition representing “at least one of A is P”.\nExample 10.26: “Most young students are tall” is an uncertain proposition\nin which the uncertain quantiﬁer Q is “most”, the uncertain subject S is\n“young students” and the uncertain predicate P is “tall”.\nTheorem 10.8 (Liu [121], Logical Equivalence Theorem) Let (Q, S, P) be\nan uncertain proposition. Then\n(Q, S, P) = (Q∗, S, ¬P)\n(10.49)\nwhere Q∗is the dual quantiﬁer of Q and ¬P is the negated predicate of P.\nProof: Note that (Q, S, P) represents “Q of S are P”. Since “Q of S are P”\nimplies “Q∗of S are not P”, we obtain (Q∗, S, ¬P). Conversely, (Q∗, S, ¬P)\nrepresents “Q∗of S are not P”. Since “Q∗of S are not P” implies “Q∗∗of S\nare P” and Q∗∗= Q, we obtain (Q, S, P). Thus (10.49) is veriﬁed.\nExample 10.27: When Q = ¬∀, we have Q∗= ∃. If S = A, then (10.49)\nbecomes the classical logic equivalence\n(¬∀, A, P) = (∃, A, ¬P).\n(10.50)\n\n\n258\nChapter 10 - Uncertain Logic\nExample 10.28: When Q = ¬∃, we have Q∗= ∀. If S = A, then (10.49)\nbecomes the classical logic equivalence\n(¬∃, A, P) = (∀, A, ¬P).\n(10.51)\n10.6\nTruth Value\nLet (Q, S, P) be an uncertain proposition. The truth value of (Q, S, P) should\nbe the uncertain measure that “Q of S are P”. That is,\nT(Q, S, P) = M{Q of S are P}.\n(10.52)\nHowever, it is impossible for us to deduce the value of M{Q of S are P} from\nthe information of Q, S and P within the framework of uncertain set theory.\nThus we need an additional formula to compose Q, S and P.\nDeﬁnition 10.8 (Liu [121]) Let (Q, S, P) be an uncertain proposition in\nwhich Q is an uncertain quantiﬁer with membership function λ, S is an un-\ncertain subject with membership function ν, and P is an uncertain predicate\nwith membership function µ. Then the truth value of (Q, S, P) with respect\nto the universe A is\nT(Q, S, P) =\nsup\n0≤ω≤1\n\n\n\nω ∧\nsup\nK⊂Sω\nλ(|K|)≥ω\ninf\na∈K µ(a) ∧\nsup\nK⊂Sω\nλ(|Sω|−|K|)≥ω\ninf\na∈K ¬µ(a)\n\n\n\nwhere Sω = {a ∈A | ν(a) ≥ω}.\nRemark 10.5: The symbol |K| represents the cardinality of the set K. For\nexample, |∅| = 0 and |{2, 5, 6}| = 3.\nRemark 10.6: Note that ¬µ is the membership function of the negated\npredicate of P, and\n¬µ(a) = 1 −µ(a).\n(10.53)\nRemark 10.7: When the subset K of individuals becomes an empty set ∅,\nwe set\ninf\na∈∅µ(a) = inf\na∈∅¬µ(a) = 1.\n(10.54)\nExercise 10.1: Assume Q is an uncertain percentage (e.g., about 70%) with\nmembership function λ, S is an uncertain subject with membership function\nν, P is an uncertain predicate with membership function µ, and A is the\nuniverse of individuals. Show that\nT(Q, S, P) = sup\n0≤ω≤1\n\n\n\nω ∧\nsup\nK⊂Sω\nλ(|K|/|Sω|)≥ω\ninf\na∈K µ(a) ∧\nsup\nK⊂Sω\nλ(1−|K|/|Sω|)≥ω\ninf\na∈K ¬µ(a)\n\n\n\n\n\nSection 10.6 - Truth Value\n259\nwhere Sω = {a ∈A | ν(a) ≥ω}.\nExercise 10.2: Assume Q is an uncertain quantiﬁer with membership func-\ntion λ, A is the universe of individuals, and P is an uncertain predicate with\nmembership function µ. Show that\nT(Q, A, P) =\nsup\n0≤ω≤1\n\n\n\nω ∧\nsup\nK⊂A\nλ(|K|)≥ω\ninf\na∈K µ(a) ∧\nsup\nK⊂A\nλ(|A|−|K|)≥ω\ninf\na∈K ¬µ(a)\n\n\n.\nExercise 10.3: Let A be the universe of individuals, and let P be an uncer-\ntain predicate with membership function µ. Show that\nT(∀, A, P) = inf\na∈A µ(a).\n(10.55)\nExercise 10.4: Let A be the universe of individuals, and let P be an uncer-\ntain predicate with membership function µ. Show that\nT(∃, A, P) = sup\na∈A\nµ(a).\n(10.56)\nExercise 10.5: Let A be the universe of individuals, and let P be an uncer-\ntain predicate with membership function µ. Show that\nT(¬∀, A, P) = 1 −inf\na∈A µ(a).\n(10.57)\nExercise 10.6: Let A be the universe of individuals, and let P be an uncer-\ntain predicate with membership function µ. Show that\nT(¬∃, A, P) = 1 −sup\na∈A\nµ(a).\n(10.58)\nTheorem 10.9 (Liu [121], Truth Value Theorem) Let (Q, S, P) be an un-\ncertain proposition in which Q is an uncertain quantiﬁer with membership\nfunction λ, S is an uncertain subject with membership function ν, and P is\nan uncertain predicate with membership function µ. Then the truth value of\n(Q, S, P) with respect to the universe A is\nT(Q, S, P) = sup\n0≤ω≤1\n(ω ∧∆(kω) ∧∆∗(k∗\nω))\n(10.59)\nwhere\nSω = {a ∈A | ν(a) ≥ω} ,\n(10.60)\nkω = min {x ∈Sω | λ(x) ≥ω} ,\n(10.61)\n∆(kω) = the kω-th largest value of {µ(a) | a ∈Sω},\n(10.62)\nk∗\nω = |Sω| −max{x ∈Sω | λ(x) ≥ω},\n(10.63)\n∆∗(k∗\nω) = the k∗\nω-th largest value of {¬µ(a) | a ∈Sω}.\n(10.64)\n\n\n260\nChapter 10 - Uncertain Logic\nProof: Since the supremum is achieved at the subset with minimum cardi-\nnality, we have\nsup\nK⊂Sω,λ(|K|)≥ω\ninf\na∈K µ(a) =\nsup\nK⊂Sω,|K|=kω\ninf\na∈K µ(a) = ∆(kω),\nsup\nK⊂Sω,λ(|Sω|−|K|)≥ω\ninf\na∈K ¬µ(a) =\nsup\nK⊂Sω,|K|=k∗\nω\ninf\na∈K ¬µ(a) = ∆∗(k∗\nω).\nThe theorem is thus veriﬁed. Please note that ∆(0) = ∆∗(0) = 1.\nExercise 10.7: Assume Q is an uncertain percentage (e.g., about 70%) with\nmembership function λ, S is an uncertain subject with membership function\nν, P is an uncertain predicate with membership function µ, and A is the\nuniverse of individuals. Show that\nT(Q, S, P) = sup\n0≤ω≤1\n(ω ∧∆(kω) ∧∆∗(k∗\nω))\n(10.65)\nwhere\nSω = {a ∈A | ν(a) ≥ω} ,\n(10.66)\nkω = min\n\u001a\nx ∈Sω\n\f\n\f λ\n\u0012 x\n|Sω|\n\u0013\n≥ω\n\u001b\n,\n(10.67)\n∆(kω) = the kω-th largest value of {µ(a) | a ∈Sω},\n(10.68)\nk∗\nω = |Sω| −max\n\u001a\nx ∈Sω\n\f\n\f λ\n\u0012 x\n|Sω|\n\u0013\n≥ω\n\u001b\n,\n(10.69)\n∆∗(k∗\nω) = the k∗\nω-th largest value of {¬µ(a) | a ∈Sω}.\n(10.70)\nExercise 10.8: Assume Q is an uncertain quantiﬁer with membership func-\ntion λ, A is the universe of individuals, and P is an uncertain predicate with\nmembership function µ. Show that\nT(Q, A, P) = sup\n0≤ω≤1\n(ω ∧∆(kω) ∧∆∗(k∗\nω))\n(10.71)\nwhere\nkω = min {x ∈A | λ(x) ≥ω} ,\n(10.72)\n∆(kω) = the kω-th largest value of {µ(a) | a ∈A},\n(10.73)\nk∗\nω = |A| −max{x ∈A | λ(x) ≥ω},\n(10.74)\n∆∗(k∗\nω) = the k∗\nω-th largest value of {¬µ(a) | a ∈A}.\n(10.75)\nExample 10.29: Assume that the daily temperatures from Monday to Sun-\nday are\n22, 23, 25, 28, 30, 32, 36\n(10.76)\n\n\nSection 10.6 - Truth Value\n261\nin centigrades. Consider an uncertain proposition\n(Q, A, P) = “two or three days are warm”.\n(10.77)\nNote that the uncertain quantiﬁer is Q = {2, 3}. We also suppose that the\nuncertain predicate P = “warm” has a membership function\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤15\n(x −15)/3,\nif 15 ≤x ≤18\n1,\nif 18 ≤x ≤24\n(28 −x)/4,\nif 24 ≤x ≤28\n0,\nif 28 ≤x.\n(10.78)\nIt is clear that Monday and Tuesday are warm with truth value 1, and\nWednesday is warm with truth value 0.75.\nBut Thursday to Sunday are\nnot “warm” at all (in fact, they are “hot”). Intuitively, the uncertain propo-\nsition “two or three days are warm” should be completely true. The truth\nvalue formula yields that the truth value is\nT(“two or three days are warm”) = 1.\n(10.79)\nThis is an intuitively expected result. In addition, we also have\nT(“two days are warm”) = 0.25,\n(10.80)\nT(“three days are warm”) = 0.75.\n(10.81)\nExample 10.30:\nAssume that in a team there are 16 sportsmen whose\nheights are\n175, 178, 178, 180, 183, 184, 186, 186\n188, 190, 192, 192, 193, 194, 195, 196\n(10.82)\nin centimeters. Consider an uncertain proposition\n(Q, A, P) = “about 70% of sportsmen are tall”.\n(10.83)\nSuppose the uncertain quantiﬁer Q = “about 70%” has a membership func-\ntion\nλ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.6\n20(x −0.6),\nif 0.6 ≤x ≤0.65\n1,\nif 0.65 ≤x ≤0.75\n20(0.8 −x),\nif 0.75 ≤x ≤0.8\n0,\nif 0.8 ≤x ≤1\n(10.84)\n\n\n262\nChapter 10 - Uncertain Logic\nand the uncertain predicate P = “tall” has a membership function\nµ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤180\n(x −180)/5,\nif 180 ≤x ≤185\n1,\nif 185 ≤x ≤195\n(200 −x)/5,\nif 195 ≤x ≤200\n0,\nif x ≥200.\n(10.85)\nThe truth value formula yields that the uncertain proposition has a truth\nvalue\nT(“about 70% of sportsmen are tall”) = 0.8.\n(10.86)\nExample 10.31: Assume that in a class there are 18 students whose ages\nand heights are\n(24, 185), (25, 190), (26, 184), (26, 170), (27, 187), (27, 188)\n(28, 160), (30, 190), (32, 185), (33, 176), (35, 185), (36, 188)\n(38, 164), (38, 178), (39, 182), (40, 186), (42, 165), (44, 170)\n(10.87)\nin years and centimeters. Consider an uncertain proposition\n(Q, S, P) = “most young students are tall”.\n(10.88)\nSuppose the uncertain quantiﬁer (percentage) Q = “most” has a membership\nfunction\nλ(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.7\n20(x −0.7),\nif 0.7 ≤x ≤0.75\n1,\nif 0.75 ≤x ≤0.85\n20(0.9 −x),\nif 0.85 ≤x ≤0.9\n0,\nif 0.9 ≤x ≤1.\n(10.89)\nNote that each individual is described by a feature data (y, z), where y rep-\nresents ages and z represents heights.\nIn this case, the uncertain subject\nS = “young students” has a membership function\nν(y) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif y ≤15\n(y −15)/5,\nif 15 ≤y ≤20\n1,\nif 20 ≤y ≤35\n(45 −y)/10,\nif 35 ≤y ≤45\n0,\nif y ≥45\n(10.90)\n\n\nSection 10.7 - Linguistic Summarizer\n263\nand the uncertain predicate P = “tall” has a membership function\nµ(z) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif z ≤180\n(z −180)/5,\nif 180 ≤z ≤185\n1,\nif 185 ≤z ≤195\n(200 −z)/5,\nif 195 ≤z ≤200\n0,\nif z ≥200.\n(10.91)\nThe truth value formula yields that the uncertain proposition has a truth\nvalue\nT(“most young students are tall”) = 0.8.\n(10.92)\n10.7\nLinguistic Summarizer\nLinguistic summary is a human language statement that is concise and easy-\nto-understand by humans. For example, “most young students are tall” is\na linguistic summary of students’ ages and heights. Thus a linguistic sum-\nmary is a special uncertain proposition whose uncertain quantiﬁer, uncertain\nsubject and uncertain predicate are linguistic terms. Uncertain logic pro-\nvides a ﬂexible means that is capable of extracting linguistic summary from\na collection of raw data.\nWhat inputs does the uncertain logic need? First, we should have some\nraw data (i.e., the individual feature data),\nA = {a1, a2, · · · , an}.\n(10.93)\nNext, we should have some linguistic terms to represent quantiﬁers, for exam-\nple, “most” and “all”. Denote them by a collection of uncertain quantiﬁers,\nQ = {Q1, Q2, · · · , Qm}.\n(10.94)\nThen, we should have some linguistic terms to represent subjects, for exam-\nple, “young students” and “old students”. Denote them by a collection of\nuncertain subjects,\nS = {S1, S2, · · · , Sn}.\n(10.95)\nLast, we should have some linguistic terms to represent predicates, for exam-\nple, “short” and “tall”. Denote them by a collection of uncertain predicates,\nP = {P1, P2, · · · , Pk}.\n(10.96)\nOne problem of data mining is to choose an uncertain quantiﬁer Q ∈Q, an\nuncertain subject S ∈S and an uncertain predicate P ∈P such that the\ntruth value of the linguistic summary “Q of S are P” to be extracted is at\nleast β, i.e.,\nT(Q, S, P) ≥β\n(10.97)\n\n\n264\nChapter 10 - Uncertain Logic\nfor the universe A = {a1, a2, · · · , an}, where β is a conﬁdence level. In order\nto solve this problem, Liu [121] proposed the following linguistic summarizer,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nFind Q, S and P\nsubject to:\nQ ∈Q\nS ∈S\nP ∈P\nT(Q, S, P) ≥β.\n(10.98)\nEach solution (Q, S, P) of the linguistic summarizer (10.98) produces a lin-\nguistic summary “Q of S are P”.\nExample 10.32: Assume that in a class there are 18 students whose ages\nand heights are\n(24, 185), (25, 190), (26, 184), (26, 170), (27, 187), (27, 188)\n(28, 160), (30, 190), (32, 185), (33, 176), (35, 185), (36, 188)\n(38, 164), (38, 178), (39, 182), (40, 186), (42, 165), (44, 170)\n(10.99)\nin years and centimeters.\nSuppose we have three linguistic terms “about\nhalf”, “most” and “all” as uncertain quantiﬁers whose membership functions\nare\nλhalf(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.4\n20(x −0.4),\nif 0.4 ≤x ≤0.45\n1,\nif 0.45 ≤x ≤0.55\n20(0.6 −x),\nif 0.55 ≤x ≤0.6\n0,\nif 0.6 ≤x ≤1,\n(10.100)\nλmost(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif 0 ≤x ≤0.7\n20(x −0.7),\nif 0.7 ≤x ≤0.75\n1,\nif 0.75 ≤x ≤0.85\n20(0.9 −x),\nif 0.85 ≤x ≤0.9\n0,\nif 0.9 ≤x ≤1,\n(10.101)\nλall(x) =\n(\n1,\nif x = 1\n0,\nif 0 ≤x < 1,\n(10.102)\nrespectively. Denote the collection of uncertain quantiﬁers by\nQ = {“about half ”, “most”,“all”}.\n(10.103)\n\n\nSection 10.7 - Linguistic Summarizer\n265\nWe also have three linguistic terms “young students”, “middle-aged students”\nand “old students” as uncertain subjects whose membership functions are\nνyoung(y) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif y ≤15\n(y −15)/5,\nif 15 ≤y ≤20\n1,\nif 20 ≤y ≤35\n(45 −y)/10,\nif 35 ≤y ≤45\n0,\nif y ≥45,\n(10.104)\nνmiddle(y) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif y ≤40\n(y −40)/5,\nif 40 ≤y ≤45\n1,\nif 45 ≤y ≤55\n(60 −y)/5,\nif 55 ≤y ≤60\n0,\nif y ≥60,\n(10.105)\nνold(y) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif y ≤55\n(y −55)/5,\nif 55 ≤y ≤60\n1,\nif 60 ≤y ≤80\n(85 −y)/5,\nif 80 ≤y ≤85\n0,\nif y ≥85,\n(10.106)\nrespectively. Denote the collection of uncertain subjects by\nS = {“young students”, “middle-aged students”, “old students”}. (10.107)\nFinally, we suppose that there are two linguistic terms “short” and “tall” as\nuncertain predicates whose membership functions are\nµshort(z) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif z ≤145\n(z −145)/5,\nif 145 ≤z ≤150\n1,\nif 150 ≤z ≤155\n(160 −z)/5,\nif 155 ≤z ≤160\n0,\nif z ≥200,\n(10.108)\nµtall(z) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif z ≤180\n(z −180)/5,\nif 180 ≤z ≤185\n1,\nif 185 ≤z ≤195\n(200 −z)/5,\nif 195 ≤z ≤200\n0,\nif z ≥200,\n(10.109)\nrespectively. Denote the collection of uncertain predicates by\nP = {“short”, “tall”}.\n(10.110)\n\n\n266\nChapter 10 - Uncertain Logic\nWe would like to extract an uncertain quantiﬁer Q ∈Q, an uncertain subject\nS ∈S and an uncertain predicate P ∈P such that the truth value of the\nlinguistic summary “Q of S are P” to be extracted is at least 0.8, i.e.,\nT(Q, S, P) ≥0.8\n(10.111)\nwhere 0.8 is a predetermined conﬁdence level.\nThe linguistic summarizer\n(10.98) yields\nQ = “most”,\nS = “young students”,\nP = “tall”\nand then extracts a linguistic summary “most young students are tall”.\n10.8\nBibliographic Notes\nBased on uncertain set theory, uncertain logic was designed by Liu [121]\nin 2011 for dealing with human language by using the truth value formula\nfor uncertain propositions. As an application of uncertain logic, Liu [121]\nalso proposed a linguistic summarizer that provides a means for extracting\nlinguistic summary from a collection of raw data.\n\n\nChapter 11\nUncertain Inference\nControl\nUncertain inference controller is a function that maps the state variables of a\nprocess under control to the action variables by using human knowledge and\nuncertain set theory. This chapter will introduce uncertain inference rule, and\nuncertain inference controller with application to inverted pendulum system.\n11.1\nUncertain Inference Rule\nLet X and Y be two concepts. Assume two rules “if X is an uncertain set ξ1\nthen Y is an uncertain set η1” and “if X is an uncertain set ξ2 then Y is an\nuncertain set η2”. From “X is a constant a”, we infer that Y is an uncertain\nset\nη∗=\nM{a ∈ξ1} · η1\nM{a ∈ξ1} + M{a ∈ξ2} +\nM{a ∈ξ2} · η2\nM{a ∈ξ1} + M{a ∈ξ2}.\n(11.1)\nThe uncertain inference rule is represented by\nRule 1: If X is ξ1 then Y is η1\nRule 2: If X is ξ2 then Y is η2\nFrom: X is a constant a\nInfer: Y is η∗determined by (11.1).\n(11.2)\nIf ξ1 and ξ2 have membership functions µ1 and µ2, respectively, then we\nimmediately have\nM{a ∈ξ1} = µ1(a),\nM{a ∈ξ2} = µ2(a).\nThus (11.1) becomes\nη∗=\nµ1(a) · η1\nµ1(a) + µ2(a) +\nµ2(a) · η2\nµ1(a) + µ2(a).\n(11.3)\n\n\n268\nChapter 11 - Uncertain Inference Control\nMultiple Antecedents\nMore generally, let X1, X2, · · · , Xm, Y be concepts. Assume rules “if X1 is ξi1\nand · · · and Xm is ξim then Y is ηi” for i = 1, 2, · · · , k. From “X1 is a1 and\n· · · and Xm is am”, we infer that Y is an uncertain set\nη∗=\nk\nX\ni=1\nci · ηi\nc1 + c2 + · · · + ck\n(11.4)\nwhere the coeﬃcients are determined by\nci = M {(a1 ∈ξi1) ∩(a2 ∈ξi2) ∩· · · ∩(am ∈ξim)}\n(11.5)\nfor i = 1, 2, · · · , k. The uncertain inference rule is represented by\nRule 1: If X1 is ξ11 and · · · and Xm is ξ1m then Y is η1\nRule 2: If X1 is ξ21 and · · · and Xm is ξ2m then Y is η2\n· · ·\nRule k: If X1 is ξk1 and · · · and Xm is ξkm then Y is ηk\nFrom: X1 is a1 and · · · and Xm is am\nInfer: Y is η∗determined by (11.4)\n(11.6)\nFor each index i with 1 ≤i ≤k, if ξi1, ξi2, · · · , ξim are independent un-\ncertain sets with membership functions µi1, µi2, · · · , µim, respectively, then\nci = M\n( m\n\\\nl=1\n(al ∈ξil)\n)\n= min\n1≤l≤m µil(al).\nThus (11.5) becomes\nci = min\n1≤l≤m µil(al),\ni = 1, 2, · · · , k.\n(11.7)\n11.2\nUncertain Inference Controller\nUncertain inference controller, proposed by Liu [118], is a function that maps\nthe inputs to the outputs based on the uncertain inference rule. Usually, an\nuncertain inference controller consists of 5 parts:\n1. inputs that are crisp data to be fed into the controller;\n2. a rule-base that contains a set of if-then rules provided by the experts;\n3. an uncertain inference rule that infers uncertain consequents from the\nuncertain antecedents;\n4. an expected value operator that converts the uncertain consequents to\ncrisp values;\n\n\nSection 11.2 - Uncertain Inference Controller\n269\nαm\n.\n.\n.\nα2\nα1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 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.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.Inference Rule\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.Rule Base\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nη∗\nn\n.\n.\n.\nη∗\n2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nη∗\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. βn = E[η∗\nn]\n.\n.\n.\nβ2 = E[η∗\n2]\nβ1 = E[η∗\n1]\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nβn\n.\n.\n.\nβ2\nβ1\nFigure 11.1: An Uncertain Inference Controller\n5. outputs that are crisp data yielded from the expected value operator.\nNow let us consider an uncertain inference controller in which there are\nm inputs α1, α2, · · · , αm and n outputs β1, β2, · · · , βn. At ﬁrst, we infer n\nuncertain sets η∗\n1, η∗\n2, · · · , η∗\nn from the m inputs α1, α2, · · · , αm by the rule-\nbase (i.e., a set of if-then rules),\nIf ξ11 and ξ12 and· · · and ξ1m then η11 and η12 and· · · and η1n\nIf ξ21 and ξ22 and· · · and ξ2m then η21 and η22 and· · · and η2n\n· · ·\nIf ξk1 and ξk2 and· · · and ξkm then ηk1 and ηk2 and· · · and ηkn\n(11.8)\nand the uncertain inference rule\nη∗\nj =\nk\nX\ni=1\nci · ηij\nc1 + c2 + · · · + ck\n(11.9)\nfor j = 1, 2, · · · , n, where the coeﬃcients are determined by\nci = M {(α1 ∈ξi1) ∩(α2 ∈ξi2) ∩· · · ∩(αm ∈ξim)}\n(11.10)\nfor i = 1, 2, · · · , k. Thus by using the expected value operator, we obtain\nβj = E[η∗\nj ]\n(11.11)\nfor j = 1, 2, · · · , n. Until now we have constructed a function from inputs\nα1, α2, · · · , αm to outputs β1, β2, · · · , βn. Write this function by f, i.e.,\n(β1, β2, · · · , βn) = f(α1, α2, · · · , αm).\n(11.12)\nThen we get an uncertain inference controller f.\nTheorem 11.1 Assume ξi1, ξi2, · · · , ξim, ηi1, ηi2, · · · , ηin are independent un-\ncertain sets with membership functions µi1, µi2, · · · , µim, νi1, νi2, · · · , νin, i =\n1, 2, · · · , k, respectively. Then the uncertain inference controller from the in-\nput (α1, α2, · · · , αm) to the output (β1, β2, · · · , βn) is\nβj =\nk\nX\ni=1\nci · E[ηij]\nc1 + c2 + · · · + ck\n(11.13)\n\n\n270\nChapter 11 - Uncertain Inference Control\nfor j = 1, 2, · · · , n, where ci are constants determined by\nci = min\n1≤l≤m µil(αl)\n(11.14)\nfor i = 1, 2, · · · , k, j = 1, 2, · · · , n, respectively.\nProof: It follows from the uncertain inference rule that the uncertain sets\nη∗\nj are\nη∗\nj =\nk\nX\ni=1\nci · ηij\nc1 + c2 + · · · + ck\nfor j = 1, 2, · · · , n. Since ηij, i = 1, 2, · · · , k, j = 1, 2, · · · , n are independent\nuncertain sets, we get the theorem immediately by the linearity of expected\nvalue operator.\nRemark 11.1: The uncertain inference controller allows the uncertain sets\nηij in the rule-base (11.8) become constants bij, i.e.,\nηij = bij\n(11.15)\nfor i = 1, 2, · · · , k and j = 1, 2, · · · , n. In this case, the uncertain inference\ncontroller (11.13) becomes\nβj =\nk\nX\ni=1\nci · bij\nc1 + c2 + · · · + ck\n(11.16)\nfor j = 1, 2, · · · , n.\nRemark 11.2: The uncertain inference controller allows the uncertain sets\nηij in the rule-base (11.8) become crisp functions hij of α1, α2, · · · , αm, i.e.,\nηij = hij(α1, α2, · · · , αm)\n(11.17)\nfor i = 1, 2, · · · , k and j = 1, 2, · · · , n. In this case, the uncertain inference\ncontroller (11.13) becomes\nβj =\nk\nX\ni=1\nci · hij(α1, α2, · · · , αm)\nc1 + c2 + · · · + ck\n(11.18)\nfor j = 1, 2, · · · , n.\nUncertain Inference Controllers are Universal Approximator\nUncertain inference controllers are capable of approximating any continuous\nfunction on a compact set (i.e., bounded and closed set) to arbitrary accuracy.\nThe following theorem shows this fact.\n\n\nSection 11.3 - Inverted Pendulum\n271\nTheorem 11.2 (Peng-Chen [188]) For any given continuous function g on a\ncompact set D ⊂ℜm and any given ε > 0, there exists an uncertain inference\ncontroller f such that\n∥f(α1, α2, · · · , αm) −g(α1, α2, · · · , αm)∥< ε\n(11.19)\nfor any (α1, α2, · · · , αm) ∈D.\nProof: Without loss of generality, we assume that the function g is a real-\nvalued function with only two variables α1 and α2, and the compact set is\na unit rectangle D = [0, 1] × [0, 1]. Since g is continuous on D and then is\nuniformly continuous, for any given number ε > 0, there is a number δ > 0\nsuch that\n|g(α1, α2) −g(α′\n1, α′\n2)| < ε\n(11.20)\nwhenever ∥(α1, α2) −(α′\n1, α′\n2)∥< δ. Let k be an integer larger than\n√\n2/δ,\nand write\nDij =\n\u001a\n(α1, α2)\n\f\n\f i −1\nk\n< α1 ≤i\nk , j −1\nk\n< α2 ≤j\nk\n\u001b\n(11.21)\nfor i, j = 1, 2, · · · , k.\nNote that {Dij| i, j = 1, 2, · · · , k} is a sequence of\ndisjoint rectangles whose “diameter” is less than δ. Deﬁne special uncertain\nsets\nξi =\n\u0012i −1\nk\n, i\nk\n\u0013\n,\ni = 1, 2, · · · , k,\n(11.22)\nηj =\n\u0012j −1\nk\n, j\nk\n\u0013\n,\nj = 1, 2, · · · , k.\n(11.23)\nThen we assume a rule-base with k × k if-then rules,\nRule ij: If ξi and ηj then g(i/k, j/k),\ni, j = 1, 2, · · · , k.\n(11.24)\nAccording to the uncertain inference rule, the corresponding uncertain infer-\nence controller from D to ℜis\nf(α1, α2) = g(i/k, j/k),\nif (α1, α2) ∈Dij, i, j = 1, 2, · · · , k.\n(11.25)\nIt follows from (11.20) that for any (α1, α2) ∈Dij ⊂D, we have\n|f(α1, α2) −g(α1, α2)| = |g(i/k, j/k) −g(α1, α2)| < ε.\n(11.26)\nThe theorem is thus veriﬁed. Hence uncertain inference controllers are uni-\nversal approximators.\nUncertain Inference Control System\nFigure 11.2 shows an uncertain inference control system consisting of an\nuncertain inference controller and a process under control. Note that t rep-\nresents time, α1(t), α2(t), · · · , αm(t) are not only the inputs of uncertain in-\nference controller but also the state variables of process under control, and\nβ1(t), β2(t), · · · , βn(t) are not only the outputs of uncertain inference con-\ntroller but also the action variables of process under control.\n\n\n272\nChapter 11 - Uncertain Inference Control\nαm(t)\n.\n.\n.\nα2(t)\nα1(t)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. Inference Rule\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nRule Base\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nη∗\nn(t)\n.\n.\n.\nη∗\n2(t)\nη∗\n1(t)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. βn(t)=E[η∗\nn(t)]\n.\n.\n.\nβ2(t)=E[η∗\n2(t)]\nβ1(t)=E[η∗\n1(t)]\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nβn(t)\n.\n.\n.\nβ2(t)\nβ1(t)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nProcess under\nControl\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nOutputs of Controller\nAction Variables of Process\nInputs of Controller\nState Variables of Process\nFigure 11.2: An Uncertain Inference Control System\n11.3\nInverted Pendulum\nInverted pendulum system is a nonlinear unstable system that is widely used\nas a benchmark for testing control algorithms. Many good techniques already\nexist for balancing inverted pendulum. Among others, Gao [57] successfully\nbalanced an inverted pendulum by the uncertain inference controller with\n5 × 5 if-then rules.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n•\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n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.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nA(t)\nF(t).\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 11.3: An Inverted Pendulum in which A(t) represents the angular\nposition and F(t) represents the force that moves the cart at time t.\nThe uncertain inference controller has two inputs (“angle” and “angular\nvelocity”) and one output (“force”). Three of them will be represented by\nuncertain sets labeled by\n“negative large”\nNL\n“negative small”\nNS\n“zero”\nZ\n“positive small”\nPS\n“positive large”\nPL\n\n\nSection 11.3 - Inverted Pendulum\n273\nThe membership functions of those uncertain sets are shown in Figures 11.4,\n11.5 and 11.6.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. (rad)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n−π/2 −π/4\n0\nπ/4\nπ/2\nNL\nNS\nZ\nPS\nPL\nFigure 11.4: Membership Functions of “Angle”\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. (rad/sec)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n−π/4 −π/8\n0\nπ/8\nπ/4\nNL\nNS\nZ\nPS\nPL\nFigure 11.5: Membership Functions of “Angular Velocity”\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. (N)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n−60\n−40\n−20\n0\n20\n40\n60\nNL\nNS\nZ\nPS\nPL\nFigure 11.6: Membership Functions of “Force”\nIntuitively, when the inverted pendulum has a large clockwise angle and\na large clockwise angular velocity, we should give it a large force to the right.\nThus we have an if-then rule,\nIf the angle is negative large\nand the angular velocity is negative large,\nthen the force is positive large.\nSimilarly, when the inverted pendulum has a large counterclockwise angle\nand a large counterclockwise angular velocity, we should give it a large force\n\n\n274\nChapter 11 - Uncertain Inference Control\nto the left. Thus we have an if-then rule,\nIf the angle is positive large\nand the angular velocity is positive large,\nthen the force is negative large.\nNote that each input or output has 5 states and each state is represented by\nan uncertain set. This implies that the rule-base contains 5 × 5 if-then rules.\nIn order to balance the inverted pendulum, the 25 if-then rules in Table 11.1\nare accepted.\nTable 11.1: Rule Base with 5 × 5 If-Then Rules\nXXXXXXXXX\nX\nangle\nvelocity\nNL\nNS\nZ\nPS\nPL\nNL\nPL\nPL\nPL\nPS\nZ\nNS\nPL\nPL\nPS\nZ\nNS\nZ\nPL\nPS\nZ\nNS\nNL\nPS\nPS\nZ\nNS\nNL\nNL\nPL\nZ\nNS\nNL\nNL\nNL\nA lot of simulation results show that the uncertain inference controller\nbased on the 25 if-then rules in Table 11.1 may balance the inverted pendulum\nsuccessfully.\n11.4\nBibliographic Notes\nThe basic uncertain inference rule was initialized by Liu [118] in 2010 by the\ntool of uncertain set theory. After that, Gao-Gao-Ralescu [52] extended the\nuncertain inference rule to the case with multiple antecedents and multiple if-\nthen rules. Based on the uncertain inference rules, Liu [118] presented the tool\nof uncertain inference controller. As an important contribution, Peng-Chen\n[188] proved that uncertain inference controllers are universal approximators\nand then demonstrated that the uncertain inference controller is a reasonable\ntool. As a successful application, Gao [57] balanced an inverted pendulum\nby using the uncertain inference controller.\n\n\nChapter 12\nUncertain Process\nThe study of uncertain process was started by Liu [114] in 2008 for modelling\nthe evolution of uncertain phenomena. This chapter will provide the concept\nof uncertain process, and introduce sample path, uncertainty distribution,\nextreme value, ﬁrst hitting time, time integral, and stationary independent\nincrement process.\n12.1\nUncertain Process\nAn uncertain process is essentially a sequence of uncertain variables indexed\nby time. A formal deﬁnition is given below.\nDeﬁnition 12.1 (Liu [114]) Let (Γ, L, M) be an uncertainty space and let T\nbe a totally ordered set (e.g. time). An uncertain process is a function Xt(γ)\nfrom T × (Γ, L, M) to the set of real numbers such that {Xt ∈B} is an event\nfor any Borel set B of real numbers at each time t.\nRemark 12.1: The above deﬁnition says Xt is an uncertain process if and\nonly if it is an uncertain variable at each time t.\nExample 12.1: Take an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = 0.6, M{γ2} = 0.4. Then\nXt(γ) =\n(\nt,\nif γ = γ1\nt + 1,\nif γ = γ2\n(12.1)\nis an uncertain process.\nExample 12.2: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Then\nXt(γ) = t −γ,\n∀γ ∈Γ\n(12.2)\n\n\n276\nChapter 12 - Uncertain Process\nis an uncertain process.\nExample 12.3: A real-valued function f(t) with respect to time t may be\nregarded as a special uncertain process on an uncertainty space (Γ, L, M),\ni.e.,\nXt(γ) = f(t),\n∀γ ∈Γ.\n(12.3)\nDeﬁnition 12.2 (Liu [114]) Let Xt be an uncertain process. Then for each\nﬁxed γ ∈Γ, the function Xt(γ) is called a sample path of Xt.\nNote that each sample path is a real-valued function of time t. Thus an\nuncertain process may also be regarded as a function from an uncertainty\nspace to a collection of sample paths.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nℜ\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 12.1: A Sample Path of Uncertain Process\nDeﬁnition 12.3 An uncertain process Xt is said to be sample-continuous if\nalmost all sample paths are continuous functions with respect to time t.\nSimilarly, an uncertain process Xt is called sample-Lipschitz-continuous\nif almost all sample paths are Lipschitz continuous functions with respect to\ntime t.\nDeﬁnition 12.4 (Liu [129]) Uncertain processes X1t, X2t, · · · , Xnt are said\nto be independent if for any positive integer k and any times t1, t2, · · · , tk,\nthe uncertain vectors\nξi = (Xit1, Xit2, · · · , Xitk),\ni = 1, 2, · · · , n\n(12.4)\nare independent, i.e., for any Borel sets B1, B2, · · · , Bn of k-dimensional real\nvectors, we have\nM\n( n\n\\\ni=1\n(ξi ∈Bi)\n)\n=\nn\n^\ni=1\nM{ξi ∈Bi}.\n(12.5)\n\n\nSection 12.2 - Uncertainty Distribution\n277\nExercise 12.1: Let X1t, X2t, · · · , Xnt be independent uncertain processes,\nand let t1, t2, · · · , tn be any times. Show that\nX1t1, X2t2, · · · , Xntn\n(12.6)\nare independent uncertain variables.\nExercise 12.2: Let Xt and Yt be independent uncertain processes. For any\ntimes t1, t2, · · · , tk and s1, s2, · · · , sm, show that\n(Xt1, Xt2, · · · , Xtk) and (Ys1, Ys2, · · · , Ysm)\n(12.7)\nare independent uncertain vectors.\nTheorem 12.1 (Liu [129]) Uncertain processes X1t, X2t, · · · , Xnt are inde-\npendent if and only if for any positive integer k, any times t1, t2, · · · , tk, and\nany Borel sets B1, B2, · · · , Bn of k-dimensional real vectors, we have\nM\n( n\n[\ni=1\n(ξi ∈Bi)\n)\n=\nn\n_\ni=1\nM{ξi ∈Bi}\n(12.8)\nwhere ξi = (Xit1, Xit2, · · · , Xitk) for i = 1, 2, · · · , n.\nProof: It follows from Theorem 3.43 that ξ1, ξ2, · · · , ξn are independent\nuncertain vectors if and only if (12.8) holds. The theorem is thus veriﬁed.\nUncertain Field\nUncertain ﬁeld is a generalization of uncertain process when the index set T\nbecomes a partially ordered set (e.g. time × space).\nDeﬁnition 12.5 (Liu [129]) Let (Γ, L, M) be an uncertainty space and let T\nbe a partially ordered set (e.g. time × space). An uncertain ﬁeld is a function\nXt(γ) from T × (Γ, L, M) to the set of real numbers such that {Xt ∈B} is\nan event for any Borel set B at each time t.\nExample 12.4:\nLet Xt and Ys be uncertain processes indexed by time\nvariable t and spatial variable s, respectively. Then\nZ(t, s) = Xt + Ys\n(12.9)\nis an uncertain ﬁeld.\n\n\n278\nChapter 12 - Uncertain Process\n12.2\nUncertainty Distribution\nAn uncertainty distribution of uncertain process is a sequence of uncertainty\ndistributions of uncertain variables indexed by time. Thus an uncertainty\ndistribution of uncertain process is a surface rather than a curve. A formal\ndeﬁnition is given below.\nDeﬁnition 12.6 (Liu [129]) The uncertainty distribution Φt(x) of an un-\ncertain process Xt is deﬁned by\nΦt(x) = M {Xt ≤x}\n(12.10)\nfor any time t and any number x.\nThat is, the uncertain process Xt has an uncertainty distribution Φt(x)\nif at each time t, the uncertain variable Xt has the uncertainty distribution\nΦt(x).\nIn other words, Φt(x) is an uncertainty distribution of uncertain\nprocess if and only if at each time t, Φt(x) is an uncertainty distribution of\nuncertain variable.\nExample 12.5: The linear uncertain process Xt ∼L(at, bt) has an uncer-\ntainty distribution,\nΦt(x) =\n\n\n\n\n\n\n\n\n\n0,\nif x ≤at\nx −at\n(b −a)t,\nif at < x ≤bt\n1,\nif x > bt.\n(12.11)\nExample 12.6:\nThe zigzag uncertain process Xt ∼Z(at, bt, ct) has an\nuncertainty distribution,\nΦt(x) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n0,\nif x ≤at\nx −at\n2(b −a)t,\nif at < x ≤bt\nx + ct −2bt\n2(c −b)t\n,\nif bt < x ≤ct\n1,\nif x > ct.\n(12.12)\nExample 12.7: The normal uncertain process Xt ∼N(et, σt) has an un-\ncertainty distribution,\nΦt(x) =\n\u0012\n1 + exp\n\u0012π(et −x)\n√\n3σt\n\u0013\u0013−1\n.\n(12.13)\n\n\nSection 12.2 - Uncertainty Distribution\n279\nExercise 12.3:\nTake an uncertainty space (Γ, L, M) to be {γ1, γ2} with\npower set and M{γ1} = 0.6, M{γ2} = 0.4. Derive the uncertainty distribu-\ntion of the uncertain process\nXt(γ) =\n(\nt,\nif γ = γ1\nt + 1,\nif γ = γ2.\n(12.14)\nExercise 12.4: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Derive the uncertainty distribution of the\nuncertain process\nXt(γ) = t −γ,\n∀γ ∈Γ.\n(12.15)\nExercise 12.5: A real-valued function f(t) with respect to time t is a special\nuncertain process. What is the uncertainty distribution of f(t)?\nExercise 12.6: Let Xt be an uncertain process with uncertainty distribution\nΦt(x), and let a and b be real numbers with a > 0. Show that aXt + b has\nan uncertainty distribution,\nΨt(x) = Φt((x −b)/a).\n(12.16)\nExercise 12.7: Let Xt be an uncertain process with continuous uncertainty\ndistribution Φt(x), and let a and b be real numbers with a < 0. Show that\naXt + b has an uncertainty distribution,\nΨt(x) = 1 −Φt((x −b)/a).\n(12.17)\nExercise 12.8: Let Xt be an uncertain process with uncertainty distribution\nΦt(x), and let f(x) be a continuous and strictly increasing function. Show\nthat f(Xt) has an uncertainty distribution\nΨt(x) = Φt(f −1(x)).\n(12.18)\nExercise 12.9: Let Xt be an uncertain process with continuous uncertainty\ndistribution Φt(x), and let f(x) be a continuous and strictly decreasing func-\ntion. Show that f(Xt) has an uncertainty distribution\nΨt(x) = 1 −Φt(f −1(x)).\n(12.19)\nRegular Uncertainty Distribution\nDeﬁnition 12.7 (Liu [129]) An uncertainty distribution Φt(x) is said to be\nregular if at each time t, it is a continuous and strictly increasing function\nwith respect to x at which 0 < Φt(x) < 1, and\nlim\nx→−∞Φt(x) = 0,\nlim\nx→+∞Φt(x) = 1.\n(12.20)\n\n\n280\nChapter 12 - Uncertain Process\nInverse Uncertainty Distribution\nDeﬁnition 12.8 (Liu [129]) Let Xt be an uncertain process with regular\nuncertainty distribution Φt(x). Then the inverse function Φ−1\nt (α) is called\nthe inverse uncertainty distribution of Xt.\nThat is, the uncertain process Xt has an inverse uncertainty distribution\nΦ−1\nt (α) if at each time t, the uncertain variable Xt has the the inverse uncer-\ntainty distribution Φ−1\nt (α). In other words, Φ−1\nt (α) is an inverse uncertainty\ndistribution of uncertain process if and only if at each time t, Φ−1\nt (α) is an\ninverse uncertainty distribution of uncertain variable.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nΦ−1\nt (α)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.α = 0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.8\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.9\nFigure 12.2: Inverse Uncertainty Distribution of Uncertain Process\nExample 12.8: The linear uncertain process Xt ∼L(at, bt) has an inverse\nuncertainty distribution,\nΦ−1\nt (α) = (1 −α)at + αbt.\n(12.21)\nExample 12.9:\nThe zigzag uncertain process Xt ∼Z(at, bt, ct) has an\ninverse uncertainty distribution,\nΦ−1\nt (α) =\n(\n(1 −2α)at + 2αbt,\nif α < 0.5\n(2 −2α)bt + (2α −1)ct,\nif α ≥0.5.\n(12.22)\nExample 12.10:\nThe normal uncertain process Xt ∼N(et, σt) has an\ninverse uncertainty distribution,\nΦ−1\nt (α) = et +\n√\n3σt\nπ\nln\nα\n1 −α.\n(12.23)\n\n\nSection 12.3 - Independent Increment Process\n281\nExercise 12.10: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure. Derive the inverse uncertainty distribution\nof the uncertain process\nXt(γ) = t −γ,\n∀γ ∈Γ.\n(12.24)\nExercise 12.11: Let Xt be an uncertain process with regular uncertainty\ndistribution Φt(x), and let a and b be real numbers. Show that (i) if a > 0,\nthen aXt + b has an inverse uncertainty distribution,\nΨ−1\nt (α) = aΦ−1\nt (α) + b;\n(12.25)\nand (ii) if a < 0, then aXt + b has an inverse uncertainty distribution,\nΨ−1\nt (α) = aΦ−1\nt (1 −α) + b.\n(12.26)\nExercise 12.12: Let Xt be an uncertain process with regular uncertainty\ndistribution Φt(x), and let f(x) be a continuous and strictly increasing func-\ntion. Show that f(Xt) has an inverse uncertainty distribution\nΨ−1\nt (α) = f(Φ−1\nt (α)).\n(12.27)\nExercise 12.13: Let Xt be an uncertain process with regular uncertainty\ndistribution Φt(x), and let f(x) be a continuous and strictly decreasing func-\ntion. Show that f(Xt) has an inverse uncertainty distribution\nΨ−1\nt (α) = f(Φ−1\nt (1 −α)).\n(12.28)\n12.3\nIndependent Increment Process\nAn independent increment process is an uncertain process that has indepen-\ndent increments. A formal deﬁnition is given below.\nDeﬁnition 12.9 (Liu [114]) An uncertain process Xt is said to have inde-\npendent increments if\nXt1, Xt2 −Xt1, Xt3 −Xt2, · · · , Xtk −Xtk−1\n(12.29)\nare independent uncertain variables where t1, t2, · · · , tk are any times with\nt1 < t2 < · · · < tk.\nThat is, an independent increment process means that its increments are\nindependent uncertain variables whenever the time intervals do not overlap.\nPlease note that the increments are also independent of the initial state.\n\n\n282\nChapter 12 - Uncertain Process\nTheorem 12.2 (Liu [129]) Let Xt be an independent increment process with\nregular uncertainty distribution Φt(x). Then for any times s and t with s < t,\nthe increment Xt −Xs has an inverse uncertainty distribution\nΨ−1(α) = Φ−1\nt (α) −Φ−1\ns (α).\n(12.30)\nProof: Since Xt is an independent increment process, Xs and Xt −Xs are\nindependent uncertain variables. It follows from\nXt = Xs + (Xt −Xs)\nthat\nΦ−1\nt (α) = Φ−1\ns (α) + Ψ−1(α).\nThe theorem is thus proved.\nRemark 12.2: It follows from (12.30) that Φ−1\nt (α) −Φ−1\ns (α) is a monotone\nincreasing function with respect to α for any times s and t with s < t. Thus\nfor any α < β, we immediately have\nΦ−1\nt (β) −Φ−1\ns (β) ≥Φ−1\nt (α) −Φ−1\ns (α).\nThat is,\nΦ−1\nt (β) −Φ−1\nt (α) ≥Φ−1\ns (β) −Φ−1\ns (α).\nTherefore, the uncertainty distribution of independent increment process has\na horn-like shape. See Figure 12.3.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nΦ−1\nt (α)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.8\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.9\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 12.3: Inverse Uncertainty Distribution of Independent Increment Pro-\ncess: A Horn-like Family of Functions of t indexed by α\n\n\nSection 12.4 - Extreme Value Theorem\n283\nTheorem 12.3 Let Xt be a sample-continuous independent increment pro-\ncess with regular uncertainty distribution Φt(x). Then for any α ∈(0, 1), we\nhave\nM{Xt ≤Φ−1\nt (α), ∀t} = α,\n(12.31)\nM{Xt > Φ−1\nt (α), ∀t} = 1 −α.\n(12.32)\nProof: It is still a conjecture.\nRemark 12.3: It is also showed that for any α ∈(0, 1), the following two\nequations are true,\nM{Xt < Φ−1\nt (α), ∀t} = α,\n(12.33)\nM{Xt ≥Φ−1\nt (α), ∀t} = 1 −α.\n(12.34)\nPlease mention that {Xt < Φ−1\nt (α), ∀t} and {Xt ≥Φ−1\nt (α), ∀t} are disjoint\nevents but not opposite. Although it is always true that\nM{Xt < Φ−1\nt (α), ∀t} + M{Xt ≥Φ−1\nt (α), ∀t} ≡1,\n(12.35)\nthe union of {Xt < Φ−1\nt (α), ∀t} and {Xt ≥Φ−1\nt (α), ∀t} does not make the\nuniversal set, and it is possible that\nM{(Xt < Φ−1\nt (α), ∀t) ∪(Xt ≥Φ−1\nt (α), ∀t)} < 1.\n(12.36)\n12.4\nExtreme Value Theorem\nThis section will present a series of extreme value theorems for sample-\ncontinuous independent increment processes.\nTheorem 12.4 (Liu [125], Extreme Value Theorem) Let Xt be a sample-\ncontinuous independent increment process with regular uncertainty distribu-\ntion Φt(x). Then the supremum\nsup\n0≤t≤s\nXt\n(12.37)\nhas an uncertainty distribution\nΨ(x) =\ninf\n0≤t≤s Φt(x);\n(12.38)\nand the inﬁmum\ninf\n0≤t≤s Xt\n(12.39)\nhas an uncertainty distribution\nΨ(x) = sup\n0≤t≤s\nΦt(x).\n(12.40)\n\n\n284\nChapter 12 - Uncertain Process\nProof: Let 0 = t1 < t2 < · · · < tn = s be a partition of the closed interval\n[0, s]. It is clear that\nXti = Xt1 + (Xt2 −Xt1) + · · · + (Xti −Xti−1)\nfor i = 1, 2, · · · , n. Since the increments\nXt1, Xt2 −Xt1, · · · , Xtn −Xtn−1\nare independent uncertain variables, it follows from Theorem 3.15 that the\nmaximum\nmax\n1≤i≤n Xti\nhas an uncertainty distribution\nmin\n1≤i≤n Φti(x).\nSince Xt is sample-continuous, we have\nmax\n1≤i≤n Xti →sup\n0≤t≤s\nXt\nand\nmin\n1≤i≤n Φti(x) →\ninf\n0≤t≤s Φt(x)\nas n →∞. Thus (12.38) is proved. Similarly, it follows from Theorem 3.15\nthat the minimum\nmin\n1≤i≤n Xti\nhas an uncertainty distribution\nmax\n1≤i≤n Φti(x).\nSince Xt is sample-continuous, we have\nmin\n1≤i≤n Xti →\ninf\n0≤t≤s Xt\nand\nmax\n1≤i≤n Φti(x) →sup\n0≤t≤s\nΦt(x)\nas n →∞. Thus (12.40) is veriﬁed.\nExample 12.11: The sample-continuity condition in Theorem 12.4 cannot\nbe removed. For example, take an uncertainty space (Γ, L, M) to be (0, 1)\nwith Borel algebra and Lebesgue measure. Deﬁne a sample-discontinuous\nuncertain process\nXt(γ) =\n(\n0,\nif γ ̸= t\n1,\nif γ = t.\n(12.41)\n\n\nSection 12.4 - Extreme Value Theorem\n285\nSince all increments are 0 almost surely, Xt is an independent increment\nprocess. On the one hand, Xt has an uncertainty distribution\nΦt(x) =\n(\n0,\nif x < 0\n1,\nif x ≥0.\n(12.42)\nOn the other hand, the supremum\nsup\n0≤t≤1\nXt(γ) ≡1\n(12.43)\nhas an uncertainty distribution\nΨ(x) =\n(\n0,\nif x < 1\n1,\nif x ≥1.\n(12.44)\nThus\nΨ(x) ̸=\ninf\n0≤t≤1 Φt(x).\n(12.45)\nTherefore, the sample-continuity condition cannot be removed.\nExercise 12.14:\nLet Xt be a sample-continuous independent increment\nprocess with regular uncertainty distribution Φt(x). Assume f is a continuous\nand strictly increasing function. Show that the supremum\nsup\n0≤t≤s\nf(Xt)\n(12.46)\nhas an uncertainty distribution\nΨ(x) =\ninf\n0≤t≤s Φt(f −1(x));\n(12.47)\nand the inﬁmum\ninf\n0≤t≤s f(Xt)\n(12.48)\nhas an uncertainty distribution\nΨ(x) = sup\n0≤t≤s\nΦt(f −1(x)).\n(12.49)\nExercise 12.15:\nLet Xt be a sample-continuous independent increment\nprocess with regular uncertainty distribution Φt(x). Assume f is a continuous\nand strictly decreasing function. Show that the supremum\nsup\n0≤t≤s\nf(Xt)\n(12.50)\nhas an uncertainty distribution\nΨ(x) = 1 −sup\n0≤t≤s\nΦt(f −1(x));\n(12.51)\n\n\n286\nChapter 12 - Uncertain Process\nand the inﬁmum\ninf\n0≤t≤s f(Xt)\n(12.52)\nhas an uncertainty distribution\nΨ(x) = 1 −inf\n0≤t≤s Φt(f −1(x)).\n(12.53)\n12.5\nFirst Hitting Time\nDeﬁnition 12.10 (Liu [125]) Let Xt be an uncertain process and let z be a\ngiven level. Then the uncertain variable\nτz =\n(\ninf\n\b\nt ≥0\n\f\n\f Xt ≥z\n\t\n,\nif z > X0\ninf\n\b\nt ≥0\n\f\n\f Xt ≤z\n\t\n,\nif z < X0\n(12.54)\nis called the ﬁrst hitting time that Xt reaches the level z.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nXt\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nz\nτz\n..........................................................................................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 12.4: First Hitting Time\nTheorem 12.5 (Liu [125]) Let Xt be a sample-continuous independent in-\ncrement process with regular uncertainty distribution Φt(x). Then the ﬁrst\nhitting time τz that Xt reaches the level z has an uncertainty distribution,\nΥ(s) =\n\n\n\n\n\n1 −inf\n0≤t≤s Φt(z),\nif z > X0\nsup\n0≤t≤s\nΦt(z),\nif z < X0.\n(12.55)\nProof: When z > X0, it follows from the deﬁnition of ﬁrst hitting time that\nτz ≤s if and only if\nsup\n0≤t≤s\nXt ≥z.\n\n\nSection 12.6 - Time Integral\n287\nThus the uncertainty distribution of τz is\nΥ(s) = M{τz ≤s} = M\n\u001a\nsup\n0≤t≤s\nXt ≥z\n\u001b\n.\nBy using the extreme value theorem, we obtain\nΥ(s) = 1 −inf\n0≤t≤s Φt(z).\nWhen z < X0, it follows from the deﬁnition of ﬁrst hitting time that\nτz ≤s if and only if\ninf\n0≤t≤s Xt ≤z.\nThus the uncertainty distribution of τz is\nΥ(s) = M{τz ≤s} = M\n\u001a\ninf\n0≤t≤s Xt ≤z\n\u001b\n= sup\n0≤t≤s\nΦt(z).\nThe theorem is veriﬁed.\nExercise 12.16:\nLet Xt be a sample-continuous independent increment\nprocess with regular uncertainty distribution Φt(x). Assume f is a continuous\nand strictly increasing function. Show that the ﬁrst hitting time τz that f(Xt)\nreaches the level z has an uncertainty distribution,\nΥ(s) =\n\n\n\n\n\n1 −inf\n0≤t≤s Φt(f −1(z)),\nif z > f(X0)\nsup\n0≤t≤s\nΦt(f −1(z)),\nif z < f(X0).\n(12.56)\nExercise 12.17:\nLet Xt be a sample-continuous independent increment\nprocess with regular uncertainty distribution Φt(x). Assume f is a continuous\nand strictly decreasing function.\nShow that the ﬁrst hitting time τz that\nf(Xt) reaches the level z has an uncertainty distribution,\nΥ(s) =\n\n\n\n\n\nsup\n0≤t≤s\nΦt(f −1(z)),\nif z > f(X0)\n1 −inf\n0≤t≤s Φt(f −1(z)),\nif z < f(X0).\n(12.57)\nExercise 12.18: Show that the sample-continuity condition in Theorem 12.5\ncannot be removed.\n12.6\nTime Integral\nThis section will give a deﬁnition of time integral that is an integral of un-\ncertain process with respect to time.\n\n\n288\nChapter 12 - Uncertain Process\nDeﬁnition 12.11 (Liu [114]) Let Xt be an uncertain process. For any par-\ntition of closed interval [a, b] with a = t1 < t2 < · · · < tk+1 = b, the mesh is\nwritten as\n∆= max\n1≤i≤k |ti+1 −ti|.\n(12.58)\nThen the time integral of Xt with respect to t is\nZ b\na\nXtdt = lim\n∆→0\nk\nX\ni=1\nXti · (ti+1 −ti)\n(12.59)\nprovided that the limit exists almost surely and is ﬁnite. In this case, the\nuncertain process Xt is said to be time integrable.\nSince Xt is an uncertain variable at each time t, the limit in (12.59) is\nalso an uncertain variable provided that the limit exists almost surely and\nis ﬁnite. Hence an uncertain process Xt is time integrable if and only if the\nlimit in (12.59) is an uncertain variable.\nTheorem 12.6 If Xt is a sample-continuous uncertain process on [a, b], then\nit is time integrable on [a, b].\nProof: Let a = t1 < t2 < · · · < tk+1 = b be a partition of the closed interval\n[a, b]. Since the uncertain process Xt is sample-continuous, almost all sample\npaths are continuous functions with respect to t. Hence the limit\nlim\n∆→0\nk\nX\ni=1\nXti(ti+1 −ti)\nexists almost surely and is ﬁnite. On the other hand, since Xt is an uncertain\nvariable at each time t, the above limit is also a measurable function. Hence\nthe limit is an uncertain variable and then Xt is time integrable.\nTheorem 12.7 If Xt is a time integrable uncertain process on [a, b], then it\nis time integrable on each subinterval of [a, b]. Moreover, if c ∈[a, b], then\nZ b\na\nXtdt =\nZ c\na\nXtdt +\nZ b\nc\nXtdt.\n(12.60)\nProof: Let [a′, b′] be a subinterval of [a, b]. Since Xt is a time integrable\nuncertain process on [a, b], for any partition\na = t1 < · · · < tm = a′ < tm+1 < · · · < tn = b′ < tn+1 < · · · < tk+1 = b,\nthe limit\nlim\n∆→0\nk\nX\ni=1\nXti(ti+1 −ti)\n\n\nSection 12.6 - Time Integral\n289\nexists almost surely and is ﬁnite. Thus the limit\nlim\n∆→0\nn−1\nX\ni=m\nXti(ti+1 −ti)\nexists almost surely and is ﬁnite. Hence Xt is time integrable on the subin-\nterval [a′, b′]. Next, for the partition\na = t1 < · · · < tm = c < tm+1 < · · · < tk+1 = b,\nwe have\nk\nX\ni=1\nXti(ti+1 −ti) =\nm−1\nX\ni=1\nXti(ti+1 −ti) +\nk\nX\ni=m\nXti(ti+1 −ti).\nNote that\nZ b\na\nXtdt = lim\n∆→0\nk\nX\ni=1\nXti(ti+1 −ti),\nZ c\na\nXtdt = lim\n∆→0\nm−1\nX\ni=1\nXti(ti+1 −ti),\nZ b\nc\nXtdt = lim\n∆→0\nk\nX\ni=m\nXti(ti+1 −ti).\nHence the equation (12.60) is proved.\nTheorem 12.8 (Linearity of Time Integral) Let Xt and Yt be time integrable\nuncertain processes on [a, b], and let α and β be real numbers. Then\nZ b\na\n(αXt + βYt)dt = α\nZ b\na\nXtdt + β\nZ b\na\nYtdt.\n(12.61)\nProof: Let a = t1 < t2 < · · · < tk+1 = b be a partition of the closed interval\n[a, b]. It follows from the deﬁnition of time integral that\nZ b\na\n(αXt + βYt)dt = lim\n∆→0\nk\nX\ni=1\n(αXti + βYti)(ti+1 −ti)\n= lim\n∆→0 α\nk\nX\ni=1\nXti(ti+1 −ti) + lim\n∆→0 β\nk\nX\ni=1\nYti(ti+1 −ti)\n= α\nZ b\na\nXtdt + β\nZ b\na\nYtdt.\nHence the equation (12.61) is proved.\n\n\n290\nChapter 12 - Uncertain Process\nTheorem 12.9 (Yao [275]) Let Xt be a sample-continuous independent in-\ncrement process with regular uncertainty distribution Φt(x). Then the time\nintegral\nYs =\nZ s\n0\nXtdt\n(12.62)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nΦ−1\nt (α)dt.\n(12.63)\nProof: For any given time s > 0, it follows from the basic property of time\nintegral that\n\u001aZ s\n0\nXtdt ≤\nZ s\n0\nΦ−1\nt (α)dt\n\u001b\n⊃{Xt ≤Φ−1\nt (α), ∀t}.\nBy using Theorem 12.3, we obtain\nM\n\u001aZ s\n0\nXtdt ≤\nZ s\n0\nΦ−1\nt (α)dt\n\u001b\n≥M{Xt ≤Φ−1\nt (α), ∀t} = α.\nSimilarly, since\n\u001aZ s\n0\nXtdt >\nZ s\n0\nΦ−1\nt (α)dt\n\u001b\n⊃{Xt > Φ−1\nt (α), ∀t},\nwe have\nM\n\u001aZ s\n0\nXtdt >\nZ s\n0\nΦ−1\nt (α)dt\n\u001b\n≥M{Xt > Φ−1\nt (α), ∀t} = 1 −α.\nIt follows from the above two inequalities and the duality axiom that\nM\n\u001aZ s\n0\nXtdt ≤\nZ s\n0\nΦ−1\nt (α)dt\n\u001b\n= α.\nThus the time integral Ys has the inverse uncertainty distribution Ψ−1\ns (α).\nExercise 12.19: Let Xt be a sample-continuous independent increment pro-\ncess with regular uncertainty distribution Φt(x), and let J(x) be a continuous\nand strictly increasing function. Show that the time integral\nYs =\nZ s\n0\nJ(Xt)dt\n(12.64)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nJ(Φ−1\nt (α))dt.\n(12.65)\n\n\nSection 12.7 - Stationary Independent Increment Process\n291\nExercise 12.20: Let Xt be a sample-continuous independent increment pro-\ncess with regular uncertainty distribution Φt(x), and let J(x) be a continuous\nand strictly decreasing function. Show that the time integral\nYs =\nZ s\n0\nJ(Xt)dt\n(12.66)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nJ(Φ−1\nt (1 −α))dt.\n(12.67)\n12.7\nStationary Independent Increment Process\nAn uncertain process Xt is said to have stationary increments if its increments\nare identically distributed uncertain variables whenever the time intervals\nhave the same length, i.e., for any given t > 0, the increments Xs+t −Xs are\nidentically distributed uncertain variables for all s > 0.\nDeﬁnition 12.12 (Liu [114]) An uncertain process is said to be a stationary\nindependent increment process if it has not only stationary increments but\nalso independent increments.\nIt is clear that a stationary independent increment process is a special\nindependent increment process.\nTheorem 12.10 Let Xt be a stationary independent increment process. Then\nfor any real numbers a and b, the uncertain process\nYt = aXt + b\n(12.68)\nis also a stationary independent increment process.\nProof: Since Xt is an independent increment process, the uncertain variables\nXt1, Xt2 −Xt1, Xt3 −Xt2, · · · , Xtk −Xtk−1\nare independent. It follows from Yt = aXt + b and Theorem 3.10 that\nYt1, Yt2 −Yt1, Yt3 −Yt2, · · · , Ytk −Ytk−1\nare also independent. That is, Yt is an independent increment process. On\nthe other hand, since Xt is a stationary increment process, the increments\nXs+t −Xs are identically distributed uncertain variables for all s > 0. Thus\nYs+t −Ys = a(Xs+t −Xs)\n\n\n292\nChapter 12 - Uncertain Process\nare also identically distributed uncertain variables for all s > 0, and Yt is a\nstationary increment process. Hence Yt is a stationary independent increment\nprocess.\nRemark 12.4: Generally speaking, a nonlinear function of stationary inde-\npendent increment process is not necessarily a stationary independent incre-\nment process. A typical example is the square of a stationary independent\nincrement process.\nTheorem 12.11 (Chen [14]) Suppose Xt is a stationary independent in-\ncrement process. Then Xt and (1 −t)X0 + tX1 are identically distributed\nuncertain variables for any time t ≥0.\nProof: We ﬁrst prove the theorem when t is a rational number. Assume t =\nq/p where p and q are irreducible integers. Let Φ be the common uncertainty\ndistribution of increments\nX1/p −X0/p, X2/p −X1/p, X3/p −X2/p, · · ·\nThen\nXt −X0 = (X1/p −X0/p) + (X2/p −X1/p) + · · · + (Xq/p −X(q−1)/p)\nhas an uncertainty distribution\nΨ(x) = Φ(x/q).\n(12.69)\nIn addition,\nt(X1 −X0) = t((X1/p −X0/p) + (X2/p −X1/p) + · · · + (Xp/p −X(p−1)/p))\nhas an uncertainty distribution\nΥ(x) = Φ(x/p/t) = Φ(x/p/(q/p)) = Φ(x/q).\n(12.70)\nIt follows from (12.69) and (12.70) that Xt−X0 and t(X1−X0) are identically\ndistributed, and so are Xt and (1 −t)X0 + tX1.\nTheorem 12.12 (Liu [129]) Let Xt be a stationary independent increment\nprocess whose initial value and increments have regular uncertainty distribu-\ntions. Then there exist two continuous and strictly increasing functions µ\nand ν such that Xt has an inverse uncertainty distribution\nΦ−1\nt (α) = µ(α) + ν(α)t.\n(12.71)\nProof: Note that X0 and X1−X0 are independent uncertain variables whose\ninverse uncertainty distributions exist and are denoted by µ(α) and ν(α), re-\nspectively. It is clear that µ(α) and ν(α) are continuous and strictly increas-\ning functions. Theorem 12.11 says that Xt and X0+(X1−X0)t are identically\n\n\nSection 12.7 - Stationary Independent Increment Process\n293\ndistributed uncertain variables. Therefore, by using the operational law, Xt\nhas the inverse uncertainty distribution Φ−1\nt (α) = µ(α)+ν(α)t. The theorem\nis veriﬁed.\nRemark 12.5: The inverse uncertainty distribution of stationary indepen-\ndent increment process is a family of linear functions of t indexed by α. See\nFigure 12.5.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nΦ−1\nt (α)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.8\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.9\nFigure 12.5: Inverse Uncertainty Distribution of Stationary Independent In-\ncrement Process\nTheorem 12.13 (Uniform Velocity) Let Xt be a stationary independent in-\ncrement process. Then\n∆Xt\n∆t\n(12.72)\nare identically distributed uncertain variables for any t and ∆t.\nProof: Without loss of generality, assume Xt has an inverse uncertainty\ndistribution\nΦ−1\nt (α) = µ(α) + ν(α)t.\n(12.73)\nIt follows from Theorem 12.2 that\n∆Xt = Xt+∆t −Xt\nhas an inverse uncertainty distribution\nΨ−1(α) = Φ−1\nt+∆t(α) −Φ−1\nt (α) = ν(α)∆t.\nThus\n∆Xt\n∆t\n∼ν(α).\n(12.74)\nIn other words, the “speed” ∆Xt/∆t are identically distributed uncertain\nvariables for any t and ∆t.\n\n\n294\nChapter 12 - Uncertain Process\nTheorem 12.14 (Liu [129]) Let µ and ν be continuous and strictly increas-\ning functions on (0, 1). Then there exists a stationary independent increment\nprocess that is sample-Lipschitz-continuous and has an inverse uncertainty\ndistribution\nΦ−1\nt (α) = µ(α) + ν(α)t.\n(12.75)\nProof: Without loss of generality, we only consider the time range of [0, 1].\nLet Q be the set of rational numbers in [0, 1].\nFor each rational number\nq ∈Q, take an uncertainty space (Γq, Lq, Mq) to be the interval (0, 1) with\nBorel algebra and Lebesgue measure, and denote the product uncertainty\nspace by (Γ, L, M). For the number 0, we deﬁne an uncertain variable\nξ0(γ) = µ(γ)\n(12.76)\non the uncertainty space (Γ0, L0, M0). Since µ(γ) is a continuous and strictly\nincreasing function with respect to γ, the inverse uncertainty distribution of\nξ0 is µ(α). For each positive rational number q ∈Q, we deﬁne an uncertain\nvariable\nξq(γ) = ν(γ)\n(12.77)\non the uncertainty space (Γq, Lq, Mq). Since ν(γ) is a continuous and strictly\nincreasing function with respect to γ, the inverse uncertainty distribution of\nξq is ν(α). For each positive integer n, we deﬁne an uncertain process\nXn\nt =\n\n\n\n\n\n\n\nξ0 + 1\nn\nk\nX\ni=1\nξ i\nn ,\nif t = k\nn\n(k = 1, 2, · · · , n)\nlinear,\notherwise.\nSince ξ’s are deﬁned on diﬀerent uncertainty spaces, they are independent\nuncertain variables. By using the operational law, Xn\nt has an inverse uncer-\ntainty distribution\nΦ−1\nt (α) = µ(α) + 1\nn\nk\nX\ni=1\nν(α) = µ(α) + ν(α)t\n(12.78)\nat\nt = k\nn,\nk = 1, 2, · · · , n.\n(12.79)\nConsider the event on the product uncertainty space (Γ, L, M) deﬁned by\n(2.34) on Page 19, i.e.,\nΛ =\n∞\n[\ni=1\nY\nq∈Q\n\u0014\n1\ni + 1,\ni\ni + 1\n\u0015\n.\n(12.80)\nNote that M{Λ} = 1, and for each γ = (γq|q ∈Q) ∈Λ, there exists a small\nnumber δ > 0 such that\nδ ≤γq ≤1 −δ,\nq ∈Q.\n(12.81)\n\n\nSection 12.7 - Stationary Independent Increment Process\n295\nThus\nµ(δ) ≤ξ0(γ0) ≤µ(1 −δ),\n(12.82)\nν(δ) ≤ξq(γq) ≤ν(1 −δ),\nq ∈Q, q ̸= 0.\n(12.83)\nFor any times s and t, we have\n|Xs(γ) −Xt(γ)| ≤L|s −t|\n(12.84)\nwhere\nL = |ν(δ)| ∨|ν(1 −δ)|.\n(12.85)\nThus Xt(γ) is Lipschitz-continuous with respect to t for each γ ∈Λ. Since\n(ξq|q ∈[0, s]) and (ξq|q ∈(s, t]) are deﬁned on diﬀerent uncertainty spaces,\nXs and Xt −Xs are independent. More generally, for any times t1, t2, · · · , tr\nwith t1 < t2 < · · · < tr, we may verify that\nXt1, Xt2 −Xt1, Xt3 −Xt2, · · · , Xtr −Xtr−1\n(12.86)\nare independent. Thus Xt has independent increments. Finally, for any given\nt > 0, it follows from Theorem 12.2 that Xs+t−Xs has an inverse uncertainty\ndistribution\nΥ−1\ns (α) = µ(α) + ν(α)(t + s) −(µ(α) + ν(α)t) = ν(α)s\n(12.87)\nfor all s > 0. Thus Xt has stationary increments. Since Xn\nt converges in\ndistribution as n →∞, the limit Xt is a stationary independent increment\nprocess that is sample-Lipschitz-continuous and has the inverse uncertainty\ndistribution Φ−1(α). The theorem is proved.\nTheorem 12.15 (Liu [120]) Let Xt be a stationary independent increment\nprocess. Then there exist two real numbers a and b such that\nE[Xt] = a + bt\n(12.88)\nfor any time t ≥0.\nProof: It follows from Theorem 12.11 that Xt and X0 + (X1 −X0)t are\nidentically distributed uncertain variables. Thus we have\nE[Xt] = E[X0 + (X1 −X0)t].\nSince X0 and X1 −X0 are independent uncertain variables, we obtain\nE[Xt] = E[X0] + E[X1 −X0]t.\nHence (12.88) holds for a = E[X0] and b = E[X1 −X0].\nTheorem 12.16 (Liu [120]) Let Xt be a stationary independent increment\nprocess with an initial value 0. Then for any times s and t, we have\nE[Xs+t] = E[Xs] + E[Xt].\n(12.89)\n\n\n296\nChapter 12 - Uncertain Process\nProof: It follows from Theorem 12.15 that there exists a real number b such\nthat E[Xt] = bt for any time t ≥0. Hence\nE[Xs+t] = b(s + t) = bs + bt = E[Xs] + E[Xt].\nTheorem 12.17 (Chen [14]) Let Xt be a stationary independent increment\nprocess with a crisp initial value X0. Then there exists a real number b such\nthat\nV [Xt] = bt2\n(12.90)\nfor any time t ≥0.\nProof:\nIt follows from Theorem 12.11 that Xt and (1 −t)X0 + tX1 are\nidentically distributed uncertain variables. Since X0 is a constant, we have\nV [Xt] = V [(1 −t)X0 + tX1] = t2V [X1].\nHence (12.90) holds for b = V [X1].\nTheorem 12.18 (Chen [14]) Let Xt be a stationary independent increment\nprocess with a crisp initial value X0. Then for any times s and t, we have\np\nV [Xs+t] =\np\nV [Xs] +\np\nV [Xt].\n(12.91)\nProof: It follows from Theorem 12.17 that there exists a real number b such\nthat V [Xt] = bt2 for any time t ≥0. Hence\np\nV [Xs+t] =\n√\nb(s + t) =\n√\nbs +\n√\nbt =\np\nV [Xs] +\np\nV [Xt].\n12.8\nBibliographic Notes\nThe study of uncertain process was started by Liu [114] in 2008 for modelling\nthe evolution of uncertain phenomena. In order to describe uncertain pro-\ncess, Liu [129] proposed the concepts of uncertainty distribution and inverse\nuncertainty distribution.\nFor independent increment process, Liu [125] presented an extreme value\ntheorem and obtained the uncertainty distribution of ﬁrst hitting time, and\nYao [275] provided a formula for calculating the inverse uncertainty distri-\nbution of time integral. For stationary independent increment process, Liu\n[120] showed that the expected value is a linear function of time, and Chen\n[14] veriﬁed that the variance is proportional to the square of time.\n\n\nChapter 13\nUncertain Renewal\nProcess\nUncertain renewal process is an uncertain process in which events occur con-\ntinuously and independently of one another in uncertain times. This chapter\nwill introduce uncertain renewal process, and provide uncertain insurance\nmodel, uncertain production model, and uncertain queueing model.\n13.1\nUncertain Renewal Process\nDeﬁnition 13.1 (Liu [114]) Let ξ1, ξ2, · · · be iid uncertain interarrival times.\nDeﬁne S0 = 0 and Sn = ξ1 + ξ2 + · · · + ξn for n ≥1. Then the uncertain\nprocess\nNt = max\nn≥0 {n | Sn ≤t}\n(13.1)\nis called an uncertain renewal process.\nIt is clear that Sn is a stationary independent increment process with re-\nspect to n. Since ξ1, ξ2, · · · denote the interarrival times of successive events,\nSn can be regarded as the waiting time until the occurrence of the nth event.\nIn this case, the uncertain renewal process Nt is the number of renewals in\n(0, t]. Note that Nt is not sample-continuous, but each sample path of Nt\nis a right-continuous and increasing step function taking only nonnegative\ninteger values. Furthermore, since the interarrival times are always assumed\nto be positive uncertain variables, the size of each jump of Nt is always 1.\nIn other words, Nt has at most one renewal at each time. In particular, Nt\ndoes not jump at time 0.\nTheorem 13.1 (Fundamental Relationship) Let Nt be an uncertain renewal\nprocess with positive uncertain interarrival times ξ1, ξ2, · · · , and Sn = ξ1 +\n\n\n298\nChapter 13 - Uncertain Renewal Process\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n4 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n3 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n2 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n1 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nξ1\nξ2\nξ3\nξ4\nS0\nS1\nS2\nS3\nS4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nNt\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 13.1: A Sample Path of Renewal Process\nξ2 + · · · + ξn. Then we have\nNt ≥n ⇔Sn ≤t\n(13.2)\nfor any time t and integer n. Furthermore, we also have\nNt ≤n ⇔Sn+1 > t.\n(13.3)\nProof: Since Nt is the largest n such that Sn ≤t, we have SNt ≤t < SNt+1.\nIf Nt ≥n, then Sn ≤SNt ≤t. Conversely, if Sn ≤t, then Sn < SNt+1\nthat implies Nt ≥n.\nThus (13.2) is veriﬁed.\nSimilarly, if Nt ≤n, then\nNt + 1 ≤n + 1 and Sn+1 ≥SNt+1 > t.\nConversely, if Sn+1 > t, then\nSn+1 > SNt that implies Nt ≤n. Thus (13.3) is veriﬁed.\nExercise 13.1: Let Nt be an uncertain renewal process with positive uncer-\ntain interarrival times ξ1, ξ2, · · · , and Sn = ξ1 + ξ2 + · · · + ξn. Show that\nM{Nt ≥n} = M{Sn ≤t},\n(13.4)\nM{Nt ≤n} = 1 −M{Sn+1 ≤t}.\n(13.5)\nTheorem 13.2 (Liu [120]) Let Nt be an uncertain renewal process with iid\npositive uncertain interarrival times ξ1, ξ2, · · · If Φ is the common regular un-\ncertainty distribution of those interarrival times, then Nt has an uncertainty\ndistribution\nΥt(x) = 1 −Φ\n\u0012\nt\n⌊x⌋+ 1\n\u0013\n,\n∀x ≥0\n(13.6)\nwhere ⌊x⌋represents the maximal integer less than or equal to x.\nProof: Note that Sn+1 has an uncertainty distribution Φ(x/(n + 1)). It\nfollows from (13.5) that\nM{Nt ≤n} = 1 −M{Sn+1 ≤t} = 1 −Φ\n\u0012\nt\nn + 1\n\u0013\n.\n\n\nSection 13.1 - Uncertain Renewal Process\n299\nSince Nt takes integer values, for any x ≥0, we have\nΥt(x) = M{Nt ≤x} = M{Nt ≤⌊x⌋} = 1 −Φ\n\u0012\nt\n⌊x⌋+ 1\n\u0013\n.\nThe theorem is veriﬁed.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n•\n•\n•\n•\n•\n•\nΥt(0)\nΥt(1)\nΥt(2)\nΥt(3)\nΥt(4)\nΥt(5)\n1\n2\n3\n4\n5\n0\nx\nΥt(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 13.2: Uncertainty Distribution Υt(x) of Renewal Process Nt\nTheorem 13.3 (Liu [120], Elementary Renewal Theorem) Let Nt be an un-\ncertain renewal process with iid positive uncertain interarrival times ξ1, ξ2, · · ·\nThen the average renewal number\nNt\nt →1\nξ1\n(13.7)\nin the sense of convergence in distribution as t →∞.\nProof: Assume Φ is the uncertainty distribution of interarrival times. For\nsimplicity, we only prove the case that Φ is regular. At ﬁrst, the uncertainty\ndistribution Υt of Nt is given by Theorem 13.2 as follows,\nΥt(x) = 1 −Φ\n\u0012\nt\n⌊x⌋+ 1\n\u0013\n.\nIt follows from the operational law that the uncertainty distribution of Nt/t\nis\nΨt(x) = 1 −Φ\n\u0012\nt\n⌊tx⌋+ 1\n\u0013\nwhere ⌊tx⌋represents the maximal integer less than or equal to tx. It is clear\nthat at each point x, we have\nlim\nt→∞Ψt(x) = 1 −Φ\n\u0012 1\nx\n\u0013\n\n\n300\nChapter 13 - Uncertain Renewal Process\nwhich is just the uncertainty distribution of 1/ξ1. Hence Nt/t converges in\ndistribution to 1/ξ1 as t →∞.\nTheorem 13.4 (Liu [120], Elementary Renewal Theorem) Let Nt be an un-\ncertain renewal process with iid positive uncertain interarrival times ξ1, ξ2, · · ·\nThen\nlim\nt→∞\nE[Nt]\nt\n= E\n\u0014 1\nξ1\n\u0015\n.\n(13.8)\nProof: Write the uncertainty distributions of Nt/t and 1/ξ1 by Ψt(x) and\nG(x), respectively. Then\nΨt(x) = 1 −Φ\n\u0012\nt\n⌊tx⌋+ 1\n\u0013\n,\nG(x) = 1 −Φ\n\u0012 1\nx\n\u0013\n.\nSince Ψt(x) ≥G(x) and Ψt(x) →G(x) as t →∞at each point x, it fol-\nlows from the Lebesgue dominated convergence theorem and the existence of\nE[1/ξ1] that\nlim\nt→∞\nE[Nt]\nt\n= lim\nt→∞\nZ +∞\n0\n(1 −Ψt(x))dx =\nZ +∞\n0\n(1 −G(x))dx = E\n\u0014 1\nξ1\n\u0015\n.\nThe theorem is proved.\nExercise 13.2: An uncertain renewal process Nt is called linear if ξ1, ξ2, · · ·\nare iid linear uncertain variables L(a, b) with a > 0. Show that\nlim\nt→∞\nE[Nt]\nt\n= ln b −ln a\nb −a\n.\n(13.9)\nExercise 13.3: An uncertain renewal process Nt is called zigzag if ξ1, ξ2, · · ·\nare iid zigzag uncertain variables Z(a, b, c) with a > 0. Show that\nlim\nt→∞\nE[Nt]\nt\n= 1\n2\n\u0012ln b −ln a\nb −a\n+ ln c −ln b\nc −b\n\u0013\n.\n(13.10)\n13.2\nUncertain Renewal Reward Process\nLet (ξ1, η1), (ξ2, η2), · · · be a sequence of pairs of uncertain variables. We\nshall interpret ηi as the rewards (or costs) associated with the i-th interarrival\ntimes ξi for i = 1, 2, · · · , respectively. An uncertain renewal reward process\nis deﬁned as the total reward earned by time t.\n\n\nSection 13.2 - Uncertain Renewal Reward Process\n301\nDeﬁnition 13.2 (Liu [120]) Let ξ1, ξ2, · · · be iid uncertain interarrival times,\nand let η1, η2, · · · be iid uncertain rewards. Then\nRt =\nNt\nX\ni=1\nηi\n(13.11)\nis called an uncertain renewal reward process, where Nt is the uncertain re-\nnewal process with uncertain interarrival times ξ1, ξ2, · · ·\nTheorem 13.5 (Liu [120]) Let Rt be an uncertain renewal reward process\nwith iid positive uncertain interarrival times ξ1, ξ2, · · · and iid positive uncer-\ntain rewards η1, η2, · · · Assume (ξ1, ξ2, · · · ) and (η1, η2, · · · ) are independent\nuncertain vectors, and those interarrival times and rewards have regular un-\ncertainty distributions Φ and Ψ, respectively. Then Rt has an uncertainty\ndistribution\nΥt(x) = max\nk≥0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0010x\nk\n\u0011\n.\n(13.12)\nHere we set Ψ(x/0) = 1 for any x ≥0.\nProof: At ﬁrst, for any index k, nonnegative time t and nonnegative number\nx, we deﬁne a number\nαk =\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0010x\nk\n\u0011\n.\n(13.13)\nThe argument breaks down into three cases. Case 1: Assume 0 < αk < 1\nfor any index k. If\nαk = 1 −Φ\n\u0012\nt\nk + 1\n\u0013\n,\nαk ≤Ψ\n\u0010x\nk\n\u0011\n,\nthen\n(k + 1)Φ−1(1 −αk) = t,\nkΨ−1(αk) ≤x.\n(13.14)\nIf\nαk < 1 −Φ\n\u0012\nt\nk + 1\n\u0013\n,\nαk = Ψ\n\u0010x\nk\n\u0011\n,\nthen k ̸= 0 and\n(k + 1)Φ−1(1 −αk) > t,\nkΨ−1(αk) = x.\n(13.15)\nTherefore, one of the alternatives (13.14) and (13.15) holds. Let us turn our\nattention to the uncertainty distribution of uncertain renewal reward process\nRt. It is easy to verify that Rt ≤x if and only if, for some nonnegative\ninteger k, the (k + 1)st reward arrives after the time t and the sum of the\nﬁrst k rewards is less than or equal to x. That is,\n{Rt ≤x} =\n∞\n[\nk=0\n k+1\nX\ni=1\nξi > t\n!\n∩\n k\nX\ni=1\nηi ≤x\n!\n.\n\n\n302\nChapter 13 - Uncertain Renewal Process\nThus the uncertain renewal reward process Rt has an uncertainty distribution\nΥt(x) = M{Rt ≤x} = M\n( ∞\n[\nk=0\n k+1\nX\ni=1\nξi > t\n!\n∩\n k\nX\ni=1\nηi ≤x\n!)\n.\nOn the one hand, by using (13.14) and (13.15), we obtain\nΥt(x) = M\n( ∞\n[\nk=0\n k+1\nX\ni=1\nξi > t\n!\n∩\n k\nX\ni=1\nηi ≤x\n!)\n≤M\n( ∞\n[\nk=0\n k+1\n[\ni=1\n(ξi > Φ−1(1 −αk))\n!\n∪\n k\n[\ni=1\n(ηi ≤Ψ−1(αk))\n!)\n= M\n( ∞\n[\ni=1\n \n∞\n[\nk=i−1\n(ξi > Φ−1(1 −αk))\n!\n∪\n ∞\n[\nk=i\n(ηi ≤Ψ−1(αk))\n!)\n≤M\n( ∞\n[\ni=1\n \nξi >\n∞\n^\nk=i−1\nΦ−1(1 −αk)\n!\n∪\n \nηi ≤\n∞\n_\nk=i\nΨ−1(αk)\n!)\n=\n∞\n_\ni=1\nM\n(\nξi >\n∞\n^\nk=i−1\nΦ−1(1 −αk)\n)\n∨M\n(\nηi ≤\n∞\n_\nk=i\nΨ−1(αk)\n)\n=\n∞\n_\ni=1\n \n∞\n_\nk=i−1\nαk\n!\n∨\n ∞\n_\nk=i\nαk\n!\n=\n∞\n_\nk=0\nαk\n=\n∞\n_\nk=0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0010x\nk\n\u0011\n.\nOn the other hand, we obtain\nΥt(x) = M\n( ∞\n[\nk=0\n k+1\nX\ni=1\nξi > t\n!\n∩\n k\nX\ni=1\nηi ≤x\n!)\n≥\n∞\n_\nk=0\nM\n( k+1\nX\ni=1\nξi > t\n!\n∩\n k\nX\ni=1\nηi ≤x\n!)\n=\n∞\n_\nk=0\nM\n(k+1\nX\ni=1\nξi > t\n)\n∧M\n( k\nX\ni=1\nηi ≤x\n)\n=\n∞\n_\nk=0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0010x\nk\n\u0011\n.\nIt follows that\nΥt(x) =\n∞\n_\nk=0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0010x\nk\n\u0011\n.\n\n\nSection 13.2 - Uncertain Renewal Reward Process\n303\nCase 2: Assume there exists at least one index k such that αk = 1. The\nproof is left as an exercise for the reader. Case 3: Assume αk < 1 for any\nindex k and there exists at least one index k such that αk = 0. The proof is\nalso left as an exercise for the reader.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\nx\nΥt(x)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n.\n.\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n.\n.\n.\n.\n..\n.\n.\n..\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\n.\n..\n.\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13.3: Uncertainty Distribution Υt(x) of Uncertain Renewal Reward\nProcess Rt in which the dashed horizontal lines are 1 −Φ(t/(k + 1)) and the\ndashed curves are Ψ(x/k) for k = 0, 1, 2, · · ·\nTheorem 13.6 (Liu [120], Renewal Reward Theorem) Let Rt be an un-\ncertain renewal reward process with iid positive uncertain interarrival times\nξ1, ξ2, · · · and iid positive uncertain rewards η1, η2, · · · Assume (ξ1, ξ2, · · · )\nand (η1, η2, · · · ) are independent uncertain vectors. Then the reward rate\nRt\nt →η1\nξ1\n(13.16)\nin the sense of convergence in distribution as t →∞.\nProof: Assume those interarrival times and rewards have uncertainty dis-\ntributions Φ and Ψ, respectively. For simplicity, we only prove the case that\nΦ and Ψ are regular.\nIt follows from Theorem 13.5 that the uncertainty\ndistribution of Rt is\nΥt(x) = max\nk≥0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0010x\nk\n\u0011\n.\nThen Rt/t has an uncertainty distribution\nΨt(x) = max\nk≥0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0012tx\nk\n\u0013\n.\n\n\n304\nChapter 13 - Uncertain Renewal Process\nWhen t →∞, we have\nΨt(x) →sup\ny≥0\n(1 −Φ(y)) ∧Ψ(xy)\nwhich is just the uncertainty distribution of η1/ξ1. Hence Rt/t converges in\ndistribution to η1/ξ1 as t →∞.\nTheorem 13.7 (Liu [120], Renewal Reward Theorem) Let Rt be an un-\ncertain renewal reward process with iid positive uncertain interarrival times\nξ1, ξ2, · · · and iid positive uncertain rewards η1, η2, · · · Assume (ξ1, ξ2, · · · )\nand (η1, η2, · · · ) are independent uncertain vectors. Then\nlim\nt→∞\nE[Rt]\nt\n= E\n\u0014η1\nξ1\n\u0015\n.\n(13.17)\nProof: It follows from Theorem 13.5 that Rt/t has an uncertainty distribu-\ntion\nFt(x) = max\nk≥0\n\u0012\n1 −Φ\n\u0012\nt\nk + 1\n\u0013\u0013\n∧Ψ\n\u0012tx\nk\n\u0013\nand η1/ξ1 has an uncertainty distribution\nG(x) = sup\ny≥0\n(1 −Φ(y)) ∧Ψ(xy).\nNote that Ft(x) →G(x) and Ft(x) ≥G(x). It follows from Lebesgue domi-\nnated convergence theorem and the existence of E[η1/ξ1] that\nlim\nt→∞\nE[Rt]\nt\n= lim\nt→∞\nZ +∞\n0\n(1 −Ft(x))dx =\nZ +∞\n0\n(1 −G(x))dx = E\n\u0014η1\nξ1\n\u0015\n.\nThe theorem is proved.\n13.3\nUncertain Insurance Model\nLiu [125] assumed that a is the initial capital of an insurance company, b is\nthe premium rate, bt is the total income up to time t, and the uncertain claim\nprocess is an uncertain renewal reward process\nRt =\nNt\nX\ni=1\nηi\n(13.18)\nwith iid uncertain interarrival times ξ1, ξ2, · · · and iid uncertain claim amounts\nη1, η2, · · · Then the capital of the insurance company at time t is\nZt = a + bt −Rt\n(13.19)\nand Zt is called an insurance risk process.\n\n\nSection 13.3 - Uncertain Insurance Model\n305\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. t\nZt\n0\na\nS1\nS2\nS3\nS4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 13.4: An Insurance Risk Process\nRuin Index\nDeﬁnition 13.3 (Liu [125]) Let Zt be an insurance risk process. Then the\nruin index is deﬁned as the uncertain measure that the capital Zt eventually\nbecomes negative, i.e.,\nRuin = M\n\u001a\ninf\nt≥0 Zt < 0\n\u001b\n.\n(13.20)\nTheorem 13.8 (Liu [125], Ruin Index Theorem) Let Zt = a + bt −Rt be\nan insurance risk process where a and b are positive numbers, and Rt is\nan uncertain renewal reward process with iid positive uncertain interarrival\ntimes ξ1, ξ2, · · · and iid positive uncertain claim amounts η1, η2, · · · Assume\n(ξ1, ξ2, · · · ) and (η1, η2, · · · ) are independent uncertain vectors, and those in-\nterarrival times and claim amounts have regular uncertainty distributions Φ\nand Ψ, respectively. Then the ruin index is\nRuin = max\nk≥1 sup\nx≥0\nΦ\n\u0010x\nk\n\u0011\n∧\n\u0012\n1 −Ψ\n\u0012a + bx\nk\n\u0013\u0013\n.\n(13.21)\nProof: At ﬁrst, we deﬁne an uncertain process indexed by index k as follows,\nYk = a + b\nk\nX\ni=1\nξi −\nk\nX\ni=1\nηi.\nIt is easy to verify that Yk is an independent increment process, and has an\nuncertainty distribution\nFk(z) = sup\nx≥0\nΦ\n\u0010x\nk\n\u0011\n∧\n\u0012\n1 −Ψ\n\u0012a + bx −z\nk\n\u0013\u0013\n.\n\n\n306\nChapter 13 - Uncertain Renewal Process\nIt follows from the extreme value theorem that\nmin\nk≥1 Yk\nhas an uncertainty distribution\nG(z) = max\nk≥1 Fk(z).\nSince Yk is just the capital Zt at the arrival time of the kth claim for each k,\nand a ruin occurs only at the arrival times, we have\nRuin = M\n\u001a\ninf\nt≥0 Zt < 0\n\u001b\n= M\n\u001a\nmin\nk≥1 Yk < 0\n\u001b\n= G(0) = max\nk≥1 Fk(0).\nThe theorem is proved.\nRuin Time\nDeﬁnition 13.4 (Liu [125]) Let Zt be an insurance risk process. Then the\nruin time is deﬁned as the ﬁrst hitting time that the capital Zt becomes neg-\native, i.e.,\nτ = inf\n\b\nt ≥0\n\f\n\f Zt < 0\n\t\n.\n(13.22)\nTheorem 13.9 (Yao-Zhou [267]) Let Zt = a + bt −Rt be an insurance\nrisk process where a and b are positive numbers, and Rt is an uncertain re-\nnewal reward process with iid positive uncertain interarrival times ξ1, ξ2, · · ·\nand iid positive uncertain claim amounts η1, η2, · · · Assume (ξ1, ξ2, · · · ) and\n(η1, η2, · · · ) are independent uncertain vectors, and those interarrival times\nand claim amounts have regular uncertainty distributions Φ and Ψ, respec-\ntively. Then the ruin time has an uncertainty distribution\nΥ(t) = max\nk≥1 sup\nx≤t\nΦ\n\u0010x\nk\n\u0011\n∧\n\u0012\n1 −Ψ\n\u0012a + bx\nk\n\u0013\u0013\n.\n(13.23)\nProof: At ﬁrst, for any index k, nonnegative time t and nonnegative number\nx, we deﬁne a number,\nαk = sup\nx≤t\nΦ\n\u0010x\nk\n\u0011\n∧\n\u0012\n1 −Ψ\n\u0012a + bx\nk\n\u0013\u0013\n.\n(13.24)\nThe argument breaks down into three cases. Case 1: Assume 0 < αk < 1\nfor any index k. Keep in mind that Φ(x/k) is an increasing function with\nrespect to x, and\n1 −Ψ\n\u0012a + bx\nk\n\u0013\n\n\nSection 13.3 - Uncertain Insurance Model\n307\nis a decreasing function with respect to x. This fact implies that\nΦ\n\u0010x\nk\n\u0011\n∧\n\u0012\n1 −Ψ\n\u0012a + bx\nk\n\u0013\u0013\nis increasing on the left side of the intersection of the two functions, and\ndecreasing on the right side. Let x∗be the supremum solution of (13.24). If\nx∗= t (i.e., t is on the left side of the intersection), then\nαk = Φ\n\u0012 t\nk\n\u0013\n,\nαk ≤1 −Ψ\n\u0012a + bt\nk\n\u0013\n.\nThat is,\nΦ−1(αk) = t\nk ,\na + bt\nk\n≤Ψ−1(1 −αk).\nThus,\nkΦ−1(αk) = t,\na + bkΦ−1(αk) −kΨ−1(1 −αk) ≤0.\n(13.25)\nIf x∗< t (i.e., t is on the right side of the intersection), then\nαk = Φ\n\u0012x∗\nk\n\u0013\n= 1 −Ψ\n\u0012a + bx∗\nk\n\u0013\n.\nThat is,\nΦ−1(αk) = x∗\nk ,\na + bx∗\nk\n= Ψ−1(1 −αk).\nThus,\nkΦ−1(αk) < t,\na + bkΦ−1(αk) −kΨ−1(1 −αk) = 0.\n(13.26)\nTherefore, one of the alternatives (13.25) and (13.26) holds. Let us turn our\nattention to the uncertainty distribution of ruin time. For any given t, the\nruin time τ ≤t if and only if, for some positive integer k, the kth claim\narrives before the time t and a ruin occurs when the kth claim arrives. That\nis,\n{τ ≤t} =\n∞\n[\nk=1\n( k\nX\ni=1\nξi ≤t, a + b\nk\nX\ni=1\nξi −\nk\nX\ni=1\nηi < 0\n)\n.\nThus the ruin time τ has an uncertainty distribution\nΥ(t) = M{τ ≤t} = M\n( ∞\n[\nk=1\n k\nX\ni=1\nξi ≤t, a + b\nk\nX\ni=1\nξi −\nk\nX\ni=1\nηi < 0\n!)\n.\n\n\n308\nChapter 13 - Uncertain Renewal Process\nOn the one hand, by using (13.25) and (13.26), we obtain\nΥ(t) = M\n( ∞\n[\nk=1\n k\nX\ni=1\nξi ≤t, a + b\nk\nX\ni=1\nξi −\nk\nX\ni=1\nηi < 0\n!)\n≥\n∞\n_\nk=1\nM\n( k\nX\ni=1\nξi ≤t, a + b\nk\nX\ni=1\nξi −\nk\nX\ni=1\nηi < 0\n)\n≥\n∞\n_\nk=1\nM\n( k\n\\\ni=1\n(ξi ≤Φ−1(αk)) ∩(ηi > Ψ−1(1 −αk))\n)\n=\n∞\n_\nk=1\nk\n^\ni=1\nM\n\b\nξi ≤Φ−1(αk)\n\t\n∧M\n\b\nηi > Ψ−1(1 −αk)\n\t\n=\n∞\n_\nk=1\nk\n^\ni=1\nαk ∧αk =\n∞\n_\nk=1\nαk.\nOn the other hand, we obtain\nΥ(t) = M\n( ∞\n[\nk=1\n k\nX\ni=1\nξi ≤t, a + b\nk\nX\ni=1\nξi −\nk\nX\ni=1\nηi < 0\n!)\n≤M\n( ∞\n[\nk=1\nk\n[\ni=1\n(ξi ≤Φ−1(αk)) ∪(ηi > Ψ−1(1 −αk))\n)\n= M\n( ∞\n[\ni=1\n∞\n[\nk=i\n(ξi ≤Φ−1(αk)) ∪(ηi > Ψ−1(1 −αk))\n)\n≤M\n( ∞\n[\ni=1\n \nξi ≤\n∞\n_\nk=i\nΦ−1(αk)\n!\n∪\n \nηi >\n∞\n^\nk=i\nΨ−1(1 −αk)\n!)\n=\n∞\n_\ni=1\nM\n(\nξi ≤\n∞\n_\nk=i\nΦ−1(αk)\n)\n∨M\n(\nηi >\n∞\n^\nk=i\nΨ−1(1 −αk)\n)\n=\n∞\n_\ni=1\n ∞\n_\nk=i\nαk\n!\n∨\n ∞\n_\nk=i\nαk\n!\n=\n∞\n_\nk=1\nαk.\nIt follows that\nΥ(t) =\n∞\n_\nk=1\nαk.\nCase 2: Assume there exists at least one index k such that αk = 1. The\nproof is left as an exercise for the reader. Case 3: Assume αk < 1 for any\nindex k and there exists at least one index k such that αk = 0. The proof is\nalso left as an exercise for the reader.\n\n\nSection 13.4 - Uncertain Production Model\n309\n13.4\nUncertain Production Model\nLio-Liu [106] assumed that a is the initial inventory level of a factory, b is\nthe demand rate, bt is the total demand up to time t, and the uncertain\nproduction process is an uncertain renewal reward process\nRt =\nNt\nX\ni=1\nηi\n(13.27)\nwith iid uncertain production times ξ1, ξ2, · · · and iid uncertain production\namounts η1, η2, · · · Then the surplus inventory of the factory at time t is\nZt = a + Rt −bt\n(13.28)\nand Zt is called a production risk process.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. t\nZt\n0\na\nS1\nS2\nS3 S4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 13.5: A Production Risk Process\nShortage Index\nDeﬁnition 13.5 (Lio-Liu [106]) Let Zt be a production risk process. Then\nthe shortage index is deﬁned as the uncertain measure that the surplus inven-\ntory Zt eventually becomes negative, i.e.,\nShortage = M\n\u001a\ninf\nt≥0 Zt < 0\n\u001b\n.\n(13.29)\nTheorem 13.10 (Lio-Liu [106], Shortage Index Theorem) Let Zt = a +\nRt −bt be a production risk process where a and b are positive numbers,\nand Rt is an uncertain renewal reward process with iid uncertain produc-\ntion times ξ1, ξ2, · · · and iid uncertain production amounts η1, η2, · · · Assume\n(ξ1, ξ2, · · · ) and (η1, η2, · · · ) are independent uncertain vectors, and those pro-\nduction times and production amounts have regular uncertainty distributions\n\n\n310\nChapter 13 - Uncertain Renewal Process\nΦ and Ψ, respectively. Then the shortage index is\nShortage = max\nk≥1 sup\nx≥0\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a\nk −1\n\u0013\n.\n(13.30)\nHere we set\nΨ\n\u0012bx −a\n0\n\u0013\n=\n(\n0,\nif bx < a\n1,\nif bx ≥a.\n(13.31)\nProof: At ﬁrst, we deﬁne an uncertain process indexed by index k as follows,\nYk = a +\nk−1\nX\ni=1\nηi −b\nk\nX\ni=1\nξi.\nIt is easy to verify that Yk is an independent increment process, and has an\nuncertainty distribution\nFk(z) = sup\nx≥0\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a + z\nk −1\n\u0013\n.\nIt follows from the extreme value theorem that\nmin\nk≥1 Yk\nhas an uncertainty distribution\nG(z) = max\nk≥1 Fk(z).\nSince the surplus inventory Zt eventually becomes negative if and only if\nYk < 0 for some k, we have\nShortage = M\n\u001a\ninf\nt≥0 Zt < 0\n\u001b\n= M\n\u001a\nmin\nk≥1 Yk < 0\n\u001b\n= G(0) = max\nk≥1 Fk(0).\nThe theorem is proved.\nShortage Time\nDeﬁnition 13.6 (Lio-Liu [106]) Let Zt be a production risk process. Then\nthe shortage time is deﬁned as the ﬁrst hitting time that the surplus inventory\nZt becomes negative, i.e.,\nτ = inf\n\b\nt ≥0\n\f\n\f Zt < 0\n\t\n.\n(13.32)\nTheorem 13.11 (Lio-Liu [106]) Let Zt = a + Rt −bt be a production risk\nprocess where a and b are positive numbers, and Rt is an uncertain renewal\n\n\nSection 13.4 - Uncertain Production Model\n311\nreward process with iid uncertain production times ξ1, ξ2, · · · and iid uncer-\ntain production amounts η1, η2, · · · Assume (ξ1, ξ2, · · · ) and (η1, η2, · · · ) are\nindependent uncertain vectors, and those production times and production\namounts are positive and have regular uncertainty distributions Φ and Ψ,\nrespectively. Then the shortage time has an uncertainty distribution\nΥ(t) = max\nk≥1 sup\nx≤t\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a\nk −1\n\u0013\n.\n(13.33)\nHere we set\nΨ\n\u0012bx −a\n0\n\u0013\n=\n(\n0,\nif bx < a\n1,\nif bx ≥a.\n(13.34)\nProof: The argument breaks down into three cases. Case 1: When bt < a,\nwe always have Zs = a + Rs −bs > 0 for any s ≤t. That is, the surplus\ninventory Zs cannot become negative before time t. Thus the shortage time\nτ > t, and\nΥ(t) = M{τ ≤t} = 0.\nOn the other hand, for any x ≤t, we have bx ≤bt < a, and then\nΨ\n\u0012bx −a\nk −1\n\u0013\n= 0\nfor each k. Thus\nmax\nk≥1 sup\nx≤t\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a\nk −1\n\u0013\n= 0.\nTherefore, (13.33) is veriﬁed.\nCase 2: When bt = a, it follows from the deﬁnition of shortage time that\nτ ≤t if and only if the ﬁrst production is not ﬁnished before time t, i.e.,\nξ1 > t. Thus\nΥ(t) = M{τ ≤t} = M{ξ1 > t} = 1 −Φ(t).\nOn the other hand, for any x ≤t, we have bx ≤bt = a, and then\nΨ\n\u0012bx −a\nk −1\n\u0013\n= 0\nfor each k ≥2. Thus\nmax\nk≥1 sup\nx≤t\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a\nk −1\n\u0013\n= 1 −Φ(t)\nsince the supremum value is achieved at k = 1 and x = t. Therefore, (13.33)\nis also veriﬁed.\n\n\n312\nChapter 13 - Uncertain Renewal Process\nCase 3: When bt > a, for any index k, nonnegative time t and nonnega-\ntive number x, we deﬁne a number,\nαk = sup\nx≤t\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a\nk −1\n\u0013\n.\n(13.35)\nThe argument breaks down into three subcases. Subcase 1: Assume 0 <\nαk < 1 for any index k. Keep in mind that 1−Φ(x/k) is a decreasing function\nwith respect to x, and\nΨ\n\u0012bx −a\nk −1\n\u0013\nis an increasing function with respect to x. This fact implies that\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0012bx −a\nk −1\n\u0013\nis increasing on the left side of the intersection of the two functions, and\ndecreasing on the right side. Assume x∗is the supremum solution of (13.35).\nIf k = 1, then x∗= a/b < t, and\nα1 = 1 −Φ\n\u0010a\nb\n\u0011\n.\nThat is,\na\nb = Φ−1(1 −α1).\nThus\na+(k −1)Ψ−1(αk) < bt,\na+(k −1)Ψ−1(αk)−bkΦ−1(1−αk) = 0. (13.36)\nIf k ≥2 and x∗= t (i.e., t is on the left side of the intersection), then\nαk ≤1 −Φ\n\u0012 t\nk\n\u0013\n,\nαk = Ψ\n\u0012bt −a\nk −1\n\u0013\n.\nThat is,\nt\nk ≤Φ−1(1 −αk),\nΨ−1(αk) = bt −a\nk −1 .\nThus\na+(k −1)Ψ−1(αk) = bt,\na+(k −1)Ψ−1(αk)−bkΦ−1(1−αk) ≤0. (13.37)\nIf k ≥2 and x∗< t (i.e., t is on the right side of the intersection), then\nαk = 1 −Φ\n\u0012x∗\nk\n\u0013\n= Ψ\n\u0012bx∗−a\nk −1\n\u0013\n.\nThat is,\nx∗\nk = Φ−1(1 −αk),\nΨ−1(αk) = bx∗−a\nk −1 .\n\n\nSection 13.4 - Uncertain Production Model\n313\nThus (13.36) holds. Therefore, one of the alternatives (13.36) and (13.37)\nholds. Let us turn our attention to the uncertainty distribution of shortage\ntime. For any given t, the shortage time τ ≤t if and only if, for some positive\ninteger k, the initial inventory level plus the ﬁrst k −1 production amounts\nis less than the total demand up to the time t and a shortage occurs before\nthe completion of the kth production. That is,\n{τ ≤t} =\n∞\n[\nk=1\n(\na +\nk−1\nX\ni=1\nηi ≤bt, a +\nk−1\nX\ni=1\nηi −b\nk\nX\ni=1\nξi < 0\n)\n.\nThus the shortage time τ has an uncertainty distribution\nΥ(t) = M{τ ≤t} = M\n( ∞\n[\nk=1\n \na +\nk−1\nX\ni=1\nηi ≤bt, a +\nk−1\nX\ni=1\nηi −b\nk\nX\ni=1\nξi < 0\n!)\n.\nOn the one hand, by using (13.36) and (13.37), we obtain\nΥ(t) = M\n( ∞\n[\nk=1\n \na +\nk−1\nX\ni=1\nηi ≤bt, a +\nk−1\nX\ni=1\nηi −b\nk\nX\ni=1\nξi < 0\n!)\n≥\n∞\n_\nk=1\nM\n(\na +\nk−1\nX\ni=1\nηi ≤bt, a +\nk−1\nX\ni=1\nηi −b\nk\nX\ni=1\nξi < 0\n)\n≥\n∞\n_\nk=1\nM\n( k−1\n\\\ni=1\nηi ≤Ψ−1(αk)\n\u0001\n!\n∩\n k\n\\\ni=1\nξi > Φ−1(1 −αk)\n\u0001\n!)\n=\n∞\n_\nk=1\n k−1\n^\ni=1\nM\n\b\nηi ≤Ψ−1(αk)\n\t\n!\n∧\n k\n^\ni=1\nM\n\b\nξi > Φ−1(1 −αk)\n\t\n!\n=\n∞\n_\nk=1\n k−1\n^\ni=1\nαk\n!\n∧\n k\n^\ni=1\nαk\n!\n=\n∞\n_\nk=1\nαk.\n\n\n314\nChapter 13 - Uncertain Renewal Process\nOn the other hand, we obtain\nΥ(t) = M\n( ∞\n[\nk=1\n \na +\nk−1\nX\ni=1\nηi ≤bt, a +\nk−1\nX\ni=1\nηi −b\nk\nX\ni=1\nξi < 0\n!)\n≤M\n( ∞\n[\nk=1\n k−1\n[\ni=1\nηi ≤Ψ−1(αk)\n\u0001\n!\n∪\n k\n[\ni=1\nξi > Φ−1(1 −αk)\n\u0001\n!)\n= M\n( ∞\n[\ni=1\n \n∞\n[\nk=i+1\nηi ≤Ψ−1(αk)\n\u0001\n!\n∪\n ∞\n[\nk=i\nξi > Φ−1(1 −αk)\n\u0001\n!)\n≤M\n( ∞\n[\ni=1\n \nηi ≤\n∞\n_\nk=i+1\nΨ−1(αk)\n!\n∪\n \nξi >\n∞\n^\nk=i\nΦ−1(1 −αk)\n!)\n=\n∞\n_\ni=1\nM\n(\nηi ≤\n∞\n_\nk=i+1\nΨ−1(αk)\n)\n∨M\n(\nξi >\n∞\n^\nk=i\nΦ−1(1 −αk)\n)\n=\n∞\n_\ni=1\n \n∞\n_\nk=i+1\nαk\n!\n∨\n ∞\n_\nk=i\nαk\n!\n=\n∞\n_\nk=1\nαk.\nIt follows that\nΥ(t) =\n∞\n_\nk=1\nαk.\nSubcase 2: Assume there exists at least one index k such that αk = 1. The\nproof is left as an exercise for the reader. Subcase 3: Assume αk < 1 for\nany index k and there exists at least one index k such that αk = 0. The proof\nis also left as an exercise for the reader.\n13.5\nUncertain Queueing Model\nYao [274] assumed that in a queueing system the customers join the queue\n(waiting line) for service with iid uncertain interarrival times denoted by\nξ1, ξ2, · · · Write S0 = 0 and\nSk = ξ1 + ξ2 + · · · + ξk,\nk ≥1.\n(13.38)\nThen the kth customers arrive at the times Sk, k = 1, 2, · · · , respectively.\nAssume η1, η2, · · · denote the iid uncertain service times for the customers,\nand the customers leave the queue after getting served. See Figure 13.6.\nBusy Period\nA busy period is a time interval during which there are always some cus-\ntomers in the queue. For any busy period, since ξ1, ξ2, · · · are iid uncertain\ninterarrival times and η1, η2, · · · are iid uncertain service times, without loss\n\n\nSection 13.5 - Uncertain Queueing Model\n315\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. t\n0\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. ξ1 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. ξ2 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nξ3 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. ξ4 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nξ5 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nξ6 .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. η1.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. η2.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. η3.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. η4.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nη5.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 13.6: An Uncertain Queueing System\nof generality, we may assume that the busy period starts when the ﬁrst cus-\ntomer arrives. We also reset the starting time of busy period to 0 in this\nsubsection. Let t be a time after the current busy period and before the\nnext one. Then there are Nt + 1 customers (including the ﬁrst one) arriving\nbefore the time t, where Nt is a renewal process with uncertain interarrival\ntimes ξ2, ξ3, · · · Since the busy period ends before the time t, the total service\ntime for those Nt + 1 customers is less than t. Thus the busy period is the\nminimum value of t such that\nNt+1\nX\ni=1\nηi < t.\n(13.39)\nThat is, the busy period can be represented by\nτ = inf\n(\nt ≥0\n\f\n\f\nNt+1\nX\ni=1\nηi < t\n)\n.\n(13.40)\nTheorem 13.12 (Yao [274]) Consider an uncertain queueing system with\niid positive uncertain interarrival times ξ1, ξ2, · · · and iid positive uncertain\nservice times η1, η2, · · · Assume (ξ1, ξ2, · · · ) and (η1, η2, · · · ) are independent\nuncertain vectors, and those interarrival times and service times have regular\nuncertainty distributions Φ and Ψ, respectively. Then the busy period τ has\nan uncertainty distribution\nΥ(t) = max\nk≥1 sup\nx≤t\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0010x\nk\n\u0011\n.\n(13.41)\nProof: At ﬁrst, for any index k, nonnegative time t and nonnegative number\nx, we deﬁne a number,\nαk = sup\nx≤t\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0010x\nk\n\u0011\n.\n(13.42)\nThe argument breaks down into three cases. Case 1: Assume 0 < αk < 1\nfor any index k. Keep in mind that\n1 −Φ\n\u0010x\nk\n\u0011\n\n\n316\nChapter 13 - Uncertain Renewal Process\nis a decreasing function with respect to x, and Ψ (x/k) is an increasing func-\ntion with respect to x. This fact implies that\n\u0010\n1 −Φ\n\u0010x\nk\n\u0011\u0011\n∧Ψ\n\u0010x\nk\n\u0011\nis increasing on the left side of the intersection of the two functions, and\ndecreasing on the right side. Let x∗be the supremum solution of (13.42). If\nx∗= t (i.e., t is on the left side of the intersection), then\nαk ≤1 −Φ\n\u0012 t\nk\n\u0013\n,\nαk = Ψ\n\u0012 t\nk\n\u0013\n.\nThat is,\nt\nk ≤Φ−1(1 −αk),\nΨ−1(αk) = t\nk .\nThus,\nΨ−1(αk) ≤Φ−1(1 −αk),\nkΨ−1(αk) = t.\n(13.43)\nIf x∗< t (i.e., t is on the right side of the intersection), then\nαk = 1 −Φ\n\u0012x∗\nk\n\u0013\n= Ψ\n\u0012x∗\nk\n\u0013\n.\nThat is,\nx∗\nk = Φ−1(1 −αk),\nΨ−1(αk) = x∗\nk .\nThus,\nΨ−1(αk) = Φ−1(1 −αk),\nkΨ−1(αk) < t.\n(13.44)\nTherefore, one of the alternatives (13.43) and (13.44) holds. Let us turn our\nattention to the uncertainty distribution of busy period τ. For any given\ntime t, the busy period τ ≤t if and only if, for some positive integer k, the\nkth customer departs before the arrival of the (k + 1)st customer and before\nthe time t. That is,\n{τ ≤t} =\n∞\n[\nk=1\n( k\nX\ni=1\nηi <\nk+1\nX\ni=2\nξi,\nk\nX\ni=1\nηi ≤t\n)\n.\nThus the uncertainty distribution of busy period τ can be calculated as\nΥ(t) = M{τ ≤t} = M\n( ∞\n[\nk=1\n k\nX\ni=1\nηi <\nk+1\nX\ni=2\nξi,\nk\nX\ni=1\nηi ≤t\n!)\n.\n\n\nSection 13.5 - Uncertain Queueing Model\n317\nOn the one hand, by using (13.43) and (13.44), we obtain\nΥ(t) = M\n( ∞\n[\nk=1\n k\nX\ni=1\nηi <\nk+1\nX\ni=2\nξi,\nk\nX\ni=1\nηi ≤t\n!)\n≥\n∞\n_\nk=1\nM\n( k\nX\ni=1\nηi <\nk+1\nX\ni=2\nξi,\nk\nX\ni=1\nηi ≤t\n)\n≥\n∞\n_\nk=1\nM\n( k\n\\\ni=1\n(ξi+1 > Φ−1(1 −αk)) ∩(ηi ≤Ψ−1(αk))\n)\n=\n∞\n_\nk=1\nk\n^\ni=1\nM\n\b\nξi+1 > Φ−1(1 −αk)\n\t\n∧M\n\b\nηi ≤Ψ−1(αk)\n\t\n=\n∞\n_\nk=1\nk\n^\ni=1\nαk ∧αk =\n∞\n_\nk=1\nαk.\nOn the other hand, we obtain\nΥ(t) = M\n( ∞\n[\nk=1\n k\nX\ni=1\nηi <\nk+1\nX\ni=2\nξi,\nk\nX\ni=1\nηi ≤t\n!)\n≤M\n( ∞\n[\nk=1\nk\n[\ni=1\n(ξi+1 > Φ−1(1 −αk)) ∪(ηi ≤Ψ−1(αk))\n)\n= M\n( ∞\n[\ni=1\n∞\n[\nk=i\n(ξi+1 > Φ−1(1 −αk)) ∪(ηi ≤Ψ−1(αk))\n)\n≤M\n( ∞\n[\ni=1\n \nξi+1 >\n∞\n^\nk=i\nΦ−1(1 −αk)\n!\n∪\n \nηi ≤\n∞\n_\nk=i\nΨ−1(αk)\n!)\n=\n∞\n_\ni=1\nM\n(\nξi+1 >\n∞\n^\nk=i\nΦ−1(1 −αk)\n)\n∨M\n(\nηi ≤\n∞\n_\nk=i\nΨ−1(αk)\n)\n=\n∞\n_\ni=1\n ∞\n_\nk=i\nαk\n!\n∨\n ∞\n_\nk=i\nαk\n!\n=\n∞\n_\nk=1\nαk.\nIt follows that\nΥ(t) =\n∞\n_\nk=1\nαk.\nCase 2: Assume there exists at least one index k such that αk = 1. The\nproof is left as an exercise for the reader. Case 3: Assume αk < 1 for any\nindex k and there exists at least one index k such that αk = 0. The proof is\nalso left as an exercise for the reader.\n\n\n318\nChapter 13 - Uncertain Renewal Process\nWaiting Time\nFor each index n, let Wn be the waiting time of the nth customer, that is, the\namount of time that the nth customer has to wait until his service begins.\nSince the ﬁrst customer does not need to wait for his service, it is clear that\nW1 = 0.\nFor any integer n ≥2, recall that the (n −1)st customer and\nnth customer arrive at the times Sn−1 and Sn, respectively. The (n −1)st\ncustomer will start his service at time Sn−1 +Wn−1, and complete his service\nat time Sn−1 + Wn−1 + ηn−1. If\nSn < Sn−1 + Wn−1 + ηn−1,\nthen the nth customer has to wait for his service and the waiting time is\nWn = Sn−1 + Wn−1 + ηn−1 −Sn = Wn−1 + ηn−1 −ξn.\nOtherwise, we get\nWn−1 + ηn−1 −ξn ≤0\nand Wn = 0. Thus we always have\nWn = (Wn−1 + ηn−1 −ξn)+\n(13.45)\nfor n ≥2.\nTheorem 13.13 (Liu-Liu [142]) Let Wn be the waiting time of the nth cus-\ntomer in an uncertain queueing system with iid positive uncertain interarrival\ntimes ξ1, ξ2, · · · and iid positive uncertain service times η1, η2, · · · Assume\n(ξ1, ξ2, · · · ) and (η1, η2, · · · ) are independent uncertain vectors, and those in-\nterarrival times and service times have regular uncertainty distributions Φ\nand Ψ, respectively. Then the uncertainty distribution of Wn (n ≥2) is\nΥn(t) = sup\nx≥0\n(1 −Φ(x)) ∧Ψ\n\u0012\nx +\nt\nn −1\n\u0013\n,\nt ≥0.\n(13.46)\nProof: Let us prove that the theorem holds for all n ≥2 by mathematical\ninduction on n. Assume n = 2. Then for any given time t ≥0, we have\nΥ2(t) = M{W2 ≤t}\n= M{W1 + η1 −ξ2 ≤t}\n= M{η1 −ξ2 ≤t}\n= sup\nx≥0\n(1 −Φ(x)) ∧Ψ(x + t).\nThis conﬁrms that the theorem holds for n = 2.\nNow assume that the\ntheorem holds for n = k with k ≥2 and let n = k + 1. For any given t ≥0,\nby using (13.45), we get\n{Wn ≤t} = {Wk + ηk −ξk+1 ≤t}.\n(13.47)\n\n\nSection 13.5 - Uncertain Queueing Model\n319\nOn the one hand, since\n[\ns≥0\n{Wk ≤s, ηk −ξk+1 ≤t −s} ⊂{Wk + ηk −ξk+1 ≤t},\n(13.48)\nwe have\nΥn(t) = M{Wn ≤t}\n≥M\n\n\n\n[\ns≥0\n(Wk ≤s, ηk −ξk+1 ≤t −s)\n\n\n\n≥sup\ns≥0\nM {Wk ≤s, ηk −ξk+1 ≤t −s}\n= sup\ns≥0\nM {Wk ≤s} ∧M {ηk −ξk+1 ≤t −s}\n= sup\nx≥0\n(1 −Φ(x)) ∧Ψ\n\u0012\nx + t\nk\n\u0013\n.\nOn the other hand, since\n[\ns≥0\n{Wk > s, ηk −ξk+1 > t −s} ⊂{Wk + ηk −ξk+1 > t},\n(13.49)\nwe have\nΥn(t) = M{Wn ≤t} = 1 −M{Wk + ηk −ξk+1 > t}\n≤1 −M\n\n\n\n[\ns≥0\n(Wk > s, ηk −ξk+1 > t −s)\n\n\n\n≤1 −sup\ns≥0\nM {Wk > s, ηk −ξk+1 > t −s}\n= 1 −sup\ns≥0\nM {Wk > s} ∧M {ηk −ξk+1 > t −s}\n= sup\nx≥0\n(1 −Φ(x)) ∧Ψ\n\u0012\nx + t\nk\n\u0013\n.\nIt follows that\nΥn(t) = sup\nx≥0\n(1 −Φ(x)) ∧Ψ\n\u0012\nx + t\nk\n\u0013\nfor any t ≥0. This implies that the theorem holds for n = k + 1. By the\nprinciple of mathematical induction, the theorem holds for all n ≥2.\nIdle Time\nFor each index n, let In denote the idle time of the system preceding the nth\ncustomer’s arrival, that is, the time interval between the departure time of\n\n\n320\nChapter 13 - Uncertain Renewal Process\nthe (n −1)st customer and the arrival time of the nth customer. Since the\nsystem is idle until the ﬁrst customer arrives, we immediately have\nI1 = ξ1.\nFor any integer n ≥2, recall that the (n −1)st customer and nth customer\narrive at the system at times Sn−1 and Sn, respectively.\nThe (n −1)st\ncustomer will start to be served at time Sn−1 + Wn−1, and depart from the\nsystem at time Sn−1 + Wn−1 + ηn−1. If\nSn > Sn−1 + Wn−1 + ηn−1,\nthen the system is idle until the nth customer arrives and the idle time of\nthe system is\nIn = Sn −(Sn−1 + Wn−1 + ηn−1) = ξn −Wn−1 −ηn−1.\nOtherwise, we get\nξn −Wn−1 −ηn−1 ≤0\nand In = 0. Thus we always have\nIn = (ξn −Wn−1 −ηn−1)+\n(13.50)\nfor n ≥2.\nTheorem 13.14 (Liu-Liu [142]) Let In be the idle time of an uncertain\nqueueing system with iid positive uncertain interarrival times ξ1, ξ2, · · · and\niid positive uncertain service times η1, η2, · · · preceding the nth customer’s\narrival. Assume (ξ1, ξ2, · · · ) and (η1, η2, · · · ) are independent uncertain vec-\ntors, and those interarrival times and service times have regular uncertainty\ndistributions Φ and Ψ, respectively. Then In, n = 2, 3, · · · have a common\nuncertainty distribution\nF(t) = sup\nx≥0\nΦ(x + t) ∧(1 −Ψ (x)) ,\nt ≥0.\n(13.51)\nProof: For any index n ≥2 and nonnegative time t, it follows from (13.50)\nthat\n{In ≤t} = {ξn −Wn−1 −ηn−1 ≤t}.\n(13.52)\nOn the one hand, since Wn−1 always takes a non-negative value, we get\n{ξn −ηn−1 ≤t} ⊂{ξn −Wn−1 −ηn−1 ≤t}.\nBy using (13.52), we obtain\nF(t) = M{In ≤t}\n≥M{ξn −ηn−1 ≤t}\n= sup\nx≥0\nΦ(x + t) ∧(1 −Ψ (x)) .\n\n\nSection 13.6 - Bibliographic Notes\n321\nOn the other hand, since\n{Wn−1 = 0, ξn −ηn−1 > t} ⊂{ξn −Wn−1 −ηn−1 > t},\nwe obtain\nF(t) = M{In ≤t}\n= 1 −M{ξn −Wn−1 −ηn−1 > t}\n≤1 −M{Wn−1 = 0, ξn −ηn−1 > t}\n= 1 −M{Wn−1 = 0} ∧M{ξn −ηn−1 > t}\n= sup\nx≥0\nΦ(x + t) ∧(1 −Ψ (x)) .\nIt follows that\nF(t) = sup\nx≥0\nΦ(x + t) ∧(1 −Ψ (x))\nfor all n ≥2. The theorem is proved.\n13.6\nBibliographic Notes\nUncertain renewal process was ﬁrst proposed by Liu [114] in 2008. Two years\nlater, Liu [120] proved some elementary renewal theorems for determining\nthe average renewal number. Liu [120] also provided an uncertain renewal\nreward process and veriﬁed some renewal reward theorems for determining\nthe long-run reward rate.\nAs the ﬁrst application of uncertain renewal process, the research of un-\ncertain insurance models was pioneered by Liu [125] in 2013 in which the\nruin index was derived.\nIn addition, Yao-Zhou [267] provided the uncer-\ntainty distribution of ruin time in 2018.\nAfter that, uncertain insurance\nmodels were further developed by Yao-Qin [263], Liu-Yang [161], and Liu-\nYang [167], among others.\nAs the second application of uncertain renewal process, the research of\nuncertain production models was started by Lio-Liu [106] in 2020 in which the\nshortage index and uncertainty distribution of shortage time were obtained.\nAfter that, uncertain production models were further developed by Lio-Jia\n[107].\nAs the third application of uncertain renewal process, uncertain queueing\nmodels were ﬁrst presented by Yao [274] in 2021 in which the uncertainty\ndistribution of busy period was derived. In addition, Liu-Liu [142] obtained\nthe uncertainty distributions of waiting time and idle time.\n\n\n\n\nChapter 14\nUncertain Calculus\nUncertain calculus is a branch of mathematics that deals with diﬀerentia-\ntion and integration of uncertain processes. This chapter will introduce Liu\nprocess, Liu integral, fundamental theorem, chain rule, change of variables,\nintegration by parts, and Fubini theorem.\n14.1\nLiu Process\nIn 2009, Liu [116] investigated a type of stationary independent increment\nprocess whose increments are normal uncertain variables. Later, this process\nwas named by the academic community as Liu process due to its importance\nand usefulness. A formal deﬁnition is given below.\nDeﬁnition 14.1 (Liu [116]) An uncertain process Ct is said to be a Liu\nprocess if\n(i) C0 = 0 and almost all sample paths are Lipschitz continuous,\n(ii) Ct has stationary and independent increments,\n(iii) every increment Cs+t −Cs is a normal uncertain variable with expected\nvalue 0 and variance t2.\nIt is clear that a Liu process Ct is a normal uncertain process with ex-\npected value 0 and variance t2, i.e., Ct ∼N(0, t). Furthermore, Ct has an\nuncertainty distribution\nΦt(x) =\n\u0012\n1 + exp\n\u0012\n−πx\n√\n3t\n\u0013\u0013−1\n(14.1)\nand an inverse uncertainty distribution\nΦ−1\nt (α) =\n√\n3t\nπ\nln\nα\n1 −α\n(14.2)\n\n\n324\nChapter 14 - Uncertain Calculus\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\n0\nΦ−1\nt (α)\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.α = 0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.8\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.9\nFigure 14.1: Inverse Uncertainty Distribution of Liu Process\nthat are homogeneous linear functions of time t for any given α. See Fig-\nure 14.1.\nA Liu process is described by three properties in the deﬁnition. Does such\nan uncertain process exist? The following theorem will answer this question.\nTheorem 14.1 (Liu [120], Existence Theorem) There exists a Liu process.\nProof: It follows from Theorem 12.14 that there exists a stationary inde-\npendent increment process Ct whose inverse uncertainty distribution is\nΦ−1\nt (α) =\n√\n3t\nπ\nln\nα\n1 −α.\nFurthermore, Ct has a Lipschitz continuous version. It is also easy to verify\nthat every increment Cs+t −Cs is a normal uncertain variable with expected\nvalue 0 and variance t2. Hence there exists a Liu process.\nTheorem 14.2 Let Ct be a Liu process. Then for each time t > 0, the ratio\nCt/t is a standard normal uncertain variable. That is,\nCt\nt ∼N(0, 1)\n(14.3)\nfor any t > 0.\nProof: Since Ct is a normal uncertain variable N(0, t), the operational law\ntells us that Ct/t has an uncertainty distribution\nΨ(x) = Φt(tx) =\n\u0012\n1 + exp\n\u0012\n−πx\n√\n3\n\u0013\u0013−1\n.\nHence Ct/t is a standard normal uncertain variable. The theorem is veriﬁed.\n\n\nSection 14.2 - Liu Integral\n325\nDeﬁnition 14.2 Let Ct be a Liu process. Then for any real numbers e and\nσ > 0, the uncertain process\nAt = et + σCt\n(14.4)\nis called an arithmetic Liu process, where e is called the drift and σ is called\nthe diﬀusion.\nIt is clear that the arithmetic Liu process At is a type of stationary in-\ndependent increment process. In addition, the arithmetic Liu process At has\na normal uncertainty distribution with expected value et and variance σ2t2,\ni.e.,\nAt ∼N(et, σt)\n(14.5)\nwhose uncertainty distribution is\nΦt(x) =\n\u0012\n1 + exp\n\u0012π(et −x)\n√\n3σt\n\u0013\u0013−1\n(14.6)\nand inverse uncertainty distribution is\nΦ−1\nt (α) = et +\n√\n3σt\nπ\nln\nα\n1 −α.\n(14.7)\nDeﬁnition 14.3 Let Ct be a Liu process. Then for any real numbers e and\nσ > 0, the uncertain process\nGt = exp(et + σCt)\n(14.8)\nis called a geometric Liu process, where e is called the log-drift and σ is called\nthe log-diﬀusion.\nNote that the geometric Liu process Gt has an uncertainty distribution\nΦt(x) =\n\u0012\n1 + exp\n\u0012π(et −ln x)\n√\n3σt\n\u0013\u0013−1\n(14.9)\nand an inverse uncertainty distribution\nΦ−1\nt (α) = exp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!\n.\n(14.10)\n14.2\nLiu Integral\nAs the most popular topic of uncertain integral, Liu integral allows us to\nintegrate an uncertain process (the integrand) with respect to Liu process\n(the integrator).\n\n\n326\nChapter 14 - Uncertain Calculus\nDeﬁnition 14.4 (Liu [116]) Let Xt be an uncertain process and let Ct be a\nLiu process. For any partition of closed interval [0, r] with\n0 = t1 < t2 < · · · < tk+1 = r,\n(14.11)\nthe mesh is written as\n∆= max\n1≤i≤k |ti+1 −ti|.\n(14.12)\nThen the Liu integral of Xt with respect to Ct is deﬁned as\nZ r\n0\nXtdCt = lim\n∆→0\nk\nX\ni=1\nXti · (Cti+1 −Cti)\n(14.13)\nprovided that there is an uncertain variable to which the above sums converge\nalmost surely as ∆→0. In this case, the uncertain process Xt is said to be\nintegrable with respect to Ct.\nExample 14.1: For any partition 0 = t1 < t2 < · · · < tk+1 = r, it follows\nfrom (14.13) that\nZ r\n0\ndCt = lim\n∆→0\nk\nX\ni=1\n(Cti+1 −Cti) ≡Cr −C0 = Cr.\nThat is,\nZ r\n0\ndCt = Cr.\n(14.14)\nExample 14.2: For any partition 0 = t1 < t2 < · · · < tk+1 = r, it follows\nfrom (14.13) that\nC2\nr =\nk\nX\ni=1\n\u0010\nC2\nti+1 −C2\nti\n\u0011\n=\nk\nX\ni=1\nCti+1 −Cti\n\u00012 + 2\nk\nX\ni=1\nCti\nCti+1 −Cti\n\u0001\n→0 + 2\nZ r\n0\nCtdCt\nas the mesh ∆→0. That is,\nZ r\n0\nCtdCt = 1\n2C2\nr.\n(14.15)\n\n\nSection 14.2 - Liu Integral\n327\nExample 14.3: For any partition 0 = t1 < t2 < · · · < tk+1 = r, it follows\nfrom (14.13) that\nrCr =\nk\nX\ni=1\nti+1Cti+1 −tiCti\n\u0001\n=\nk\nX\ni=1\nCti+1(ti+1 −ti) +\nk\nX\ni=1\nti(Cti+1 −Cti)\n→\nZ r\n0\nCtdt +\nZ r\n0\ntdCt\nas the mesh ∆→0. That is,\nZ r\n0\nCtdt +\nZ r\n0\ntdCt = rCr.\n(14.16)\nTheorem 14.3 If Xt is a sample-continuous uncertain process on [0, r], then\nit is integrable with respect to Ct on [0, r].\nProof: Let 0 = t1 < t2 < · · · < tk+1 = r be a partition of the closed interval\n[0, r]. Since the uncertain process Xt is sample-continuous, almost all sample\npaths are continuous functions with respect to t. Hence the limit\nlim\n∆→0\nk\nX\ni=1\nXti(Cti+1 −Cti)\nexists almost surely and is ﬁnite. On the other hand, since Xt and Ct are\nuncertain variables at each time t, the above limit is also a measurable func-\ntion. Hence the limit is an uncertain variable and then Xt is integrable with\nrespect to Ct.\nTheorem 14.4 If Xt is an integrable uncertain process on [0, r], then it is\nintegrable on each subinterval of [0, r]. Moreover, if s ∈[0, r], then\nZ r\n0\nXtdCt =\nZ s\n0\nXtdCt +\nZ r\ns\nXtdCt.\n(14.17)\nProof: Let [a, b] be a subinterval of [0, r]. Since Xt is an integrable uncertain\nprocess on [0, r], for any partition\n0 = t1 < · · · < tm = a < tm+1 < · · · < tn = b < tn+1 < · · · < tk+1 = r,\nthe limit\nlim\n∆→0\nk\nX\ni=1\nXti(Cti+1 −Cti)\n\n\n328\nChapter 14 - Uncertain Calculus\nexists almost surely and is ﬁnite. Thus the limit\nlim\n∆→0\nn−1\nX\ni=m\nXti(Cti+1 −Cti)\nexists almost surely and is ﬁnite. Hence Xt is integrable on the subinterval\n[a, b]. Next, for the partition\n0 = t1 < · · · < tm = s < tm+1 < · · · < tk+1 = r,\nwe have\nk\nX\ni=1\nXti(Cti+1 −Cti) =\nm−1\nX\ni=1\nXti(Cti+1 −Cti) +\nk\nX\ni=m\nXti(Cti+1 −Cti).\nNote that\nZ r\n0\nXtdCt = lim\n∆→0\nk\nX\ni=1\nXti(Cti+1 −Cti),\nZ s\n0\nXtdCt = lim\n∆→0\nm−1\nX\ni=1\nXti(Cti+1 −Cti),\nZ r\ns\nXtdCt = lim\n∆→0\nk\nX\ni=m\nXti(Cti+1 −Cti).\nHence the equation (14.17) is proved.\nTheorem 14.5 (Linearity of Liu Integral) Let Xt and Yt be integrable un-\ncertain processes on [0, r], and let α and β be real numbers. Then\nZ r\n0\n(αXt + βYt)dCt = α\nZ r\n0\nXtdCt + β\nZ r\n0\nYtdCt.\n(14.18)\nProof: Let 0 = t1 < t2 < · · · < tk+1 = r be a partition of the closed interval\n[0, r]. It follows from the deﬁnition of Liu integral that\nZ r\n0\n(αXt + βYt)dCt = lim\n∆→0\nk\nX\ni=1\n(αXti + βYti)(Cti+1 −Cti)\n= lim\n∆→0 α\nk\nX\ni=1\nXti(Cti+1 −Cti) + lim\n∆→0 β\nk\nX\ni=1\nYti(Cti+1 −Cti)\n= α\nZ r\n0\nXtdCt + β\nZ r\n0\nYtdCt.\nHence the equation (14.18) is proved.\n\n\nSection 14.3 - Differential\n329\nTheorem 14.6 Let f(t) be an integrable function with respect to t. Then\nthe Liu integral\nZ r\n0\nf(t)dCt\n(14.19)\nis a normal uncertain variable at each time r, and\nZ r\n0\nf(t)dCt ∼N\n\u0012\n0,\nZ r\n0\n|f(t)|dt\n\u0013\n.\n(14.20)\nProof: Since the increments of Ct are stationary and independent normal\nuncertain variables, for any partition of closed interval [0, r] with 0 = t1 <\nt2 < · · · < tk+1 = r, it follows from Theorem 3.14 that\nk\nX\ni=1\nf(ti)(Cti+1 −Cti) ∼N\n \n0,\nk\nX\ni=1\n|f(ti)|(ti+1 −ti)\n!\n.\nThat is, the sum is also a normal uncertain variable. Since f is an integrable\nfunction, we have\nk\nX\ni=1\n|f(ti)|(ti+1 −ti) →\nZ r\n0\n|f(t)|dt\nas the mesh ∆→0. Hence we obtain\nZ r\n0\nf(t)dCt = lim\n∆→0\nk\nX\ni=1\nf(ti)(Cti+1 −Cti) ∼N\n\u0012\n0,\nZ r\n0\n|f(t)|dt\n\u0013\n.\nThe theorem is proved.\nExercise 14.1: Let r be a given time with r > 0. Show that the Liu integral\nZ r\n0\ntdCt\n(14.21)\nis a normal uncertain variable N(0, r2/2) and has an uncertainty distribution\nΦr(x) =\n\u0012\n1 + exp\n\u0012\n−2πx\n√\n3r2\n\u0013\u0013−1\n.\n(14.22)\n14.3\nDiﬀerential\nDeﬁnition 14.5 (Liu [116], Chen-Ralescu [17] and Ye [279]) Let Ct be a Liu\nprocess and let Zt be an uncertain process. If there exist sample-continuous\nuncertain processes µt and σt such that\nZt = Z0 +\nZ t\n0\nµsds +\nZ t\n0\nσsdCs\n(14.23)\n\n\n330\nChapter 14 - Uncertain Calculus\nfor any t ≥0, then Zt is called a general Liu process with drift µt and\ndiﬀusion σt. Furthermore, Zt has a diﬀerential\ndZt = µtdt + σtdCt.\n(14.24)\nExample 14.4: It follows from the equation (14.14) that Liu process Ct can\nbe written as\nCt =\nZ t\n0\ndCs.\nThus Ct is a general Liu process with drift 0 and diﬀusion 1, and has a\ndiﬀerential dCt.\nExample 14.5: It follows from the equation (14.15) that C2\nt can be written\nas\nC2\nt = 2\nZ t\n0\nCsdCs.\nThus C2\nt is a general Liu process with drift 0 and diﬀusion 2Ct, and has a\ndiﬀerential\nd(C2\nt ) = 2CtdCt.\nExample 14.6: It follows from the equation (14.16) that tCt can be written\nas\ntCt =\nZ t\n0\nCsds +\nZ t\n0\nsdCs.\nThus tCt is a general Liu process with drift Ct and diﬀusion t, and has a\ndiﬀerential\nd(tCt) = Ctdt + tdCt.\nTheorem 14.7 (Ye [279]) Almost all sample paths of general Liu process\nare locally Lipschitz continuous.\nProof: Let Zt be a general Liu process with drift µt and diﬀusion σt. Then\nwe immediately have\nZt = Z0 +\nZ t\n0\nµsds +\nZ t\n0\nσsdCs.\n(14.25)\nLet [a, b] be any bounded interval on [0, +∞), and ﬁx t1 and t2 with a ≤t1 <\nt2 ≤b. For any partition of the interval [t1, t2] with t1 = s1 < s2 < · · · <\nsk+1 = t2, the mesh is written as\n∆= max\n1≤i≤k |si+1 −si|,\n\n\nSection 14.4 - Fundamental Theorem\n331\nand for each γ ∈Γ, we have\nZ t2\nt1\nσs(γ)dCs(γ) = lim\n∆→0\nk\nX\ni=1\nσsi(γ)(Csi+1(γ) −Csi(γ)).\nIf Ct(γ) is a Lipschitz continuous function with respect to t, then there exists\na Lipschitz constant L(γ) such that\n|Csi+1(γ) −Csi(γ)| ≤L(γ)(si+1 −si),\ni = 1, 2, · · · , k.\nThus\n\f\n\f\n\f\n\f\n\f\nk\nX\ni=1\nσsi(γ)(Csi+1(γ) −Csi(γ))\n\f\n\f\n\f\n\f\n\f ≤\nk\nX\ni=1\n|σsi(γ)||Csi+1(γ) −Csi(γ)|\n≤L(γ)\nk\nX\ni=1\n|σsi(γ)|(si+1 −si).\nLetting ∆→0, we get\n\f\n\f\n\f\n\f\nZ t2\nt1\nσs(γ)dCs(γ)\n\f\n\f\n\f\n\f ≤L(γ)\nZ t2\nt1\n|σs(γ)|ds.\n(14.26)\nIf µt(γ) and σt(γ) are continuous functions with respect to t, then there exists\na positive number A(γ) such that\n|µt(γ)| ≤A(γ),\n|σt(γ)| ≤A(γ),\n∀t ∈[a, b].\n(14.27)\nIt follows from (14.25), (14.26) and (14.27) that\n|Zt1(γ) −Zt2(γ)| =\n\f\n\f\n\f\n\f\nZ t2\nt1\nµs(γ)ds +\nZ t2\nt1\nσs(γ)dCs(γ)\n\f\n\f\n\f\n\f\n≤\nZ t2\nt1\n|µs(γ)|ds + L(γ)\nZ t2\nt1\n|σs(γ)|ds\n≤\nZ t2\nt1\nA(γ)ds + L(γ)\nZ t2\nt1\nA(γ)ds\n= A(γ)(1 + L(γ))|t1 −t2|.\nHence Zt(γ) is locally Lipschitz continuous due to the arbitrariness of [a, b].\nSince almost all paths of µt and σt are continuous, and almost all paths of\nCt are Lipschitz continuous, we conclude that almost all paths of general Liu\nprocess are locally Lipschitz continuous. The theorem is proved.\n\n\n332\nChapter 14 - Uncertain Calculus\n14.4\nFundamental Theorem\nTheorem 14.8 (Liu [116], Fundamental Theorem of Uncertain Calculus)\nLet Ct be a Liu process, and let h(t, c) be a continuously diﬀerentiable func-\ntion. Then Zt = h(t, Ct) has a diﬀerential\ndZt = ∂h\n∂t (t, Ct)dt + ∂h\n∂c (t, Ct)dCt.\n(14.28)\nProof: This proof was provided by Ye [279]. For any given γ ∈Γ, note that\nZt(γ) = h(t, Ct(γ)).\nIf Ct(γ) is a Lipschitz continuous function with respect to t, then by using\nYe Lemma (Page 459), we obtain\nZt(γ) = Z0(γ) +\nZ t\n0\n∂h\n∂s (s, Cs(γ))ds +\nZ t\n0\n∂h\n∂c (s, Cs(γ))dCs(γ).\nSince almost all sample paths of Ct are Lipschitz continuous, we get\nZt = Z0 +\nZ t\n0\n∂h\n∂s (s, Cs)ds +\nZ t\n0\n∂h\n∂c (s, Cs)dCs,\na.s.\nThus the diﬀerential form (14.28) is proved.\nExample 14.7: Let us calculate the diﬀerential of tCt. In this case, we have\nh(t, c) = tc whose partial derivatives are\n∂h\n∂t (t, c) = c,\n∂h\n∂c (t, c) = t.\nIt follows from the fundamental theorem of uncertain calculus that\nd(tCt) = Ctdt + tdCt.\n(14.29)\nThus tCt is a general Liu process with drift Ct and diﬀusion t.\nExample 14.8: Let us calculate the diﬀerential of the arithmetic Liu process\nAt = et+σCt. In this case, we have h(t, c) = et+σc whose partial derivatives\nare\n∂h\n∂t (t, c) = e,\n∂h\n∂c (t, c) = σ.\nIt follows from the fundamental theorem of uncertain calculus that\ndAt = edt + σdCt.\n(14.30)\nThus At is a general Liu process with drift e and diﬀusion σ.\n\n\nSection 14.5 - Chain Rule\n333\nExample 14.9: Let us calculate the diﬀerential of the geometric Liu process\nGt = exp(et+σCt). In this case, we have h(t, c) = exp(et+σc) whose partial\nderivatives are\n∂h\n∂t (t, c) = eh(t, c),\n∂h\n∂c (t, c) = σh(t, c).\nIt follows from the fundamental theorem of uncertain calculus that\ndGt = eGtdt + σGtdCt.\n(14.31)\nThus Gt is a general Liu process with drift eGt and diﬀusion σGt.\nExercise 14.2: (Ye [279]) Let C1t, C2t, · · · , Cnt be Liu processes, and let\nh(t, c1, c2, · · · , cn) be a continuously diﬀerentiable function. Then\nZt = h(t, C1t, C2t, · · · , Cnt)\nhas a diﬀerential\ndZt = ∂h\n∂t (t, C1t, C2t, · · · , Cnt)dt +\nn\nX\ni=1\n∂h\n∂ci\n(t, C1t, C2t, · · · , Cnt)dCit.\nExercise 14.3: (Ye [279]) Let X1t, X2t, · · · , Xnt be general Liu processes,\nand let h(x1, x2, · · · , xn) be a continuously diﬀerentiable function. Then\nZt = h(X1t, X2t, · · · , Xnt)\nhas a diﬀerential\ndZt =\nn\nX\ni=1\n∂h\n∂xi\n(X1t, X2t, · · · , Xnt)dXit.\n14.5\nChain Rule\nTheorem 14.9 (Liu [116], Chain Rule) Let f(c) be a continuously diﬀeren-\ntiable function. Then f(Ct) has a diﬀerential\ndf(Ct) = f ′(Ct)dCt.\n(14.32)\nProof: Since f(c) is a continuously diﬀerentiable function, we immediately\nhave\n∂\n∂tf(c) = 0,\n∂\n∂cf(c) = f ′(c).\nThe equation (14.32) follows from the fundamental theorem of uncertain\ncalculus,\ndf(Ct) = ∂\n∂tf(Ct)dt + ∂\n∂cf(Ct)dCt.\n\n\n334\nChapter 14 - Uncertain Calculus\nExample 14.10: Let us calculate the diﬀerential of C2\nt . In this case, we\nhave f(c) = c2 and f ′(c) = 2c. It follows from the chain rule that\ndC2\nt = 2CtdCt.\n(14.33)\nExample 14.11: Let us calculate the diﬀerential of sin(Ct). In this case,\nwe have f(c) = sin(c) and f ′(c) = cos(c). It follows from the chain rule that\nd sin(Ct) = cos(Ct)dCt.\n(14.34)\nExample 14.12: Let us calculate the diﬀerential of exp(Ct). In this case,\nwe have f(c) = exp(c) and f ′(c) = exp(c). It follows from the chain rule that\nd exp(Ct) = exp(Ct)dCt.\n(14.35)\n14.6\nChange of Variables\nTheorem 14.10 (Liu [116], Change of Variables) Let f be a continuously\ndiﬀerentiable function. Then for any s > 0, we have\nZ s\n0\nf ′(Ct)dCt = f(Cs) −f(C0).\n(14.36)\nProof: Since f is a continuously diﬀerentiable function, it follows from the\nchain rule that\ndf(Ct) = f ′(Ct)dCt.\nThis formula implies that\nf(Cs) = f(C0) +\nZ s\n0\nf ′(Ct)dCt.\nHence the theorem is veriﬁed.\nExample 14.13: Since the function f ′(c) = c has an antiderivative f(c) =\nc2/2, it follows from the change of variables of integral that\nZ s\n0\nCtdCt = 1\n2C2\ns −1\n2C2\n0 = 1\n2C2\ns.\nExample 14.14: Since the function f ′(c) = c2 has an antiderivative f(c) =\nc3/3, it follows from the change of variables of integral that\nZ s\n0\nC2\nt dCt = 1\n3C3\ns −1\n3C3\n0 = 1\n3C3\ns.\nExample 14.15: Since the function f ′(c) = exp(c) has an antiderivative\nf(c) = exp(c), it follows from the change of variables of integral that\nZ s\n0\nexp(Ct)dCt = exp(Cs) −exp(C0) = exp(Cs) −1.\n\n\nSection 14.7 - Integration by Parts\n335\n14.7\nIntegration by Parts\nTheorem 14.11 (Liu [116], Integration by Parts) Suppose Xt and Yt are\ngeneral Liu processes. Then\nd(XtYt) = YtdXt + XtdYt.\n(14.37)\nProof: Since h(x, y) = xy is a continuously diﬀerentiable function, and\n∂h\n∂x(x, y) = y,\n∂h\n∂y (x, y) = x,\nthe equation (14.37) follows from the fundamental theorem of uncertain cal-\nculus,\ndh(Xt, Yt) = ∂h\n∂x(Xt, Yt)dXt + ∂h\n∂y (Xt, Yt)dYt.\nExample 14.16: In order to illustrate the integration by parts, let us cal-\nculate the diﬀerential of\nZt = exp(t)C2\nt .\nIn this case, we deﬁne\nXt = exp(t),\nYt = C2\nt .\nThen\ndXt = exp(t)dt,\ndYt = 2CtdCt.\nIt follows from the integration by parts that\ndZt = exp(t)C2\nt dt + 2 exp(t)CtdCt.\nExample 14.17: The integration by parts may also calculate the diﬀerential\nof\nZt = sin(t + 1)\nZ t\n0\nsdCs.\nIn this case, we deﬁne\nXt = sin(t + 1),\nYt =\nZ t\n0\nsdCs.\nThen\ndXt = cos(t + 1)dt,\ndYt = tdCt.\nIt follows from the integration by parts that\ndZt =\n\u0012Z t\n0\nsdCs\n\u0013\ncos(t + 1)dt + sin(t + 1)tdCt.\n\n\n336\nChapter 14 - Uncertain Calculus\nExample 14.18: Let f and g be continuously diﬀerentiable functions. It is\nclear that\nZt = f(t)g(Ct)\nis an uncertain process. In order to calculate the diﬀerential of Zt, we deﬁne\nXt = f(t),\nYt = g(Ct).\nThen\ndXt = f ′(t)dt,\ndYt = g′(Ct)dCt.\nIt follows from the integration by parts that\ndZt = f ′(t)g(Ct)dt + f(t)g′(Ct)dCt.\n14.8\nFubini Theorem\nFubini theorem allows us to interchange the order of iterated Liu integrals\nunder the sample-continuity condition of the integrand.\nTheorem 14.12 (Zhang-Liu [300], Fubini Theorem) Suppose Xt and Ys are\ngeneral Liu processes. If Z(t, s) is a sample-continuous uncertain ﬁeld, then\nZ a\n0\nZ b\n0\nZ(t, s)dYsdXt =\nZ b\n0\nZ a\n0\nZ(t, s)dXtdYs,\na.s.\n(14.38)\nProof: For any ﬁxed γ ∈Γ, since Z(t, s; γ) is a continuous function, and\nXt(γ) is a locally Lipschitz continuous function with respect to t, the integral\nF(s; γ) =\nZ a\n0\nZ(t, s; γ)dXt(γ)\nexists for each s. Similarly, the integral\nG(t; γ) =\nZ b\n0\nZ(t, s; γ)dYs(γ)\nexists for each t. Furthermore, F(s; γ) and G(t; γ) are also continuous func-\ntions of s and t, respectively. Thus the integrals\nZ b\n0\nF(s; γ)dYs(γ)\nand\nZ a\n0\nG(t; γ)dXt(γ)\nexist. It follows from the deﬁnition of integral that for any small number\nε > 0, there exist partitions\n0 = t1 < t2 < · · · < tn+1 = a\nand\n0 = s1 < s2 < · · · < sm+1 = b\n\n\nSection 14.8 - Fubini Theorem\n337\nsuch that\n\f\n\f\n\f\n\f\n\fF(sj; γ) −\nn\nX\ni=1\nZ(ti, sj; γ)(Xti+1(γ) −Xti(γ))\n\f\n\f\n\f\n\f\n\f < ε, j = 1, 2, · · · , m,\n\f\n\f\n\f\n\f\n\f\n\f\nG(ti; γ) −\nm\nX\nj=1\nZ(ti, sj; γ)(Ysj+1(γ) −Ysj(γ))\n\f\n\f\n\f\n\f\n\f\n\f\n< ε, i = 1, 2, · · · , n,\n\f\n\f\n\f\n\f\n\f\n\f\nZ b\n0\nF(s; γ)dYs(γ) −\nm\nX\nj=1\nF(sj; γ)(Ysj+1(γ) −Ysj(γ))\n\f\n\f\n\f\n\f\n\f\n\f\n< ε,\n\f\n\f\n\f\n\f\n\f\nZ a\n0\nG(t; γ)dXt(γ) −\nn\nX\ni=1\nG(ti; γ)(Xti+1(γ) −Xti(γ))\n\f\n\f\n\f\n\f\n\f < ε.\nTherefore, we have\n\f\n\f\n\f\n\f\n\f\nZ a\n0\nG(t; γ)dXt(γ) −\nZ b\n0\nF(s; γ)dYs(γ)\n\f\n\f\n\f\n\f\n\f\n< 2ε +\n\f\n\f\n\f\n\f\n\f\n\f\nn\nX\ni=1\nG(ti; γ)(Xti+1(γ) −Xti(γ)) −\nm\nX\nj=1\nF(sj; γ)(Ysj+1(γ) −Ysj(γ))\n\f\n\f\n\f\n\f\n\f\n\f\n< 2ε + ε\nn\nX\ni=1\n\f\n\fXti+1(γ) −Xti(γ)\n\f\n\f + ε\nm\nX\nj=1\n\f\n\fYsj+1(γ) −Ysj(γ)\n\f\n\f\n< (2 + aL(γ) + bL(γ)) ε\nwhere L(γ) is a Lipschitz constant of Xt(γ) and Ys(γ) on [0, a] and [0, b],\nrespectively. Letting ε →0, we obtain\nZ a\n0\nG(t; γ)dXt(γ) =\nZ b\n0\nF(s; γ)dYs(γ).\nThe theorem is proved.\nExercise 14.4: Suppose Xt and Ys are general Liu processes. If Z(t, s) is a\nsample-continuous uncertain ﬁeld, then\nZ r\n0\nZ s\n0\nZ(t, s)dXtdYs =\nZ r\n0\nZ r\nt\nZ(t, s)dYsdXt,\na.s.\n(14.39)\nTheorem 14.13 (Zhang-Liu [300]) Let Ct be a Liu process, and let h(t, s)\nbe a continuously diﬀerentiable function. Then\nZt =\nZ t\n0\nh(t, s)dCs\n(14.40)\n\n\n338\nChapter 14 - Uncertain Calculus\nhas a diﬀerential\ndZt =\nZ t\n0\n∂h\n∂t (t, s)dCsdt + h(t, t)dCt.\n(14.41)\nProof: Since h(t, s) is a continuously diﬀerentiable function, for any t and\ns, we have\nh(t, s) = h(0, s) +\nZ t\n0\n∂h\n∂t (r, s)dr.\nThus\nZt =\nZ t\n0\nh(0, s)dCs +\nZ t\n0\nZ t\n0\n∂h\n∂t (r, s)drdCs.\nIt follows from Fubini theorem that\nZt =\nZ t\n0\nh(0, s)dCs +\nZ t\n0\nZ t\n0\n∂h\n∂t (r, s)dCsdr\n=\nZ t\n0\nh(0, s)dCs +\nZ t\n0\nZ r\n0\n∂h\n∂t (r, s)dCsdr +\nZ t\n0\nZ t\nr\n∂h\n∂t (r, s)dCsdr\n=\nZ t\n0\nh(0, s)dCs +\nZ t\n0\nZ r\n0\n∂h\n∂t (r, s)dCsdr +\nZ t\n0\nZ s\n0\n∂h\n∂t (r, s)drdCs\n=\nZ t\n0\n\u0012\nh(0, s) +\nZ s\n0\n∂h\n∂t (r, s)dr\n\u0013\ndCs +\nZ t\n0\nZ r\n0\n∂h\n∂t (r, s)dCsdr\n=\nZ t\n0\nh(s, s)dCs +\nZ t\n0\nZ r\n0\n∂h\n∂t (r, s)dCsdr.\nHence (14.41) is proved.\nExercise 14.5: Let Ct be a Liu process. Show that the uncertain process\nZt =\nZ t\n0\n(t −s)dCs\n(14.42)\nhas a diﬀerential\ndZt = Ctdt.\n(14.43)\nExercise 14.6: Let Ct be a Liu process, and let α be a constant with α > 1.\nShow that the uncertain process\nZt =\nZ t\n0\n(t −s)αdCs\n(14.44)\nhas a diﬀerential\ndZt = α\nZ t\n0\n(t −s)α−1dCsdt.\n(14.45)\n\n\nSection 14.9 - Bibliographic Notes\n339\n14.9\nBibliographic Notes\nUncertain calculus was pioneered by Liu [114] in 2008 where Liu integral was\ninvented that allows us to integrate an uncertain process with respect to Liu\nprocess. One year later, Liu [116] presented a fundamental theorem of un-\ncertain calculus from which the techniques of chain rule, change of variables,\nand integration by parts were derived. This theorem was generalized to the\nmultifactor version by Chen [12] in 2011, and was rigorously proved by Ye\n[279] in 2021. In addition, Zhang-Liu [300] provided a Fubini theorem that\ngives conditions for interchanging the order of iterated Liu integrals.\nMultivariate uncertain calculus was developed by Ye [285] where the par-\ntial derivatives of uncertain ﬁeld were created. Meanwhile, higher-order un-\ncertain calculus was initialized by Zhang-Liu [300] where the higher-order\nderivatives of uncertain process were rigorously deﬁned.\n\n\n\n\nChapter 15\nUncertain Diﬀerential\nEquation\nUncertain diﬀerential equation is a type of diﬀerential equation involving un-\ncertain processes. This chapter will discuss the existence, uniqueness and\nstability of solutions of uncertain diﬀerential equations, and introduce Yao-\nChen formula that represents the solution of an uncertain diﬀerential equation\nby a family of solutions of ordinary diﬀerential equations. On the basis of\nthis formula, some formulas to calculate extreme value, ﬁrst hitting time, and\ntime integral of solution will be provided, and an Euler method for solving\nuncertain diﬀerential equations will also be documented. Assume an uncer-\ntain process follows an uncertain diﬀerential equation and some realizations\nof this process are observed. In order to make a connection between uncer-\ntain diﬀerential equation and observed data, this chapter will introduce the\nconcept of residual. Based on those residuals, uncertain hypothesis test will\nbe employed to determine whether an uncertain diﬀerential equation ﬁts the\nobserved data, and the method of moments will be presented to estimate\nthe unknown parameters in an uncertain diﬀerential equation that ﬁts the\nobserved data as much as possible. Finally, some real-world examples are\ndocumented.\n15.1\nUncertain Diﬀerential Equation\nDeﬁnition 15.1 (Liu [114]) Suppose f and g are continuous functions, and\nCt is a Liu process. Then\ndXt = f(t, Xt)dt + g(t, Xt)dCt\n(15.1)\nis called an uncertain diﬀerential equation. A solution is an uncertain process\nXt that satisﬁes (15.1) identically in t.\n\n\n342\nChapter 15 - Uncertain Differential Equation\nRemark 15.1: The uncertain diﬀerential equation (15.1) means the solution\nXt meets the uncertain integral equation\nXt = X0 +\nZ t\n0\nf(s, Xs)ds +\nZ t\n0\ng(s, Xs)dCs.\n(15.2)\nTheorem 15.1 Let ut and vt be two continuous functions of t. Then the\nuncertain diﬀerential equation\ndXt = utdt + vtdCt\n(15.3)\nhas a solution\nXt = X0 +\nZ t\n0\nusds +\nZ t\n0\nvsdCs.\n(15.4)\nProof: This theorem is essentially the deﬁnition of uncertain diﬀerential or\na direct deduction of the fundamental theorem of uncertain calculus.\nExample 15.1: Let a and b be constants. Consider the uncertain diﬀerential\nequation\ndXt = adt + bdCt.\n(15.5)\nIt follows from Theorem 15.1 that the solution is\nXt = X0 +\nZ t\n0\nads +\nZ t\n0\nbdCs.\nThat is,\nXt = X0 + at + bCt.\n(15.6)\nTheorem 15.2 Let ut and vt be two continuous functions of t. Then the\nuncertain diﬀerential equation\ndXt = utXtdt + vtXtdCt\n(15.7)\nhas a solution\nXt = X0 exp\n\u0012Z t\n0\nusds +\nZ t\n0\nvsdCs\n\u0013\n.\n(15.8)\nProof: At ﬁrst, the original uncertain diﬀerential equation is equivalent to\ndXt\nXt\n= utdt + vtdCt.\nIt follows from the fundamental theorem of uncertain calculus that\nd ln Xt = dXt\nXt\n= utdt + vtdCt\nand then\nln Xt = ln X0 +\nZ t\n0\nusds +\nZ t\n0\nvsdCs.\n\n\nSection 15.1 - Uncertain Differential Equation\n343\nTherefore the uncertain diﬀerential equation has the solution (15.8).\nExample 15.2: Let e and σ be constants. Consider the uncertain diﬀerential\nequation\ndXt = eXtdt + σXtdCt.\n(15.9)\nIt follows from Theorem 15.2 that the solution is\nXt = X0 exp\n\u0012Z t\n0\neds +\nZ t\n0\nσdCs\n\u0013\n.\nThat is,\nXt = X0 exp (et + σCt) .\n(15.10)\nLinear Uncertain Diﬀerential Equation\nTheorem 15.3 (Chen-Liu [8]) Let u1t, u2t, v1t, v2t be continuous functions\nof t. Then the linear uncertain diﬀerential equation\ndXt = (u1tXt + u2t)dt + (v1tXt + v2t)dCt\n(15.11)\nhas a solution\nXt = Ut\n\u0012\nX0 +\nZ t\n0\nu2s\nUs\nds +\nZ t\n0\nv2s\nUs\ndCs\n\u0013\n(15.12)\nwhere\nUt = exp\n\u0012Z t\n0\nu1sds +\nZ t\n0\nv1sdCs\n\u0013\n.\n(15.13)\nProof: At ﬁrst, we deﬁne two uncertain processes Ut and Vt via uncertain\ndiﬀerential equations,\ndUt = u1tUtdt + v1tUtdCt,\ndVt = u2t\nUt\ndt + v2t\nUt\ndCt.\nIt follows from the integration by parts that\nd(UtVt) = VtdUt + UtdVt = (u1tUtVt + u2t)dt + (v1tUtVt + v2t)dCt.\nThat is, the uncertain process Xt = UtVt is a solution of the uncertain\ndiﬀerential equation (15.11). Note that\nUt = U0 exp\n\u0012Z t\n0\nu1sds +\nZ t\n0\nv1sdCs\n\u0013\n,\nVt = V0 +\nZ t\n0\nu2s\nUs\nds +\nZ t\n0\nv2s\nUs\ndCs.\nTaking U0 = 1 and V0 = X0, we get the solution (15.12). The theorem is\nproved.\n\n\n344\nChapter 15 - Uncertain Differential Equation\nExample 15.3: Let m, a, σ be constants with a ̸= 0.\nConsider a linear\nuncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt.\n(15.14)\nAt ﬁrst, we have\nUt = exp\n\u0012Z t\n0\n(−a)ds +\nZ t\n0\n0dCs\n\u0013\n= exp(−at).\nIt follows from Theorem 15.3 that the solution is\nXt = exp(−at)\n\u0012\nX0 +\nZ t\n0\nm exp(as)ds +\nZ t\n0\nσ exp(as)dCs\n\u0013\n.\nThat is,\nXt = m\na + exp(−at)\n\u0010\nX0 −m\na\n\u0011\n+ σ exp(−at)\nZ t\n0\nexp(as)dCs.\n(15.15)\nExample 15.4: Let m, a, σ be constants. Consider a linear uncertain dif-\nferential equation\ndXt = (m −aXt)dt + σXtdCt.\n(15.16)\nAt ﬁrst, we have\nUt = exp\n\u0012Z t\n0\n(−a)ds +\nZ t\n0\nσdCs\n\u0013\n= exp(−at + σCt).\nIt follows from Theorem 15.3 that the solution is\nXt = exp(−at + σCt)\n\u0012\nX0 +\nZ t\n0\nm exp(as −σCs)ds +\nZ t\n0\n0dCs\n\u0013\n.\nThat is,\nXt = exp(−at + σCt)\n\u0012\nX0 + m\nZ t\n0\nexp(as −σCs)ds\n\u0013\n.\n(15.17)\nNonlinear Uncertain Diﬀerential Equations\nTheorem 15.4 (Liu [151]) Let f be a continuous function of two variables,\nand let σt be a continuous function of t.\nThen the uncertain diﬀerential\nequation\ndXt = f(t, Xt)dt + σtXtdCt\n(15.18)\nhas a solution\nXt = Y −1\nt\nZt\n(15.19)\n\n\nSection 15.1 - Uncertain Differential Equation\n345\nwhere\nYt = exp\n\u0012\n−\nZ t\n0\nσsdCs\n\u0013\n(15.20)\nand Zt is the solution of the uncertain diﬀerential equation\ndZt = Ytf(t, Y −1\nt\nZt)dt\n(15.21)\nwith initial value Z0 = X0.\nProof: At ﬁrst, by using the chain rule, the uncertain process Yt has an\nuncertain diﬀerential\ndYt = −exp\n\u0012\n−\nZ t\n0\nσsdCs\n\u0013\nσtdCt = −YtσtdCt.\nIt follows from the integration by parts that\nd(XtYt) = XtdYt + YtdXt = −XtYtσtdCt + Ytf(t, Xt)dt + YtσtXtdCt.\nThat is,\nd(XtYt) = Ytf(t, Xt)dt.\nDeﬁning Zt = XtYt, we obtain Xt = Y −1\nt\nZt and dZt = Ytf(t, Y −1\nt\nZt)dt.\nFurthermore, since Y0 = 1, the initial value Z0 is just X0. The theorem is\nthus veriﬁed.\nExample 15.5: Let k and σ be constants with k ̸= 1. Consider the uncertain\ndiﬀerential equation\ndXt = Xk\nt dt + σXtdCt.\n(15.22)\nAt ﬁrst, we have\nYt = exp\n\u0012\n−\nZ t\n0\nσdCs\n\u0013\n= exp(−σCt)\nand Zt satisﬁes the uncertain diﬀerential equation,\ndZt = exp(−σCt)(exp(σCt)Zt)kdt = exp((k −1)σCt)Zk\nt dt.\nSince k ̸= 1, we have\ndZ1−k\nt\n= (1 −k)Z−k\nt\ndZt = (1 −k) exp((k −1)σCt)dt.\nIt follows from the fundamental theorem of uncertain calculus that\nZ1−k\nt\n= Z1−k\n0\n+ (1 −k)\nZ t\n0\nexp((k −1)σCs)ds.\nSince the initial value Z0 is just X0, we have\nZt =\n\u0012\nX1−k\n0\n+ (1 −k)\nZ t\n0\nexp((k −1)σCs)ds\n\u00131/(1−k)\n.\n\n\n346\nChapter 15 - Uncertain Differential Equation\nTheorem 15.4 says the uncertain diﬀerential equation (15.22) has a solution\nXt = Y −1\nt\nZt, i.e.,\nXt = exp(σCt)\n\u0012\nX1−k\n0\n+ (1 −k)\nZ t\n0\nexp((k −1)σCs)ds\n\u00131/(1−k)\n.\nTheorem 15.5 (Liu [151]) Let g be a continuous function of two variables,\nand let αt be a continuous function of t.\nThen the uncertain diﬀerential\nequation\ndXt = αtXtdt + g(t, Xt)dCt\n(15.23)\nhas a solution\nXt = Y −1\nt\nZt\n(15.24)\nwhere\nYt = exp\n\u0012\n−\nZ t\n0\nαsds\n\u0013\n(15.25)\nand Zt is the solution of the uncertain diﬀerential equation\ndZt = Ytg(t, Y −1\nt\nZt)dCt\n(15.26)\nwith initial value Z0 = X0.\nProof: At ﬁrst, by using the chain rule, the uncertain process Yt has an\nuncertain diﬀerential\ndYt = −exp\n\u0012\n−\nZ t\n0\nαsds\n\u0013\nαtdt = −Ytαtdt.\nIt follows from the integration by parts that\nd(XtYt) = XtdYt + YtdXt = −XtYtαtdt + YtαtXtdt + Ytg(t, Xt)dCt.\nThat is,\nd(XtYt) = Ytg(t, Xt)dCt.\nDeﬁning Zt = XtYt, we obtain Xt = Y −1\nt\nZt and dZt = Ytg(t, Y −1\nt\nZt)dCt.\nFurthermore, since Y0 = 1, the initial value Z0 is just X0. The theorem is\nthus veriﬁed.\nExample 15.6: Let α and k be constants with k ̸= 1. Consider the uncertain\ndiﬀerential equation\ndXt = αXtdt + Xk\nt dCt.\n(15.27)\nAt ﬁrst, we have\nYt = exp\n\u0012\n−\nZ t\n0\nαds\n\u0013\n= exp(−αt)\n\n\nSection 15.1 - Uncertain Differential Equation\n347\nand Zt satisﬁes the uncertain diﬀerential equation,\ndZt = exp(−αt)(exp(αt)Zt)kdCt = exp((k −1)αt)Zk\nt dCt.\nSince k ̸= 1, we have\ndZ1−k\nt\n= (1 −k)Z−k\nt\ndZt = (1 −k) exp((k −1)αt)dCt.\nIt follows from the fundamental theorem of uncertain calculus that\nZ1−k\nt\n= Z1−k\n0\n+ (1 −k)\nZ t\n0\nexp((k −1)αs)dCs.\nSince the initial value Z0 is just X0, we have\nZt =\n\u0012\nX1−k\n0\n+ (1 −k)\nZ t\n0\nexp((k −1)αs)dCs\n\u00131/(1−k)\n.\nTheorem 15.5 says the uncertain diﬀerential equation (15.27) has a solution\nXt = Y −1\nt\nZt, i.e.,\nXt = exp(αt)\n\u0012\nX1−k\n0\n+ (1 −k)\nZ t\n0\nexp((k −1)αs)dCs\n\u00131/(1−k)\n.\nTheorem 15.6 (Yao [255]) Let f be a continuous function of two variables,\nand let σt be a continuous function of t.\nThen the uncertain diﬀerential\nequation\ndXt = f(t, Xt)dt + σtdCt\n(15.28)\nhas a solution\nXt = Yt + Zt\n(15.29)\nwhere\nYt =\nZ t\n0\nσsdCs\n(15.30)\nand Zt is the solution of the uncertain diﬀerential equation\ndZt = f(t, Yt + Zt)dt\n(15.31)\nwith initial value Z0 = X0.\nProof: At ﬁrst, Yt has an uncertain diﬀerential dYt = σtdCt. It follows that\nd(Xt −Yt) = dXt −dYt = f(t, Xt)dt + σtdCt −σtdCt.\nThat is,\nd(Xt −Yt) = f(t, Xt)dt.\n\n\n348\nChapter 15 - Uncertain Differential Equation\nDeﬁning Zt = Xt −Yt, we obtain Xt = Yt + Zt and dZt = f(t, Yt + Zt)dt.\nFurthermore, since Y0 = 0, the initial value Z0 is just X0. The theorem is\nproved.\nExample 15.7: Let α and σ be constants with α ̸= 0. Consider the uncer-\ntain diﬀerential equation\ndXt = α exp(Xt)dt + σdCt.\n(15.32)\nAt ﬁrst, we have\nYt =\nZ t\n0\nσdCs = σCt\nand Zt satisﬁes the uncertain diﬀerential equation,\ndZt = α exp(σCt + Zt)dt.\nSince α ̸= 0, we have\nd exp(−Zt) = −exp(−Zt)dZt = −α exp(σCt)dt.\nIt follows from the fundamental theorem of uncertain calculus that\nexp(−Zt) = exp(−Z0) −α\nZ t\n0\nexp(σCs)ds.\nSince the initial value Z0 is just X0, we have\nZt = X0 −ln\n\u0012\n1 −α\nZ t\n0\nexp(X0 + σCs)ds\n\u0013\n.\nHence\nXt = X0 + σCt −ln\n\u0012\n1 −α\nZ t\n0\nexp(X0 + σCs)ds\n\u0013\n.\nTheorem 15.7 (Yao [255]) Let g be a continuous function of two variables,\nand let αt be a continuous function of t.\nThen the uncertain diﬀerential\nequation\ndXt = αtdt + g(t, Xt)dCt\n(15.33)\nhas a solution\nXt = Yt + Zt\n(15.34)\nwhere\nYt =\nZ t\n0\nαsds\n(15.35)\nand Zt is the solution of the uncertain diﬀerential equation\ndZt = g(t, Yt + Zt)dCt\n(15.36)\nwith initial value Z0 = X0.\n\n\nSection 15.1 - Uncertain Differential Equation\n349\nProof: The uncertain process Yt has an uncertain diﬀerential dYt = αtdt. It\nfollows that\nd(Xt −Yt) = dXt −dYt = αtdt + g(t, Xt)dCt −αtdt.\nThat is,\nd(Xt −Yt) = g(t, Xt)dCt.\nDeﬁning Zt = Xt −Yt, we obtain Xt = Yt + Zt and dZt = g(t, Yt + Zt)dCt.\nFurthermore, since Y0 = 0, the initial value Z0 is just X0. The theorem is\nproved.\nExample 15.8: Let α and σ be constants with σ ̸= 0. Consider the uncertain\ndiﬀerential equation\ndXt = αdt + σ exp(Xt)dCt.\n(15.37)\nAt ﬁrst, we have\nYt =\nZ t\n0\nαds = αt\nand Zt satisﬁes the uncertain diﬀerential equation,\ndZt = σ exp(αt + Zt)dCt.\nSince σ ̸= 0, we have\nd exp(−Zt) = −exp(−Zt)dZt = −σ exp(αt)dCt.\nIt follows from the fundamental theorem of uncertain calculus that\nexp(−Zt) = exp(−Z0) −σ\nZ t\n0\nexp(αs)dCs.\nSince the initial value Z0 is just X0, we have\nZt = X0 −ln\n\u0012\n1 −σ\nZ t\n0\nexp(X0 + αs)dCs\n\u0013\n.\nHence\nXt = X0 + αt −ln\n\u0012\n1 −σ\nZ t\n0\nexp(X0 + αs)dCs\n\u0013\n.\nExistence and Uniqueness Theorem\nTheorem 15.8 (Chen-Liu [8], Existence and Uniqueness Theorem) The un-\ncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt\n(15.38)\n\n\n350\nChapter 15 - Uncertain Differential Equation\nhas a unique solution if the coeﬃcients f(t, x) and g(t, x) satisfy the linear\ngrowth condition\n|f(t, x)| + |g(t, x)| ≤L(1 + |x|),\n∀x ∈ℜ, t ≥0\n(15.39)\nand Lipschitz condition\n|f(t, x) −f(t, y)| + |g(t, x) −g(t, y)| ≤L|x −y|,\n∀x, y ∈ℜ, t ≥0\n(15.40)\nfor some constant L. Moreover, the solution is sample-continuous.\nProof: We ﬁrst prove the existence of solution by a successive approximation\nmethod. Deﬁne X(0)\nt\n= X0, and\nX(n)\nt\n= X0 +\nZ t\n0\nf\n\u0010\ns, X(n−1)\ns\n\u0011\nds +\nZ t\n0\ng\n\u0010\ns, X(n−1)\ns\n\u0011\ndCs\nfor n = 1, 2, · · · and write\nD(n)\nt\n(γ) = max\n0≤s≤t\n\f\n\f\n\fX(n+1)\ns\n(γ) −X(n)\ns\n(γ)\n\f\n\f\n\f\nfor each γ ∈Γ. It follows from the linear growth condition and Lipschitz\ncondition that\nD(0)\nt (γ) = max\n0≤s≤t\n\f\n\f\n\f\n\f\nZ s\n0\nf(v, X0)dv +\nZ s\n0\ng(v, X0)dCv(γ)\n\f\n\f\n\f\n\f\n≤\nZ t\n0\n|f(v, X0)| dv + Kγ\nZ t\n0\n|g(v, X0)| dv\n≤(1 + |X0|)L(1 + Kγ)t\nwhere Kγ is the Lipschitz constant to the sample path Ct(γ). In fact, by\nusing the induction method, we may verify\nD(n)\nt\n(γ) ≤(1 + |X0|)Ln+1(1 + Kγ)n+1\n(n + 1)!\ntn+1\nfor each n. This means that, for each γ ∈Γ, the sequence X(n)\nt\n(γ) converges\nuniformly on any given time interval as n →∞. Write the limit by Xt(γ)\nthat is just a solution of the uncertain diﬀerential equation because\nXt = X0 +\nZ t\n0\nf(s, Xs)ds +\nZ t\n0\ng(s, Xs)dCs.\nNext we prove that the solution is unique. Assume that both Xt and X∗\nt\nare solutions of the uncertain diﬀerential equation. Then for each γ ∈Γ, it\nfollows from the linear growth condition and Lipschitz condition that\n|Xt(γ) −X∗\nt (γ)| ≤L(1 + Kγ)\nZ t\n0\n|Xv(γ) −X∗\nv(γ)|dv.\n\n\nSection 15.1 - Uncertain Differential Equation\n351\nBy using Gronwall inequality, we obtain\n|Xt(γ) −X∗\nt (γ)| ≤0 · exp(L(1 + Kγ)t) = 0.\nHence Xt = X∗\nt . The uniqueness is veriﬁed. Finally, for each γ ∈Γ, we have\n|Xt(γ) −Xr(γ)| =\n\f\n\f\n\f\n\f\nZ t\nr\nf(s, Xs(γ))ds +\nZ t\nr\ng(s, Xs(γ))dCs(γ)\n\f\n\f\n\f\n\f →0\nas r →t. Thus Xt is sample-continuous and the theorem is proved.\nStability\nDeﬁnition 15.2 (Liu [116]) An uncertain diﬀerential equation is said to be\nstable if for any two solutions Xt and Yt, we have\nlim\n|X0−Y0|→0 M{|Xt −Yt| < ε for all t ≥0} = 1\n(15.41)\nfor any given number ε > 0.\nExample 15.9: In order to illustrate the concept of stability, let us consider\nthe uncertain diﬀerential equation\ndXt = adt + bdCt.\n(15.42)\nIt is clear that two solutions with initial values X0 and Y0 are\nXt = X0 + at + bCt,\nYt = Y0 + at + bCt.\nThen for any given number ε > 0, we have\nlim\n|X0−Y0|→0 M{|Xt −Yt| < ε for all t ≥0} =\nlim\n|X0−Y0|→0 M{|X0 −Y0| < ε} = 1.\nHence the uncertain diﬀerential equation (15.42) is stable.\nExample 15.10: Some uncertain diﬀerential equations are not stable. For\nexample, consider\ndXt = Xtdt + bdCt.\n(15.43)\nIt is clear that two solutions with diﬀerent initial values X0 and Y0 are\nXt = exp(t)X0 + b exp(t)\nZ t\n0\nexp(−s)dCs,\nYt = exp(t)Y0 + b exp(t)\nZ t\n0\nexp(−s)dCs.\n\n\n352\nChapter 15 - Uncertain Differential Equation\nThen for any given number ε > 0, we have\nlim\n|X0−Y0|→0 M{|Xt −Yt| < ε for all t ≥0}\n=\nlim\n|X0−Y0|→0 M{exp(t)|X0 −Y0| < ε for all t ≥0} = 0.\nHence the uncertain diﬀerential equation (15.43) is unstable.\nTheorem 15.9 (Yao-Gao-Gao [251], Stability Theorem) The uncertain dif-\nferential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt\n(15.44)\nis stable if the coeﬃcients f(t, x) and g(t, x) satisfy the linear growth condition\n|f(t, x)| + |g(t, x)| ≤K(1 + |x|),\n∀x ∈ℜ, t ≥0\n(15.45)\nfor some constant K and strong Lipschitz condition\n|f(t, x) −f(t, y)| + |g(t, x) −g(t, y)| ≤L(t)|x −y|,\n∀x, y ∈ℜ, t ≥0 (15.46)\nfor some bounded and integrable function L(t) on [0, +∞).\nProof: Since L(t) is bounded on [0, +∞), there is a constant R such that\nL(t) ≤R for any t. Then the strong Lipschitz condition (15.46) implies the\nfollowing Lipschitz condition,\n|f(t, x) −f(t, y)| + |g(t, x) −g(t, y)| ≤R|x −y|,\n∀x, y ∈ℜ, t ≥0. (15.47)\nIt follows from the linear growth condition (15.45), the Lipschitz condition\n(15.47) and the existence and uniqueness theorem that the uncertain diﬀer-\nential equation (15.44) has a unique solution. Let Xt and Yt be two solutions\nwith initial values X0 and Y0, respectively. Then for each γ, we have\nd|Xt(γ) −Yt(γ)|\n≤|f(t, Xt(γ)) −f(t, Yt(γ))|dt + |g(t, Xt(γ)) −g(t, Yt(γ))||dCt(γ)|\n≤L(t)|Xt(γ) −Yt(γ)|dt + L(t)K(γ)|Xt(γ) −Yt(γ)|dt\n= L(t)(1 + K(γ))|Xt(γ) −Yt(γ)|dt\nwhere K(γ) is the Lipschitz constant of the sample path Ct(γ). It follows\nthat\n|Xt(γ) −Yt(γ)| ≤|X0 −Y0| exp\n\u0012\n(1 + K(γ))\nZ +∞\n0\nL(s)ds\n\u0013\n.\nThus for any given ε > 0, we always have\nM{|Xt −Yt| < ε for all t ≥0}\n≥M\n\u001a\n|X0 −Y0| exp\n\u0012\n(1 + K(γ))\nZ +∞\n0\nL(s)ds\n\u0013\n< ε\n\u001b\n.\n\n\nSection 15.2 - α-Path\n353\nSince\nM\n\u001a\n|X0 −Y0| exp\n\u0012\n(1 + K(γ))\nZ +∞\n0\nL(s)ds\n\u0013\n< ε\n\u001b\n→1\nas |X0 −Y0| →0, we obtain\nlim\n|X0−Y0|→0 M{|Xt −Yt| < ε for all t ≥0} = 1.\nHence the uncertain diﬀerential equation is stable.\nExercise 15.1: Suppose u1t, u2t, v1t, v2t are bounded and continuous func-\ntions of t such that\nZ +∞\n0\n|u1t|dt < +∞,\nZ +∞\n0\n|v1t|dt < +∞.\n(15.48)\nShow that the linear uncertain diﬀerential equation\ndXt = (u1tXt + u2t)dt + (v1tXt + v2t)dCt\n(15.49)\nis stable.\n15.2\nα-Path\nDeﬁnition 15.3 (Yao-Chen [254]) Let α be a number between 0 and 1. An\nuncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt\n(15.50)\nis said to have an α-path Xα\nt if it solves the corresponding ordinary diﬀeren-\ntial equation\ndXα\nt = f(t, Xα\nt )dt + |g(t, Xα\nt )|Φ−1(α)dt\n(15.51)\nwhere Φ−1(α) is the inverse standard normal uncertainty distribution, i.e.,\nΦ−1(α) =\n√\n3\nπ ln\nα\n1 −α.\n(15.52)\nRemark 15.2: Note that each α-path Xα\nt is a real-valued function of time\nt, but is not necessarily one of sample paths. Furthermore, all α-paths are\ncontinuous functions with respect to time t.\nExample 15.11:\nAssume a and b are constants with b > 0.\nThen the\nuncertain diﬀerential equation\ndXt = adt + bdCt\n(15.53)\n\n\n354\nChapter 15 - Uncertain Differential Equation\nhas an α-path Xα\nt that solves the corresponding ordinary diﬀerential equation\ndXα\nt = adt +\n√\n3b\nπ\nln\nα\n1 −αdt.\n(15.54)\nThus\nXα\nt = X0 +\nZ t\n0\nads +\nZ t\n0\n√\n3b\nπ\nln\nα\n1 −αds.\nThat is,\nXα\nt = X0 + at +\n√\n3bt\nπ\nln\nα\n1 −α.\n(15.55)\nExercise 15.2: Assume e and σ are constants with σ > 0. Show that the\nuncertain diﬀerential equation\ndXt = eXtdt + σXtdCt\n(15.56)\nhas an α-path\nXα\nt = X0 exp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!\n(15.57)\nprovided that X0 > 0.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nXα\nt\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.3\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.4\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.5\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.6\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.7\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.8\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. α = 0.9\nFigure 15.1: A Spectrum of α-Paths of dXt = aXtdt + bXtdCt\nExercise 15.3: Assume m, a and σ are constants with a ̸= 0 and σ > 0.\nShow that the uncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt\n(15.58)\n\n\nSection 15.3 - Yao-Chen Formula\n355\nhas an α-path\nXα\nt = m\na + exp(−at)\n\u0010\nX0 −m\na\n\u0011\n+\n√\n3σ(1 −exp(−at))\nπa\nln\nα\n1 −α.\n(15.59)\nExercise 15.4: Assume m, a and σ are constants with m > 0 and σ > 0.\nShow that the uncertain diﬀerential equation\ndXt = (m −aXt)dt + σXtdCt\n(15.60)\nhas an α-path\nXα\nt = exp(−ν(α)t)X0 + m(1 −exp(−ν(α)t))\nν(α)\n(15.61)\nwhere\nν(α) = a −\n√\n3σ\nπ\nln\nα\n1 −α\n(15.62)\nprovided that X0 > 0. Note that Xα\nt = X0 + mt when ν(α) = 0.\nTheorem 15.10 Let f and g be continuous functions with g ̸= 0, and let\nXα\nt be the α-path of the uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.63)\nrespectively. Then Xα\nt is a strictly increasing function with respect to α at\neach time t > 0.\nProof: Let α and β be two numbers with 0 < α < β < 1, and let Φ−1 be\nthe inverse standard normal uncertainty distribution. Then the α-path Xα\nt\nand the β-path Xβ\nt solve the ordinary diﬀerential equations\ndYt = f(t, Yt)dt + |g(t, Yt)|Φ−1(α)dt,\nY0 = X0\nand\ndYt = f(t, Yt)dt + |g(t, Yt)|Φ−1(β)dt,\nY0 = X0,\nrespectively. Since Φ−1(α) < Φ−1(β) and g ̸= 0, we have\nf(t, Yt) + |g(t, Yt)|Φ−1(α) < f(t, Yt) + |g(t, Yt)|Φ−1(β)\nfor any (t, Yt). It follows from the comparison theorem1 that\nXα\nt < Xβ\nt ,\n∀t > 0.\nThe theorem is veriﬁed.\n1Comparison Theorem:\nLet h(t, x) and H(t, x) be continuous functions such that\nh(t, x) < H(t, x) for any (t, x). If zt and Zt solve the ordinary diﬀerential equations\ndYt = h(t, Yt)dt,\nY0 = X0\nand\ndYt = H(t, Yt)dt,\nY0 = X0,\nrespectively, then zt < Zt for any t > 0.\n\n\n356\nChapter 15 - Uncertain Differential Equation\n15.3\nYao-Chen Formula\nYao-Chen formula relates uncertain diﬀerential equations and ordinary dif-\nferential equations, just like that Feynman-Kac formula relates stochastic\ndiﬀerential equations and partial diﬀerential equations.\nTheorem 15.11 (Yao-Chen Formula [254]) Let Xt and Xα\nt be the solution\nand α-path of the uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.64)\nrespectively. Then\nM{Xt ≤Xα\nt , ∀t} = α,\n(15.65)\nM{Xt > Xα\nt , ∀t} = 1 −α.\n(15.66)\nProof: At ﬁrst, for each α-path Xα\nt , we divide the time interval into two\nparts,\nT + =\n\b\nt\n\f\n\f g (t, Xα\nt ) ≥0\n\t\n,\nT −=\n\b\nt\n\f\n\f g (t, Xα\nt ) < 0\n\t\n.\nIt is obvious that T + ∩T −= ∅and T + ∪T −= [0, +∞). Write\nΛ+\n1 =\n\u001a\nγ\n\f\n\f dCt(γ)\ndt\n≤Φ−1(α) for any t ∈T +\n\u001b\n,\nΛ−\n1 =\n\u001a\nγ\n\f\n\f dCt(γ)\ndt\n≥Φ−1(1 −α) for any t ∈T −\n\u001b\nwhere Φ−1 is the inverse standard normal uncertainty distribution. Since T +\nand T −are disjoint sets and Ct has independent increments, we get\nM{Λ+\n1 } = α,\nM{Λ−\n1 } = α,\nM{Λ+\n1 ∩Λ−\n1 } = α.\nFor any γ ∈Λ+\n1 ∩Λ−\n1 , we always have\ng(t, Xt(γ))dCt(γ)\ndt\n≤|g(t, Xα\nt )|Φ−1(α), ∀t.\nHence Xt(γ) ≤Xα\nt for all t and\nM{Xt ≤Xα\nt , ∀t} ≥M{Λ+\n1 ∩Λ−\n1 } = α.\n(15.67)\nOn the other hand, let us deﬁne\nΛ+\n2 =\n\u001a\nγ\n\f\n\f dCt(γ)\ndt\n> Φ−1(α) for any t ∈T +\n\u001b\n,\nΛ−\n2 =\n\u001a\nγ\n\f\n\f dCt(γ)\ndt\n< Φ−1(1 −α) for any t ∈T −\n\u001b\n.\n\n\nSection 15.3 - Yao-Chen Formula\n357\nSince T + and T −are disjoint sets and Ct has independent increments, we\nobtain\nM{Λ+\n2 } = 1 −α,\nM{Λ−\n2 } = 1 −α,\nM{Λ+\n2 ∩Λ−\n2 } = 1 −α.\nFor any γ ∈Λ+\n2 ∩Λ−\n2 , we always have\ng(t, Xt(γ))dCt(γ)\ndt\n> |g(t, Xα\nt )|Φ−1(α), ∀t.\nHence Xt(γ) > Xα\nt for all t and\nM{Xt > Xα\nt , ∀t} ≥M{Λ+\n2 ∩Λ−\n2 } = 1 −α.\n(15.68)\nNote that {Xt ≤Xα\nt , ∀t} and {Xt ̸≤Xα\nt , ∀t} are opposite events with each\nother. By using the duality axiom, we obtain\nM{Xt ≤Xα\nt , ∀t} + M{Xt ̸≤Xα\nt , ∀t} = 1.\nIt follows from {Xt > Xα\nt , ∀t} ⊂{Xt ̸≤Xα\nt , ∀t} and monotonicity theorem\nthat\nM{Xt ≤Xα\nt , ∀t} + M{Xt > Xα\nt , ∀t} ≤1.\n(15.69)\nThus (15.65) and (15.66) follow from (15.67), (15.68) and (15.69) immedi-\nately.\nRemark 15.3: Please mention that {Xt ≤Xα\nt , ∀t} and {Xt > Xα\nt , ∀t}\nare disjoint events but not opposite. That is, their union does not make the\nuniversal set. However, we always have\nM{Xt ≤Xα\nt , ∀t} + M{Xt > Xα\nt , ∀t} ≡1.\n(15.70)\nRemark 15.4: It is also showed that for any α ∈(0, 1), the following two\nequations are true,\nM{Xt < Xα\nt , ∀t} = α,\n(15.71)\nM{Xt ≥Xα\nt , ∀t} = 1 −α.\n(15.72)\nUncertainty Distribution of Solution\nTheorem 15.12 (Yao-Chen [254]) Let Xt and Xα\nt be the solution and α-\npath of the uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.73)\nrespectively. Then Xt has an inverse uncertainty distribution\nΨ−1\nt (α) = Xα\nt .\n(15.74)\n\n\n358\nChapter 15 - Uncertain Differential Equation\nProof: Note that {Xt ≤Xα\nt } ⊃{Xs ≤Xα\ns , ∀s} holds for each t. By using\nthe monotonicity theorem and Yao-Chen formula, we obtain\nM{Xt ≤Xα\nt } ≥M{Xs ≤Xα\ns , ∀s} = α.\n(15.75)\nSimilarly, we also have\nM{Xt > Xα\nt } ≥M{Xs > Xα\ns , ∀s} = 1 −α.\n(15.76)\nSince {Xt ≤Xα\nt } and {Xt > Xα\nt } are opposite events for each t, the duality\naxiom makes\nM{Xt ≤Xα\nt } + M{Xt > Xα\nt } = 1.\n(15.77)\nIt follows from (15.75), (15.76) and (15.77) that M{Xt ≤Xα\nt } = α. Thus\nΨ−1\nt (α) = Xα\nt is the inverse uncertainty distribution of Xt.\nExercise 15.5: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively, and let J be a continuous and strictly\nincreasing function. Show that J(Xt) has an inverse uncertainty distribution\nΨ−1\nt (α) = J(Xα\nt ).\n(15.78)\nExercise 15.6: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively, and let J be a continuous and strictly\ndecreasing function. Show that J(Xt) has an inverse uncertainty distribution\nΨ−1\nt (α) = J(X1−α\nt\n).\n(15.79)\nExpected Value of Solution\nTheorem 15.13 (Yao-Chen [254]) Let Xt and Xα\nt be the solution and α-\npath of the uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.80)\nrespectively. Then\nE[Xt] =\nZ 1\n0\nXα\nt dα.\n(15.81)\nProof: Yao-Chen formula says that Xt has an inverse uncertainty distribu-\ntion Ψ−1\nt (α) = Xα\nt . It follows from Theorem 3.24 that (15.81) holds.\nExercise 15.7: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively, and let J be a continuous and monotone\n(increasing or decreasing) function. Show that\nE[J(Xt)] =\nZ 1\n0\nJ(Xα\nt )dα.\n(15.82)\n\n\nSection 15.3 - Yao-Chen Formula\n359\nExtreme Value of Solution\nTheorem 15.14 (Yao [252]) Let Xt and Xα\nt be the solution and α-path of\nthe uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.83)\nrespectively. Then for any time s > 0, the supremum\nsup\n0≤t≤s\nXt\n(15.84)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) = sup\n0≤t≤s\nXα\nt ;\n(15.85)\nand the inﬁmum\ninf\n0≤t≤s Xt\n(15.86)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\ninf\n0≤t≤s Xα\nt .\n(15.87)\nProof:\nFor any given time s > 0, it follows from the basic property of\nextreme value that\n\u001a\nsup\n0≤t≤s\nXt ≤sup\n0≤t≤s\nXα\nt\n\u001b\n⊃{Xt ≤Xα\nt , ∀t}.\nBy using Yao-Chen formula, we obtain\nM\n\u001a\nsup\n0≤t≤s\nXt ≤sup\n0≤t≤s\nXα\nt\n\u001b\n≥M{Xt ≤Xα\nt , ∀t} = α.\n(15.88)\nSimilarly, we have\nM\n\u001a\nsup\n0≤t≤s\nXt > sup\n0≤t≤s\nXα\nt\n\u001b\n≥M{Xt > Xα\nt , ∀t} = 1 −α.\n(15.89)\nIt follows from (15.88), (15.89) and the duality axiom that\nM\n\u001a\nsup\n0≤t≤s\nXt ≤sup\n0≤t≤s\nXα\nt\n\u001b\n= α\n(15.90)\nwhich proves (15.85). Next, it follows from the basic property of extreme\nvalue that\n\u001a\ninf\n0≤t≤s Xt ≤\ninf\n0≤t≤s Xα\nt\n\u001b\n⊃{Xt ≤Xα\nt , ∀t}.\n\n\n360\nChapter 15 - Uncertain Differential Equation\nBy using Yao-Chen formula, we obtain\nM\n\u001a\ninf\n0≤t≤s Xt ≤\ninf\n0≤t≤s Xα\nt\n\u001b\n≥M{Xt ≤Xα\nt , ∀t} = α.\n(15.91)\nSimilarly, we have\nM\n\u001a\ninf\n0≤t≤s Xt >\ninf\n0≤t≤s Xα\nt\n\u001b\n≥M{Xt > Xα\nt , ∀t} = 1 −α.\n(15.92)\nIt follows from (15.91), (15.92) and the duality axiom that\nM\n\u001a\ninf\n0≤t≤s Xt ≤\ninf\n0≤t≤s Xα\nt\n\u001b\n= α\n(15.93)\nwhich proves (15.87). The theorem is thus veriﬁed.\nExercise 15.8: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively.\nAssume J is a continuous and strictly\nincreasing function. For any time s > 0, show that the supremum\nsup\n0≤t≤s\nJ(Xt)\n(15.94)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) = sup\n0≤t≤s\nJ(Xα\nt );\n(15.95)\nand the inﬁmum\ninf\n0≤t≤s J(Xt)\n(15.96)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\ninf\n0≤t≤s J(Xα\nt ).\n(15.97)\nExercise 15.9: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively.\nAssume J is a continuous and strictly\ndecreasing function. For any time s > 0, show that the supremum\nsup\n0≤t≤s\nJ(Xt)\n(15.98)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) = sup\n0≤t≤s\nJ(X1−α\nt\n);\n(15.99)\nand the inﬁmum\ninf\n0≤t≤s J(Xt)\n(15.100)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\ninf\n0≤t≤s J(X1−α\nt\n).\n(15.101)\n\n\nSection 15.3 - Yao-Chen Formula\n361\nFirst Hitting Time of Solution\nTheorem 15.15 (Yao [252]) Let Xt and Xα\nt be the solution and α-path of\nthe uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.102)\nrespectively.\nThen for any given level z, the ﬁrst hitting time τz that Xt\nreaches z has an uncertainty distribution\nΨ(s) =\n\n\n\n\n\n\n\n\n\n1 −inf\n\u001a\nα\n\f\n\f sup\n0≤t≤s\nXα\nt ≥z\n\u001b\n,\nif z > X0\nsup\n\u001a\nα\n\f\n\f\ninf\n0≤t≤s Xα\nt ≤z\n\u001b\n,\nif z < X0.\n(15.103)\nProof: At ﬁrst, assume z > X0 and write\nα0 = inf\n\u001a\nα\n\f\n\f sup\n0≤t≤s\nXα\nt ≥z\n\u001b\n.\nThen\nsup\n0≤t≤s\nXα0\nt\n= z,\n{τz ≤s} =\n\u001a\nsup\n0≤t≤s\nXt ≥z\n\u001b\n⊃{Xt ≥Xα0\nt , ∀t},\n{τz > s} =\n\u001a\nsup\n0≤t≤s\nXt < z\n\u001b\n⊃{Xt < Xα0\nt , ∀t}.\nBy using Yao-Chen formula, we obtain\nM{τz ≤s} ≥M{Xt ≥Xα0\nt , ∀t} = 1 −α0,\nM{τz > s} ≥M{Xt < Xα0\nt , ∀t} = α0.\nIt follows from M{τz ≤s} + M{τz > s} = 1 that M{τz ≤s} = 1 −α0. Hence\nthe ﬁrst hitting time τz has an uncertainty distribution\nΨ(s) = M{τz ≤s} = 1 −α0 = 1 −inf\n\u001a\nα\n\f\n\f sup\n0≤t≤s\nXα\nt ≥z\n\u001b\n.\nSimilarly, assume z < X0 and write\nα0 = sup\n\u001a\nα\n\f\n\f\ninf\n0≤t≤s Xα\nt ≤z\n\u001b\n.\nThen\ninf\n0≤t≤s Xα0\nt\n= z,\n\n\n362\nChapter 15 - Uncertain Differential Equation\n{τz ≤s} =\n\u001a\ninf\n0≤t≤s Xt ≤z\n\u001b\n⊃{Xt ≤Xα0\nt , ∀t},\n{τz > s} =\n\u001a\ninf\n0≤t≤s Xt > z\n\u001b\n⊃{Xt > Xα0\nt , ∀t}.\nBy using Yao-Chen formula, we obtain\nM{τz ≤s} ≥M{Xt ≤Xα0\nt , ∀t} = α0,\nM{τz > s} ≥M{Xt > Xα0\nt , ∀t} = 1 −α0.\nIt follows from M{τz ≤s} + M{τz > s} = 1 that M{τz ≤s} = α0. Hence\nthe ﬁrst hitting time τz has an uncertainty distribution\nΨ(s) = M{τz ≤s} = α0 = sup\n\u001a\nα\n\f\n\f\ninf\n0≤t≤s Xα\nt ≤z\n\u001b\n.\nThe theorem is veriﬁed.\nExercise 15.10: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively.\nAssume J is a continuous and strictly\nincreasing function. For any given level z, show that the ﬁrst hitting time τz\nthat J(Xt) reaches z has an uncertainty distribution\nΨ(s) =\n\n\n\n\n\n\n\n\n\n1 −inf\n\u001a\nα\n\f\n\f sup\n0≤t≤s\nJ(Xα\nt ) ≥z\n\u001b\n,\nif z > J(X0)\nsup\n\u001a\nα\n\f\n\f\ninf\n0≤t≤s J(Xα\nt ) ≤z\n\u001b\n,\nif z < J(X0).\n(15.104)\nExercise 15.11: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively.\nAssume J is a continuous and strictly\ndecreasing function. For any given level z, show that the ﬁrst hitting time τz\nthat J(Xt) reaches z has an uncertainty distribution\nΨ(s) =\n\n\n\n\n\n\n\n\n\nsup\n\u001a\nα\n\f\n\f sup\n0≤t≤s\nJ(Xα\nt ) ≥z\n\u001b\n,\nif z > J(X0)\n1 −inf\n\u001a\nα\n\f\n\f\ninf\n0≤t≤s J(Xα\nt ) ≤z\n\u001b\n,\nif z < J(X0).\n(15.105)\nTime Integral of Solution\nTheorem 15.16 (Yao [252]) Let Xt and Xα\nt be the solution and α-path of\nthe uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\n(15.106)\n\n\nSection 15.4 - Numerical Solution\n363\nrespectively. Then for any time s > 0, the time integral\nZ s\n0\nXtdt\n(15.107)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nXα\nt dt.\n(15.108)\nProof: For any given time s > 0, it follows from the basic property of time\nintegral that\n\u001aZ s\n0\nXtdt ≤\nZ s\n0\nXα\nt dt\n\u001b\n⊃{Xt ≤Xα\nt , ∀t}.\nBy using Yao-Chen formula, we obtain\nM\n\u001aZ s\n0\nXtdt ≤\nZ s\n0\nXα\nt dt\n\u001b\n≥M{Xt ≤Xα\nt , ∀t} = α.\n(15.109)\nSimilarly, we have\nM\n\u001aZ s\n0\nXtdt >\nZ s\n0\nXα\nt dt\n\u001b\n≥M{Xt > Xα\nt , ∀t} = 1 −α.\n(15.110)\nIt follows from (15.109), (15.110) and the duality axiom that\nM\n\u001aZ s\n0\nXtdt ≤\nZ s\n0\nXα\nt dt\n\u001b\n= α.\n(15.111)\nThe theorem is thus veriﬁed.\nExercise 15.12: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively.\nAssume J is a continuous and strictly\nincreasing function. For any time s > 0, show that the time integral\nZ s\n0\nJ(Xt)dt\n(15.112)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nJ(Xα\nt )dt.\n(15.113)\nExercise 15.13: Let Xt and Xα\nt be the solution and α-path of an uncertain\ndiﬀerential equation, respectively.\nAssume J is a continuous and strictly\ndecreasing function. For any time s > 0, show that the time integral\nZ s\n0\nJ(Xt)dt\n(15.114)\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nJ(X1−α\nt\n)dt.\n(15.115)\n\n\n364\nChapter 15 - Uncertain Differential Equation\n15.4\nNumerical Solution\nIt is almost impossible to ﬁnd analytic solutions for general uncertain diﬀer-\nential equations. This fact provides a motivation to design some numerical\nsolution methods for uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt.\n(15.116)\nIn order to do so, a key point is to obtain a spectrum of α-paths of the\nuncertain diﬀerential equation. For this purpose, Yao-Chen [254] designed\nan Euler method:\nStep 1. Fix α on (0, 1).\nStep 2. Solve dXα\nt = f(t, Xα\nt )dt+|g(t, Xα\nt )|Φ−1(α)dt by any method of or-\ndinary diﬀerential equation and obtain the α-path Xα\nt , for example,\nby using the recursion formula\nXα\nti+1 = Xα\nti + f(ti, Xα\nti)h + |g(ti, Xα\nti)|Φ−1(α)h\n(15.117)\nwhere Φ−1 is the inverse standard normal uncertainty distribution\nand h is the step length.\nStep 3. The α-path Xα\nt is obtained.\nExercise 15.14: Employ Euler method to solve the uncertain diﬀerential\nequation\ndXt = (t −Xt)dt + tXtdCt,\nX0 = 1.\n(15.118)\n(i) Plot the uncertainty distribution of the solution Xt at t = 2. (ii) Calculate\nthe expected value of (3−X2)+. (iii) Plot the uncertainty distribution of the\nextreme value\nsup\n0≤t≤2\n(1 + t)Xt.\n(15.119)\n(iv) Plot the uncertainty distribution of the ﬁrst hitting time that exp(Xt)\nreaches 5. (v) Plot the uncertainty distribution of the time integral\nZ 2\n0\nexp(−t)(Xt −1)+dt.\n(15.120)\n15.5\nResidual\nAssume an uncertain process follows an uncertain diﬀerential equation and\nsome realizations of this process are observed. In order to make a connec-\ntion between uncertain diﬀerential equation and observed data, Liu-Liu [144]\nproposed the concept of residual. Consider an uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt\n(15.121)\n\n\nSection 15.5 - Residual\n365\nwhere f and g are known continuous functions, and Ct is a Liu process.\nAssume\nxt1, xt2, · · · , xtn\n(15.122)\nare observed values of the uncertain process Xt at the times t1, t2, · · · , tn\nwith t1 < t2 < · · · < tn, respectively. For each i (2 ≤i ≤n), let us ﬁrst solve\nthe updated uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt,\nXti−1 = xti−1\n(15.123)\nwhere xti−1 is set as the initial value at the initial time ti−1. The uncer-\ntainty distribution of Xti is thus obtained and represented by Φti. Note that\nΦti(Xti) is always a linear uncertain variable L(0, 1). Substitute Xti with\nthe corresponding observed value xti, and write\nεi = Φti(xti).\n(15.124)\nThen εi may be regarded as a sample of the linear uncertain variable Φti(Xti).\nIn other words, εi is always a sample of linear uncertainty distribution L(0, 1),\ni.e.,\nεi ∼L(0, 1).\n(15.125)\nDeﬁnition 15.4 (Liu-Liu [144]) For each i (2 ≤i ≤n), the term εi deﬁned\nby (15.124) is called the ith residual of the uncertain diﬀerential equation\n(15.121) corresponding to the observed data (15.122).\nExample 15.12: Assume xt1, xt2, · · · , xtn are observed values of some un-\ncertain process Xt that follows the uncertain diﬀerential equation\ndXt = µXtdt + σXtdCt\n(15.126)\nwhere µ and σ are constants. For each i (2 ≤i ≤n), we solve the updated\nuncertain diﬀerential equation\ndXt = µXtdt + σXtdCt,\nXti−1 = xti−1\n(15.127)\nand obtain the uncertainty distribution of Xti as follows,\nΦti(x) =\n\u0012\n1 + exp\n\u0012π(µ(ti −ti−1) −ln x + ln xti−1)\n√\n3σ(ti −ti−1)\n\u0013\u0013−1\n.\n(15.128)\nIt follows from Deﬁnition 15.4 that the ith residual is\nεi =\n\u0012\n1 + exp\n\u0012π(µ(ti −ti−1) −ln xti + ln xti−1)\n√\n3σ(ti −ti−1)\n\u0013\u0013−1\n.\n(15.129)\n\n\n366\nChapter 15 - Uncertain Differential Equation\nExample 15.13: Assume xt1, xt2, · · · , xtn are observed values of some un-\ncertain process Xt that follows the uncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt\n(15.130)\nwhere m, a and σ are constants. For each i (2 ≤i ≤n), we solve the updated\nuncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt,\nXti−1 = xti−1\n(15.131)\nand obtain the uncertainty distribution of Xti as follows,\nΦti(x) =\n\u0012\n1 + exp\n\u0012π((axti−1 −m) exp(a(ti−1 −ti)) + m −ax)\n√\n3σ(1 −exp(a(ti−1 −ti)))\n\u0013\u0013−1\n.\nIt follows from Deﬁnition 15.4 that the ith residual is\nεi =\n\u0012\n1 + exp\n\u0012π((axti−1 −m) exp(a(ti−1 −ti)) + m −axti)\n√\n3σ(1 −exp(a(ti−1 −ti)))\n\u0013\u0013−1\n.\nFor the general uncertain diﬀerential equation (15.121), there do not exist\nexplicit formulas like (15.129). In order to calculate each residual εi (2 ≤i ≤\nn) numerically, Liu-Liu [144] suggested a bisection method:\nStep 0. Set l = 0, r = 1 and a precision δ = 0.0001.\nStep 1. Set α = (l + r)/2.\nStep 2. Employ Euler method to calculate Xα\nti of the updated uncertain\ndiﬀerential equation (15.123).\nStep 3. If Xα\nti < xti, then l = α. Otherwise, r = α.\nStep 4. If |l −r| > δ, then go to Step 1.\nStep 5. Output εi = (l + r)/2.\nExample 15.14: Let us use the above algorithm to calculate the residuals\nof the uncertain diﬀerential equation\ndXt = (t −Xt)dt + XtdCt\n(15.132)\ncorresponding to the 8 observed data\nt\n0.00\n1.22\n2.13\n3.52\n4.63\n5.98\n7.87\n9.00\nXt\n1.00\n0.96\n2.54\n6.52\n5.60\n5.21\n5.00\n7.95\non the time horizon from 0 to 9. A run of the algorithm shows that the 7\nresiduals are\n0.611, 0.868, 0.862, 0.575, 0.467, 0.300, 0.554.\n(15.133)\n\n\nSection 15.6 - Uncertain Hypothesis Test\n367\nExercise 15.15: Calculate all residuals of the uncertain diﬀerential equation\ndXt = ln Xtdt + 2XtdCt\n(15.134)\ncorresponding to the 8 observed data\nt\n0.00\n1.97\n3.21\n4.86\n5.65\n7.47\n7.94\n9.00\nXt\n2.00\n21.4\n45.2\n112\n78.5\n157\n189\n21.0\non the time horizon from 0 to 9.\n15.6\nUncertain Hypothesis Test\nOne of the core problems in practice is how to test whether an uncertain dif-\nferential equation ﬁts the observed data of some uncertain process. Consider\nan uncertain diﬀerential equation\ndXt = f(t, Xt)dt + g(t, Xt)dCt\n(15.135)\nwhere f and g are known continuous functions, and Ct is a Liu process.\nAssume\nxt1, xt2, · · · , xtn\n(15.136)\nare observed values of some uncertain process Xt at the times t1, t2, · · · , tn\nwith t1 < t2 < · · · < tn, respectively. Using Deﬁnition 15.4, we may produce\nn −1 residuals\nε2, ε3, · · · , εn\n(15.137)\nof the uncertain diﬀerential equation (15.135) corresponding to the observed\ndata (15.136).\nIn order to test whether the uncertain diﬀerential equation (15.135) ﬁts\nthe observed data (15.136), we should test whether the linear uncertainty\ndistribution L(0, 1) ﬁts the n −1 residuals ε2, ε3, · · · , εn, i.e.,\nε2, ε3, · · · , εn ∼L(0, 1).\n(15.138)\nIn order to do so, Ye-Liu [280][282] suggested using uncertain hypothesis test.\nGiven a signiﬁcance level α (e.g. 0.05), it follows from Theorem 4.4 that the\ntest is\nW =\n\u001a\n(z2, z3, · · · , zn) : there are more than α of indexes i’s\nwith 2 ≤i ≤n such that zi < α\n2 or zi > 1 −α\n2\n\u001b\n.\nIf the vector of the n −1 residuals ε2, ε3, · · · , εn belongs to the test W, i.e.,\n(ε2, ε3, · · · , εn) ∈W,\n(15.139)\n\n\n368\nChapter 15 - Uncertain Differential Equation\nthen the uncertain diﬀerential equation (15.135) is not a good ﬁt to the\nobserved data (15.136). If\n(ε2, ε3, · · · , εn) ̸∈W,\n(15.140)\nthen the uncertain diﬀerential equation (15.135) is a good ﬁt to the observed\ndata (15.136).\nExercise 15.16: Employ the uncertain hypothesis test to determine whether\nthe uncertain diﬀerential equation\ndXt = Xtdt + 2XtdCt\n(15.141)\nﬁts the 30 observed data\nt\n0.00\n0.12\n0.18\n0.30\n0.39\n0.51\n0.63\n0.72\n0.87\n0.93\nXt\n1.00\n1.09\n1.35\n1.30\n1.75\n1.28\n1.75\n2.80\n2.30\n2.54\nt\n1.02\n1.08\n1.23\n1.35\n1.47\n1.59\n1.74\n1.89\n2.04\n2.16\nXt\n2.17\n2.80\n2.31\n3.22\n2.51\n3.77\n3.49\n4.38\n4.29\n4.93\nt\n2.28\n2.40\n2.49\n2.61\n2.70\n2.76\n2.91\n3.00\n3.06\n3.12\nXt\n5.23\n5.46\n6.49\n6.89\n7.76\n8.22\n8.45\n9.22\n9.50\n9.94\non the time horizon from 0 to 3.12.\nExercise 15.17: Employ the uncertain hypothesis test to determine whether\nthe uncertain diﬀerential equation\ndXt = (5 −Xt)dt + XtdCt\n(15.142)\nﬁts the 30 observed data\nt\n0.00\n0.20\n0.40\n0.60\n0.80\n1.00\n1.30\n1.60\n1.70\n2.00\nXt\n2.00\n2.75\n3.59\n4.45\n5.48\n6.97\n6.63\n8.35\n8.82\n15.3\nt\n2.30\n2.80\n3.30\n3.50\n3.80\n4.30\n4.60\n4.90\n5.40\n5.90\nXt\n12.4\n14.1\n14.9\n12.7\n9.81\n7.85\n5.27\n4.35\n3.64\n3.71\nt\n6.20\n6.50\n6.80\n7.30\n7.60\n7.90\n8.30\n8.70\n9.20\n9.50\nXt\n3.12\n2.75\n3.44\n3.16\n2.49\n2.46\n2.90\n2.99\n4.19\n4.23\non the time horizon from 0 to 9.5.\n15.7\nParameter Estimation\nOne of the core problems in practice is how to estimate the unknown param-\neters in an uncertain diﬀerential equation that ﬁts the observed data of some\nuncertain process as much as possible.\nConsider an uncertain diﬀerential\nequation\ndXt = f(t, Xt; θ)dt + g(t, Xt; θ)dCt\n(15.143)\n\n\nSection 15.8 - Real-Life Examples\n369\nwhere f and g are known continuous functions but θ is an unknown vector\nof parameters. Assume\nxt1, xt2, · · · , xtn\n(15.144)\nare observed values of the uncertain process Xt at the times t1, t2, · · · , tn\nwith t1 < t2 < · · · < tn, respectively.\nIn order to estimate the unknown vector θ of parameters in the uncer-\ntain diﬀerential equation (15.143) based on the observed data (15.144), the\nmethod of moments was proposed by Yao-Liu [273] and revised by Liu-Liu\n[144]. For each given θ, we may produce n −1 residuals\nε2(θ), ε3(θ), · · · , εn(θ)\n(15.145)\nof the uncertain diﬀerential equation (15.143) corresponding to the observed\ndata (15.144). Note that ε2(θ), ε3(θ), · · · , εn(θ) should follow the linear un-\ncertainty distribution L(0, 1), i.e.,\nε2(θ), ε3(θ), · · · , εn(θ) ∼L(0, 1).\n(15.146)\nFor each positive integer k, the kth sample moment of the n −1 residuals\nε2(θ), ε3(θ), · · · , εn(θ) is\n1\nn −1\nn\nX\ni=2\nεk\ni (θ),\nand the kth population moment of the linear uncertainty distribution L(0, 1)\nis\n1\nk + 1.\nThe moment estimate θ is then obtained by equating the ﬁrst p sample mo-\nments with the corresponding ﬁrst p population moments, where p is the\nnumber of unknown parameters.\nIn other words, the moment estimate θ\nshould solve the system of equations,\n1\nn −1\nn\nX\ni=2\nεk\ni (θ) =\n1\nk + 1,\nk = 1, 2, · · · , p.\n(15.147)\nRemark 15.5: Sometimes the system of equations (15.147) has no solu-\ntion. In this case, it is suggested to use other methods, e.g., the maximum\nlikelihood estimation (Liu-Liu [143][147]) and the method of least squares\n(Liu-Liu [148]).\n15.8\nReal-Life Examples\nThis section will provide some real-life examples to illustrate the tool of\nuncertain diﬀerential equation.\n\n\n370\nChapter 15 - Uncertain Differential Equation\nExample 15.15: (Liu-Liu [144]) Alibaba is a world-famous internet com-\npany. Table 15.1 shows Alibaba stock prices (weekly average) in US dollars\nfrom January 1, 2019 to June 30, 2020 reported by Nasdaq.\nTable 15.1: Alibaba Stock Prices (Weekly Average) in US Dollars from Jan-\nuary 1, 2019 to June 30, 2020 reported by Nasdaq\n136.03\n148.96\n153.60\n154.81\n163.82\n168.87\n168.02\n172.37\n183.66\n181.75\n180.61\n180.60\n178.81\n181.47\n186.75\n185.84\n186.66\n189.48\n181.26\n173.52\n158.78\n151.91\n152.29\n160.19\n165.34\n168.65\n174.62\n167.96\n173.66\n177.36\n170.18\n158.32\n165.39\n173.44\n169.48\n175.59\n177.25\n179.81\n173.23\n167.59\n166.89\n173.91\n172.04\n177.25\n183.93\n184.89\n184.77\n196.49\n198.28\n202.65\n209.51\n215.24\n215.45\n219.58\n226.68\n219.38\n208.58\n218.73\n219.46\n218.32\n206.71\n209.29\n196.41\n181.17\n186.91\n189.86\n196.70\n206.91\n207.81\n201.74\n195.80\n202.03\n212.23\n202.45\n215.42\n219.26\n221.62\n222.85\nLet i = 1, 2, · · · , 78 represent the weeks from January 1, 2019 to June 30,\n2020, and denote the stock prices in Table 15.1 by\nx1, x2, · · · , x78.\n(15.148)\nAssume Xt is an uncertain process that represents Alibaba stock price and\nfollows the uncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt\n(15.149)\nwhere m, a and σ are unknown parameters. For any ﬁxed parameters m, a, σ\nand i (2 ≤i ≤78), we solve the updated uncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt,\nXi−1 = xi−1\n(15.150)\nand obtain the ith residual\nεi(m, a, σ) =\n\u0012\n1 + exp\n\u0012π((axi−1 −m) exp(−a) + m −axi)\n√\n3σ(1 −exp(−a))\n\u0013\u0013−1\n.\nSince the number of unknown parameters is 3 and the ﬁrst three moments of\nthe linear uncertainty distribution L(0, 1) are 1/2, 1/3 and 1/4, the system\n\n\nSection 15.8 - Real-Life Examples\n371\nof equations (15.147) becomes\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1\n77\n78\nX\ni=2\nεi(m, a, σ) = 1\n2\n1\n77\n78\nX\ni=2\nε2\ni (m, a, σ) = 1\n3\n1\n77\n78\nX\ni=2\nε3\ni (m, a, σ) = 1\n4\nwhose root is\nm = 45.9292,\na = 0.2404,\nσ = 8.6308.\nThus we obtain an uncertain stock model,\ndXt = (45.9292 −0.2404Xt)dt + 8.6308dCt\n(15.151)\nwhere Xt represents Alibaba stock price. Finally, let us test whether the\nuncertain stock model (15.151) ﬁts the stock prices x1, x2, · · · , x78. That is,\nwe should test whether the linear uncertainty distribution L(0, 1) ﬁts the 77\nresiduals\nεi(45.9292, 0.2404, 8.6308), i = 2, 3, · · · , 78.\n(15.152)\nSee Figure 15.2. Given a signiﬁcance level α = 0.05, it follows from α × 77 =\n3.85 and Theorem 4.4 that the test is\nW =\nn\n(z2, z3, · · · , z78) : there are at least 4 of indexes i’s with\n2 ≤i ≤78 such that zi < 0.025 or zi > 0.975\no\n.\nSince only ε21, ε32, ε75 ̸∈[0.025, 0.975], we have (ε2, ε3, · · · , ε78) ̸∈W. Thus\nthe uncertain stock model (15.151) is a good ﬁt to the stock prices x1, x2, · · · ,\nx78.\nExample 15.16:\n(Ye-Liu [282]) Table 15.2 shows US dollar to Chinese\nyuan (USD-CNY) exchange rates (weekly average) in Forex Capital Markets\n(FXCM) from October 1, 2019 to June 30, 2021.\nLet i = 1, 2, · · · , 91 represent the weeks from October 1, 2019 to June 30,\n2021, and denote the exchange rates in Table 15.2 by\nx1, x2, · · · , x91.\n(15.153)\nAssume Xt is an uncertain process that represents USD-CNY exchange rate\nand follows the uncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt\n(15.154)\n\n\n372\nChapter 15 - Uncertain Differential Equation\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 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15.2: Residual Plot of Uncertain Stock Model (15.151) Corresponding\nto Alibaba Stock Prices. Since the frequency is far from being stable, the\nresiduals are regarded as uncertain variables rather than random variables.\nThis is the reason why we use uncertain diﬀerential equation rather than\nstochastic diﬀerential equation. See Section B.5.\nwhere m, a and σ are unknown parameters. For any ﬁxed parameters m, a, σ\nand i (2 ≤i ≤91), we solve the updated uncertain diﬀerential equation\ndXt = (m −aXt)dt + σdCt,\nXi−1 = xi−1\n(15.155)\nand obtain the ith residual\nεi(m, a, σ) =\n\u0012\n1 + exp\n\u0012π((axi−1 −m) exp(−a) + m −axi)\n√\n3σ(1 −exp(−a))\n\u0013\u0013−1\n.\nSince the number of unknown parameters is 3 and the ﬁrst three moments of\nthe linear uncertainty distribution L(0, 1) are 1/2, 1/3 and 1/4, the system\nof equations (15.147) becomes\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1\n90\n91\nX\ni=2\nεi(m, a, σ) = 1\n2\n1\n90\n91\nX\ni=2\nε2\ni (m, a, σ) = 1\n3\n1\n90\n91\nX\ni=2\nε3\ni (m, a, σ) = 1\n4\nwhose root is\nm = 1.4448,\na = 0.2136,\nσ = 0.0775.\n\n\nSection 15.9 - Bibliographic Notes\n373\nTable 15.2: USD-CNY Exchange Rates (Weekly Average) in Forex Capital\nMarkets from October 1, 2019 to June 30, 2021\n7.1145\n7.0821\n7.0679\n7.0526\n7.0096\n7.0188\n7.0376\n7.0289\n7.0441\n7.0245\n7.0009\n6.9990\n6.9715\n6.9368\n6.8804\n6.9173\n6.9847\n6.9975\n6.9839\n7.0327\n7.0169\n6.9486\n6.9822\n7.0573\n7.1059\n7.1030\n7.0757\n7.0663\n7.0945\n7.0969\n7.1147\n7.1106\n7.1327\n7.1536\n7.1111\n7.0721\n7.0813\n7.0770\n7.0725\n7.0135\n6.9970\n7.0075\n7.0011\n6.9618\n6.9473\n6.9194\n6.8921\n6.8422\n6.8399\n6.7845\n6.8071\n6.7946\n6.7359\n6.7253\n6.6694\n6.7085\n6.6473\n6.6171\n6.5659\n6.5726\n6.5514\n6.5257\n6.5221\n6.5310\n6.5131\n6.4592\n6.4712\n6.4851\n6.4759\n6.4664\n6.4293\n6.4434\n6.4724\n6.4931\n6.5084\n6.5043\n6.5325\n6.5756\n6.5556\n6.5360\n6.4977\n6.4753\n6.4587\n6.4413\n6.4379\n6.3855\n6.3890\n6.3943\n6.4416\n6.4716\n6.4713\nThus we obtain an uncertain currency model,\ndXt = (1.4448 −0.2136Xt)dt + 0.0775dCt\n(15.156)\nwhere Xt represents USD-CNY exchange rate. Finally, let us test whether\nthe uncertain currency model (15.156) ﬁts the exchange rates x1, x2, · · · , x91.\nThat is, we should test whether the linear uncertainty distribution L(0, 1) ﬁts\nthe 90 residuals\nεi(1.4448, 0.2136, 0.0775), i = 2, 3, · · · , 91.\n(15.157)\nSee Figure 15.3. Given a signiﬁcance level α = 0.05, it follows from α × 90 =\n4.5 and Theorem 4.4 that the test is\nW =\nn\n(z2, z3, · · · , z78) : there are at least 5 of indexes i’s with\n2 ≤i ≤91 such that zi < 0.025 or zi > 0.975\no\n.\nSince all residuals are in [0.025, 0.975], we have (ε2, ε3, · · · , ε91) ̸∈W. Thus\nthe uncertain currency model (15.156) is a good ﬁt to the exchange rates\nx1, x2, · · · , x91.\n15.9\nBibliographic Notes\nUncertain diﬀerential equation is a type of diﬀerential equation involving\nuncertain processes. The study of uncertain diﬀerential equation was pio-\nneered by Liu [114] in 2008. This work was immediately followed by many\n\n\n374\nChapter 15 - Uncertain Differential Equation\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 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\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\n0.5\n1\n2\n30\n60\n91\nFigure 15.3: Residual Plot of Uncertain Currency Model (15.156) Corre-\nsponding to USD-CNY Exchange Rates.\nSince the frequency is far from\nbeing stable, the residuals are regarded as uncertain variables rather than\nrandom variables. This is the reason why we use uncertain diﬀerential equa-\ntion rather than stochastic diﬀerential equation. See Section B.5.\nresearchers. Nowadays, uncertain diﬀerential equation has achieved fruitful\nresults in both theory and practice.\nThe existence and uniqueness theorem of solution of uncertain diﬀerential\nequation was ﬁrst proved by Chen-Liu [8] under linear growth condition and\nLipschitz condition. Later, the theorem was veriﬁed again by Gao [58] under\nlocal linear growth condition and local Lipschitz condition.\nThe ﬁrst concept of stability (i.e., stability in measure) of uncertain diﬀer-\nential equation was presented by Liu [116], and some stability theorems were\nproved by Yao-Gao-Gao [251]. Following that, diﬀerent types of stability of\nuncertain diﬀerential equations were explored, for example, stability in mean\n(Yao-Ke-Sheng [258]), stability in moment (Sheng-Wang [200]), stability in\ndistribution (Yang-Ni-Zhang [240]), almost sure stability (Liu-Ke-Fei [136]),\nand exponential stability (Sheng-Gao [204]).\nAs an important contribution, Yao-Chen [254] showed that the solution of\nan uncertain diﬀerential equation can be represented by a family of solutions\nof ordinary diﬀerential equations, thus relating uncertain diﬀerential equa-\ntions and ordinary diﬀerential equations. On the basis of Yao-Chen formula,\nYao [252] presented some formulas to calculate extreme value, ﬁrst hitting\ntime, and time integral of solution of uncertain diﬀerential equation. Fur-\nthermore, some numerical solution methods for uncertain diﬀerential equa-\ntions were designed, including Euler method (Yao-Chen [254]), Runge-Kutta\nmethod (Yang-Shen [236]), Milne method (Gao [41]), Adams method (Yang-\nRalescu [238]), and Hamming method (Zhang-Gao-Huang [305]).\nAssume an uncertain process follows an uncertain diﬀerential equation\nand some realizations of this process are observed. In order to make a con-\n\n\nSection 15.9 - Bibliographic Notes\n375\nnection between uncertain diﬀerential equation and observed data, Liu-Liu\n[144] proposed the concept of residual and designed an algorithm to calculate\nthe residuals of uncertain diﬀerential equation corresponding to the observed\ndata.\nWith the help of residuals, Ye-Liu [282] suggested an uncertain hypothesis\ntest to determine whether or not an uncertain diﬀerential equation ﬁts the\nobserved data.\nFor practical purpose, it is extremely interesting to estimate the unknown\nparameters in an uncertain diﬀerential equation that ﬁts the observed data as\nmuch as possible. In order to solve this problem, the method of moments was\nproposed by Yao-Liu [273] and Liu-Liu [144], the maximum likelihood esti-\nmation was developed by Liu-Liu [143][147], and the method of least squares\nwas investigated by Liu-Liu [148].\nAs a supplement, some nonparametric\nestimations were also suggested, including Legendre polynomials approxima-\ntion (He-Zhu-Gu [68]), Hermite polynomials approximation (Li-Yang [96]),\nNadaraya-Watson estimation (Li-Xia [97]), and cubic spline method (Shi-\nZhao-Sheng [208].\nUncertain diﬀerential equation has been extended in many directions.\nUncertain partial diﬀerential equation was ﬁrst investigated by Yang-Yao\n[241], eventually created by Yang-Liu [234] and rigorously deﬁned by Ye\n[285]. Higher-order uncertain diﬀerential equation was ﬁrst discussed by Yao\n[266] and rigorously deﬁned by Zhang-Liu [300]. More generally, higher-order\nuncertain partial diﬀerential equation was tentatively explored by Zhu [319].\nUncertain diﬀerential equation has been successfully applied in many\nﬁelds such as Alibaba stock price (Liu-Liu [144]), China’s birth rate (Ye-\nZheng [283]), China’s population (Liu [145] and Yang-Liu [234]), crude oil\nprice (Xie-Gao [230]), currency exchange rate (Ye-Liu [282]), interest rate\n(Yang-Ke [247]), and pendulum (Xie [232]).\n\n\n\n\nChapter 16\nUncertain Finance\nThis chapter will introduce uncertain stock model, uncertain interest rate\nmodel, and uncertain currency model by using the tool of uncertain diﬀer-\nential equation. Based on the fair price principle, this chapter will also price\nEuropean options, American options, Asian options, zero-coupon bond, in-\nterest rate ceiling, and interest rate ﬂoor.\n16.1\nUncertain Stock Model\nAssume that there exists a bond and a stock in the ﬁnancial market. Let Xt\nbe the bond price at time t, and let r be the interest rate. Then\ndXt\ndt\n= rXt.\n(16.1)\nThat is, the bond price Xt follows the ordinary diﬀerential equation,\ndXt = rXtdt.\n(16.2)\nLet Yt be the stock price at time t whose growth rate is\ne + σ · “noise”\n(16.3)\nwhere e and σ are constants. Take the mathematical interpretation of the\n“noise” term as\n“noise” = dCt\ndt\n(16.4)\nwhere Ct is a Liu process. Then the stock price Yt follows the uncertain\ndiﬀerential equation,\ndYt\ndt =\n\u0012\ne + σ dCt\ndt\n\u0013\nYt,\n(16.5)\ni.e.,\ndYt = eYtdt + σYtdCt.\n(16.6)\n\n\n378\nChapter 16 - Uncertain Finance\nBased on the above analysis, in 2009 Liu [116] ﬁrst presented an uncertain\nstock model in which the bond price Xt and the stock price Yt are determined\nby\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.7)\nwhere r is the riskless interest rate, e is the log-drift, σ is the log-diﬀusion,\nand Ct is a Liu process.\n16.2\nEuropean Options\nThis section will price European call and put options for the ﬁnancial market\ndetermined by uncertain stock models.\nEuropean Call Option\nDeﬁnition 16.1 A European call option is a contract that gives the holder\nthe right to buy a stock at an expiration time s for a strike price K.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nt\nY0\nYt\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\ns\nYs .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nK .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure 16.1: Payoﬀ(Ys −K)+ from European Call Option\nLet fc represent the price of this contract. Then the investor pays fc for\nbuying the contract at time 0, and has a payoﬀ(Ys −K)+ at time s since\nthe option is rationally exercised if and only if Ys > K. See Figure 16.1.\nConsidering the time value of money resulted from the bond, the present\nvalue of the payoﬀis exp(−rs)(Ys −K)+. Thus the net return of the investor\nat time 0 is\n−fc + exp(−rs)(Ys −K)+.\n(16.8)\nOn the other hand, the bank receives fc for selling the contract at time 0,\nand pays (Ys −K)+ at the expiration time s. Thus the net return of the\nbank at the time 0 is\nfc −exp(−rs)(Ys −K)+.\n(16.9)\n\n\nSection 16.2 - European Options\n379\nThe fair price of this contract should make the investor and the bank have\nan identical expected return (we will call it fair price principle1 hereafter),\ni.e.,\n−fc + exp(−rs)E[(Ys −K)+] = fc −exp(−rs)E[(Ys −K)+].\n(16.10)\nThus fc = exp(−rs)E[(Ys −K)+]. That is, the European call option price is\njust the expected present value of the payoﬀ.\nDeﬁnition 16.2 (Liu [116]) Assume a European call option has a strike\nprice K and an expiration time s. Then the European call option price is\nfc = exp(−rs)E[(Ys −K)+]\n(16.11)\nwhere Ys is the stock price at time s, and r is the riskless interest rate.\nTheorem 16.1 (Liu [116]) Consider a general uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = F(t, Yt)dt + G(t, Yt)dCt\n(16.12)\nwhere r is a constant, and F and G are continuous functions. Assume a\nEuropean call option has a strike price K and an expiration time s. Then\nthe European call option price is\nfc = exp(−rs)\nZ 1\n0\n(Y α\ns −K)+dα\n(16.13)\nwhere Y α\ns is the α-path of the corresponding uncertain diﬀerential equation.\nProof: It follows from Theorem 15.12 that the stock price Ys has an inverse\nuncertainty distribution\nΦ−1\ns (α) = Y α\ns .\nThus (Ys −K)+ has an inverse uncertainty distribution\nΨ−1\ns (α) = (Y α\ns −K)+.\nBy using (16.11) and the expected value formula, we get\nfc = exp(−rs)E[(Ys −K)+]\n= exp(−rs)\nZ 1\n0\n(Y α\ns −K)+ dα.\n1Fair price principle does not meet no arbitrage principle (i.e., there are never oppor-\ntunities to make risk-free proﬁt). In fact, I do not agree with no arbitrage principle since\nit may lead to unreasonable results.\n\n\n380\nChapter 16 - Uncertain Finance\nThe European call option price formula (16.13) is veriﬁed.\nExample 16.1: (Liu [116]) Consider an uncertain stock model in which the\nbond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.14)\nwhere r, e and σ are constants with σ > 0. Assume a European call option\nhas a strike price K and an expiration time s. Note that the uncertain stock\nmodel (16.14) has an α-path\nY α\ns = Y0 exp\n \nes +\n√\n3σs\nπ\nln\nα\n1 −α\n!\n.\nIt follows from Theorem 16.1 that the European call option price is\nfc = exp(−rs)\nZ 1\n0\n(Y α\ns −K)+dα\n= exp(−rs)\nZ 1\n0\n \nY0 exp\n \nes +\n√\n3σs\nπ\nln\nα\n1 −α\n!\n−K\n!+\ndα.\nThat is,\nfc = exp(−rs)\nZ 1\n0\n \nY0 exp\n \nes +\n√\n3σs\nπ\nln\nα\n1 −α\n!\n−K\n!+\ndα.\n(16.15)\nExercise 16.1: (Xie-Zhang-Jia [231]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σdCt\n(16.16)\nwhere r, m, a and σ are constants with a ̸= 0 and σ > 0. Assume a European\ncall option has a strike price K and an expiration time s. Show that the\nEuropean call option price is\nfc =\n√\n3σ\nπa exp(−rs)(1 −exp(−as)) ln\n\u0012\n1 + exp\n\u0012\nπa(K −µ)\n√\n3σ(exp(−as) −1)\n\u0013\u0013\nwhere\nµ = m\na + exp(−as)\n\u0010\nY0 −m\na\n\u0011\n.\n\n\nSection 16.2 - European Options\n381\nExercise 16.2: (Jia-Xie-Zhang [83]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σYtdCt\n(16.17)\nwhere r, m, a and σ are constants with m > 0 and σ > 0. Assume a European\ncall option has a strike price K and an expiration time s. Show that the\nEuropean call option price is\nfc = exp(−rs)\nZ 1\n0\n\u0012\nexp(−ν(α)s)Y0 −(exp(−ν(α)s) −1)m\nν(α)\n−K\n\u0013+\ndα\nwhere\nν(α) = a +\n√\n3σ\nπ\nln\nα\n1 −α.\nEuropean Put Option\nDeﬁnition 16.3 A European put option is a contract that gives the holder\nthe right to sell a stock at an expiration time s for a strike price K.\nLet fp represent the price of this contract. Then the investor pays fp for\nbuying the contract at time 0, and has a payoﬀ(K −Ys)+ at time s since\nthe option is rationally exercised if and only if Ys < K. Considering the time\nvalue of money resulted from the bond, the present value of the payoﬀis\nexp(−rs)(K −Ys)+. Thus the net return of the investor at time 0 is\n−fp + exp(−rs)(K −Ys)+.\n(16.18)\nOn the other hand, the bank receives fp for selling the contract at time 0,\nand pays (K −Ys)+ at the expiration time s. Thus the net return of the\nbank at the time 0 is\nfp −exp(−rs)(K −Ys)+.\n(16.19)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−fp + exp(−rs)E[(K −Ys)+] = fp −exp(−rs)E[(K −Ys)+].\n(16.20)\nThus fp = exp(−rs)E[(K −Ys)+]. That is, the European put option price is\njust the expected present value of the payoﬀ.\nDeﬁnition 16.4 (Liu [116]) Assume a European put option has a strike\nprice K and an expiration time s. Then the European put option price is\nfp = exp(−rs)E[(K −Ys)+]\n(16.21)\nwhere Ys is the stock price at time s, and r is the riskless interest rate.\n\n\n382\nChapter 16 - Uncertain Finance\nTheorem 16.2 (Liu [116]) Consider a general uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = F(t, Yt)dt + G(t, Yt)dCt\n(16.22)\nwhere r is a constant, and F and G are continuous functions. Assume a\nEuropean put option has a strike price K and an expiration time s. Then the\nEuropean put option price is\nfp = exp(−rs)\nZ 1\n0\n(K −Y α\ns )+dα\n(16.23)\nwhere Y α\ns is the α-path of the corresponding uncertain diﬀerential equation.\nProof: It follows from Theorem 15.12 that the stock price Ys has an inverse\nuncertainty distribution\nΦ−1\ns (α) = Y α\ns .\nThus (K −Ys)+ has an inverse uncertainty distribution\nΨ−1\ns (α) = (K −Y 1−α\ns\n)+.\nTherefore, by using (16.21), the expected value formula and the change of\nvariables of integral, we get\nfp = exp(−rs)E[(K −Ys)+]\n= exp(−rs)\nZ 1\n0\nK −Y 1−α\ns\n\u0001+ dα\n= exp(−rs)\nZ 1\n0\n(K −Y α\ns )+ dα.\nThe European put option price formula (16.23) is veriﬁed.\nExample 16.2: (Liu [116]) Consider an uncertain stock model in which the\nbond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.24)\nwhere r, e and σ are constants with σ > 0. Assume a European put option\nhas a strike price K and an expiration time s. Note that the uncertain stock\nmodel (16.24) has an α-path\nY α\ns = Y0 exp\n \nes +\n√\n3σs\nπ\nln\nα\n1 −α\n!\n.\n\n\nSection 16.3 - American Options\n383\nIt follows from Theorem 16.2 that the European put option price is\nfp = exp(−rs)\nZ 1\n0\n(K −Y α\ns )+dα\n= exp(−rs)\nZ 1\n0\n \nK −Y0 exp\n \nes +\n√\n3σs\nπ\nln\nα\n1 −α\n!!+\ndα.\nThat is,\nfp = exp(−rs)\nZ 1\n0\n \nK −Y0 exp\n \nes +\n√\n3σs\nπ\nln\nα\n1 −α\n!!+\ndα.\n(16.25)\nExercise 16.3: (Xie-Zhang-Jia [231]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σdCt\n(16.26)\nwhere r, m, a and σ are constants with a ̸= 0 and σ > 0. Assume a European\nput option has a strike price K and an expiration time s. Show that the\nEuropean put option price is\nfp =\n√\n3σ\nπa exp(−rs)(1 −exp(−as)) ln\n\u0012\n1 + exp\n\u0012\nπa(µ −K)\n√\n3σ(exp(−as) −1)\n\u0013\u0013\nwhere\nµ = m\na + exp(−as)\n\u0010\nY0 −m\na\n\u0011\n.\nExercise 16.4: (Jia-Xie-Zhang [83]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σYtdCt\n(16.27)\nwhere r, m, a and σ are constants with m > 0 and σ > 0. Assume a European\ncall option has a strike price K and an expiration time s. Show that the\nEuropean put option price is\nfp = exp(−rs)\nZ 1\n0\n\u0012\nK −exp(−ν(α)s)Y0 + (exp(−ν(α)s) −1)m\nν(α)\n\u0013+\ndα\nwhere\nν(α) = a +\n√\n3σ\nπ\nln\nα\n1 −α.\n\n\n384\nChapter 16 - Uncertain Finance\n16.3\nAmerican Options\nThis section will price American call and put options for the ﬁnancial market\ndetermined by uncertain stock models.\nAmerican Call Option\nDeﬁnition 16.5 An American call option is a contract that gives the holder\nthe right to buy a stock at any time prior to an expiration time s for a strike\nprice K.\nLet fc represent the price of this contract. Then the net return of the\ninvestor at time 0 is\n−fc + sup\n0≤t≤s\nexp(−rt)(Yt −K)+,\n(16.28)\nand the net return of the bank at the time 0 is\nfc −sup\n0≤t≤s\nexp(−rt)(Yt −K)+.\n(16.29)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−fc + E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(Yt −K)+\n\u0015\n= fc −E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(Yt −K)+\n\u0015\n.\nThus the American call option price is just the expected present value of the\npayoﬀ.\nDeﬁnition 16.6 (Chen [9]) Assume an American call option has a strike\nprice K and an expiration time s. Then the American call option price is\nfc = E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(Yt −K)+\n\u0015\n(16.30)\nwhere Yt is the stock price, and r is the riskless interest rate.\nTheorem 16.3 (Chen [9]) Consider a general uncertain stock model in which\nthe bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = F(t, Yt)dt + G(t, Yt)dCt\n(16.31)\nwhere r is a constant, and F and G are continuous functions. Assume an\nAmerican call option has a strike price K and an expiration time s. Then\nthe American call option price is\nfc =\nZ 1\n0\nsup\n0≤t≤s\nexp(−rt)(Y α\nt −K)+dα\n(16.32)\nwhere Y α\nt\nis the α-path of the corresponding uncertain diﬀerential equation.\n\n\nSection 16.3 - American Options\n385\nProof: It follows from Theorem 15.12 that the stock price Yt has an inverse\nuncertainty distribution\nΦ−1\nt (α) = Y α\nt .\nSince exp(−rt)(Yt −K)+ is an increasing function of Yt, it follows from\nTheorem 15.14 that the extreme value\nsup\n0≤t≤s\nexp(−rt)(Yt −K)+\nhas an inverse uncertainty distribution\nΨ−1\ns (α) = sup\n0≤t≤s\nexp(−rt)(Y α\nt −K)+.\nTherefore, by using (16.30) and the expected value formula, we get\nfc = E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(Yt −K)+\n\u0015\n=\nZ 1\n0\nsup\n0≤t≤s\nexp(−rt)(Y α\nt −K)+dα.\nThe American call option price formula (16.32) is veriﬁed.\nExercise 16.5: (Zhang-Jia-Xie [301]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.33)\nwhere r, e and σ are constants with σ > 0. Assume an American call option\nhas a strike price K and an expiration time s. Show that the American call\noption price is\nfc =\nZ 1\n0\nH(α)dα\nwhere\nH(α) =\n(\nh(0, α) ∨h(τ(α), α) ∨h(s, α),\nif 0 < τ(α) < s\nh(0, α) ∨h(s, α),\notherwise,\nh(t, α) = exp(−rt)\n \nexp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!\nY0 −K\n!+\n,\nτ(α) =\n1\nν(α) ln\nrK\n(r −ν(α))Y0\n,\nν(α) = e +\n√\n3σ\nπ\nln\nα\n1 −α.\n\n\n386\nChapter 16 - Uncertain Finance\nExercise 16.6: (Xie-Zhang-Jia [231]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σdCt\n(16.34)\nwhere r, m, a and σ are constants with a ̸= 0 and σ > 0. Assume an American\ncall option has a strike price K and an expiration time s. Show that the\nAmerican call option price is\nfc =\nZ 1\n0\nH(α)dα\nwhere\nH(α) =\n(\nh(0, α) ∨h(τ(α), α) ∨h(s, α),\nif 0 < τ(α) < s\nh(0, α) ∨h(s, α),\notherwise,\nh(t, α) = exp(−rt)\n\u0012\nexp(−at)Y0 −(exp(−at) −1)ν(α)\na\n−K\n\u0013+\n,\nτ(α) = 1\na ln (r + a)(ν(α) −aY0)\nr(ν(α) −aK)\n,\nν(α) = m +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.7: (Jia-Xie-Zhang [83]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σYtdCt\n(16.35)\nwhere r, m, a and σ are constants with m > 0 and σ > 0.\nAssume an\nAmerican call option has a strike price K and an expiration time s. Show\nthat the American call option price is\nfc =\nZ 1\n0\nH(α)dα\nwhere\nH(α) =\n(\nh(0, α) ∨h(τ(α), α) ∨h(s, α),\nif 0 < τ(α) < s\nh(0, α) ∨h(s, α),\notherwise,\nh(t, α) = exp(−rt)\n\u0012\nexp(−ν(α)t)Y0 −(exp(−ν(α)t) −1)m\nν(α)\n−K\n\u0013+\n,\nτ(α) =\n1\nν(α) ln (r + ν(α))(m −ν(α)Y0)\nr(m −ν(α)K)\n,\nν(α) = a +\n√\n3σ\nπ\nln\nα\n1 −α.\n\n\nSection 16.3 - American Options\n387\nAmerican Put Option\nDeﬁnition 16.7 An American put option is a contract that gives the holder\nthe right to sell a stock at any time prior to an expiration time s for a strike\nprice K.\nLet fp represent the price of this contract. Then the net return of the\ninvestor at time 0 is\n−fp + sup\n0≤t≤s\nexp(−rt)(K −Yt)+,\n(16.36)\nand the net return of the bank at the time 0 is\nfp −sup\n0≤t≤s\nexp(−rt)(K −Yt)+.\n(16.37)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−fp + E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(K −Yt)+\n\u0015\n= fp −E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(K −Yt)+\n\u0015\n.\nThus the American put option price is just the expected present value of the\npayoﬀ.\nDeﬁnition 16.8 (Chen [9]) Assume an American put option has a strike\nprice K and an expiration time s. Then the American put option price is\nfp = E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(K −Yt)+\n\u0015\n(16.38)\nwhere Yt is the stock price, and r is the riskless interest rate.\nTheorem 16.4 (Chen [9]) Consider a general uncertain stock model in which\nthe bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = F(t, Yt)dt + G(t, Yt)dCt\n(16.39)\nwhere r is a constant, and F and G are continuous functions. Assume an\nAmerican put option has a strike price K and an expiration time s. Then\nthe American put option price is\nfp =\nZ 1\n0\nsup\n0≤t≤s\nexp(−rt)(K −Y α\nt )+dα\n(16.40)\nwhere Y α\nt\nis the α-path of the corresponding uncertain diﬀerential equation.\n\n\n388\nChapter 16 - Uncertain Finance\nProof: It follows from Theorem 15.12 that the stock price Yt has an inverse\nuncertainty distribution\nΦ−1\nt (α) = Y α\nt .\nSince exp(−rt)(K −Yt)+ is a decreasing function of Yt, it follows from The-\norem 15.14 that the extreme value\nsup\n0≤t≤s\nexp(−rt)(K −Yt)+\nhas an inverse uncertainty distribution\nΨ−1\ns (α) = sup\n0≤t≤s\nexp(−rt)(K −Y 1−α\nt\n)+.\nTherefore, by using (16.38), the expected value formula and the change of\nvariables of integral, we get\nfc = E\n\u0014\nsup\n0≤t≤s\nexp(−rt)(Yt −K)+\n\u0015\n=\nZ 1\n0\nsup\n0≤t≤s\nexp(−rt)(K −Y 1−α\nt\n)+dα\n=\nZ 1\n0\nsup\n0≤t≤s\nexp(−rt)(K −Y α\nt )+dα.\nThe American put option price formula (16.40) is veriﬁed.\nExercise 16.8: (Zhang-Jia-Xie [301]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.41)\nwhere r, e and σ are constants with σ > 0. Assume an American put option\nhas a strike price K and an expiration time s. Show that the American put\noption price is\nfp =\nZ 1\n0\nH(α)dα\nwhere\nH(α) =\n(\nh(0, α) ∨h(τ(α), α) ∨h(s, α),\nif 0 < τ(α) < s\nh(0, α) ∨h(s, α),\notherwise,\nh(t, α) = exp(−rt)\n \nK −exp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!\nY0\n!+\n,\n\n\nSection 16.3 - American Options\n389\nτ(α) =\n1\nν(α) ln\nrK\n(r −ν(α))Y0\n,\nν(α) = e +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.9: (Xie-Zhang-Jia [231]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σdCt\n(16.42)\nwhere r, m, a and σ are constants with a ̸= 0 and σ > 0. Assume an American\nput option has a strike price K and an expiration time s. Show that the\nAmerican put option price is\nfp =\nZ 1\n0\nH(α)dα\nwhere\nH(α) =\n(\nh(0, α) ∨h(τ(α), α) ∨h(s, α),\nif 0 < τ(α) < s\nh(0, α) ∨h(s, α),\notherwise,\nh(t, α) = exp(−rt)\n\u0012\nK −exp(−at)Y0 + (exp(−at) −1)ν(α)\na\n\u0013+\n,\nτ(α) = 1\na ln (r + a)(ν(α) −aY0)\nr(ν(α) −aK)\n,\nν(α) = m +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.10: (Jia-Xie-Zhang [83]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σYtdCt\n(16.43)\nwhere r, m, a and σ are constants with m > 0 and σ > 0.\nAssume an\nAmerican put option has a strike price K and an expiration time s. Show\nthat the American put option price is\nfp =\nZ 1\n0\nH(α)dα\n\n\n390\nChapter 16 - Uncertain Finance\nwhere\nH(α) =\n(\nh(0, α) ∨h(τ(α), α) ∨h(s, α),\nif 0 < τ(α) < s\nh(0, α) ∨h(s, α),\notherwise,\nh(t, α) = exp(−rt)\n\u0012\nK −exp(−ν(α)t)Y0 + (exp(−ν(α)t) −1)m\nν(α)\n\u0013+\n,\nτ(α) =\n1\nν(α) ln (r + ν(α))(m −ν(α)Y0)\nr(m −ν(α)K)\n,\nν(α) = a +\n√\n3σ\nπ\nln\nα\n1 −α.\n16.4\nAsian Options\nThis section will price Asian call and put options for the ﬁnancial market\ndetermined by uncertain stock models.\nAsian Call Option\nDeﬁnition 16.9 An Asian call option is a contract whose payoﬀat the ex-\npiration time s is\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\n(16.44)\nwhere K is a strike price.\nLet fc represent the price of this contract. Then the investor pays fc for\nbuying the contract at time 0, and has a payoﬀ\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\n(16.45)\nat time s. Considering the time value of money resulted from the bond, the\npresent value of the payoﬀis\nexp(−rs)\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\n.\n(16.46)\nThus the net return of the investor at time 0 is\n−fc + exp(−rs)\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\n.\n(16.47)\nOn the other hand, the bank receives fc for selling the contract at time 0,\nand pays\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\n(16.48)\n\n\nSection 16.4 - Asian Options\n391\nat the expiration time s. Thus the net return of the bank at the time 0 is\nfc −exp(−rs)\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\n.\n(16.49)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−fc + exp(−rs)E\n\"\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+#\n= fc −exp(−rs)E\n\"\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+#\n.\n(16.50)\nThus the Asian call option price is just the expected present value of the\npayoﬀ.\nDeﬁnition 16.10 (Sun-Chen [211]) Assume an Asian call option has a strike\nprice K and an expiration time s. Then the Asian call option price is\nfc = exp(−rs)E\n\"\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+#\n(16.51)\nwhere Yt is the stock price, and r is the riskless interest rate.\nTheorem 16.5 (Sun-Chen [211]) Consider a general uncertain stock model\nin which the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = F(t, Yt)dt + G(t, Yt)dCt\n(16.52)\nwhere r is a constant, and F and G are continuous functions. Assume an\nAsian call option has a strike price K and an expiration time s. Then the\nAsian call option price is\nfc = exp(−rs)\nZ 1\n0\n\u00121\ns\nZ s\n0\nY α\nt dt −K\n\u0013+\ndα\n(16.53)\nwhere Y α\nt\nis the α-path of the corresponding uncertain diﬀerential equation.\nProof: It follows from Theorem 15.12 that the stock price Yt has an inverse\nuncertainty distribution\nΦ−1\nt (α) = Y α\nt .\nIn addition, Theorem 15.16 tells us that the time integral\nZ s\n0\nYtdt\n\n\n392\nChapter 16 - Uncertain Finance\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nY α\nt dt.\nThus\n\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+\nhas an inverse uncertainty distribution\nΥ−1\ns (α) =\n\u00121\ns\nZ s\n0\nY α\nt dt −K\n\u0013+\n.\nTherefore, by using (16.51) and the expected value formula, we get\nfc = exp(−rs)E\n\"\u00121\ns\nZ s\n0\nYtdt −K\n\u0013+#\n= exp(−rs)\nZ 1\n0\n\u00121\ns\nZ s\n0\nY α\nt dt −K\n\u0013+\ndα.\nThe Asian call option price formula (16.53) is veriﬁed.\nExercise 16.11: (Zhang-Jia-Xie [301]) Consider an uncertain stock model\nin which the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.54)\nwhere r, e and σ are constants with σ > 0. Assume an Asian call option has\na strike price K and an expiration time s. Show that the Asian call option\nprice is\nfc = exp(−rs)\nZ 1\n0\n\u0012(exp(ν(α)s) −1)Y0\nν(α)s\n−K\n\u0013+\ndα\nwhere\nν(α) = e +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.12: (Xie-Zhang-Jia [231]) Consider an uncertain stock model\nin which the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σdCt\n(16.55)\n\n\nSection 16.4 - Asian Options\n393\nwhere r, m, a and σ are constants with a ̸= 0 and σ > 0. Assume an Asian\ncall option has a strike price K and an expiration time s. Show that the\nAsian call option price is\nfc = exp(−rs)\nZ 1\n0\n\u0012(exp(−as) −1)(ν(α) −aY0)\na2s\n+ ν(α)\na\n−K\n\u0013+\ndα\nwhere\nν(α) = m +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.13: (Jia-Xie-Zhang [83]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σYtdCt\n(16.56)\nwhere r, m, a and σ are constants with m > 0 and σ > 0. Assume an Asian\ncall option has a strike price K and an expiration time s. Show that the\nAsian call option price is\nfc = exp(−rs)\nZ 1\n0\n\u0012(exp(−ν(α)s) −1)(m −ν(α)Y0) + ν(α)ms\nν2(α)s\n−K\n\u0013+\ndα\nwhere\nν(α) = a +\n√\n3σ\nπ\nln\nα\n1 −α.\nAsian Put Option\nDeﬁnition 16.11 An Asian put option is a contract whose payoﬀat the\nexpiration time s is\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\n(16.57)\nwhere K is a strike price.\nLet fp represent the price of this contract. Then the investor pays fp for\nbuying the contract at time 0, and has a payoﬀ\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\n(16.58)\nat time s. Considering the time value of money resulted from the bond, the\npresent value of the payoﬀis\nexp(−rs)\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\n.\n(16.59)\n\n\n394\nChapter 16 - Uncertain Finance\nThus the net return of the investor at time 0 is\n−fp + exp(−rs)\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\n.\n(16.60)\nOn the other hand, the bank receives fp for selling the contract at time 0,\nand pays\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\n(16.61)\nat the expiration time s. Thus the net return of the bank at the time 0 is\nfp −exp(−rs)\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\n.\n(16.62)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−fp + exp(−rs)E\n\"\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+#\n= fp −exp(−rs)E\n\"\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+#\n.\n(16.63)\nThus the Asian put option price should be the expected present value of the\npayoﬀ.\nDeﬁnition 16.12 (Sun-Chen [211]) Assume an Asian put option has a strike\nprice K and an expiration time s. Then the Asian put option price is\nfp = exp(−rs)E\n\"\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+#\n(16.64)\nwhere Yt is the stock price, and r is the riskless interest rate.\nTheorem 16.6 (Sun-Chen [211]) Consider a general uncertain stock model\nin which the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = F(t, Yt)dt + G(t, Yt)dCt\n(16.65)\nwhere r is a constant, and F and G are continuous functions. Assume an\nAsian put option has a strike price K and an expiration time s. Then the\nAsian put option price is\nfp = exp(−rs)\nZ 1\n0\n\u0012\nK −1\ns\nZ s\n0\nY α\nt dt\n\u0013+\ndα\n(16.66)\nwhere Y α\nt\nis the α-path of the corresponding uncertain diﬀerential equation.\n\n\nSection 16.4 - Asian Options\n395\nProof: It follows from Theorem 15.12 that the stock price Yt has an inverse\nuncertainty distribution\nΦ−1\nt (α) = Y α\nt .\nIn addition, Theorem 15.16 tells us that the time integral\nZ s\n0\nYtdt\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nY α\nt dt.\nThus\n\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+\nhas an inverse uncertainty distribution\nΥ−1\ns (α) =\n\u0012\nK −1\ns\nZ s\n0\nY 1−α\nt\ndt\n\u0013+\n.\nTherefore, by using (16.64), the expected value formula and the change of\nvariables of integral, we get\nfp = exp(−rs)E\n\"\u0012\nK −1\ns\nZ s\n0\nYtdt\n\u0013+#\n= exp(−rs)\nZ 1\n0\n\u0012\nK −1\ns\nZ s\n0\nY 1−α\nt\ndt\n\u0013+\ndα\n= exp(−rs)\nZ 1\n0\n\u0012\nK −1\ns\nZ s\n0\nY α\nt dt\n\u0013+\ndα.\nThe Asian put option price formula (16.66) is veriﬁed.\nExercise 16.14: (Zhang-Jia-Xie [301]) Consider an uncertain stock model\nin which the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = eYtdt + σYtdCt\n(16.67)\nwhere r, e and σ are constants with σ > 0. Assume an Asian put option has\na strike price K and an expiration time s. Show that the Asian put option\nprice is\nfp = exp(−rs)\nZ 1\n0\n\u0012\nK −(exp(ν(α)s) −1)Y0\nν(α)s\n\u0013+\ndα\n\n\n396\nChapter 16 - Uncertain Finance\nwhere\nν(α) = e +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.15: (Xie-Zhang-Jia [231]) Consider an uncertain stock model\nin which the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σdCt\n(16.68)\nwhere r, m, a and σ are constants with a ̸= 0 and σ > 0. Assume an Asian\nput option has a strike price K and an expiration time s. Show that the\nAsian put option price is\nfp = exp(−rs)\nZ 1\n0\n\u0012\nK −(exp(−as) −1)(ν(α) −aY0)\na2s\n−ν(α)\na\n\u0013+\ndα\nwhere\nν(α) = m +\n√\n3σ\nπ\nln\nα\n1 −α.\nExercise 16.16: (Jia-Xie-Zhang [83]) Consider an uncertain stock model in\nwhich the bond price Xt and the stock price Yt are determined by\n(\ndXt = rXtdt\ndYt = (m −aYt)dt + σYtdCt\n(16.69)\nwhere r, m, a and σ are constants with m > 0 and σ > 0. Assume an Asian\nput option has a strike price K and an expiration time s. Show that the\nAsian put option price is\nfp = exp(−rs)\nZ 1\n0\n\u0012\nK −(exp(−ν(α)s) −1)(m −ν(α)Y0) + ν(α)ms\nν2(α)s\n\u0013+\ndα\nwhere\nν(α) = a +\n√\n3σ\nπ\nln\nα\n1 −α.\n16.5\nUncertain Interest Rate Model\nReal interest rates do not remain unchanged. Chen-Gao [18] assumed that\nthe interest rate Xt follows an uncertain diﬀerential equation and presented\nan uncertain interest rate model,\ndXt = (m −aXt)dt + σdCt\n(16.70)\nwhere m, a and σ are constants, and Ct is a Liu process.\n\n\nSection 16.5 - Uncertain Interest Rate Model\n397\nZero-Coupon Bond\nA zero-coupon bond is a bond bought at a price lower than its face value that\nis the amount it promises to pay at the maturity date. For simplicity, we\nassume the face value is always 1 dollar.\nLet f represent the price of this zero-coupon bond. Then the investor\npays f for buying it at time 0, and receives 1 dollar at the maturity date s.\nSince the interest rate is Xt, the present value of 1 dollar is\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\n.\n(16.71)\nThus the net return of the investor at time 0 is\n−f + exp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\n.\n(16.72)\nOn the other hand, the bank receives f for selling the zero-coupon bond at\ntime 0, and pays 1 dollar at the maturity date s. Thus the net return of the\nbank at the time 0 is\nf −exp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\n.\n(16.73)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−f + E\n\u0014\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\u0015\n= f −E\n\u0014\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\u0015\n(16.74)\nThus the price of the zero-coupon bond is just the expected present value of\nits face value.\nDeﬁnition 16.13 (Chen-Gao [18]) Let Xt be the uncertain interest rate.\nThen the price of a zero-coupon bond with a maturity date s is\nf = E\n\u0014\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\u0015\n.\n(16.75)\nTheorem 16.7 (Jiao-Yao [86]) Assume the uncertain interest rate Xt fol-\nlows the uncertain diﬀerential equation\ndXt = F(t, Xt)dt + G(t, Xt)dCt\n(16.76)\nwhere F and G are continuous functions. Then the price of a zero-coupon\nbond with maturity date s is\nf =\nZ 1\n0\nexp\n\u0012\n−\nZ s\n0\nXα\nt dt\n\u0013\ndα\n(16.77)\nwhere Xα\nt is the α-path of the corresponding uncertain diﬀerential equation.\n\n\n398\nChapter 16 - Uncertain Finance\nProof: It follows from the time integral of solution of uncertain diﬀerential\nequation that\nZ s\n0\nXtdt\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\nXα\nt dt.\nThus\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013\nhas an inverse uncertainty distribution\nΥ−1\ns (α) = exp\n\u0012\n−\nZ s\n0\nX1−α\nt\ndt\n\u0013\n.\nBy using (16.75), the expected value formula and the change of variables of\nintegral, we get (16.77).\nExercise 16.17: Consider an uncertain interest rate model in which the\ninterest rate Xt is determined by\ndXt = eXtdt + σXtdCt\n(16.78)\nwhere e and σ are constants with σ > 0. Show that the price of a zero-coupon\nbond with maturity date s is\nf =\nZ 1\n0\nexp\n \n−X0\nZ s\n0\nexp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!\ndt\n!\ndα.\n(16.79)\nInterest Rate Ceiling\nAn interest rate ceiling is a derivative contract in which the borrower will not\npay any more than a predetermined level of interest on his loan. Assume K\nis the maximum interest rate and s is the maturity date. For simplicity, we\nalso assume the amount of loan is always 1 dollar.\nLet f represent the price of this contract. Then the borrower pays f for\nbuying the contract at time 0, and has a payoﬀ\nexp\n\u0012Z s\n0\nXtdt\n\u0013\n−exp\n\u0012Z s\n0\nXt ∧Kdt\n\u0013\n(16.80)\nat the maturity date s. Considering the time value of money, the present\n\n\nSection 16.5 - Uncertain Interest Rate Model\n399\nvalue of the payoﬀis\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013 \u0012\nexp\n\u0012Z s\n0\nXtdt\n\u0013\n−exp\n\u0012Z s\n0\nXt ∧Kdt\n\u0013\u0013\n= 1 −exp\n\u0012\n−\nZ s\n0\nXtdt +\nZ s\n0\nXt ∧Kdt\n\u0013\n= 1 −exp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\n.\nThus the net return of the borrower at time 0 is\n−f + 1 −exp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\n.\n(16.81)\nSimilarly, we may verify that the net return of the bank at the time 0 is\nf −1 + exp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\n.\n(16.82)\nIt follows from the fair price principle that the price of this contract should\nmake the borrower and the bank have an identical expected return, i.e.,\n−f+1−E\n\u0014\nexp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\u0015\n= f−1+E\n\u0014\nexp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\u0015\n.\nThus we have the following deﬁnition of the price of interest rate ceiling.\nDeﬁnition 16.14 (Zhang-Ralescu-Liu [311]) Assume an interest rate ceiling\nhas a maximum interest rate K and a maturity date s. Then the price of the\ninterest rate ceiling is\nf = 1 −E\n\u0014\nexp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\u0015\n.\n(16.83)\nTheorem 16.8 (Zhang-Ralescu-Liu [311]) Assume the uncertain interest\nrate Xt follows the uncertain diﬀerential equation\ndXt = F(t, Xt)dt + G(t, Xt)dCt\n(16.84)\nwhere F and G are continuous functions. Then the price of the interest rate\nceiling with a maximum interest rate K and a maturity date s is\nf = 1 −\nZ 1\n0\nexp\n\u0012\n−\nZ s\n0\n(Xα\nt −K)+dt\n\u0013\ndα\n(16.85)\nwhere Xα\nt is the α-path of the corresponding uncertain diﬀerential equation.\n\n\n400\nChapter 16 - Uncertain Finance\nProof: It follows from the time integral of solution of uncertain diﬀerential\nequation that\nZ s\n0\n(Xt −K)+dt\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\n(Xα\nt −K)+dt.\nThus\nexp\n\u0012\n−\nZ s\n0\n(Xt −K)+dt\n\u0013\nhas an inverse uncertainty distribution\nΥ−1\ns (α) = exp\n\u0012\n−\nZ s\n0\n(X1−α\nt\n−K)+dt\n\u0013\n.\nBy using (16.83), the expected value formula and the change of variables of\nintegral, we get (16.85).\nExercise 16.18: Consider an uncertain interest rate model in which the\ninterest rate Xt is determined by\ndXt = eXtdt + σXtdCt\n(16.86)\nwhere e and σ are constants with σ > 0. Show that the price of the interest\nrate ceiling with a maximum interest rate K and a maturity date s is\nf = 1 −\nZ 1\n0\nexp\n\n−\nZ s\n0\n \nX0 exp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!\n−K\n!+\ndt\n\ndα.\nInterest Rate Floor\nAn interest rate ﬂoor is a derivative contract in which the investor will not\nreceive any less than a predetermined level of interest on his investment.\nAssume K is the minimum interest rate and s is the maturity date. For\nsimplicity, we also assume the amount of investment is always 1 dollar.\nLet f represent the price of this contract. Then the investor pays f for\nbuying the contract at time 0, and has a payoﬀ\nexp\n\u0012Z s\n0\nXt ∨Kdt\n\u0013\n−exp\n\u0012Z s\n0\nXtdt\n\u0013\n(16.87)\n\n\nSection 16.5 - Uncertain Interest Rate Model\n401\nat the maturity date s. Considering the time value of money, the present\nvalue of the payoﬀis\nexp\n\u0012\n−\nZ s\n0\nXtdt\n\u0013 \u0012\nexp\n\u0012Z s\n0\nXt ∨Kdt\n\u0013\n−exp\n\u0012Z s\n0\nXtdt\n\u0013\u0013\n= exp\n\u0012\n−\nZ s\n0\nXtdt +\nZ s\n0\nXt ∨Kdt\n\u0013\n−1\n= exp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\n−1.\nThus the net return of the investor at time 0 is\n−f + exp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\n−1.\n(16.88)\nSimilarly, we may verify that the net return of the bank at the time 0 is\nf −exp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\n+ 1.\n(16.89)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−f +E\n\u0014\nexp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\u0015\n−1 = f −E\n\u0014\nexp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\u0015\n+1.\nThus we have the following deﬁnition of the price of interest rate ﬂoor.\nDeﬁnition 16.15 (Zhang-Ralescu-Liu [311]) Assume an interest rate ﬂoor\nhas a minimum interest rate K and a maturity date s. Then the price of the\ninterest rate ﬂoor is\nf = E\n\u0014\nexp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\u0015\n−1.\n(16.90)\nTheorem 16.9 (Zhang-Ralescu-Liu [311]) Assume the uncertain interest\nrate Xt follows the uncertain diﬀerential equation\ndXt = F(t, Xt)dt + G(t, Xt)dCt\n(16.91)\nwhere F and G are continuous functions. Then the price of the interest rate\nﬂoor with a minimum interest rate K and a maturity date s is\nf =\nZ 1\n0\nexp\n\u0012Z s\n0\n(K −Xα\nt )+dt\n\u0013\ndα −1\n(16.92)\nwhere Xα\nt is the α-path of the corresponding uncertain diﬀerential equation.\n\n\n402\nChapter 16 - Uncertain Finance\nProof: It follows from the time integral of solution of uncertain diﬀerential\nequation that\nZ s\n0\n(K −Xt)+dt\nhas an inverse uncertainty distribution\nΨ−1\ns (α) =\nZ s\n0\n(K −X1−α\nt\n)+dt.\nThus\nexp\n\u0012Z s\n0\n(K −Xt)+dt\n\u0013\nhas an inverse uncertainty distribution\nΥ−1\ns (α) = exp\n\u0012Z s\n0\n(K −X1−α\nt\n)+dt\n\u0013\n.\nBy using (16.90), the expected value formula and the change of variables of\nintegral, we get (16.92).\nExercise 16.19: Consider an uncertain interest rate model in which the\ninterest rate Xt is determined by\ndXt = eXtdt + σXtdCt\n(16.93)\nwhere e and σ are constants with σ > 0. Show that the price of the interest\nrate ﬂoor with a minimum interest rate K and a maturity date s is\nf =\nZ 1\n0\nexp\n\n\nZ s\n0\n \nK −X0 exp\n \net +\n√\n3σt\nπ\nln\nα\n1 −α\n!!+\ndt\n\ndα −1.\n16.6\nUncertain Currency Model\nLiu-Chen-Ralescu [155] assumed that the exchange rate follows an uncertain\ndiﬀerential equation and proposed an uncertain currency model,\n\n\n\n\n\n\n\ndXt = uXtdt\n(Domestic Currency)\ndYt = vYtdt\n(Foreign Currency)\ndZt = eZtdt + σZtdCt\n(Exchange Rate)\n(16.94)\nwhere Xt represents the domestic currency with domestic interest rate u, Yt\nrepresents the foreign currency with foreign interest rate v, and Zt repre-\nsents the exchange rate that is domestic currency price of one unit of foreign\ncurrency at time t.\n\n\nSection 16.6 - Uncertain Currency Model\n403\nEuropean Currency Option\nDeﬁnition 16.16 A European currency option is a contract that gives the\nholder the right to exchange one unit of foreign currency at an expiration\ntime s for K units of domestic currency.\nSuppose that the price of this contract is f in domestic currency. Then\nthe investor pays f for buying the contract at time 0, and receives (Zs −K)+\nin domestic currency at the expiration time s. Thus the net return of the\ninvestor at time 0 is\n−f + exp(−us)(Zs −K)+.\n(16.95)\nOn the other hand, the bank receives f for selling the contract at time 0, and\npays (1 −K/Zs)+ in foreign currency at the expiration time s. Thus the net\nreturn of the bank at the time 0 is\nf −exp(−vs)Z0(1 −K/Zs)+.\n(16.96)\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−f + exp(−us)E[(Zs −K)+] = f −exp(−vs)Z0E[(1 −K/Zs)+]. (16.97)\nThus the European currency option price is given by the deﬁnition below.\nDeﬁnition 16.17 (Liu-Chen-Ralescu [155]) Assume a European currency\noption has a strike price K and an expiration time s. Then the European\ncurrency option price is\nf = 1\n2 exp(−us)E[(Zs −K)+] + 1\n2 exp(−vs)Z0E[(1 −K/Zs)+].\n(16.98)\nTheorem 16.10 (Liu-Chen-Ralescu [155]) Consider a general uncertain cur-\nrency model in which the domestic currency Xt, the foreign currency Yt and\nthe exchange rate Zt are determined by\n\n\n\n\n\n\n\ndXt = uXtdt\ndYt = vYtdt\ndZt = F(t, Zt)dt + G(t, Zt)dCt\n(16.99)\nwhere F and G are continuous functions. Assume a European currency op-\ntion has a strike price K and an expiration time s.\nThen the European\ncurrency option price is\nf = 1\n2\nZ 1\n0\nexp(−us)(Zα\ns −K)+ + exp(−vs)Z0(1 −K/Zα\ns )+\u0001\ndα\n(16.100)\nwhere Zα\nt is the α-path of the corresponding uncertain diﬀerential equation.\n\n\n404\nChapter 16 - Uncertain Finance\nProof: It follows from Theorem 15.12 that the exchange rate Zs has an\ninverse uncertainty distribution\nΦ−1\ns (α) = Zα\ns .\nSince (Zs −K)+ and (1 −K/Zs)+ are increasing functions with respect to\nZs, they have inverse uncertainty distributions\nΨ−1\ns (α) = (Zα\ns −K)+,\nΥ−1\ns (α) = (1 −K/Zα\ns )+,\nrespectively. By using (16.98) and the expected value formula, we get the\nresult.\nExercise 16.20: (Liu-Chen-Ralescu [155]) Consider the uncertain currency\nmodel\n\n\n\n\n\n\n\ndXt = uXtdt\n(Domestic Currency)\ndYt = vYtdt\n(Foreign Currency)\ndZt = eZtdt + σZtdCt\n(Exchange Rate)\n(16.101)\nwhere u, v, e and σ are constants with σ > 0. Assume a European currency\noption has a strike price K and an expiration time s. Show that the European\ncurrency option price is\nf = 1\n2 exp(−us)\nZ 1\n0\n \nZ0 exp\n \nes + σs\n√\n3\nπ\nln\nα\n1 −α\n!\n−K\n!+\ndα\n+1\n2 exp(−vs)\nZ 1\n0\n \nZ0 −K/ exp\n \nes + σs\n√\n3\nπ\nln\nα\n1 −α\n!!+\ndα.\nAmerican Currency Option\nDeﬁnition 16.18 An American currency option is a contract that gives the\nholder the right to exchange one unit of foreign currency at any time prior\nto an expiration time s for K units of domestic currency.\nSuppose that the price of this contract is f in domestic currency. Then\nthe net return of the investor at time 0 is\n−f + sup\n0≤t≤s\nexp(−ut)(Zt −K)+,\n(16.102)\nand the net return of the bank at time 0 is\nf −sup\n0≤t≤s\nexp(−vt)Z0(1 −K/Zt)+.\n(16.103)\n\n\nSection 16.6 - Uncertain Currency Model\n405\nIt follows from the fair price principle that the price of this contract should\nmake the investor and the bank have an identical expected return, i.e.,\n−f + E\n\u0014\nsup\n0≤t≤s\nexp(−ut)(Zt −K)+\n\u0015\n= f −E\n\u0014\nsup\n0≤t≤s\nexp(−vt)Z0(1 −K/Zt)+\n\u0015\n.\n(16.104)\nThus the American currency option price is given by the deﬁnition below.\nDeﬁnition 16.19 (Liu-Chen-Ralescu [155]) Assume an American currency\noption has a strike price K and an expiration time s. Then the American\ncurrency option price is\nf = 1\n2E\n\u0014\nsup\n0≤t≤s\nexp(−ut)(Zt −K)+\n\u0015\n+ 1\n2E\n\u0014\nsup\n0≤t≤s\nexp(−vt)Z0(1 −K/Zt)+\n\u0015\n.\nTheorem 16.11 (Liu-Chen-Ralescu [155]) Consider a general uncertain cur-\nrency model in which the domestic currency Xt, the foreign currency Yt and\nthe exchange rate Zt are determined by\n\n\n\n\n\n\n\ndXt = uXtdt\ndYt = vYtdt\ndZt = F(t, Zt)dt + G(t, Zt)dCt\n(16.105)\nwhere F and G are continuous functions. Assume an American currency\noption has a strike price K and an expiration time s. Then the American\ncurrency option price is\nf = 1\n2\nZ 1\n0\n\u0012\nsup\n0≤t≤s\nexp(−ut)(Zα\nt −K)+ + sup\n0≤t≤s\nexp(−vt)Z0(1 −K/Zα\nt )+\n\u0013\ndα\nwhere Zα\nt is the α-path of the corresponding uncertain diﬀerential equation.\nProof: It follows from Theorem 15.12 that the exchange rate Zt has an\ninverse uncertainty distribution\nΦ−1\nt (α) = Zα\nt .\nSince exp(−ut)(Zt−K)+ and exp(−vt)Z0(1−K/Zt)+ are increasing functions\nwith respect to Zt, it follows from the extreme value theorem of solution of\nuncertain diﬀerential equation that\nsup\n0≤t≤s\nexp(−ut)(Zt −K)+\nand\nsup\n0≤t≤s\nexp(−vt)Z0(1 −K/Zt)+\n\n\n406\nChapter 16 - Uncertain Finance\nhave inverse uncertainty distributions\nΨ−1\ns (α) = sup\n0≤t≤s\nexp(−ut)(Zα\nt −K)+,\nΥ−1\ns (α) = sup\n0≤t≤s\nexp(−vt)Z0(1 −K/Zα\nt )+,\nrespectively. By using Deﬁnition 16.19 and the expected value formula, we\nget the result.\nExercise 16.21: (Liu-Chen-Ralescu [155]) Consider the uncertain currency\nmodel\n\n\n\n\n\n\n\ndXt = uXtdt\n(Domestic Currency)\ndYt = vYtdt\n(Foreign Currency)\ndZt = eZtdt + σZtdCt\n(Exchange Rate)\n(16.106)\nwhere u, v, e and σ are constants with σ > 0. Assume an American currency\noption has a strike price K and an expiration time s. Show that the American\ncurrency option price is\nf = 1\n2\nZ 1\n0\nsup\n0≤t≤s\nexp(−ut)\n \nZ0 exp\n \net + σt\n√\n3\nπ\nln\nα\n1 −α\n!\n−K\n!+\ndα\n+1\n2\nZ 1\n0\nsup\n0≤t≤s\nexp(−vt)\n \nZ0 −K/ exp\n \net + σt\n√\n3\nπ\nln\nα\n1 −α\n!!+\ndα.\n16.7\nBibliographic Notes\nStochastic ﬁnance theory assumes that stock prices follow stochastic diﬀer-\nential equations. However, this preassumption was challenged by Liu [124] in\nwhich a convincing paradox was presented to show why real stock prices are\nimpossible to follow any stochastic diﬀerential equations. In order to support\nthis viewpoint, numerous empirical examples were reported, including stock\nprice (Liu-Liu [144]), currency exchange rate (Ye-Liu [282]), and interest rate\n(Yang-Ke [247]), among others.\nAs an alternative to stochastic ﬁnance theory, uncertain ﬁnance theory\nassumes that stock prices follow uncertain diﬀerential equations. The study\nof uncertain ﬁnance theory was started by Liu [116] in 2009 in which an un-\ncertain stock model was proposed, and European option price formulas were\nprovided. After that, numerous ﬁnancial derivatives were actively investi-\ngated, including American option (Chen [9]), Asian option (Sun-Chen [211]),\nbarrier option (Yao-Qin [276]), equity swap (Yu-Yang-Lei [288]), and equity\nwarrant (Shokrollahi [209]).\nUncertain diﬀerential equations were used to simulate ﬂoating interest\nrate by Chen-Gao [18] in 2013. Following that, Jiao-Yao [86] presented a\n\n\nSection 16.7 - Bibliographic Notes\n407\nprice formula of zero-coupon bond, Zhang-Ralescu-Liu [311] discussed the\nvaluation of interest rate ceiling and ﬂoor, and Xiao-Zhang-Fu [228] valuated\nthe interest rate swap.\nUncertain diﬀerential equations were employed to model currency ex-\nchange rate by Liu-Chen-Ralescu [155] in 2015 in which some currency option\nprice formulas were derived for the uncertain currency markets. Afterwards,\nuncertain currency models were also actively investigated among others by\nLiu [130], Shen-Yao [199], Wang-Ning [220], and Zhang-Gao-Fu [306].\n\n\n\n\nAppendix A\nChance Theory\nUncertain random variable was initialized by Liu [152] in 2013 for modelling\ncomplex systems with not only uncertainty but also randomness. This ap-\npendix will introduce the concepts of chance measure, uncertain random vari-\nable, chance distribution, operational law, expected value, variance, and law\nof large numbers. This appendix will also solve the choice problem in Ellsberg\nexperiment by uncertainty theory, probability theory and chance theory.\nA.1\nChance Measure\nLet (Γ, L, M) be an uncertainty space and let (Ω, A, Pr) be a probability\nspace.\nThen the product (Γ, L, M) × (Ω, A, Pr) is called a chance space.\nEssentially, it is another triplet,\n(Γ × Ω, L × A, M × Pr)\n(A.1)\nwhere Γ × Ωis the universal set, L × A is the product σ-algebra, and M × Pr\nis the product measure.\nThe universal set Γ × Ωis clearly the set of all ordered pairs of the form\n(γ, ω), where γ ∈Γ and ω ∈Ω. That is,\nΓ × Ω= {(γ, ω) | γ ∈Γ, ω ∈Ω} .\n(A.2)\nNote that Γ × Ωcan be understood as a rectangular coordinate system if Γ\nis understood as the horizontal axis and Ωis understood as the vertical axis.\nThe product σ-algebra L×A is the smallest σ-algebra containing measurable\nrectangles of the form Λ×A, where Λ ∈L and A ∈A. Each element in L×A\nis called an event in the chance space. What is the product measure M × Pr\nfor an event Θ? We will call M × Pr chance measure and represent it by\nCh{Θ}.\n\n\n410\nAppendix A - Chance Theory\nDeﬁnition A.1 (Liu [152]) Let (Γ, L, M)×(Ω, A, Pr) be a chance space, and\nlet Θ ∈L × A be an event. Then the chance measure of Θ is deﬁned as\nCh{Θ} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θ} ≥x} dx.\n(A.3)\nRemark A.1: Note that M{γ ∈Γ | (γ, ω) ∈Θ} is just the uncertain measure\nof cross section of Θ at ω. Since M{γ ∈Γ | (γ, ω) ∈Θ} can be regarded\nas a function from the probability space (Ω, A, Pr) to [0, 1], it is a random\nvariable. Thus the chance measure Ch{Θ} is just the expected value (i.e.,\naverage value) of this random variable.\nExercise A.1: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure, and take a probability space (Ω, A, Pr) to be\nalso (0, 1) with Borel algebra and Lebesgue measure. Then\nΘ = {(γ, ω) ∈Γ × Ω| γ + ω ≤1}\n(A.4)\nis an event in the chance space (Γ, L, M) × (Ω, A, Pr). Show that\nCh{Θ} = 1\n2.\n(A.5)\nExercise A.2: Take an uncertainty space (Γ, L, M) to be (0, 1) with Borel\nalgebra and Lebesgue measure, and take a probability space (Ω, A, Pr) to be\nalso (0, 1) with Borel algebra and Lebesgue measure. Then\nΘ =\n\b\n(γ, ω) ∈Γ × Ω| (γ −0.5)2 + (ω −0.5)2 < 0.52\t\n(A.6)\nis an event in the chance space (Γ, L, M) × (Ω, A, Pr). Show that\nCh{Θ} = π\n4 .\n(A.7)\nTheorem A.1 (Liu [152]) Let (Γ, L, M)×(Ω, A, Pr) be a chance space. Then\nCh{Λ × A} = M{Λ} × Pr{A}\n(A.8)\nfor any Λ ∈L and any A ∈A. Furthermore, we have\nCh{∅} = 0,\nCh{Γ × Ω} = 1.\n(A.9)\nProof: Let us ﬁrst prove the identity (A.8). For any real number x ∈(0, 1],\nif M{Λ} ≥x, then\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Λ × A} ≥x} = Pr{A}.\nIf M{Λ} < x, then\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Λ × A} ≥x} = Pr{∅} = 0.\n\n\nSection A.1 - Chance Measure\n411\nThus\nCh{Λ × A} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Λ × A} ≥x} dx\n=\nZ M{Λ}\n0\nPr{A}dx +\nZ 1\nM{Λ}\n0dx\n= M{Λ} × Pr{A}.\nFurthermore, it follows from (A.8) that\nCh{∅} = M{∅} × Pr{∅} = 0,\nCh{Γ × Ω} = M{Γ} × Pr{Ω} = 1.\nThe theorem is thus veriﬁed.\nTheorem A.2 (Liu [152], Monotonicity Theorem) The chance measure is\na monotone increasing set function. That is, for any events Θ1 and Θ2 with\nΘ1 ⊂Θ2, we have\nCh{Θ1} ≤Ch{Θ2}.\n(A.10)\nProof:\nSince Θ1 and Θ2 are two events with Θ1 ⊂Θ2, for each ω, we\nimmediately have\n{γ ∈Γ | (γ, ω) ∈Θ1} ⊂{γ ∈Γ | (γ, ω) ∈Θ2}\nand\nM{γ ∈Γ | (γ, ω) ∈Θ1} ≤M{γ ∈Γ | (γ, ω) ∈Θ2}.\nThus\nCh{Θ1} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θ1} ≥x} dx\n≤\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θ2} ≥x} dx\n= Ch{Θ2}.\nThat is, Ch{Θ} is a monotone increasing function with respect to Θ. The\ntheorem is thus veriﬁed.\nTheorem A.3 (Liu [152], Duality Theorem) The chance measure is self-\ndual. That is, for any event Θ, we have\nCh{Θ} + Ch{Θc} = 1.\n(A.11)\n\n\n412\nAppendix A - Chance Theory\nProof: Since both uncertain measure and probability measure are self-dual,\nwe have\nCh{Θ} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θ} ≥x} dx\n=\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θc} ≤1 −x} dx\n=\nZ 1\n0\n(1 −Pr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θc} > 1 −x}) dx\n= 1 −\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θc} > x} dx\n= 1 −Ch{Θc}.\nThat is, Ch{Θ} + Ch{Θc} = 1, i.e., the chance measure is self-dual.\nTheorem A.4 (Hou [69], Subadditivity Theorem) The chance measure is\nsubadditive. That is, for any countable sequence of events Θ1, Θ2, · · · , we\nhave\nCh\n( ∞\n[\ni=1\nΘi\n)\n≤\n∞\nX\ni=1\nCh{Θi}.\n(A.12)\nProof: At ﬁrst, it follows from the subadditivity of uncertain measure that\nM\n(\nγ ∈Γ | (γ, ω) ∈\n∞\n[\ni=1\nΘi\n)\n≤\n∞\nX\ni=1\nM{γ ∈Γ | (γ, ω) ∈Θi}.\nThus\nCh\n( ∞\n[\ni=1\nΘi\n)\n=\nZ 1\n0\nPr\n(\nω ∈Ω| M\n(\nγ ∈Γ | (γ, ω) ∈\n∞\n[\ni=1\nΘi\n)\n≥x\n)\ndx\n≤\nZ +∞\n0\nPr\n(\nω ∈Ω|\n∞\nX\ni=1\nM{γ ∈Γ | (γ, ω) ∈Θi} ≥x\n)\ndx\n=\n∞\nX\ni=1\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈Θi} ≥x} dx\n=\n∞\nX\ni=1\nCh{Θi}.\nThat is, the chance measure is subadditive.\n\n\nSection A.2 - Uncertain Random Variable\n413\nA.2\nUncertain Random Variable\nDeﬁnition A.2 (Liu [152]) An uncertain random variable is a function ξ\nfrom a chance space (Γ, L, M) × (Ω, A, Pr) to the set of real numbers such\nthat {ξ ∈B} is an event in L × A for any Borel set B of real numbers.\nRemark A.2: An uncertain random variable ξ(γ, ω) degenerates to a ran-\ndom variable if it does not vary with γ. Thus a random variable is a special\nuncertain random variable.\nRemark A.3: An uncertain random variable ξ(γ, ω) degenerates to an un-\ncertain variable if it does not vary with ω. Thus an uncertain variable is a\nspecial uncertain random variable.\nTheorem A.5 Let ξ1, ξ2, · · ·, ξn be uncertain random variables on the chance\nspace (Γ, L, M) × (Ω, A, Pr), and let f be a measurable function. Then\nξ = f(ξ1, ξ2, · · · , ξn)\n(A.13)\nis an uncertain random variable determined by\nξ(γ, ω) = f(ξ1(γ, ω), ξ2(γ, ω), · · · , ξn(γ, ω))\n(A.14)\nfor all (γ, ω) ∈Γ × Ω.\nProof: Since ξ1, ξ2, · · · , ξn are uncertain random variables, we know that\nthey are measurable functions on the chance space, and ξ = f(ξ1, ξ2, · · · , ξn)\nis also a measurable function. Hence ξ is an uncertain random variable.\nExample A.1: A random variable η plus an uncertain variable τ makes an\nuncertain random variable ξ, i.e.,\nξ(γ, ω) = η(ω) + τ(γ)\n(A.15)\nfor all (γ, ω) ∈Γ × Ω.\nExample A.2: A random variable η times an uncertain variable τ makes\nan uncertain random variable ξ, i.e.,\nξ(γ, ω) = η(ω) · τ(γ)\n(A.16)\nfor all (γ, ω) ∈Γ × Ω.\nTheorem A.6 (Liu [152]) Let ξ be an uncertain random variable on the\nchance space (Γ, L, M) × (Ω, A, Pr), and let B be a Borel set of real numbers.\nThen {ξ ∈B} is an uncertain random event with chance measure\nCh{ξ ∈B} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | ξ(γ, ω) ∈B} ≥x} dx.\n(A.17)\n\n\n414\nAppendix A - Chance Theory\nProof: Since {ξ ∈B} is an event in the chance space, the equation (A.17)\nfollows from Deﬁnition A.1 immediately.\nRemark A.4: If the uncertain random variable degenerates to a random\nvariable η, then Ch{η ∈B} = Ch{Γ × (η ∈B)} = M{Γ} × Pr{η ∈B} =\nPr{η ∈B}. That is,\nCh{η ∈B} = Pr{η ∈B}.\n(A.18)\nRemark A.5: If the uncertain random variable degenerates to an uncertain\nvariable τ, then Ch{τ ∈B} = Ch{(τ ∈B) × Ω} = M{τ ∈B} × Pr{Ω} =\nM{τ ∈B}. That is,\nCh{τ ∈B} = M{τ ∈B}.\n(A.19)\nTheorem A.7 (Liu [152]) Let ξ be an uncertain random variable. Then the\nchance measure Ch{ξ ∈B} is a monotone increasing function of B and\nCh{ξ ∈∅} = 0,\nCh{ξ ∈ℜ} = 1.\n(A.20)\nProof: Let B1 and B2 be Borel sets of real numbers with B1 ⊂B2. Then\nwe immediately have {ξ ∈B1} ⊂{ξ ∈B2}. It follows from the monotonicity\nof chance measure that\nCh{ξ ∈B1} ≤Ch{ξ ∈B2}.\nHence Ch{ξ ∈B} is a monotone increasing function of B. Furthermore, we\nhave\nCh{ξ ∈∅} = Ch{∅} = 0,\nCh{ξ ∈ℜ} = Ch{Γ × Ω} = 1.\nThe theorem is veriﬁed.\nTheorem A.8 (Liu [152]) Let ξ be an uncertain random variable. Then for\nany Borel set B of real numbers, we have\nCh{ξ ∈B} + Ch{ξ ∈Bc} = 1.\n(A.21)\nProof: It follows from {ξ ∈B}c = {ξ ∈Bc} and the duality of chance\nmeasure immediately.\nA.3\nChance Distribution\nDeﬁnition A.3 (Liu [152]) Let ξ be an uncertain random variable. Then\nits chance distribution is deﬁned by\nΦ(x) = Ch{ξ ≤x}\n(A.22)\nfor any x ∈ℜ.\n\n\nSection A.3 - Chance Distribution\n415\nExample A.3: As a special uncertain random variable, the chance distri-\nbution of a random variable η is just its probability distribution, that is,\nΦ(x) = Ch{η ≤x} = Pr{η ≤x}.\n(A.23)\nExample A.4: As a special uncertain random variable, the chance distri-\nbution of an uncertain variable τ is just its uncertainty distribution, that\nis,\nΦ(x) = Ch{τ ≤x} = M{τ ≤x}.\n(A.24)\nTheorem A.9 (Liu [152], Chance Inversion Theorem) Let ξ be an uncertain\nrandom variable with chance distribution Φ. Then for any real number x, we\nhave\nCh{ξ ≤x} = Φ(x),\nCh{ξ > x} = 1 −Φ(x).\n(A.25)\nProof: The equation Ch{ξ ≤x} = Φ(x) follows from the deﬁnition of chance\ndistribution immediately. By using the duality of chance measure, we get\nCh{ξ > x} = 1 −Ch{ξ ≤x} = 1 −Φ(x).\nTheorem A.10 (Liu [152], Suﬃcient and Necessary Condition for Chance\nDistribution) A real-valued function Φ(x) on ℜis a chance distribution if and\nonly if it is a monotone increasing function satisfying\n0 ≤Φ(x) ≤1,\n(A.26)\nΦ(x) ̸≡0,\n(A.27)\nΦ(x) ̸≡1,\n(A.28)\nΦ(x0) = 1 if Φ(x) = 1 for any x > x0.\n(A.29)\nProof: Suppose Φ is a chance distribution of some uncertain random variable\nξ. For any points x1 and x2 with x1 < x2, by using the monotonicity theorem,\nwe have\nΦ(x1) = Ch{ξ ≤x1} ≤Ch{ξ ≤x2} = Φ(x2).\nThus Φ is a monotone increasing function. For any point x, since\n0 ≤Ch{ξ ≤x} ≤1,\nwe have 0 ≤Φ(x) ≤1. By using the chance inversion theorem and subaddi-\ntivity theorem, we have\n1 = Ch{ξ ∈ℜ} = Ch\n( ∞\n[\nn=1\n(ξ ≤n)\n)\n≤\n∞\nX\nn=1\nCh{ξ ≤n} =\n∞\nX\nn=1\nΦ(n).\n\n\n416\nAppendix A - Chance Theory\nThus Φ(x) ̸≡0. Similarly, we have\n1 = Ch{ξ ∈ℜ} = Ch\n( ∞\n[\nn=1\n(ξ > −n)\n)\n≤\n∞\nX\nn=1\nCh{ξ > −n} =\n∞\nX\nn=1\n(1 −Φ(−n)).\nThus Φ(x) ̸≡1.\nFurthermore, let x0 be a given point.\nIf Φ(x) = 1 for\nany x > x0, then by using the chance inversion theorem and subadditivity\ntheorem, we obtain\n1 −Φ(x0) = Ch{ξ > x0}\n= Ch\n( ∞\n[\ni=1\n\u0012\nξ > x0 + 1\ni\n\u0013)\n≤\n∞\nX\ni=1\nCh\n\u001a\nξ > x0 + 1\ni\n\u001b\n=\n∞\nX\ni=1\n\u0012\n1 −Φ\n\u0012\nx0 + 1\ni\n\u0013\u0013\n= 0.\nThus Φ(x0) = 1 and (A.29) is veriﬁed.\nConversely, suppose that Φ is a monotone increasing function satisfying\n(A.26) to (A.29). It follows from Theorem 3.5 that there is an uncertain\nvariable whose uncertainty distribution is just Φ(x). Since an uncertain vari-\nable is a special uncertain random variable, we know that Φ is a chance\ndistribution.\nA.4\nOperational Law\nAssume η1, η2, · · · , ηm are independent random variables with probability\ndistributions Ψ1, Ψ2, · · · , Ψm, and τ1, τ2, · · · , τn are independent uncertain\nvariables with uncertainty distributions Υ1, Υ2, · · ·, Υn, respectively. What\nis the chance distribution of the uncertain random variable\nξ = f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn)?\n(A.30)\nThis section will provide an operational law to answer this question.\nTheorem A.11 (Liu [153]) Let η1, η2, · · · , ηm be independent random vari-\nables with probability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be\nindependent uncertain variables with uncertainty distributions Υ1, Υ2, · · ·, Υn,\nrespectively. If f is a measurable function, then the uncertain random vari-\nable\nξ = f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn)\n(A.31)\nhas a chance distribution\nΦ(x) =\nZ\nℜmF(x; y1, y2, · · · , ym)dΨ1(y1)dΨ2(y2) · · · dΨm(ym)\n(A.32)\n\n\nSection A.4 - Operational Law\n417\nwhere\nF(x; y1, y2, · · · , ym) = M{f(y1, y2, · · · , ym, τ1, τ2, · · · , τn) ≤x}\n(A.33)\nis the uncertainty distribution of f(y1, y2, · · · , ym, τ1, τ2, · · · , τn) for any real\nnumbers y1, y2, · · · , ym, and is determined by Υ1, Υ2, · · · , Υn.\nProof: It follows from Theorem A.6 that the uncertain random variable ξ\nhas a chance distribution\nΦ(x) =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | ξ(γ, ω) ≤x} ≥r} dr\n=\nZ 1\n0\nPr {ω ∈Ω| M{f(η1(ω), · · · , ηm(ω), τ1, · · · , τn) ≤x} ≥r} dr\n=\nZ\nℜmM{f(y1, y2, · · ·, ym, τ1, τ2, · · ·, τn) ≤x}dΨ1(y1) · · · dΨm(ym)\n=\nZ\nℜmF(x; y1, y2, · · ·, ym)dΨ1(y1)dΨ2(y2) · · · dΨm(ym).\nThe theorem is thus veriﬁed.\nExercise A.3:\nLet η1, η2, · · · , ηm be independent random variables with\nprobability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be indepen-\ndent uncertain variables with uncertainty distributions Υ1, Υ2, · · · , Υn, re-\nspectively. Show that the sum\nξ = η1 + η2 + · · · + ηm + τ1 + τ2 + · · · + τn\n(A.34)\nhas a chance distribution\nΦ(x) =\nZ +∞\n−∞\nΥ(x −y)dΨ(y)\n(A.35)\nwhere\nΨ(y) =\nZ\ny1+y2+···+ym≤y\ndΨ1(y1)dΨ2(y2) · · · dΨm(ym)\n(A.36)\nis the probability distribution of η1 + η2 + · · · + ηm, and\nΥ(z) =\nsup\nz1+z2+···+zn=z Υ1(z1) ∧Υ2(z2) ∧· · · ∧Υn(zn)\n(A.37)\nis the uncertainty distribution of τ1 + τ2 + · · · + τn.\nExercise A.4:\nLet η1, η2, · · · , ηm be independent positive random vari-\nables with probability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn\n\n\n418\nAppendix A - Chance Theory\nbe independent positive uncertain variables with uncertainty distributions\nΥ1, Υ2, · · · , Υn, respectively. Show that the product\nξ = η1η2 · · · ηmτ1τ2 · · · τn\n(A.38)\nhas a chance distribution\nΦ(x) =\nZ +∞\n0\nΥ(x/y)dΨ(y)\n(A.39)\nwhere\nΨ(y) =\nZ\ny1y2···ym≤y\ndΨ1(y1)dΨ2(y2) · · · dΨm(ym)\n(A.40)\nis the probability distribution of η1η2 · · · ηm, and\nΥ(z) =\nsup\nz1z2···zn=z Υ1(z1) ∧Υ2(z2) ∧· · · ∧Υn(zn)\n(A.41)\nis the uncertainty distribution of τ1τ2 · · · τn.\nExercise A.5:\nLet η1, η2, · · · , ηm be independent random variables with\nprobability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be indepen-\ndent uncertain variables with uncertainty distributions Υ1, Υ2, · · · , Υn, re-\nspectively. Show that the minimum\nξ = η1 ∧η2 ∧· · · ∧ηm ∧τ1 ∧τ2 ∧· · · ∧τn\n(A.42)\nhas a chance distribution\nΦ(x) = Ψ(x) + Υ(x) −Ψ(x)Υ(x)\n(A.43)\nwhere\nΨ(x) = 1 −(1 −Ψ1(x))(1 −Ψ2(x)) · · · (1 −Ψm(x))\n(A.44)\nis the probability distribution of η1 ∧η2 ∧· · · ∧ηm, and\nΥ(x) = Υ1(x) ∨Υ2(x) ∨· · · ∨Υn(x)\n(A.45)\nis the uncertainty distribution of τ1 ∧τ2 ∧· · · ∧τn.\nExercise A.6:\nLet η1, η2, · · · , ηm be independent random variables with\nprobability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be indepen-\ndent uncertain variables with uncertainty distributions Υ1, Υ2, · · · , Υn, re-\nspectively. Show that the maximum\nξ = η1 ∨η2 ∨· · · ∨ηm ∨τ1 ∨τ2 ∨· · · ∨τn\n(A.46)\nhas a chance distribution\nΦ(x) = Ψ(x)Υ(x)\n(A.47)\n\n\nSection A.4 - Operational Law\n419\nwhere\nΨ(x) = Ψ1(x)Ψ2(x) · · · Ψm(x)\n(A.48)\nis the probability distribution of η1 ∨η2 ∨· · · ∨ηm, and\nΥ(x) = Υ1(x) ∧Υ2(x) ∧· · · ∧Υn(x)\n(A.49)\nis the uncertainty distribution of τ1 ∨τ2 ∨· · · ∨τn.\nTheorem A.12 (Liu [153]) Let η1, η2, · · · , ηm be independent random vari-\nables with probability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be\nindependent uncertain variables with regular uncertainty distributions Υ1, Υ2,\n· · · , Υn, respectively. Assume f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn) is continuous,\nstrictly increasing with respect to τ1, τ2, · · · , τk and strictly decreasing with\nrespect to τk+1, τk+2, · · · , τn. Then the uncertain random variable\nξ = f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn)\n(A.50)\nhas a chance distribution\nΦ(x) =\nZ\nℜmF(x; y1, y2, · · · , ym)dΨ1(y1)dΨ2(y2) · · · dΨm(ym)\n(A.51)\nwhere F(x; y1, y2, · · · , ym) is the root α of the equation\nf(y1, y2, · · · , ym, Υ−1\n1 (α), · · · , Υ−1\nk (α), Υ−1\nk+1(1 −α), · · · , Υ−1\nn (1 −α)) = x.\nProof: Since F(x; y1, y2, · · · , ym) = M{f(y1, y2, · · · , ym, τ1, τ2, · · · , τn) ≤x}\nis just the root α of the equation\nf(y1, y2, · · · , ym, Υ−1\n1 (α), · · · , Υ−1\nk (α), Υ−1\nk+1(1 −α), · · · , Υ−1\nn (1 −α)) = x,\nwe get the result by Theorem A.11.\nOperational Law for Boolean System\nTheorem A.13 (Liu [153]) Assume η1, η2, · · · , ηm are independent Boolean\nrandom variables, i.e.,\nηi =\n(\n1 with probability measure ai\n0 with probability measure 1 −ai\n(A.52)\nfor i = 1, 2, · · · , m, and τ1, τ2, · · · , τn are independent Boolean uncertain\nvariables, i.e.,\nτj =\n(\n1 with uncertain measure bj\n0 with uncertain measure 1 −bj\n(A.53)\n\n\n420\nAppendix A - Chance Theory\nfor j = 1, 2, · · · , n. If f is a Boolean function, then\nξ = f(η1, · · · , ηm, τ1, · · · , τn)\n(A.54)\nis a Boolean uncertain random variable such that\nCh{ξ = 1} =\nX\n(x1,··· ,xm)∈{0,1}m\n m\nY\ni=1\nµi(xi)\n!\nf ∗(x1, · · · , xm)\n(A.55)\nwhere\nf ∗(x1, · · · , xm) =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nf(x1,··· ,xm,y1,··· ,yn)=1\nmin\n1≤j≤n νj(yj),\nif\nsup\nf(x1,··· ,xm,y1,··· ,yn)=1\nmin\n1≤j≤n νj(yj) < 0.5\n1 −\nsup\nf(x1,··· ,xm,y1,··· ,yn)=0\nmin\n1≤j≤n νj(yj),\nif\nsup\nf(x1,··· ,xm,y1,··· ,yn)=1\nmin\n1≤j≤n νj(yj) ≥0.5,\n(A.56)\nµi(xi) =\n(\nai,\nif xi = 1\n1 −ai,\nif xi = 0\n(i = 1, 2, · · · , m),\n(A.57)\nνj(yj) =\n(\nbj,\nif yj = 1\n1 −bj,\nif yj = 0\n(j = 1, 2, · · · , n).\n(A.58)\nProof: At ﬁrst, when (x1, · · · , xm) is given, f(x1, · · · , xm, τ1, · · · , τn) is es-\nsentially a Boolean function of uncertain variables. It follows from the oper-\national law of uncertain variables that\nM{f(x1, · · · , xm, τ1, · · · , τn) = 1} = f ∗(x1, · · · , xm)\nthat is determined by (A.56). On the other hand, it follows from the opera-\ntional law of uncertain random variables that\nCh{ξ = 1} =\nX\n(x1,··· ,xm)∈{0,1}m\n m\nY\ni=1\nµi(xi)\n!\nM{f(x1, · · · , xm, τ1, · · · , τn) = 1}.\nThus (A.55) is veriﬁed.\nRemark A.6: When the uncertain variables disappear, the operational law\nbecomes\nPr{ξ = 1} =\nX\n(x1,x2,··· ,xm)∈{0,1}m\n m\nY\ni=1\nµi(xi)\n!\nf(x1, x2, · · · , xm).\n(A.59)\n\n\nSection A.5 - Expected Value\n421\nRemark A.7: When the random variables disappear, the operational law\nbecomes\nM{ξ = 1} =\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nsup\nf(y1,y2,··· ,yn)=1\nmin\n1≤j≤n νj(yj),\nif\nsup\nf(y1,y2,··· ,yn)=1\nmin\n1≤j≤n νj(yj) < 0.5\n1 −\nsup\nf(y1,y2,··· ,yn)=0\nmin\n1≤j≤n νj(yj),\nif\nsup\nf(y1,y2,··· ,yn)=1\nmin\n1≤j≤n νj(yj) ≥0.5.\n(A.60)\nExercise A.7: Let η1, η2, · · · , ηm be independent Boolean random variables\ndeﬁned by (A.52) and let τ1, τ2, · · · , τn be independent Boolean uncertain\nvariables deﬁned by (A.53). Then the minimum\nξ = η1 ∧η2 ∧· · · ∧ηm ∧τ1 ∧τ2 ∧· · · ∧τn\n(A.61)\nis a Boolean uncertain random variable. Show that\nCh{ξ = 1} = a1a2 · · · am(b1 ∧b2 ∧· · · ∧bn).\n(A.62)\nExercise A.8: Let η1, η2, · · · , ηm be independent Boolean random variables\ndeﬁned by (A.52) and let τ1, τ2, · · · , τn be independent Boolean uncertain\nvariables deﬁned by (A.53). Then the maximum\nξ = η1 ∨η2 ∨· · · ∨ηm ∨τ1 ∨τ2 ∨· · · ∨τn\n(A.63)\nis a Boolean uncertain random variable. Show that\nCh{ξ = 1} = 1 −(1 −a1)(1 −a2) · · · (1 −am)(1 −b1 ∨b2 ∨· · · ∨bn). (A.64)\nA.5\nExpected Value\nDeﬁnition A.4 (Liu [152]) Let ξ be an uncertain random variable. Then\nits expected value is deﬁned by\nE[ξ] =\nZ +∞\n0\nCh{ξ ≥x}dx −\nZ 0\n−∞\nCh{ξ ≤x}dx\n(A.65)\nprovided that at least one of the two integrals is ﬁnite.\nTheorem A.14 (Liu [152]) Let ξ be an uncertain random variable with\nchance distribution Φ. Then\nE[ξ] =\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx.\n(A.66)\n\n\n422\nAppendix A - Chance Theory\nProof:\nIt follows from the chance inversion theorem that for almost all\nnumbers x, we have Ch{ξ ≥x} = 1−Φ(x) and Ch{ξ ≤x} = Φ(x). By using\nthe deﬁnition of expected value operator, we obtain\nE[ξ] =\nZ +∞\n0\nCh{ξ ≥x}dx −\nZ 0\n−∞\nCh{ξ ≤x}dx\n=\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx.\nThus we obtain the equation (A.66).\nTheorem A.15 Let ξ be an uncertain random variable with regular chance\ndistribution Φ. Then\nE[ξ] =\nZ 1\n0\nΦ−1(α)dα.\n(A.67)\nProof: Since α = Φ(x) and x = Φ−1(α) represent the same curve in the\nrectangular coordinate system (x, α), we have\nZ +∞\n0\n(1 −Φ(x))dx =\nZ 1\nΦ(0)\nΦ−1(α)dα\n(A.68)\nbecause the two integrals make an identical acreage. Similarly, we also have\nZ 0\n−∞\nΦ(x)dx = −\nZ Φ(0)\n0\nΦ−1(α)dα.\n(A.69)\nIt follows from Theorem A.14, (A.68) and (A.69) that the expected value is\nE[ξ] =\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx\n=\nZ 1\nΦ(0)\nΦ−1(α)dα +\nZ Φ(0)\n0\nΦ−1(α)dα\n=\nZ 1\n0\nΦ−1(α)dα.\nThe theorem is proved.\nTheorem A.16 (Liu [153]) Let η1, η2, · · · , ηm be independent random vari-\nables with probability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be\nindependent uncertain variables with uncertainty distributions Υ1, Υ2, · · ·, Υn,\nrespectively. If f is a measurable function, then\nξ = f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn)\n(A.70)\n\n\nSection A.5 - Expected Value\n423\nhas an expected value\nE[ξ] =\nZ\nℜm G(y1, y2, · · · , ym)dΨ1(y1)dΨ2(y2) · · · dΨm(ym)\n(A.71)\nwhere\nG(y1, y2, · · · , ym) = E[f(y1, y2, · · · , ym, τ1, τ2, · · · , τn)]\n(A.72)\nis the expected value of the uncertain variable f(y1, y2, · · · , ym, τ1, τ2, · · · , τn)\nfor any real numbers y1, y2, · · · , ym, and is determined by Υ1, Υ2, · · · , Υn.\nProof:\nFor simplicity, we only prove the case m = n = 2.\nWrite the\nuncertainty distribution of f(y1, y2, τ1, τ2) by F(x; y1, y2) for any real numbers\ny1 and y2. Then\nE[f(y1, y2, τ1, τ2)] =\nZ +∞\n0\n(1 −F(x; y1, y2))dx −\nZ 0\n−∞\nF(x; y1, y2)dx.\nOn the other hand, the uncertain random variable ξ = f(η1, η2, τ1, τ2) has a\nchance distribution\nΦ(x) =\nZ\nℜ2 F(x; y1, y2)dΨ1(y1)dΨ2(y2).\nIt follows from Theorem A.14 and Fubini theorem that\nE[ξ] =\nZ +∞\n0\n(1 −Φ(x))dx −\nZ 0\n−∞\nΦ(x)dx\n=\nZ +∞\n0\n\u0012\n1 −\nZ\nℜ2 F(x; y1, y2)dΨ1(y1)dΨ2(y2)\n\u0013\ndx\n−\nZ 0\n−∞\nZ\nℜ2 F(x; y1, y2)dΨ1(y1)dΨ2(y2)dx\n=\nZ\nℜ2\n\u0012Z +∞\n0\n(1 −F(x; y1, y2))dx −\nZ 0\n−∞\nF(x; y1, y2)dx\n\u0013\ndΨ1(y1)dΨ2(y2)\n=\nZ\nℜ2 E[f(y1, y2, τ1, τ2)]dΨ1(y1)dΨ2(y2).\nThus the theorem is proved.\nExercise A.9: Let η be a random variable and let τ be an uncertain variable.\nShow that\nE[η + τ] = E[η] + E[τ]\n(A.73)\nand\nE[ητ] = E[η]E[τ].\n(A.74)\n\n\n424\nAppendix A - Chance Theory\nTheorem A.17 (Liu [153]) Let η1, η2, · · · , ηm be independent random vari-\nables with probability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn be\nindependent uncertain variables with regular uncertainty distributions Υ1, Υ2,\n· · · , Υn, respectively. If f(η1, · · · , ηm, τ1, · · · , τn) is a continuous and strictly\nincreasing function (or strictly decreasing function) with respect to τ1, · · · , τn,\nthen the expected function\nE[f(η1, · · · , ηm, τ1, · · · , τn)]\n(A.75)\nis equal to\nZ\nℜm\nZ 1\n0\nf(y1, · · · , ym, Υ−1\n1 (α), · · · , Υ−1\nn (α))dαdΨ1(y1) · · · dΨm(ym).\nProof: Since f(y1, · · · , ym, τ1, · · · , τn) is a continuous and strictly increasing\nfunction (or strictly decreasing function) with respect to τ1, · · · , τn, we have\nE[f(y1, · · · , ym, τ1, · · · , τn)] =\nZ 1\n0\nf(y1, · · · , ym, Υ−1\n1 (α), · · · , Υ−1\nn (α))dα.\nIt follows from Theorem A.16 that the result holds.\nRemark A.8: If f(η1, · · · , ηm, τ1, · · · , τn) is continuous, strictly increasing\nwith respect to τ1, · · · , τk and strictly decreasing with respect to τk+1, · · · , τn,\nthen the integrand in the formula of expected value should be replaced with\nf(y1, · · · , ym, Υ−1\n1 (α), · · · , Υ−1\nk (α), Υ−1\nk+1(1 −α), · · · , Υ−1\nn (1 −α)).\nExercise A.10: Let η be a random variable with probability distribution\nΨ, and let τ be an uncertain variable with regular uncertainty distribution\nΥ. Show that\nE[η ∨τ] =\nZ\nℜ\nZ 1\n0\ny ∨Υ−1(α)\n\u0001\ndαdΨ(y)\n(A.76)\nand\nE[η ∧τ] =\nZ\nℜ\nZ 1\n0\ny ∧Υ−1(α)\n\u0001\ndαdΨ(y).\n(A.77)\nTheorem A.18 (Liu [153], Linearity of Expected Value Operator) Assume\nη1 and η2 are random variables (not necessarily independent), τ1 and τ2 are\nindependent uncertain variables, and f1 and f2 are measurable functions.\nThen\nE[f1(η1, τ1) + f2(η2, τ2)] = E[f1(η1, τ1)] + E[f2(η2, τ2)].\n(A.78)\n\n\nSection A.6 - Variance\n425\nProof: Since τ1 and τ2 are independent uncertain variables, for any real\nnumbers y1 and y2, the functions f1(y1, τ1) and f2(y2, τ2) are also independent\nuncertain variables. Thus\nE[f1(y1, τ1) + f2(y2, τ2)] = E[f1(y1, τ1)] + E[f2(y2, τ2)].\nLet Ψ1 and Ψ2 be the probability distributions of random variables η1 and\nη2, respectively. Then we have\nE[f1(η1, τ1) + f2(η2, τ2)]\n=\nZ\nℜ2 E[f1(y1, τ1) + f2(y2, τ2)]dΨ1(y1)dΨ2(y2)\n=\nZ\nℜ2(E[f1(y1, τ1)] + E[f2(y2, τ2)])dΨ1(y1)dΨ2(y2)\n=\nZ\nℜ\nE[f1(y1, τ1)]dΨ1(y1) +\nZ\nℜ\nE[f2(y2, τ2)]dΨ2(y2)\n= E[f1(η1, τ1)] + E[f2(η2, τ2)].\nThe theorem is proved.\nExercise A.11: Assume η1 and η2 are random variables, and τ1 and τ2 are\nindependent uncertain variables. Show that\nE[η1 ∨τ1 + η2 ∧τ2] = E[η1 ∨τ1] + E[η2 ∧τ2].\n(A.79)\nA.6\nVariance\nDeﬁnition A.5 (Liu [152]) Let ξ be an uncertain random variable with ﬁnite\nexpected value e. Then the variance of ξ is\nV [ξ] = E[(ξ −e)2].\n(A.80)\nSince (ξ −e)2 is a nonnegative uncertain random variable, we also have\nV [ξ] =\nZ +∞\n0\nCh{(ξ −e)2 ≥x}dx.\n(A.81)\nTheorem A.19 (Liu [152]) If ξ is an uncertain random variable with ﬁnite\nexpected value, a and b are real numbers, then\nV [aξ + b] = a2V [ξ].\n(A.82)\nProof: Let e be the expected value of ξ. Then aξ + b has an expected value\nae + b. Thus the variance is\nV [aξ + b] = E[(aξ + b −(ae + b))2] = E[a2(ξ −e)2] = a2V [ξ].\nThe theorem is veriﬁed.\n\n\n426\nAppendix A - Chance Theory\nTheorem A.20 (Liu [152]) Let ξ be an uncertain random variable with ex-\npected value e. Then V [ξ] = 0 if and only if Ch{ξ = e} = 1.\nProof: We ﬁrst assume V [ξ] = 0. It follows from the equation (A.81) that\nZ +∞\n0\nCh{(ξ −e)2 ≥x}dx = 0\nwhich implies Ch{(ξ −e)2 ≥x} = 0 for any x > 0. Hence we have\nCh{(ξ −e)2 = 0} = 1.\nThat is, Ch{ξ = e} = 1. Conversely, assume Ch{ξ = e} = 1. Then we\nimmediately have Ch{(ξ −e)2 = 0} = 1 and Ch{(ξ −e)2 ≥x} = 0 for any\nx > 0. Thus\nV [ξ] =\nZ +∞\n0\nCh{(ξ −e)2 ≥x}dx = 0.\nThe theorem is proved.\nHow to Obtain Variance from Distributions?\nLet ξ be an uncertain random variable with expected value e. If we only\nknow its chance distribution Φ, then the variance\nV [ξ] =\nZ +∞\n0\nCh{(ξ −e)2 ≥x}dx\n=\nZ +∞\n0\nCh{(ξ ≥e + √x) ∪(ξ ≤e −√x)}dx\n≤\nZ +∞\n0\n(Ch{ξ ≥e + √x} + Ch{ξ ≤e −√x})dx\n=\nZ +∞\n0\n(1 −Φ(e + √x) + Φ(e −√x))dx.\nThus we have the following stipulation.\nStipulation A.1 (Guo-Wang [63]) Let ξ be an uncertain random variable\nwith chance distribution Φ and ﬁnite expected value e. Then\nV [ξ] =\nZ +∞\n0\n(1 −Φ(e + √x) + Φ(e −√x))dx.\n(A.83)\nTheorem A.21 (Sheng-Yao [201]) Let ξ be an uncertain random variable\nwith regular chance distribution Φ and ﬁnite expected value e. Then\nV [ξ] =\nZ 1\n0\n(Φ−1(α) −e)2dα.\n(A.84)\n\n\nSection A.6 - Variance\n427\nProof: Since α = Φ(e + √x) on (0, +∞) and x = (Φ−1(α) −e)2 on (Φ(e), 1)\nrepresent the same curve in the rectangular coordinate system (x, α), we have\nZ +∞\n0\n(1 −Φ(e + √x))dx =\nZ 1\nΦ(e)\n(Φ−1(α) −e)2dα\n(A.85)\nbecause the two integrals make an identical acreage. Since α = Φ(e −√x)\non (0, +∞) and x = (Φ−1(α) −e)2 on (0, Φ(e)) represent the same curve, we\nhave\nZ +∞\n0\nΦ(e −√x)dx =\nZ Φ(e)\n0\n(Φ−1(α) −e)2dα.\n(A.86)\nIt follows from Stipulation A.1, (A.85) and (A.86) that the variance is\nV [ξ] =\nZ +∞\n0\n(1 −Φ(e + √x) + Φ(e −√x))dx\n=\nZ 1\nΦ(e)\n(Φ−1(α) −e)2dα +\nZ Φ(e)\n0\n(Φ−1(α) −e)2dα\n=\nZ 1\n0\n(Φ−1(α) −e)2dα.\nThe theorem is proved.\nTheorem A.22 (Guo-Wang [63]) Let η1, η2, · · · , ηm be independent random\nvariables with probability distributions Ψ1, Ψ2, · · · , Ψm, and let τ1, τ2, · · · , τn\nbe independent uncertain variables with regular uncertainty distributions Υ1,\nΥ2, · · · , Υn, respectively. Assume f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn) is contin-\nuous, strictly increasing with respect to τ1, τ2, · · · , τk and strictly decreasing\nwith respect to τk+1, τk+2, · · · , τn. Then\nξ = f(η1, η2, · · · , ηm, τ1, τ2, · · · , τn)\n(A.87)\nhas a variance\nV [ξ] =\nZ\nℜm\nZ +∞\n0\n(1 −F(e + √x; y1, y2, · · · , ym)\n+F(e −√x; y1, y2, · · · , ym))dxdΨ1(y1)dΨ2(y2) · · · Ψm(ym)\nwhere F(x; y1, y2, · · · , ym) is the root α of the equation\nf(y1, y2, · · · , ym, Υ−1\n1 (α), · · · , Υ−1\nk (α), Υ−1\nk+1(1 −α), · · · , Υ−1\nn (1 −α)) = x.\nProof: It follows from the operational law of uncertain random variables\nthat ξ has a chance distribution\nΦ(x) =\nZ\nℜm F(x; y1, y2, · · · , ym)dΨ1(y1)dΨ2(y2) · · · Ψm(ym)\n\n\n428\nAppendix A - Chance Theory\nwhere F(x; y1, y2, · · · , ym) is the uncertainty distribution of the uncertain\nvariable f(y1, y2, · · · , ym, τ1, τ2, · · · , τn). Thus the theorem follows Stipula-\ntion A.1 immediately.\nExercise A.12: Let η be a random variable with probability distribution\nΨ, and let τ be an uncertain variable with uncertainty distribution Υ. Show\nthat the sum\nξ = η + τ\n(A.88)\nhas a variance\nV [ξ] =\nZ +∞\n−∞\nZ +∞\n0\n(1 −Υ(e + √x −y) + Υ(e −√x −y))dxdΨ(y).\n(A.89)\nA.7\nLaw of Large Numbers\nTheorem A.23 (Yao-Gao [264], Law of Large Numbers) Let η1, η2, · · · be\niid random variables with a common probability distribution Ψ, and let τ1, τ2,\n· · · be iid uncertain variables. Assume f is a strictly monotone function.\nThen\nSn = f(η1, τ1) + f(η2, τ2) + · · · + f(ηn, τn)\n(A.90)\nis a sequence of uncertain random variables and\nSn\nn →\nZ +∞\n−∞\nf(y, τ1)dΨ(y)\n(A.91)\nin the sense of convergence in distribution as n →∞.\nProof: According to the deﬁnition of convergence in distribution, it suﬃces\nto prove\nlim\nn→∞Ch\n\u001aSn\nn ≤\nZ +∞\n−∞\nf(y, z)dΨ(y)\n\u001b\n= M\n\u001aZ +∞\n−∞\nf(y, τ1)dΨ(y) ≤\nZ +∞\n−∞\nf(y, z)dΨ(y)\n\u001b\n(A.92)\nfor any real number z (i.e., continuous point) with\nlim\nw→z M\n\u001aZ +∞\n−∞\nf(y, τ1)dΨ(y) ≤\nZ +∞\n−∞\nf(y, w)dΨ(y)\n\u001b\n= M\n\u001aZ +∞\n−∞\nf(y, τ1)dΨ(y) ≤\nZ +∞\n−∞\nf(y, z)dΨ(y)\n\u001b\n.\nThe argument breaks into two cases. Case 1: Assume f(y, z) is strictly in-\ncreasing with respect to z. Let Υ denote the common uncertainty distribution\nof τ1, τ2, · · · It is clear that\nM{f(y, τ1) ≤f(y, z)} = M{τ1 ≤z} = Υ(z)\n\n\nSection A.7 - Law of Large Numbers\n429\nfor any real numbers y and z. Thus we have\nM\n\u001aZ +∞\n−∞\nf(y, τ1)dΨ(y) ≤\nZ +∞\n−∞\nf(y, z)dΨ(y)\n\u001b\n= Υ(z).\n(A.93)\nIn addition, since f(η1, z), f(η2, z), · · · are a sequence of iid random variables,\nthe law of large numbers for random variables tells us that\nf(η1, z) + f(η2, z) + · · · + f(ηn, z)\nn\n→\nZ +∞\n−∞\nf(y, z)dΨ(y),\na.s.\nas n →∞. Thus\nlim\nn→∞Ch\n\u001aSn\nn ≤\nZ +∞\n−∞\nf(y, z)dΨ(y)\n\u001b\n= Υ(z).\n(A.94)\nIt follows from (A.93) and (A.94) that (A.92) holds. Case 2: Assume f(y, z)\nis strictly decreasing with respect to z. Then −f(y, z) is strictly increasing\nwith respect to z. By using Case 1, we obtain\nlim\nn→∞Ch\n\u001a\n−Sn\nn < −z\n\u001b\n= M\n\u001a\n−\nZ +∞\n−∞\nf(y, τ1)dΨ(y) < −z\n\u001b\n.\nThat is,\nlim\nn→∞Ch\n\u001aSn\nn > z\n\u001b\n= M\n\u001aZ +∞\n−∞\nf(y, τ1)dΨ(y) > z\n\u001b\n.\nIt follows from the duality property that\nlim\nn→∞Ch\n\u001aSn\nn ≤z\n\u001b\n= M\n\u001aZ +∞\n−∞\nf(y, τ1)dΨ(y) ≤z\n\u001b\n.\nThe theorem is thus proved.\nExercise A.13: Let η1, η2, · · · be iid random variables, and let τ1, τ2, · · · be\niid uncertain variables. Deﬁne\nSn = (η1 + τ1) + (η2 + τ2) + · · · + (ηn + τn).\n(A.95)\nShow that\nSn\nn →E[η1] + τ1\n(A.96)\nin the sense of convergence in distribution as n →∞. Especially, if\nSn = τ1 + τ2 + · · · + τn,\n(A.97)\nthen\nSn\nn →τ1\n(A.98)\n\n\n430\nAppendix A - Chance Theory\nin the sense of convergence in distribution as n →∞.\nExercise A.14:\nLet η1, η2, · · · be iid positive random variables, and let\nτ1, τ2, · · · be iid positive uncertain variables. Deﬁne\nSn = η1τ1 + η2τ2 + · · · + ηnτn.\n(A.99)\nShow that\nSn\nn →E[η1]τ1\n(A.100)\nin the sense of convergence in distribution as n →∞.\nA.8\nEllsberg Experiment\nAssume an urn contains 30 red balls and 60 other balls that are either black\nor yellow in unknown proportion. One ball is randomly drawn from the urn.\nConsider the following two options:\nA: You receive $30 if the drawn ball is red;\nB: You receive $30 if the drawn ball is black1.\nWhat is your choice between A and B? Through a lot of surveys, Ellsberg [36]\nshowed that most people strictly prefer A to B since they prefer gambling on\na known number of balls to gambling on an unknown number. However, the\nchoice problem is clearly a scientiﬁc one. Does it make sense to vote on such a\nscientiﬁc problem? Deﬁnitely no! What we need is a scientiﬁc method rather\nthan public opinion. Therefore, in order to solve the choice problem, Liu [134]\npioneered a rigorous mathematical solution comprehensively by uncertainty\ntheory, probability theory and chance theory, and concluded that we should\nbe indiﬀerent between the two options.\nAt ﬁrst, all balls are virtually numbered from 1 to 90 in order of ﬁrst\nblack, then yellow and ﬁnally red. Take an uncertainty space (Γ, L, M) to be\n{0, 1, 2, · · · , 60} with power set and uncertain measure\nM{Λ} = |Λ|\n61\n(A.101)\nwhere |Λ| represents the cardinality of Λ. Since the composition of black and\nyellow balls is completely unknown and can be any integer pair among\n(0, 60), (1, 59), (2, 58), · · · , (58, 2), (59, 1), (60, 0)\nwith equal belief degrees, we can treat the number of black balls as an un-\ncertain variable\nξ(γ) = γ,\n(A.102)\n1The color is arbitrarily chosen by you from black and yellow.\n\n\nSection A.8 - Ellsberg Experiment\n431\nand then the number of yellow balls is another uncertain variable\nη(γ) = 60 −γ.\n(A.103)\nIt is easy to verify that ξ and η are identically distributed uncertain variables\nwith\nξ = i with belief degree 1\n61,\ni = 0, 1, 2, · · · , 60,\n(A.104)\nη = j with belief degree 1\n61,\nj = 0, 1, 2, · · · , 60,\n(A.105)\nand ξ +η ≡60. Thus the formulations (A.102) and (A.103) are indeed “fair”\nfor both black and yellow balls. Therefore, the black balls are numbered from\n1 to γ, the yellow balls are numbered from γ + 1 to 60, and the red balls are\nnumbered from 61 to 90.\nTake a probability space (Ω, A, Pr) to be {1, 2, · · · , 90} with power set\nand probability measure\nPr{Λ} = |Λ|\n90 .\n(A.106)\nThus drawing one ball in an equally likely manner from the urn is equivalent\nto sampling one ω from the probability space (Ω, A, Pr).\nSince drawing a ball from the urn is a mixture of uncertainty (unknown\nnumber of balls) and randomness (randomly drawing a ball), it has to be\nrepresented by an event in the chance space\n(Γ, L, M) × (Ω, A, Pr).\n(A.107)\nEspecially, a red ball is drawn if and only if ω ≥61. Thus drawing a red ball\nis represented by the event,\n“red” = {(γ, ω) ∈Γ × Ω| ω ≥61}.\n(A.108)\nA black ball is drawn if and only if ω ≤γ. Thus drawing a black ball is\nrepresented by the event,\n“black” = {(γ, ω) ∈Γ × Ω| ω ≤γ}.\n(A.109)\nA yellow ball is drawn if and only if ω > γ and ω ≤60. Thus drawing a\nyellow ball is represented by the event,\n“yellow” = {(γ, ω) ∈Γ × Ω| γ < ω ≤60}.\n(A.110)\nSee Figure A.1.\n\n\n432\nAppendix A - Chance Theory\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΓ\nΩ\n90\n60\n1\n0\n60\n“red”\n“yellow”\n“black”\nFigure A.1: Three Events: “red”, “black” and “yellow”\nIt follows from Deﬁnition A.1 that the chance measure of drawing a red\nball is\nCh{“red”} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈“red”} ≥x} dx\n=\nZ 1\n0\nPr {ω ∈{61, 62, · · · , 90} | M{0, 1, · · · , 60} ≥x} dx\n=\nZ 1\n0\nPr\n\u001a\nω ∈{61, 62, · · · , 90} | 61\n61 ≥x\n\u001b\ndx\n=\nZ 1\n0\nPr {61, 62, · · · , 90} dx\n=\nZ 1\n0\n30\n90dx\n= 1\n3,\n\n\nSection A.8 - Ellsberg Experiment\n433\nthe chance measure of drawing a black ball is\nCh{“black”} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈“black”} ≥x} dx\n=\nZ 1\n0\nPr {ω ∈{1, 2, · · · , 60} | M{ω, ω + 1, · · · , 60} ≥x} dx\n=\nZ 1\n0\nPr\n\u001a\nω ∈{1, 2, · · · , 60} | 61 −ω\n61\n≥x\n\u001b\ndx\n=\n60\nX\nk=0\nZ\nk+1\n61\nk\n61\nPr\n\u001a\nω ∈{1, 2, · · · , 60} | 61 −ω\n61\n≥x\n\u001b\ndx\n=\n60\nX\nk=0\nZ\nk+1\n61\nk\n61\nPr {1, 2, · · · , 60 −k} dx\n=\n60\nX\nk=0\nZ\nk+1\n61\nk\n61\n60 −k\n90\ndx\n= 1\n3,\nand the chance measure of drawing a yellow ball is\nCh{“yellow”} =\nZ 1\n0\nPr {ω ∈Ω| M{γ ∈Γ | (γ, ω) ∈“yellow”} ≥x} dx\n=\nZ 1\n0\nPr {ω ∈{1, 2, · · · , 60} | M{0, 1, · · · , ω −1} ≥x} dx\n=\nZ 1\n0\nPr\nn\nω ∈{1, 2, · · · , 60} | ω\n61 ≥x\no\ndx\n=\n60\nX\nk=0\nZ\nk+1\n61\nk\n61\nPr\nn\nω ∈{1, 2, · · · , 60} | ω\n61 ≥x\no\ndx\n=\n60\nX\nk=0\nZ\nk+1\n61\nk\n61\nPr {k + 1, k + 2, · · · , 60} dx\n=\n60\nX\nk=0\nZ\nk+1\n61\nk\n61\n60 −k\n90\ndx\n= 1\n3.\nNow we are ready to solve the choice problem. The income of A is an\n\n\n434\nAppendix A - Chance Theory\nuncertain random variable\nA(γ, ω) =\n(\n30,\nif (γ, ω) ∈“red”\n0,\notherwise\n(A.111)\nwhose expected value is\nE[A] = 30 × Ch{“red”} + 0 × (1 −Ch{“red”}) = 10.\n(A.112)\nThe income of B is an uncertain random variable\nB(γ, ω) =\n(\n30,\nif (γ, ω) ∈“black”\n0,\notherwise\n(A.113)\nwhose expected value is\nE[B] = 30 × Ch{“black”} + 0 × (1 −Ch{“black”}) = 10.\n(A.114)\nIt follows that\nE[A] = E[B].\n(A.115)\nTherefore, Liu [134] concluded that we should be indiﬀerent between A and B.\nSimulation experiments also veriﬁed this conclusion. It is thus unreasonable\nto strictly prefer A to B.\nA New Problem\nIn order to further explore this issue, Liu [134] revised the choice problem as\nfollows: What is your choice if B is replaced with\nC: You receive $31 if the drawn ball is black?\nThrough a lot of surveys, Liu [134] showed that most people continue to\nprefer A to C. However, the income of C is an uncertain random variable\nC(γ, ω) =\n(\n31,\nif (γ, ω) ∈“black”\n0,\notherwise\n(A.116)\nwhose expected value is\nE[C] = 31 × Ch{“black”} + 0 × (1 −Ch{“black”}) = 31\n3 .\n(A.117)\nIt follows that\nE[A] < E[C].\n(A.118)\nTherefore, Liu [134] concluded that we should prefer C to A. This conclusion\nwas also conﬁrmed through simulation experiments. It is thus unreasonable\nto prefer A to C.\n\n\nSection A.8 - Ellsberg Experiment\n435\nExercise A.15: An urn contains 30 red balls and 60 other balls that are\neither black or yellow in unknown proportion. One ball is randomly drawn\nfrom the urn. Consider the following three options:\nA: You receive $30 if the drawn ball is black;\nB: You receive $b if the drawn ball is black;\nC: You receive $y if the drawn ball is black;\nwhere b and y are the numbers of black and yellow balls in the urn, respec-\ntively. Through surveys, Eliaz-Ortoleva [34] showed that 36% of people bet\non A, 52% bet on B, and 12% bet on C. What is your choice among them if\nuncertainty theory, probability theory and chance theory are comprehensively\nused? Hint: The incomes of A, B and C are uncertain random variables,\nA(γ, ω) =\n(\n30,\nif (γ, ω) ∈“black”\n0,\notherwise,\n(A.119)\nB(γ, ω) =\n(\nγ,\nif (γ, ω) ∈“black”\n0,\notherwise,\n(A.120)\nC(γ, ω) =\n(\n60 −γ,\nif (γ, ω) ∈“black”\n0,\notherwise,\n(A.121)\nwhere “black” = {(γ, ω) ∈Γ × Ω| ω ≤γ} on the chance space (A.107).\n(Please refer to Liu-Qin [149].)\nExercise A.16: An urn contains 30 red balls and 60 other balls that are\neither black or yellow in unknown proportion. Two balls are randomly drawn\nfrom the urn.\n(i) How likely is it that the two drawn balls are red?\n(ii) How likely is it that the two drawn balls are black?\nHint: Take an uncertainty space (Γ, L, M) to be {0, 1, 2, · · · , 60} with power\nset and uncertain measure\nM{Λ} = |Λ|\n|Γ| ,\nand take a probability space (Ω, A, Pr) to be {(i, j) | i, j = 1, 2, · · · , 90, i ̸= j}\nwith power set and probability measure\nPr{Λ} = |Λ|\n|Ω|.\nThen\n“Two drawn balls are red” =\n\b\n(γ, ω1, ω2) ∈Γ × Ω\n\f\n\f ω1 ∧ω2 ≥61\n\t\n,\n\n\n436\nAppendix A - Chance Theory\n“Two drawn balls are black” =\n\b\n(γ, ω1, ω2) ∈Γ × Ω\n\f\n\f ω1 ∨ω2 ≤γ\n\t\n.\nExercise A.17: An urn contains 30 red balls and 60 other balls that are\neither black or yellow in unknown proportion.\nThree balls are randomly\ndrawn from the urn. What is the most probable color distribution among\n3-0-0 (three drawn balls are of the same color), 2-1-0 (only two drawn balls\nare of the same color), and 1-1-1 (three drawn balls are of diﬀerent colors)?\nPlease justify your answer. (Please refer to Lio-Cheng [109].)\nA.9\nBibliographic Notes\nIn many cases, uncertainty and randomness simultaneously appear in a com-\nplex system. In order to describe this phenomenon, uncertain random vari-\nable was initialized by Liu [152] in 2013 with the concepts of chance measure\nand chance distribution. As an important contribution, Liu [153] presented\nan operational law of uncertain random variables. Furthermore, Yao-Gao\n[264], Gao-Sheng [42] and Gao-Ralescu [49] veriﬁed some laws of large num-\nbers for uncertain random variables.\nIn order to model optimization problems with not only uncertainty but\nalso randomness, uncertain random programming was founded by Liu [153]\nin 2013. As extensions, Zhou-Yang-Wang [314] proposed uncertain random\nmultiobjective programming for optimizing multiple, noncommensurable and\nconﬂicting objectives, Qin [190] proposed uncertain random goal program-\nming in order to satisfy as many goals as possible in the order speciﬁed, and\nKe-Su-Ni [88] proposed uncertain random multilevel programming for study-\ning decentralized decision systems in which the leader and followers may have\ntheir own decision variables and objective functions. After that, uncertain\nrandom programming was developed steadily and applied widely.\nIn order to quantify the risk of uncertain random systems, Liu-Ralescu\n[154] invented the tool of uncertain random risk analysis in 2014. Further-\nmore, the value-at-risk methodology was presented by Liu-Ralescu [156], and\nthe expected loss methodology was investigated by Liu-Ralescu [158] for deal-\ning with uncertain random systems.\nFor dealing with uncertain random systems, Wen-Kang [225] presented\nthe tool of uncertain random reliability analysis and deﬁned the reliability\nindex in 2016. After that, uncertain random reliability analysis was studied\nby Gao-Yao [44] and Zhang-Kang-Wen [302].\nAssuming some edges exist with some degrees in probability measure and\nothers exist with some degrees in uncertain measure, Liu [128] deﬁned the\nconcept of uncertain random graph and analyzed the connectivity index in\n2014. After that, Zhang-Peng-Li [296] and Chen-Peng-Rao-Rosyida [7] dis-\ncussed the Euler index and cycle index of uncertain random graph, respec-\ntively.\nAssuming some weights are random variables and others are uncertain\n\n\nSection A.9 - Bibliographic Notes\n437\nvariables, Liu [128] initialized the concept of uncertain random network and\ndiscussed the shortest path problem in 2014. Following that, uncertain ran-\ndom network was explored by many researchers. For example, Sheng-Gao\n[202] investigated the maximum ﬂow problem, and Sheng-Qin-Shi [205] dealt\nwith the minimum spanning tree problem of uncertain random network.\nIn order to deal with uncertain random phenomenon evolving in time,\nGao-Yao [39] presented an uncertain random process in the light of chance\ntheory in 2015. Gao-Yao [39] also proposed an uncertain random renewal\nprocess. As extensions, Yao-Zhou [265][269] and Yao [271] discussed an un-\ncertain random renewal reward process, and Yao-Gao [260] investigated an\nuncertain random alternating renewal process.\n\n\n\n\nAppendix B\nFrequently Asked\nQuestions\nThis appendix will answer some frequently asked questions related to uncer-\ntainty theory. This appendix will also show that none of fuzzy set theory,\ninterval analysis, rough set theory and grey system is a consistent mathe-\nmatical system. Finally, the evolution history of the term uncertainty will be\nsummarized.\nB.1\nWhat is belief degree?\nBelief degrees are familiar to all of us. The object of belief is an event (i.e.,\na proposition). For example, “the sun will rise tomorrow”, “it will be sunny\nnext week”, and “John is a young man” are all instances of object of belief.\nA belief degree represents the strength with which you believe the event will\nhappen. If you completely believe the event will happen, then your belief\ndegree is 1 (complete belief). If you think it is completely impossible, then\nyour belief degree is 0 (complete disbelief).\nGenerally, you will assign a\nnumber between 0 and 1 to the belief degree for each event because you can\nbe neither in more belief than “complete belief ” nor in more disbelief than\n“complete disbelief ”. The higher the belief degree is, the more strongly you\nbelieve the event will happen.\nThe belief degree of an event may also be interpreted as the fair betting\nratio (price/stake) for the event. Assume a bet oﬀers $1 if the event happens\nand nothing otherwise. What price of this bet do you think is reasonable? If\nyou think the bet is worth $1, then your belief degree of this event is 100%; if\nyou think the bet is worth nothing, then your belief degree is 0%; and if you\nthink the bet is worth 60¢, then your belief degree is 60%. Here the word\n“fair” means you are willing to either buy or sell this bet at this price.\nThe belief degree depends heavily on the personal knowledge and pref-\n\n\n440\nAppendix B - Frequently Asked Questions\nerence concerning the event. When the personal knowledge and preference\nchange, the belief degree changes too1.\nFor example, let us consider my\nbirthday. Someone who does not know me would be only 8% (i.e., 1/12) sure\nthat I was born in February. Some friends of mine might be 80% sure for it\nsince they shared my birthday cake last year. However, my mother can be\n100% sure that I was born in February. Diﬀerent people hold diﬀerent belief\ndegrees due to their diﬀerent knowledge and preference.\nPerhaps some readers may ask which belief degree is correct. I have to\nsay that all belief degrees are wrong, but some are useful. Through a lot\nof surveys, Kahneman and Tversky [87] showed that human beings usually\noverweight unlikely events. From another side, Liu [130] showed that human\nbeings usually estimate a much wider range of values than the object actually\ntakes. This conservatism of human beings makes the belief degrees deviate\nfar from the frequency. Thus all belief degrees are wrong compared with its\nfrequency. However, it cannot be denied that those belief degrees are indeed\nhelpful for decision making. A belief degree becomes “correct” only when it\ncoincides with the frequency. However, usually we cannot make it to that.\nB.2\nWhat is the diﬀerence between probability theory\nand uncertainty theory?\nThe diﬀerence between probability theory and uncertainty theory does not lie\nin whether the measures are additive or not, but how the product measures\nare deﬁned. The product probability measure is the multiplication of the\nprobability measures of individual events, i.e.,\nPr{Λ1 × Λ2} = Pr{Λ1} × Pr{Λ2},\n(B.1)\nwhile the product uncertain measure is the minimum of the uncertain mea-\nsures of individual events, i.e.,\nM{Λ1 × Λ2} = M{Λ1} ∧M{Λ2}\n(B.2)\nwhere Λ1 and Λ2 are events from diﬀerent spaces. See Figure B.1.\nIn other words, the diﬀerence between probability theory and uncertainty\ntheory is that the former assumes the joint probability measure of indepen-\ndent events is the multiplication of probability measures of individual events,\ni.e.,\nPr{Λ1 ∩Λ2} = Pr{Λ1} × Pr{Λ2},\n(B.3)\nwhile the latter assumes the joint uncertain measure of independent events\nis the minimum of uncertain measures of individual events, i.e.,\nM{Λ1 ∩Λ2} = M{Λ1} ∧M{Λ2}\n(B.4)\nwhere Λ1 and Λ2 are independent events.\n1In contrast, frequency does not change with the personal knowledge and preference.\n\n\nSection B.4 - Stochastic Equations of Mathematical Physics\n441\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ1 × Λ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ1.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nΛ2\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n..................\n..................\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nFigure B.1: Events Λ1, Λ2 and Λ1 × Λ2\nB.3\nHow do we distinguish between randomness and\nuncertainty in practice?\nRandomness is anything that follows the laws of probability theory (i.e.,\nthe three axioms of probability theory plus product probability theorem),\nwhile uncertainty is anything that follows the laws of uncertainty theory\n(i.e., the four axioms of uncertainty theory). In other words, frequency is\nthe empirical basis of probability theory, while belief degree is the empirical\nbasis of uncertainty theory.\nOf course, we can distinguish between randomness and uncertainty by the\nabove deﬁnitions. However, in practice, we can quickly distinguish between\nthem in the following way: In order to use probability theory or uncertainty\ntheory, for any quantity, we must produce a distribution function in advance.\nIf the distribution function is close enough to the frequency, then it can be\ntreated as randomness. Otherwise, it has to be treated as uncertainty.\nMost people believe that probability distribution is easy to obtain from\nthe historical data, and then we should use probability theory. However, the\ndistribution function obtained in most practical problems is, unfortunately,\nnot close enough to the frequency. In this case, we should regard it as an\nuncertainty distribution and then use uncertainty theory.\nB.4\nWhy is stochastic diﬀerential equation not suitable\nfor modelling physical systems?\nIn 1827 Robert Brown observed irregular movement of pollen particles sus-\npended in liquid. This movement is now known as Brownian motion. In\n1923 Norbert Wiener modeled Brownian motion by the following stochastic\nprocess.\nDeﬁnition B.1 (Wiener [227]) A stochastic process Wt is called a Wiener\nprocess if\n(i) W0 = 0 and almost all sample paths are continuous (but non-Lipschitz),\n\n\n442\nAppendix B - Frequently Asked Questions\n(ii) Wt has stationary and independent increments, and\n(iii) every increment Ws+t −Ws is a normal random variable with expected\nvalue 0 and variance t.\nIn 1940s Kiyoshi Ito invented stochastic calculus with respect to Wiener\nprocess and stochastic diﬀerential equation driven by Wiener process. After\nthat, stochastic diﬀerential equation was applied to physical systems such as\nheat conduction, string vibration, spring vibration, and ﬂuid ﬂow.\nWhy does a pollen particle not follow Wiener process?\nMany people believe that the irregular movement of pollen particles sus-\npended in liquid follows a Wiener process Wt. Since Wt represents the posi-\ntion of pollen particle at time t, the speed over the time interval [t, t + ∆t]\nis\n∆Wt\n∆t\n= Wt+∆t −Wt\n∆t\n∼N(0, ∆t)\n∆t\n= N\n\u0012\n0, 1\n∆t\n\u0013\n,\na normal random variable with expected value 0 and variance 1/∆t.\nLet\nK = 299,792,458 meters per second (the speed of light) and ∆t = 10−25\nseconds. Then\nPr\n\u001a\f\n\f\n\f\n\f\n∆Wt\n∆t\n\f\n\f\n\f\n\f > K\n\u001b\n= 2\nZ ∞\nK\n√\n∆t\n√\n2π exp\n\u0012\n−∆tx2\n2\n\u0013\ndx > 99.99%.\nThis means the pollen particle moves at a superluminal speed. It is speeding!\nWhy does spring vibration not follow any stochastic diﬀerential\nequation?\nLet us consider a mass that is hanging from a spring. Assume Xt is the\nposition of a mass at time t, and\nsin t + dWt\ndt\nis the external force with white noise, where Wt is a Wiener process. Newton’s\nsecond law states that a force acting on a body is equal to the acceleration\nof that body times its mass. Hooke’s law states that for a spring the force\nand distance are proportional to each other. From the two laws we may get,\nfor example, a stochastic diﬀerential equation\nd2Xt\ndt2\n+ 2dXt\ndt + 2Xt = sin t + dWt\ndt .\nOne solution is\nXt =\nZ t\n0\nexp(s −t) sin(t −s) sin sds +\nZ t\n0\nexp(s −t) sin(t −s)dWs.\n\n\nSection B.5 - Challenge to Stochastic Finance Theory\n443\nThus the speed of the mass over the time interval [t, t + ∆t] is\n∆Xt\n∆t\n= Xt+∆t −Xt\n∆t\n∼N\n\u0012\na, b\n∆t\n\u0013\n,\na normal random variable with expected value a and variance b/∆t, where\na ≈\nZ t\n0\nexp(s −t)(cos(t −s) −sin(t −s)) sin(s)ds,\nb ≈\nZ t\n0\nexp(2(s −t))\nsin(2(t −s)) −2 sin2(t −s)\n\u0001\nds\nprovided that ∆t is suﬃciently small. Let K = 299,792,458 meters per second\n(the speed of light) and ∆t = 10−27 seconds. Then\nPr\n\u001a\f\n\f\n\f\n\f\n∆Xt\n∆t\n\f\n\f\n\f\n\f > K\n\u001b\n> 99.99%\nat time t = 1. This means the mass moves at a superluminal speed. There-\nfore, spring vibration does not follow any stochastic diﬀerential equation.\nB.5\nWhy is stochastic diﬀerential equation not suitable\nfor modelling ﬁnancial markets?\nThe origin of stochastic ﬁnance theory can be traced to Louis Bachelier’s\ndoctoral dissertation Th´\neorie de la Speculation in 1900.\nHowever, Bache-\nlier’s work had little impact for more than a half century. After Kiyosi Ito\ninvented stochastic calculus [70] in 1944 and stochastic diﬀerential equation\n[71] in 1951, stochastic ﬁnance theory was well developed among others by\nSamuelson [195], Black-Scholes [3] and Merton [175] during the 1960s and\n1970s. Traditionally, stochastic ﬁnance theory presumes that the stock price\n(including interest rate and currency exchange rate) follows Ito’s stochastic\ndiﬀerential equation. Is it really reasonable? In fact, this widely accepted\npresumption was challenged among others by Liu [124] in 2013.\nExample B.1: (Liu [124]) Assume the stock price Xt follows the stochastic\ndiﬀerential equation,\ndXt = eXtdt + σXtdWt\n(B.5)\nwhere Wt is a Wiener process. The solution Xt is a geometric Wiener process,\nXt = X0 exp((e −σ2/2)t + σWt)\n(B.6)\nfrom which we derive\nWt = ln Xt −ln X0 −(e −σ2/2)t\nσ\n(B.7)\n\n\n444\nAppendix B - Frequently Asked Questions\nwhose increment is\n∆Wt = ln Xt+∆t −ln Xt −(e −σ2/2)∆t\nσ\n.\n(B.8)\nWrite\nA = −(e −σ2/2)∆t\nσ\n.\n(B.9)\nNote that the stock price Xt is actually a step function of time with a ﬁnite\nnumber of jumps although it looks like a curve. During a ﬁxed period (e.g.\none week), without loss of generality, we assume that Xt is observed to have\n100 jumps. Now we divide the period into 10,000 equal intervals. Then we\nmay observe 10,000 samples of Xt. It follows from (B.8) that ∆Wt has 10,000\nsamples that consist of 9,900 A’s and 100 other numbers:\nA, A, · · · , A\n|\n{z\n},\nB, C, · · · , Z.\n|\n{z\n}\n9900\n100\n(B.10)\nIt is obvious that nobody can believe that those 10,000 samples follow a\nnormal probability distribution with expected value 0 and variance ∆t. This\nfact is in contradiction with the property of Wiener process that the increment\n∆Wt is a normal random variable. Therefore, the real stock price Xt does\nnot follow the stochastic diﬀerential equation.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n99%\nFigure B.2: Normal density function (curve) cannot approximate to the rel-\native frequency (histogram). Hence it is impossible that the real stock price\nXt follows any Ito’s stochastic diﬀerential equation.\nPerhaps some people think that the stock price does behave like a geomet-\nric Wiener process (or Ornstein-Uhlenbeck process) in macroscopy although\nthey recognize the paradox in microscopy.\nHowever, as the very core of\nstochastic ﬁnance theory, Ito’s calculus is just built on the microscopic struc-\nture (i.e., the diﬀerential dWt) of Wiener process rather than macroscopic\nstructure.\n\n\nSection B.5 - Challenge to Stochastic Finance Theory\n445\nOn the basis of the above paradox, personally I do not think Ito’s calculus\ncan play the essential tool of ﬁnance theory because Ito’s stochastic diﬀeren-\ntial equation is impossible to model stock price. As a substitute, uncertain\ncalculus may be a potential mathematical foundation of ﬁnance theory. We\nwill have a theory of uncertain ﬁnance if the stock price, interest rate and\nexchange rate are assumed to follow uncertain diﬀerential equations.\nExample B.2: (Liu-Liu [144]) Let us reconsider Alibaba stock prices (weekly\naverage) from January 1, 2019 to June 30, 2020. See Table 15.1 on Page 370.\nLet i = 1, 2, · · · , 78 represent the weeks from January 1, 2019 to June 30,\n2020, and denote the stock prices in Table 15.1 by\nx1, x2, · · · , x78.\n(B.11)\nAssume Xt is a stochastic process that represents Alibaba stock price and\nfollows the stochastic diﬀerential equation\ndXt = (m −aXt)dt + σdWt\n(B.12)\nwhere m, a and σ are unknown parameters. For any ﬁxed parameters m, a, σ\nand i (2 ≤i ≤78), we solve the updated stochastic diﬀerential equation\ndXt = (m −aXt)dt + σdWt,\nXi−1 = xi−1\n(B.13)\nand ﬁnd that Xi is a normal random variable with expected value\nei = m\na +\n\u0010\nxi−1 −m\na\n\u0011\nexp(−a)\n(B.14)\nand variance\nv2 = σ2\n2a (1 −exp(−2a)) .\n(B.15)\nThus the probability distribution function of the normal random variable Xi\nis\nΦi(x) =\n1\nv\n√\n2π\nZ x\n−∞\nexp\n\u0012\n−(y −ei)2\n2v2\n\u0013\ndy\n(B.16)\nand Φi(Xi) is always a uniform random variable U(0, 1). Substitute Xi with\nthe corresponding observed value xi, and write\nεi(m, a, σ) = Φi(xi).\n(B.17)\nThen εi(m, a, σ) is always a sample of uniform probability distribution U(0, 1)\nand called the ith residual of the stochastic diﬀerential equation (B.12) cor-\nresponding to the observed data (B.11). For each positive integer k, the k-th\nsample moment of the 77 residuals ε2(m, a, σ), ε3(m, a, σ), · · · , ε78(m, a, σ) is\n1\n77\n78\nX\ni=2\nεk\ni (m, a, σ),\n\n\n446\nAppendix B - Frequently Asked Questions\nand the k-th population moment of the uniform probability distribution\nU(0, 1) is\n1\nk + 1.\nSince the number of unknown parameters in the stochastic diﬀerential equa-\ntion is 3, the moment estimate (m, a, σ) is obtained by equating the ﬁrst 3\nsample moments to the corresponding ﬁrst 3 population moments. In other\nwords, the moment estimate (m, a, σ) should solve the system of equations,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1\n77\n78\nX\ni=2\nεi(m, a, σ) = 1\n2\n1\n77\n78\nX\ni=2\nε2\ni (m, a, σ) = 1\n3\n1\n77\n78\nX\ni=2\nε3\ni (m, a, σ) = 1\n4\nwhose root is\nm = 45.8790,\na = 0.2401,\nσ = 8.0366.\nThus we obtain a stochastic stock model,\ndXt = (45.8790 −0.2401Xt)dt + 8.0366dWt\n(B.18)\nwhere Xt represents Alibaba stock price. Does the stochastic stock model\n(B.18) ﬁt the stock prices x1, x2, · · · , x78? In order to answer this question,\nlet us consider the 77 residuals\nεi(45.8790, 0.2401, 8.0366), i = 2, 3, · · · , 78.\n(B.19)\nSee Figure B.3. It is clear that the residuals are far from frequency stability.\nThus they cannot be regarded as random variables, let alone follow the uni-\nform probability distribution U(0, 1). Thus the stochastic stock model (B.18)\ndoes not ﬁt the stock prices x1, x2, · · · , x78. In fact, it is impossible for us to\nﬁnd a stochastic diﬀerential equation that is suitable for modelling Alibaba\nstock price.\nExample B.3: (Ye-Liu [282]) Let us reconsider USD-CNY exchange rates\n(weekly average) from October 1, 2019 to June 30, 2021. See Table 15.2 on\nPage 373. Let i = 1, 2, · · · , 91 represent the weeks from October 1, 2019 to\nJune 30, 2021, and denote the exchange rates in Table 15.2 by\nx1, x2, · · · , x91.\n(B.20)\n\n\nSection B.5 - Challenge to Stochastic Finance Theory\n447\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\ni\nε\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n•\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\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B.3: Residual Plot of Stochastic Stock Model (B.18) Corresponding\nto Alibaba Stock Prices. Since the frequency is far from being stable, the\nresiduals cannot be regarded as random variables, let alone follow the uniform\nprobability distribution U(0, 1). Thus stochastic diﬀerential equation is not\nsuitable for modelling Alibaba stock price.\nAssume Xt is a stochastic process that represents USD-CNY exchange rate\nand follows the stochastic diﬀerential equation\ndXt = (m −aXt)dt + σdWt\n(B.21)\nwhere m, a and σ are unknown parameters. For any ﬁxed parameters m, a, σ\nand i (2 ≤i ≤91), we solve the updated stochastic diﬀerential equation\ndXt = (m −aXt)dt + σdWt,\nXi−1 = xi−1\n(B.22)\nand ﬁnd that Xi is a normal random variable with expected value\nei = m\na +\n\u0010\nxi−1 −m\na\n\u0011\nexp(−a)\n(B.23)\nand variance\nv2 = σ2\n2a (1 −exp(−2a)) .\n(B.24)\nThus the probability distribution function of the normal random variable Xi\nis\nΦi(x) =\n1\nv\n√\n2π\nZ x\n−∞\nexp\n\u0012\n−(y −ei)2\n2v2\n\u0013\ndy\n(B.25)\nand Φi(Xi) is always a uniform random variable U(0, 1). Substitute Xi with\nthe corresponding observed value xi, and write\nεi(m, a, σ) = Φi(xi).\n(B.26)\n\n\n448\nAppendix B - Frequently Asked Questions\nThen εi(m, a, σ) is always a sample of uniform probability distribution U(0, 1)\nand called the ith residual of the stochastic diﬀerential equation (B.21) cor-\nresponding to the observed data (B.20). For each positive integer k, the k-th\nsample moment of the 90 residuals ε2(m, a, σ), ε3(m, a, σ), · · · , ε91(m, a, σ) is\n1\n90\n91\nX\ni=2\nεk\ni (m, a, σ),\nand the k-th population moment of the uniform probability distribution\nU(0, 1) is\n1\nk + 1.\nSince the number of unknown parameters in the stochastic diﬀerential equa-\ntion is 3, the moment estimate (m, a, σ) is obtained by equating the ﬁrst 3\nsample moments to the corresponding ﬁrst 3 population moments. In other\nwords, the moment estimate (m, a, σ) should solve the system of equations,\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1\n90\n91\nX\ni=2\nεi(m, a, σ) = 1\n2\n1\n90\n91\nX\ni=2\nε2\ni (m, a, σ) = 1\n3\n1\n90\n91\nX\ni=2\nε3\ni (m, a, σ) = 1\n4\nwhose root is\nm = 1.4896,\na = 0.2202,\nσ = 0.0731.\nThus we obtain a stochastic currency model,\ndXt = (1.4896 −0.2202Xt)dt + 0.0731dWt\n(B.27)\nwhere Xt represents USD-CNY exchange rate. Does the stochastic currency\nmodel (B.27) ﬁt the exchange rates x1, x2, · · · , x91? In order to answer this\nquestion, let us consider the 90 residuals\nεi(1.4896, 0.2202, 0.0731), i = 2, 3, · · · , 91.\n(B.28)\nSee Figure B.4. It is clear that the residuals are far from frequency stabil-\nity. Thus they cannot be regarded as random variables, let alone follow the\nuniform probability distribution U(0, 1). Thus the stochastic currency model\n(B.27) does not ﬁt the exchange rates x1, x2, · · · , x91. In fact, it is impossible\nfor us to ﬁnd a stochastic diﬀerential equation that is suitable for modelling\nUSD-CNY exchange rate.\n\n\nSection B.7 - Fuzzy set theory is wrong\n449\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. 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\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n0\n0.5\n1\n2\n30\n60\n91\nFigure B.4: Residual Plot of Stochastic Currency Model (B.27) Correspond-\ning to USD-CNY Exchange Rates.\nSince the frequency is far from being\nstable, the residuals cannot be regarded as random variables, let alone fol-\nlow the uniform probability distribution U(0, 1). Thus stochastic diﬀerential\nequation is not suitable for modelling USD-CNY exchange rate.\nB.6\nWhat is the diﬀerence between uncertainty theory\nand possibility theory?\nThe essential diﬀerence between uncertainty theory (Liu [113]) and possibility\ntheory (Zadeh [291]) is that the former holds\nM{Λ1 ∪Λ2} = M{Λ1} ∨M{Λ2}\n(B.29)\nonly for independent events Λ1 and Λ2, and the latter holds\nPos{Λ1 ∪Λ2} = Pos{Λ1} ∨Pos{Λ2}\n(B.30)\nfor any events Λ1 and Λ2 no matter if they are independent or not.\nA lot of surveys showed that the measure of the union of events is usually\ngreater than the maximum of the measures of individual events when they are\nnot independent. This fact states that human brains do not behave fuzziness.\nBoth uncertainty theory and possibility theory attempt to model belief\ndegrees, where the former uses the tool of uncertain measure and the latter\nuses the tool of possibility measure. Thus they are complete competitors.\nB.7\nWhy do I think fuzzy set theory is wrong?\nA fuzzy set is deﬁned by its membership function µ which assigns to each\nelement x a real number µ(x) in the interval [0, 1], where the value of µ(x)\nrepresents the grade of membership of x in the fuzzy set. This deﬁnition was\ngiven by Zadeh [290] in 1965. Since then, fuzzy set theory has been spread\n\n\n450\nAppendix B - Frequently Asked Questions\nbroadly. Although I strongly respect Professor LotﬁZadeh’s achievements, I\nhave to declare that fuzzy set theory is not consistent in mathematics.\nA very strange phenomenon in the fuzzy world is that diﬀerent people have\ndiﬀerent fuzzy set theories. Even so, we have to admit that every version of\nfuzzy set theory contains at least the following four items. The ﬁrst one is a\nfuzzy set ξ with membership function µ. The next one is a complement set\nξc with membership function\nλ(x) = 1 −µ(x).\n(B.31)\nThe third one is a possibility measure deﬁned by the three axioms,\nPos{Ω} = 1 for the universal set Ω,\n(B.32)\nPos{∅} = 0 for the empty set ∅,\n(B.33)\nPos{Λ1 ∪Λ2} = Pos{Λ1} ∨Pos{Λ2} for any events Λ1 and Λ2.\n(B.34)\nAnd the fourth one is a relation between membership function and possibility\nmeasure (Zadeh [291]),\nµ(x) = Pos{x ∈ξ}.\n(B.35)\nNow for any point x, it is clear that {x ∈ξ} and {x ∈ξc} are opposite\nevents2, and then\n{x ∈ξ} ∪{x ∈ξc} = Ω.\n(B.36)\nOn the one hand, by using the possibility axioms, we have\nPos{x ∈ξ} ∨Pos{x ∈ξc} = Pos{Ω} = 1.\n(B.37)\nOn the other hand, by using the relation (B.35), we have\nPos{x ∈ξ} = µ(x),\n(B.38)\nPos{x ∈ξc} = 1 −µ(x).\n(B.39)\nIt follows from (B.37), (B.38) and (B.39) that\nµ(x) ∨(1 −µ(x)) = 1.\n(B.40)\nHence\nµ(x) = 0 or 1.\n(B.41)\nThis result shows that the membership function µ can only be an indicator\nfunction of crisp set. In other words, only crisp sets can simultaneously satisfy\n(B.31)∼(B.35). In this sense, fuzzy set theory collapses mathematically to\n2Perhaps some fuzzists insist that {x ∈ξ} and {x ∈ξc} are not opposite.\nHere I\nwould like to advise them not to think so because it is in contradiction with ξc having the\nmembership function λ(x) = 1 −µ(x).\n\n\nSection B.8 - Fuzzy variable cannot model any quantity\n451\nclassical set theory.\nThat is, fuzzy set theory is nothing but classical set\ntheory.\nFurthermore, it seems both in theory and practice that inclusion relation\nbetween fuzzy sets has to be needed. Thus fuzzy set theory also assumes a\nformula (Zadeh [291]),\nPos{ξ ⊂B} = sup\nx∈B\nµ(x)\n(B.42)\nfor any crisp set B. Now consider two crisp intervals [1, 2] and [2, 3]. It is\ncompletely inacceptable in mathematical community that [1, 2] is included in\n[2, 3], i.e., the inclusion relation\n[1, 2] ⊂[2, 3]\n(B.43)\nis 100% wrong.\nNote that [1, 2] is a special fuzzy set whose membership\nfunction is\nµ(x) =\n(\n1,\nif 1 ≤x ≤2\n0,\notherwise.\n(B.44)\nIt follows from the formula (B.42) that\nPos{[1, 2] ⊂[2, 3]} = sup\nx∈[2,3]\nµ(x) = 1.\n(B.45)\nThat is, fuzzy set theory says that [1, 2] ⊂[2, 3] is 100% right. Are you willing\nto accept this result? If not, then fuzzy set theory is not acceptable.\nPerhaps some fuzzists may argue that they never use possibility measure\nin fuzzy set theory. Here I would like to remind them that the membership\ndegree µ(x) is just the possibility measure that the fuzzy set ξ contains the\npoint x (i.e., x belongs to ξ).\nPlease also keep in mind that we cannot\ndistinguish fuzzy set from random set (Robbins [193] and Matheron [173])\nand uncertain set (Liu [118]) if the underlying measures are not available.\nFrom the above discussion, we can see that fuzzy set theory is not self-\nconsistent in mathematics and may lead to wrong results in practice. There-\nfore, I would like to conclude that fuzzy set theory cannot be called mathe-\nmatics. Can we improve fuzzy set theory? Yes, we can. But the change is so\nbig that I have to give the revision a new name called uncertain set theory.\nSee Chapter 9.\nB.8\nWhy is fuzzy variable not suitable for modelling\nanything in the real world?\nA fuzzy variable is a function from a possibility space to the set of real\nnumbers (Nahmias [178]). Fuzzists think that fuzzy variable is a suitable tool\nfor modelling some quantity. Is it really true? Unfortunately, the answer is\n\n\n452\nAppendix B - Frequently Asked Questions\nnegative. For example, you may think my height is a fuzzy variable ξ, and\nassign it a membership function,\nµ(x) =\n\n\n\n\n\n\n\n\n\n0,\nif x ≤1.6\n(x −1.6)/0.1,\nif 1.6 < x ≤1.7\n(1.8 −x)/0.1,\nif 1.7 < x ≤1.8\n0,\nif x > 1.8\n(B.46)\nthat is just the triangular fuzzy variable (1.6, 1.7, 1.8) in meters. Please do\nnot argue why such a membership function is chosen because it is not impor-\ntant for the focus of debate. Based on the membership function µ and the\ndeﬁnition of possibility measure\nPos{ξ ∈B} = sup\nx∈B\nµ(x),\n(B.47)\nit is easy for us to infer that\nPos{“my height” = 1.7m} = 1\n(B.48)\nand\nPos{“my height” ̸= 1.7m} = 1\n(B.49)\nby setting B = {1.7} and B = {1.7}c, respectively. Thus we immediately\nconclude the following three propositions:\n(a) my height is “exactly 1.7m” with possibility measure 1,\n(b) my height is “not 1.7m” with possibility measure 1,\n(c) “exactly 1.7m” is as possible as “not 1.7m”.\nThe ﬁrst proposition says you are 100% sure that my height is “exactly 1.7m”,\nneither less nor more. What a coincidence it should be! It is doubtless that\nnobody is so naive to expect that “exactly 1.7m” is the true value of my\nheight.\nThe second proposition sounds good.\nThe third proposition says\n“exactly 1.7m” and “not 1.7m” have the same possibility measure. Thus you\nhave to regard them “equally likely”. Consider a bet:\nYou get $100 if my height is exactly 1.7m, and pay $100 otherwise\n(i.e., my height is not 1.7m).\nDo you think the bet is fair? If not, then “exactly 1.7m” is not as possible\nas “not 1.7m”, i.e., the conclusion (c) is unacceptable. In fact, it is obvious\nthat “exactly 1.7m” is almost impossible compared with “not 1.7m”. This\nparadox shows that those quantities like my height cannot be quantiﬁed by\npossibility measure. Therefore, fuzzy variable is not suitable for modelling\nany quantity.\n\n\nSection B.9 - How to Handle Interval Numbers?\n453\nB.9\nHow do we handle interval numbers by uncertainty\ntheory?\nIn practice, information is sometimes only given by lower and upper bounds\ndue to the imprecise observations or estimations by human beings. For exam-\nple, “I think your height is between 1.6 and 1.8 meters”. From this statement,\nwe may infer the following conclusions:\n(i) Your height is not exactly known to us;\n(ii) The true value of your height is on the interval [1.6, 1.8];\n(iii) All numbers on the interval [1.6, 1.8] are equally likely.\nThis type of information is called interval-valued. In order to describe interval-\nvalued information, we deﬁne an interval number as a number equally dis-\ntributed on a speciﬁed interval. Using this concept, my statement becomes\n“I think your height is an interval number [1.6, 1.8]”. Hence how to rationally\nhandle interval numbers is an important topic in science and engineering.\nIn uncertainty theory, an interval number [a, b] is regarded as a linear\nuncertain variable (written as L(a, b) in this book) with uncertainty distri-\nbution\nΦ(x) = x −a\nb −a ,\nif a ≤x ≤b\n(B.50)\nand inverse uncertainty distribution\nΦ−1(α) = (1 −α)a + αb,\n0 < α < 1.\n(B.51)\nOperational Law: Let [a1, b1], [a2, b2], · · · , [an, bn] be independent inter-\nval numbers. Assume f(x1, x2, · · · , xn) is strictly increasing with respect to\nx1, x2, · · · , xm and strictly decreasing with xm+1, xm+2, · · · , xn. It follows\nfrom Theorem 3.18 (i.e., operational law of uncertain variables) that\nξ = f([a1, b1], [a2, b2], · · · , [an, bn])\n(B.52)\nhas an inverse uncertainty distribution\nΨ−1(α) = f(Φ−1\n1 (α), · · · , Φ−1\nm (α), Φ−1\nm+1(1 −α), · · · , Φ−1\nn (1 −α))\n(B.53)\nwhere\nΦ−1\ni (α) = (1 −α)ai + αbi\n(B.54)\nfor i = 1, 2, · · · , n. Note that ξ determined by (B.52) is an uncertain variable,\nbut not necessarily an interval number.\nAddition: Let [a1, b1] and [a2, b2] be independent interval numbers. It fol-\nlows from the operational law that the addition [a1, b1]+[a2, b2] has an inverse\nuncertainty distribution\nΨ−1(α) = ((1 −α)a1 + αb1) + ((1 −α)a2 + αb2)\n= (1 −α)(a1 + a2) + α(b1 + b2)\n(B.55)\n\n\n454\nAppendix B - Frequently Asked Questions\nthat happens to be an interval number [a1 + a2, b1 + b2], i.e.,\n[a1, b1] + [a2, b2] = [a1 + b1, a2 + b2].\n(B.56)\nSubtraction: Let [a1, b1] and [a2, b2] be independent interval numbers. It\nfollows from the operational law that the subtraction [a1, b1] −[a2, b2] has an\ninverse uncertainty distribution\nΨ−1(α) = ((1 −α)a1 + αb1) −(αa2 + (1 −α)b2)\n= (1 −α)(a1 −b2) + α(b1 −a2)\n(B.57)\nthat happens to be an interval number [a1 −b2, b1 −a2], i.e.,\n[a1, b1] −[a2, b2] = [a1 −b2, b1 −a2].\n(B.58)\nScalar Multiplication: Let [a, b] be an interval number, and let k be a\nscalar number. It follows from the operational law that the scalar product\nk · [a, b] has an inverse uncertainty distribution\nΨ−1(α) =\n(\n(1 −α)(ka) + α(kb),\nif k ≥0\n(1 −α)(kb) + α(ka),\nif k < 0\n(B.59)\nthat happens to be an interval number, and\nk · [a, b] =\n(\n[ka, kb],\nif k ≥0\n[kb, ka],\nif k < 0.\n(B.60)\nLinear Function: Let [a1, b1], [a2, b2], · · · , [an, bn] be independent interval\nnumbers. It follows from addition, subtraction and scalar multiplication of\ninterval numbers that the linear function\nk1 · [a1, b1] + k2 · [a2, b2] + · · · + kn · [an, bn]\n(B.61)\nhappens to be an interval number.\nMultiplication: Let [a1, b1] and [a2, b2] be independent interval numbers\nwith a1 ≥0 and a2 ≥0. It follows from the operational law that the multi-\nplication [a1, b1] × [a2, b2] has an inverse uncertainty distribution\nΨ−1(α) = ((1 −α)a1 + αb1) × ((1 −α)a2 + αb2)\n(B.62)\nthat is no longer an interval number, i.e.,\n[a1, b1] × [a2, b2] ̸= [a1 × a2, b1 × b2].\n(B.63)\n\n\nSection B.10 - Interval Analysis, Rough Set and Grey System\n455\nDivision: Let [a1, b1] and [a2, b2] be independent interval numbers with a1 ≥\n0 and a2 > 0. It follows from the operational law that the division [a1, b1] ÷\n[a2, b2] has an inverse uncertainty distribution\nΨ−1(α) = ((1 −α)a1 + αb1) ÷ (αa2 + (1 −α)b2)\n(B.64)\nthat is no longer an interval number, i.e.,\n[a1, b1] ÷ [a2, b2] ̸= [a1 ÷ b2, b1 ÷ a2].\n(B.65)\nRanking Method: Let [a, b] be an interval number, and let c be a constant.\nIt follows from Theorem 3.2 (i.e., measure inversion theorem) that the belief\ndegree of [a, b] being less than or equal to c is\nM{[a, b] ≤c} =\n\n\n\n\n\n\n\n\n\n0,\nif c < a\nc −a\nb −a,\nif a ≤c ≤b\n1,\nif c > b,\n(B.66)\nand the belief degree of [a, b] being greater than or equal to c is\nM{[a, b] ≥c} =\n\n\n\n\n\n\n\n\n\n0,\nif b < c\nb −c\nb −a,\nif a ≤c ≤b\n1,\nif a > c.\n(B.67)\nFor any independent interval numbers [a1, b1] and [a2, b2], it follows from the\nsubtraction of interval numbers that\nM{[a1, b1] ≤[a2, b2]} = M{[a1 −b2, b1 −a2] ≤0}.\n(B.68)\nBy using (B.66), we get\nM{[a1, b1] ≤[a2, b2]} =\n\n\n\n\n\n\n\n\n\n0,\nif a1 > b2\n1,\nif a2 > b1\nb2 −a1\nb1 −a1 + b2 −a2\n,\notherwise.\n(B.69)\nExpected Value: It follows from Theorem 3.24 (i.e., expected value oper-\nator) that the interval number [a, b] has an expected value\nE[a, b] = a + b\n2\n.\n(B.70)\nVariance: It follows from Theorem 3.29 that the interval number [a, b] has\na variance\nV [a, b] = (b −a)2\n12\n.\n(B.71)\n\n\n456\nAppendix B - Frequently Asked Questions\nB.10\nWhy do I think none of interval analysis, rough set\ntheory and grey system is self-consistent in math-\nematics?\nInterval analysis (Moore [176]), rough set theory (Pawlak [182]) and grey\nsystem (Deng [30]) declare that they are also able to handle interval numbers,\nand each of them contains the following three assumptions:\n(i) [a1, b1] + [a2, b2] = [a1 + a2, b1 + b2],\n(ii) [a1, b1] × [a2, b2] = [a1 × a2, b1 × b2],\nif a1 ≥0, a2 ≥0,\n(iii) π{[a, b] ≤c} = (c −a)/(b −a),\nif a ≤c ≤b\nwhere π{[a, b] ≤c} represents the possibility that the interval number [a, b]\nis less than or equal to a constant c. Although engineers like this type of\nmathematical system very much, unfortunately, there does not exist any\nmathematical system that simultaneously contains (i), (ii) and (iii) since the\nthree items are inconsistent. For this reason, none of interval analysis, rough\nset theory and grey system is a consistent mathematical system.\nIn order to show the inconsistence, let us consider two interval numbers\n[0, 1] and [0, 1]. It follows from items (i) and (iii) that\n[0, 1] + [0, 1] = [0, 2]\nand\n0.5 = π{[0, 2] ≤1}\n= π{[0, 1] + [0, 1] ≤1}\n= π{(x, y) | x + y ≤1, x ≥0, y ≥0}.\nOn the other hand, it follows from (ii) and (iii) that\n[0, 1] × [0, 1] = [0, 1]\nand\n0.4 = π{[0, 1] ≤0.4}\n= π{[0, 1] × [0, 1] ≤0.4}\n= π{(x, y) | xy ≤0.4, 0 ≤x ≤1, 0 ≤y ≤1}.\nAs a summary, from (i), (ii) and (iii) we derive the following two equations:\nπ{(x, y) | x + y ≤1, x ≥0, y ≥0\n|\n{z\n}\nΛ\n} = 0.5,\n(B.72)\nπ{(x, y) | xy ≤0.4, 0 ≤x ≤1, 0 ≤y ≤1\n|\n{z\n}\n∆\n} = 0.4.\n(B.73)\nThat is, π{Λ} > π{∆}. However, unfortunately, Λ ⊂∆. See Figure B.5.\nThis contradiction shows that a mathematical system is not consistent if\n\n\nSection B.11 - What is Uncertainty?\n457\nit simultaneously contains items (i), (ii) and (iii). Hence none of interval\nanalysis, rough set theory and grey system is consistent in mathematics.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n. .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\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+ y = 1\nxy = 0.4\nFigure B.5: Events Λ (Lower Triangle) and ∆(Lower Triangle + Middle\nStrip) determined in (B.72) and (B.73)\nB.11\nHow did “uncertainty” evolve over the past 100\nyears?\nAfter the word “randomness” was used to represent probabilistic phenomena,\nKnight (1921) and Keynes (1936) started to use the word “uncertainty” to\nrepresent non-probabilistic phenomena. The academic community also calls\nit Knightian uncertainty, Keynesian uncertainty, or true uncertainty. Unfor-\ntunately, they did not invent any mathematical theory to deal with uncertain\nphenomena. This disadvantage makes uncertainty in the sense of Knight and\nKeynes not able to become a scientiﬁc terminology. Even so, we have to rec-\nognize that they made a great process to break the monopoly of probability\ntheory.\nAn inﬂuential exploration by Zadeh (1965) was fuzzy set theory that was\nwidely said to be successfully applied in many areas of our life. However,\nfuzzy set theory has neither evolved as a mathematical system nor become a\nsuitable tool in practice. The main mistake of fuzzy set theory is based on\nthe wrong assumption that the possibility measure of the union of events is\nthe maximum of the possibility measures of the individual events no matter\nif they are independent or not. Of course, the extensions of fuzzy set theory\nfail to become a consistent mathematical system, too.\nIn addition, interval analysis (Moore, 1966), rough set theory (Pawlak,\n1982) and grey system (Deng, 1982) each have also a signiﬁcant eﬀect on\nengineering and management. However, unfortunately, none of them is self-\nconsistent in mathematics.\n\n\n458\nAppendix B - Frequently Asked Questions\nThe latest development was uncertainty theory founded by Liu (2007).\nNowadays, uncertainty theory has become a branch of mathematics that is\nnot only a formal study of an abstract structure (i.e., uncertainty space) but\nalso applicable to modelling uncertain phenomena. Uncertainty is deﬁned as\nanything that follows the laws of uncertainty theory. From then on, “uncer-\ntainty” became a scientiﬁc terminology on the basis of uncertainty theory.\n\n\nAppendix C\nYe Lemma\nLemma C.1 (Ye [279]) Let ut be a Lipschitz continuous function, and let\nh(t, u) be a continuously diﬀerentiable function. Then\nh(t, ut) = h(0, u0) +\nZ t\n0\n∂h\n∂s (s, us)ds +\nZ t\n0\n∂h\n∂u(s, us)dus\n(C.1)\nfor any t > 0.\nProof: Fix t > 0. For any partition of closed interval [0, t] with 0 = t1 <\nt2 < · · · < tk+1 = t, the mesh is written as\n∆= max\n1≤i≤k |ti+1 −ti|.\nThen the two integrals in (C.1) are deﬁned as\nZ t\n0\n∂h\n∂s (s, us)ds = lim\n∆→0\nk\nX\ni=1\n∂h\n∂s (ti, uti)(ti+1 −ti),\n(C.2)\nZ t\n0\n∂h\n∂u(s, us)dus = lim\n∆→0\nk\nX\ni=1\n∂h\n∂u(ti, uti)(uti+1 −uti).\n(C.3)\nStep 1: Since us is a Lipschitz continuous function, there exists a Lips-\nchitz constant L such that\n|uti+1 −uti| ≤L|ti+1 −ti|,\ni = 1, 2, · · · , k.\n(C.4)\nIt is also obvious that there is a bounded interval A such that us ∈A for any\ns ∈[0, t].\nStep 2: Since h(s, u) is a continuously diﬀerentiable function, the partial\nderivatives\n∂h\n∂s (s, u)\nand\n∂h\n∂u(s, u)\n\n\n460\nAppendix C - Ye Lemma\nare uniformly continuous on the bounded interval [0, t] × A. Thus, for any\ngiven number ε > 0, there exists a corresponding number δ > 0 such that\n\f\n\f\n\f\n\f\n∂h\n∂s (s1, u1) −∂h\n∂s (s2, u2)\n\f\n\f\n\f\n\f ≤ε,\n\f\n\f\n\f\n\f\n∂h\n∂u(s1, u1) −∂h\n∂u(s2, u2)\n\f\n\f\n\f\n\f ≤ε\n(C.5)\nprovided that (s1, u1), (s2, u2) ∈[0, t] × A, |s1 −s2| < δ and |u1 −u2| < δL.\nStep 3: When ∆< δ, we have |ti+1 −ti| < δ and |uti+1 −uti| < δL,\ni = 1, 2, · · · , k. It follows from (C.5) that\n\f\n\f\n\f\n\f\n∂h\n∂s (s, u) −∂h\n∂s (ti, uti)\n\f\n\f\n\f\n\f ≤ε,\n\f\n\f\n\f\n\f\n∂h\n∂u(s, u) −∂h\n∂u(ti, uti)\n\f\n\f\n\f\n\f ≤ε\n(C.6)\nprovided that s ∈[ti, ti+1] and u ∈[uti, uti+1], i = 1, 2, · · · , k, respectively.\nStep 4: For each i with 1 ≤i ≤k, it follows from the mean value theorem\nthat there exist two numbers s ∈[ti, ti+1] and u ∈[uti, uti+1] such that\nh(ti+1, uti+1) −h(ti, uti) = ∂h\n∂s (s, u)(ti+1 −ti) + ∂h\n∂u(s, u)(uti+1 −uti).\nOn the one hand, by using (C.6), we obtain\nh(ti+1, uti+1) −h(ti, uti) ≤∂h\n∂s (ti, uti)(ti+1 −ti) + ε|ti+1 −ti|\n+∂h\n∂u(ti, uti)(uti+1 −uti) + ε|uti+1 −uti|.\nBy using (C.4), we obtain\nh(ti+1, uti+1) −h(ti, uti) ≤∂h\n∂s (ti, uti)(ti+1 −ti)\n+∂h\n∂u(ti, uti)(uti+1 −uti) + ε(1 + L)|ti+1 −ti|.\n(C.7)\nOn the other hand, by using (C.6), we obtain\nh(ti+1, uti+1) −h(ti, uti) ≥∂h\n∂s (ti, uti)(ti+1 −ti) −ε|ti+1 −ti|\n+∂h\n∂u(ti, uti)(uti+1 −uti) −ε|uti+1 −uti|.\nBy using (C.4), we obtain\nh(ti+1, uti+1) −h(ti, uti) ≥∂h\n∂s (ti, uti)(ti+1 −ti)\n+∂h\n∂u(ti, uti)(uti+1 −uti) −ε(1 + L)|ti+1 −ti|.\n(C.8)\n\n\nSection C.0 - What is Uncertainty?\n461\nStep 5: On the one hand, it follows from (C.7) that\nh(t, ut) −h(0, u0) ≡\nk\nX\ni=1\n(h(ti+1, uti+1) −h(ti, uti))\n≤\nk\nX\ni=1\n\u0012∂h\n∂s (ti, uti)(ti+1 −ti) + ∂h\n∂u(ti, uti)(uti+1 −uti) + ε(1 + L)|ti+1 −ti|\n\u0013\n=\nk\nX\ni=1\n∂h\n∂s (ti, uti)(ti+1 −ti) +\nk\nX\ni=1\n∂h\n∂u(ti, uti)(uti+1 −uti) + ε(1 + L)t.\nLetting ∆→0 and ε →0, we have\nh(t, ut) −h(0, u0) ≤\nZ t\n0\n∂h\n∂s (s, us)ds +\nZ t\n0\n∂h\n∂u(s, us)dus.\n(C.9)\nOn the other hand, it follows from (C.8) that\nh(t, ut) −h(0, u0) ≡\nk\nX\ni=1\n(h(ti+1, uti+1) −h(ti, uti))\n≥\nk\nX\ni=1\n\u0012∂h\n∂s (ti, uti)(ti+1 −ti) + ∂h\n∂u(ti, uti)(uti+1 −uti) −ε(1 + L)|ti+1 −ti|\n\u0013\n=\nk\nX\ni=1\n∂h\n∂s (ti, uti)(ti+1 −ti) +\nk\nX\ni=1\n∂h\n∂u(ti, uti)(uti+1 −uti) −ε(1 + L)t.\nLetting ∆→0 and ε →0, we have\nh(t, ut) −h(0, u0) ≥\nZ t\n0\n∂h\n∂s (s, us)ds +\nZ t\n0\n∂h\n∂u(s, us)dus.\n(C.10)\nStep 6: It follows from (C.9) and (C.10) that (C.1) holds. 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b, c, d)\ntrapezoidal uncertain set\nE\nexpected value\nV\nvariance\nH\nentropy\nXt, Yt, Zt\nuncertain processes\nCt\nLiu process\nNt\nrenewal process\nQ\nuncertain quantiﬁer\n(Q, S, P)\nuncertain proposition\n∀\nuniversal quantiﬁer\n∃\nexistential quantiﬁer\n∨\nmaximum operator\n∧\nminimum operator\n¬\nnegation symbol\nPr\nprobability measure\n(Ω, A, Pr)\nprobability space\nCh\nchance measure\n∅\nthe empty set\nℜ\nthe set of real numbers\n|A|\ncardinality of set A\niid\nindependent and identically distributed\n\n\nIndex\nα-path, 353\nalternative hypothesis, 110\nAmerican option, 384\nAsian option, 390\nbelief degree, 439\nbetting ratio, 439\nBoolean function, 70\nBoolean uncertain variable, 70\nbridge system, 176\nbusy period, 314\nchain rule, 333\nchance distribution, 414\nchance inversion theorem, 415\nchance measure, 410\nchange of variables, 334\nChen-Ralescu theorem, 181\ncomplement of uncertain set, 224\nconditional uncertain measure, 23\nconﬁdence interval, 125, 134\ncontainment, 232\nconvergence almost surely, 95\nconvergence in distribution, 95\nconvergence in mean, 95\nconvergence in measure, 95\ncurrency option, 403\nDe Morgan’s law, 200\ndiﬀusion, 325, 330\ndistance, 88, 242\ndisturbance term, 116, 131\ndrift, 325, 330\ndual quantiﬁer, 252\nduality axiom, 7\nEllsberg experiment, 430\nempirical uncertainty distribution, 103\nentropy, 89, 243\nEuler method, 364\nEuropean option, 378\nevent, 7\nexpected loss, 170\nexpected value, 78, 234, 421\nextreme value theorem, 58, 283\nfair price principle, 379\nfeasible solution, 141\nﬁrst hitting time, 286, 361\nforecast value, 123, 134\nfrequency, 4\nFubini theorem, 336\nfundamental theorem of calculus, 332\nfuzzy set, 449\ngoal programming, 158\nhypothetical syllogism, 189\nidle time, 319\nimaginary inclusion, 235\ninclusion, 231\nindependence, 20, 50, 217\nindependent increment, 281\nindicator function, 103\nindividual feature data, 247\ninference rule, 267\nintegration by parts, 335\ninterest rate ceiling, 398\ninterest rate ﬂoor, 400\nintersection of uncertain sets, 222\ninterval number, 453\ninverse membership function, 215\ninverse uncertainty distribution, 46\ninverted pendulum, 272\nLaplace criterion, 1\nlaw of contradiction, xiv, 199\nlaw of excluded middle, xiv, 199\nlaw of large numbers, 428\nlaw of truth conservation, xiv\nLebesgue measure, 9\nlinear uncertain variable, 44\nlinguistic summarizer, 263\nLiu integral, 325\nLiu process, 323\nlogical equivalence theorem, 257\n\n\nIndex\n483\nloss function, 161\nmachine scheduling problem, 146\nmaximum entropy principle, 94\nmaximum likelihood estimation, 139\nmaximum uncertainty principle, xiv\nmeasure inversion formula, 201\nmeasure inversion theorem, 33\nmedian, 45\nmembership function, 201\nmethod of least squares, 109\nmethod of moments, 105, 369\nmodus ponens, 187\nmodus tollens, 188\nmoment, 86\nmonotonicity theorem, 10\nmultilevel programming, 159\nmultiobjective programming, 156\nNash equilibrium, 159\nnegated quantiﬁer, 250\nnonempty uncertain set, 196\nnormal uncertain variable, 45\nnormality axiom, 7\nnull hypothesis, 110\noperational law, 53, 219, 416\noptimal solution, 142\noption pricing, 378\nparallel system, 162\nPareto solution, 157\npossibility measure, 450\npower set, 9\nproduct axiom, 12\nproduct uncertain measure, 12\nproject scheduling problem, 153\nrandomness, deﬁnition of, 441\nregular membership function, 210\nregular uncertainty distribution, 46\nrejection region, 111\nreliability index, 174\nrenewal process, 297\nrenewal reward process, 300\nresidual, 118, 132, 365\nrisk index, 163\nruin index, 305\nruin time, 306\nrule-base, 269\nsample path, 276\nseries system, 161\nshortage index, 309\nshortage time, 310\nstability, 351\nStackelberg-Nash equilibrium, 160\nstandby system, 162\nstationary increment, 291\nstrictly decreasing function, 60\nstrictly increasing function, 53\nstrictly monotone function, 62\nstructural risk analysis, 165\nstructure function, 173\nsubadditivity axiom, 7\ntime integral, 288, 362\ntotally ordered uncertain set, 197\ntrapezoidal uncertain set, 207\ntriangular uncertain set, 207\ntruth value, 179, 258\nuncertain calculus, 323\nuncertain currency model, 402\nuncertain diﬀerential equation, 341\nuncertain entailment, 185\nuncertain ﬁeld, 277\nuncertain ﬁnance, 377\nuncertain hypothesis test, 110, 367\nuncertain inference control, 268\nuncertain inference rule, 267\nuncertain insurance model, 304\nuncertain integral, 325\nuncertain interest rate model, 396\nuncertain logic, 247\nuncertain measure, 8\nuncertain process, 275\nuncertain production model, 309\nuncertain programming, 141\nuncertain proposition, 177, 257\nuncertain quantiﬁer, 248\nuncertain queueing model, 314\nuncertain random variable, 413\nuncertain regression analysis, 116\nuncertain reliability analysis, 174\nuncertain renewal process, 297\nuncertain risk analysis, 161\nuncertain sequence, 95\nuncertain set, 193\nuncertain statistics, 103\nuncertain stock model, 378\nuncertain time series analysis, 131\nuncertain variable, 27\nuncertain vector, 100\n\n\n484\nIndex\nuncertainty, deﬁnition of, 441\nuncertainty distribution, 30, 278\nuncertainty space, 11\nuncertainty theory, xi\nunion of uncertain sets, 220\nurn problem, 1, 430\nvalue-at-risk, 169\nvariance, 83, 425\nvehicle routing problem, 149\nwaiting time, 318\nWiener process, 441\nYao-Chen formula, 356\nYe lemma, 459\nzero-coupon bond, 397\nzigzag uncertain variable, 45\n\n\nBaoding Liu\nUncertainty Theory\nSomething is called random if its frequency of occurrence is known. Other-\nwise, it is called uncertain. The outcome of tossing a coin is an example of\nrandomness since the frequency that the coin will come up heads is known.\nThe outcome of a falling cake is an example of uncertainty since the frequency\nthat the cake will land butter-side down is unknown. In order to rationally\ndeal with those phenomena, there exist two mathematical systems, one is\nprobability theory and the other is uncertainty theory. Probability theory is\na branch of mathematics concerned with the analysis of random phenomena,\nwhile uncertainty theory is a branch of mathematics concerned with the anal-\nysis of uncertain phenomena. In order to use them to handle some quantity\n(e.g., stock price) in practice, the ﬁrst action is to produce a distribution\nfunction representing the possibility that the quantity falls into the left side\nof the current point. If you believe the distribution function is close enough\nto the future frequency, then you should use probability theory. Otherwise,\nyou have to use uncertainty theory. Numerous empirical studies show that\nthe real world is far from frequency stability. This fact makes the distribu-\ntion function obtained in practice usually deviate from the future frequency\neven when numerous observed data are available, and consequently provides\na motivation to learn and use uncertainty theory.\nThis is an introductory textbook on uncertainty theory, uncertain statistics,\nuncertain programming, uncertain risk analysis, uncertain reliability anal-\nysis, uncertain set, uncertain logic, uncertain inference, uncertain process,\nuncertain calculus, and uncertain diﬀerential equation. This textbook also\nshows applications of uncertainty theory to scheduling, logistics, data mining,\ncontrol, and ﬁnance.\nAxioms of Uncertainty Theory\nAxiom 1. (Normality Axiom) M{Γ} = 1 for the universal set Γ.\nAxiom 2. (Duality Axiom) M{Λ} + M{Λc} = 1 for any event Λ.\nAxiom 3. (Subadditivity Axiom) For every countable sequence of events Λ1,\nΛ2, · · · , we have\nM\n( ∞\n[\ni=1\nΛi\n)\n≤\n∞\nX\ni=1\nM{Λi}.\nAxiom 4. (Product Axiom) Let (Γk, Lk, Mk) be uncertainty spaces for k =\n1, 2, · · · The product uncertain measure M is an uncertain measure satisfying\nM\n( ∞\nY\nk=1\nΛk\n)\n=\n∞\n^\nk=1\nMk{Λk}\nwhere Λk are arbitrarily chosen events from Lk for k = 1, 2, · · · , respectively.","difficulty":"easy","domain":"Single-Document QA","length":"long","question":"Construct an uncertain variable with linear uncertainty distribution L(1,5)","sub_domain":"Academic"}

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