aa r X i v : . [ ec on . T H ] F e b A Theory of Choice Bracketing under Risk ∗ Mu Zhang † February 16, 2021
Abstract
Aggregating risks from multiple sources can be complex and demanding, anddecision makers usually adopt heuristics to simplify the decision process. Thispaper axiomatizes two such heuristics, narrow bracketing and correlation neglect, byrelaxing the standard independence axiom in the expected utility benchmark. Ourrepresentation theorem allows for either narrow bracketing, or correlation neglect, orboth of them. The flexibility of our framework allows for applications in various setups.For example, we accommodate the experimental evidence on narrow bracketing andrisk aversion over small gambles with background risk. In intertemporal choices, weshow how our framework unifies three seemingly distinct models in the literatureand introduce a new model that can satisfy many desirable normative propertiesin time preferences simultaneously, including indifference to temporal resolution ofuncertainty, dynamic consistency and separation of time and risk preferences. Onespecial class of the model shares the same predictions as Epstein and Zin (1989)in macroeconomics and finance applications, and is immune to the critique inEpstein, Farhi, and Strzalecki (2014). ∗ I am deeply indebted to Faruk Gul, Wolfgang Pesendorfer and Pietro Ortoleva for their invaluableadvice, encouragement and support throughout this project. I have greatly benefited from discussions withXiaosheng Mu and Rui Tang. I also thank Roland Bénabou, Modibo Camara, Sylvain Chassang, XiaoyuCheng, Francesco Fabbri, Shaowei Ke, Alessandro Lizzeri, Dan McGee, Evgenii Safonov, Ludvig Sinander,João Thereze, Can Urgan, Leeat Yariv and seminar participants at Princeton Microeconomic Theory StudentLunch Seminar for helpful comments and discussions. † Department of Economics, Princeton University, [email protected] Introduction
Decision makers in the real world usually face multiple risky decision problems. For instance,an investor might need to take care of her investment accounts simultaneously in differentfinancial markets, ranging from stocks, bonds, to bitcoins; a worker has to decide on howmuch to save for retirement every month; a manager could be responsible for multiple plantsor sectors and make customized plans for each of them. One implicit assumption of the long-standing focus on single choice problems in theories of risk preferences is that agents canrationally aggregate and assess risks and consequences in multiple sources. However, multi-source risk is naturally more complex and challenging than single-source risk. As a result,people usually adopt heuristics to simplify their decision-making process, among which tworepresentative examples are narrow bracketing and correlation neglect.Narrow bracketing, formalized by Thaler (1985) and Read, Loewenstein, and Rabin(1999), describes the situation where a decision maker (DM) faced with multiple choiceproblems tends to choose an option in each decision without full regard to other de-cisions. This is a simplifying heuristic as searching for a local optimum is generallyless costly than searching for a global optimum. As the building block of manybehavioral models, narrow bracketing helps to explain financial anomalies like the equitypremium puzzle (Benartzi and Thaler, 1995) and the stock market participation puzzle(Barberis, Huang, and Thaler, 2006). It can also make a complex model tractable byassuming agents optimize each decisions in isolation (Barberis, Jin, and Wang, 2020).Besides strong experimental evidence (Tversky and Kahneman, 1981, Rabin and Weizsäcker,2009, Ellis and Freeman, 2020), narrow bracketing itself serves as an important assumptionin many experiments. For example, show-up fees are ubiquitous but almost no experimentson choices under risk take them into account when estimating the subjects’ risk attitudes.Also, evidence on loss aversion is only valid if we ignore all wealth accrued from decisionsoutside of the laboratory.The second commonly used heuristic is correlation neglect. Given the prevalence ofcorrelated choices, agents tend to ignore the interdependence among decision problemsand treat them as if they are independent. This simplifies the decision process as onlymarginal distributions need to be considered. Experimental evidence of correlation neglecthas been found in various setups including belief formation (Enke and Zimmermann, 2019),2ortfolio allocation (Eyster and Weizsacker, 2016, Kallir and Sonsino, 2009) and schoolchoice (Rees-Jones, Shorrer, and Tergiman, 2020). It is also an important element in manybehavioral models. For instance, Ortoleva and Snowberg (2015) uses correlation neglect asthe micro foundation of overconfidence in political behavior.Despite the popularity of these heuristics in behavioral and experimental economics, theyhave received little attention in the choice-theoretic literature. Actually, they are typicallyinterpreted as behavioral or “irrational” biases and supposed to deviate drastically fromthe standard axiomatic framework for choice under risk. Also, in most applications, onlyone of them is incorporated and it is usually coupled with other behavioral factors like lossaversion. However, recent findings urge for a better understanding of the two heuristics. Takechoice bracketing as our major example. Empirically, through a novel revealed preferencetest, Ellis and Freeman (2020) find that most subjects are best described as either narrowor broad bracketing, even if intermediate cases are allowed. They suggest that narrowbracketing might be better viewed as a “heuristic” instead of a “bias”, and it may occurwhen agents “are unaware of how to broadly bracket, or or are unaware that broad bracketingcan lead to notably higher payoffs, or choose to employ to simplify their decision-making”.Theoretically, based on an impossibility result, Mu, Pomatto, Strack, and Tamuz (2020)suggest that “theories that do not account for narrow framing · · · cannot explain commonlyobserved choices among risky alternatives.”To our best knowledge, this paper is the first to provide a choice-theoretic foundationfor narrow bracketing and correlation neglect. We consider the preference of a DM overlotteries of two-dimensional outcome profiles P . An outcome profile can be interpretedas the consequences of two decision problems, such as simultaneous monetary gambles,intertemporal choices and consumption choices involving multiple goods. We will labeleach dimension as a source of risk and call the marginal distribution in some source as a marginal lottery . We start with the benchmark where the preference admits an expectedutility (EU) representation, which is characterized by the von Neumann–Morgenstern(vNM) independence axiom. Then we model narrow bracketing and correlation neglect asrelaxations of the independence axiom. This approach has the following advantages. First,we can argue that narrow bracketing and correlation neglect are not more behavioral orexotic than other commonly accepted non-EU theories in the literature like certainty effectand reference dependence. Second, we can compare different utility representations involving3he heuristics in a unified framework and understand how much they deviate from the EUbenchmark. Finally, we decouple the effects of narrow bracketing and correlation neglectfrom other behavioral factors like loss aversion. We admit that restricting the benchmark toEU limits the explanatory power of the model, but it is a nice starting point.In our model, narrow bracketing and correlation neglect are closed related heuristics butthey differ in the following sense. A narrow bracketer acts if she can perceive the correlationamong different sources in each lottery correctly, but she is optimizing in each single source inisolation. In contrast, a DM who ignores correlation understands how to aggregate outcomesand optimizes globally but she misperceives the interdependence of risks in difference sources.Our main results are two representation theorems. We first assume that the DM alwaysignores the correlation among two sources of risk. This is consistent with the experimentaldesigns for testing narrow bracketing where risks in different decision problems are resolvedindependently (Rabin and Weizsäcker, 2009, Ellis and Freeman, 2020). We call a lotterywith independent marginals as a product lottery and denote the set of all product lotteriesas ˆ P ⊂ P . Correlation neglect implies that we can just focus on the preference restricted toproduct lotteries. The EU benchmark in this case becomes the
EU with correlation neglect(EU-CN) model with the representation V EU − CN ( P ) = X x,y w ( x, y ) P ( x ) P ( y ) , ∀ P ∈ P . By comparison, a DM exhibits narrow bracketing will evaluate the marginal lotteries inisolation by first taking the certainty equivalents of them. The corresponding
NarrowBracketing (NB) representation is V NB ( P ) = w ( CE v ( P ) , CE v ( P )) , ∀ P ∈ P . A DM might only narrowly bracket marginal lotteries in one source instead both sourcesas NB. For instance, she might adopt a backward induction evaluation process where shefirst reduces the marginal lottery in source 2 to its certainty equivalent and then evaluatesthe risk in source 1 using expected utility. This is called backward induction bracketing withcorrelation neglect (BIB-CN) representation where V BIB − CN ( P ) = X x w ( x, CE v ( P )) P ( x ) , ∀ P ∈ P . forward induction bracketing with correlation neglect (FIB-CN) representation: V F IB − CN ( P ) = X y w ( CE v ( P ) , y ) P ( y ) , ∀ P ∈ P . We then provide an axiomatic foundation for those representations. Besides technicalaxioms including weak order, monotonicity and continuity, we consider two relaxations of thevNM independence axiom on the set of product lotteries. Axiom Conditional Independencestates that the DM satisfies the vNM independence axiom when she only needs to makedecisions in one source. That is, our model coincides with the EU benchmark when there isonly one decision problem. Axiom Weak Multilinear Independence is our central axiom. Itstates that the independence property holds only if the two product lotteries that are goingto be mixed should be “similar” enough in the sense that they should agree on the marginallottery in one source, and their marginal lotteries in the other source should be indifferentto the DM if evaluated narrowly. This axiom states that the independence property holdslocally for a source if the DM does not narrowly frame risks in that source and hence narrowbracketing can be interpreted as a violation of the vNM independence axiom. We name thecombination of the two axioms as Axiom Weak Independence.Our axiomatization also allows for representations that agree with NB on part of thedomain and with BIB-CN or FIB-CN on the rest of the domain. We denote them as GBIB-CN and GFIB-CN representations and the concrete functional forms are included in Section3.3. Representations like BIB-CN, FIB-CN and NB are special cases of GBIB-CN andGFIB-CN. Then Theorem 1 states that a preference satisfies correlation neglect, AxiomWeak Independence and some technical axioms if and only if it admits a representationamong the three classes of models: EU-CN, GBIB-CN and GFIB-CN.Then we discard the correlation neglect assumption and allow the DM to correctly perceivethe interdependence of risks in different sources. We generalize previous models to backwardinduction bracketing (BIB) and forward induction bracketing (FIB) . For instance, the BIBrepresentation is V BIB ( P ) = X x w ( x, CE v ( P | x )) P ( x ) , ∀ P ∈ P , where P is the marginal lottery of P in source 1 and P | x is the conditional lottery of P insource 2 given outcome x in source 1. We weaken correlation neglect to Axiom Correlation5ensitivity, which states that independence holds if the correlation structures are not affectedafter the mixture. Our characterization results Theorem 2 and Corollary 1 extend Theorem 1by allowing for representations EU, BIB and FIB.Unlike many decision theory papers that only involve a single representation, our resultscharacterize several seemingly distinct and extreme functional forms. We interpret thedistinction as an important and necessary feature of our framework instead of a drawback,since our goal is to model choice bracketing and correlation neglect as simplifying andintuitive heuristics, while many “intermediate” functional forms in our setup would beeither complicated or hard to interpret. For instance, our model excludes the “partialnarrow bracketing” representation in the literature (e.g., Barberis, Huang, and Thaler, 2006,Rabin and Weizsäcker, 2009, Ellis and Freeman, 2020), which features a weighted average ofthe broad bracketing representation (EU) and NB . We will justify our approach in two ways.Empirically, Ellis and Freeman (2020) show very few (around 5%) of their subjects are bestclassified to partial narrow bracketing. This suggests that incorporating such intermediatecases might not necessarily have larger explanatory power in practice. From the theoreticalpoint of view, the weighted average of the broad bracketing utility and the narrow bracketingutility arguably more complex and involving than the computation of either representation.This contradicts with our interpretation that choice bracketing should simplify the evaluationprocess compared to the EU benchmark.We then apply our model to different economic setups. When we interpret thetwo sources of risks as simultaneous and independent monetary gambles, then our NBmodel can be used to accommodate the experimental evidence about choice bracketingin Tversky and Kahneman (1981) and Rabin and Weizsäcker (2009). When we interpretsource 1 as the background risk and source 2 as the gamble at hand, then our frameworkis suitable to study risk preferences with background risk, or more specifically, riskaversion over small gambles following Rabin (2000), Barberis, Huang, and Thaler (2006)and Mu, Pomatto, Strack, and Tamuz (2020). When a DM admits a NB representation,then she will ignores the effect of the background risk and hence can exhibit enough riskaversion over small gambles without inducing unrealistic risk aversion over large gambles.Finally, when the two sources are interpreted as intertemporal choices in two periods, then The NB representation in those papers differ from ours. Please refer to Section 6.1 for a detaileddiscussion.
Related Literature.
The most similar work to ours is Vorjohann (2020), where the author7ndependently develops a choice-theoretic model for choice bracketing within the expectedutility framework. In the model, each DM is endowed with a broad preference and a narrowpreference, both of which are EU and can be observed or identified in the experiment. Sheprovides two axioms to connect the two preferences by interpreting narrow bracketing asdeviations from broad bracketing by ignoring correlation and changing the EU index. Ourpaper differ from Vorjohann (2020) in three aspects. First, in our framework, a DM onlyhas one preference and we try identify where she is subject to choice bracketing and/orcorrelation neglect from her choice data over lotteries. Second, Vorjohann (2020) maintainsthe EU paradigm, while we interpret choice bracketing and correlation neglect as deviationsfrom the EU benchmark. Actually, in the case with two sources, her representation ofnarrow bracketing lies in the intersection of our EU and NB representations. Finally, weallow for more general forms of choice bracketing and separation between choice bracketingand correlation neglect, while Vorjohann (2020) regards correlation neglect as an ingredientof choice bracketing.Another related strand of literature provides an alternative explanation for narrowbracketing in the DM’s consumption behavior by assuming limited attention to price orpreference shocks. For instance, Kőszegi and Matějka (2020) show that an rationallyinattentive consumer with imperfect information about the shocks would exhibit mentalbudgeting and naiver diversification. Lian (2020) proposes a theory of narrow thinking wherethe DM makes each decision with imperfect information of other decision problems. As aresult, the optimization problem is equivalent to solving an incomplete information, commoninterest game played by multiple selves. In contrast, we adopt an choice-theoretic approachto axiomatize choice bracketing directly and model it as a simplifying heuristic to deal withmulti-source risks. Also, our model can be applied in simple settings without shocks to pricesor preferences like experiments on choices under objective risk (Rabin and Weizsäcker, 2009,Ellis and Freeman, 2020). Hence the two approaches are complementary to each other.
We use a version of the standard expected utility framework with a two-dimensional outcomespace X = X × X , where X i is the set of outcomes in source i ∈ { , } . Throughout thepaper, we assume that X i is a nontrivial closed interval on the real line R that includes 0,8hat is, for each i = 1 , X i = [ c i , c i ] ∩ R , where c i > c i , 0 ∈ [ c i , c i ] and c i , c i ∈ R ∪{−∞ , + ∞} .Note that we allow for both bounded and unbounded outcome spaces. We call ( x , x ) ∈ X an outcome profile and x i is the outcome in source i for i ∈ { , } . A positive outcome canbe interpreted as a gain , while a negative one is a loss . For each set A ⊆ R , we denote A o as the relative interior of A with respect to R .A (joint) lottery is a probability distribution over X with a finite support. Denote P asset of all lotteries endowed with the topology of weak convergence and the standard mixtureoperation. For each lottery P ∈ P , the marginal lottery of P in source 1 is denoted as P ∈ L ( X ) such that for each x ∈ X , P ( x ) = P x ∈ X P ( x , x ). P represents themarginal risk of P in source 1. The marginal lottery P of P in source 2 can be definedsimilarly. Sometimes we will call a marginal lottery p ∈ L ( X ) ∪ L ( X ) a single-source lottery and call a lottery P ∈ P a multi-source lottery. An important subclass of lotteries isthe set of product lotteries ˆ P = L ( X ) × L ( X ) ( P where the marginal lotteries of eachproduct lottery is independent from each other. Easy to see that for each lottery P , the tupleof marginal lotteries ( P , P ) is a product lottery. As is discussed in the introduction, manyexperiments on narrow bracketing (e.g., Rabin and Weizsäcker (2009) and Ellis and Freeman(2020)) stress that risks in different gambles are resolved independently and hence the setof product lotteries ˆ P might be a more appropriate domain than the set of all lotteries P .For each lottery P ∈ P and marginal lottery p ∈ L ( X ) ∪ L ( X ), we denote supp( P ) := { ( x , x ) ∈ X × X : P ( x , x ) > } and supp( p ) := { x ∈ X ∪ X : p ( x ) > } . When thereis no confusion, we write the degenerate marginal lottery δ x as x for x ∈ X ∪ X .The primitive of our analysis is a binary relation % on P . We define the narrow preferencein source % as the restriction of % on L ( X ) × { } , that is, p % q if and only if( p, % ( q, δ , the comparison oflotteries ( p,
0) and ( q,
0) can be interpreted as the comparison of p and q in source 1 as ifsource 2 does not exist. Symmetrically, we denote % as the narrow preference in source 2.These notions will prove useful in our utility representations involving choice bracketing inthe next section.It is worthwhile to mention that our framework can accommodate many different economicapplications, depending on our interpretations of the two sources of outcomes. For example,in lab experiments on choices under risk, they can represent money or tokens in two differentgambles; in individual portfolio choices, they can represent account balances on the stock9arket and the bitcoin market respectively; in intertemporal choices, they can representconsumptions in two different periods; in consumption choices with multiple goods, theycan be the expenditures on food and clothing respectively. We will elaborate more on thoseexamples in Section 6. In this section, we introduce different decision rules adopted by a DM faced with two-source risk. We start with the expected utility model as the benchmark. As is discussed inthe introduction, people in practice usually deviate from the benchmark systematically byadopting some simplifying heuristics. In the following we will focus on two such heuristics:choice bracketing and correlation neglect.
For each function f : X → R or f : X i → R for some i = 1 ,
2, we say f is regular if itis continuous, strictly monotone and bounded. If the domain of f is compact (i.e., when X = [0 , c ] × [0 , c ]), then boundedness is implied by continuity and hence redundant. Thedefinition of an expected utility representation is standard. Definition 1 (EU) . Let % be a binary relation on P and let w : R → R be a regularfunction. The utility index w is an expected utility (EU) representation of % if % isrepresented by V EU : P → R , which is defined by V EU ( P ) = X x,y w ( x, y ) P ( x, y ) . We first consider choice bracketing, where the DM might evaluate risks in different sourcesin isolation and hence make choices in some source without regard to lotteries in the othersource. We start with the case where decisions in both sources are made separately. Foreach i ∈ { , } and each regular function f : X i → R , we denote the certainty equivalent of p ∈ L ( X i ) under f as CE f ( p ) = f − ( P x f ( x ) p ( x )). As X i is a closed interval and f is10trictly monotone and continuous, the certainty equivalent is well-defined and CE f ( p ) ∈ X i for each p ∈ L ( X i ). The definition of narrow bracketing representation is as follows. Definition 2 (NB) . Let % be a binary relation on P and let w : X × X → R , v i : X i → R , i = 1 , , be regular functions. The tuple ( w, v , v ) is a (fully) narrow bracketing (NB) representation of % if % is represented by V NB : P → R , which is defined by V NB ( P ) = w ( CE v ( P ) , CE v ( P )) . Intuitively, v i is the EU index of narrow preference % i for i = 1 , w representsthe DM’s preference in the absence of risk. If % admits a NB representation, then theDM evaluates each lottery by first reducing the marginal lotteries in both sources to theircertainty equivalents under source-sensitive utility indices. Hence it captures the idea thatchoices made in source 1 are independent of alternatives in source 2 and vice versa.Then we consider the case where the DM adopts partial narrow bracketing and choicesmade in one source are independent of alternatives in the other source while the reversefails. For each lottery P ∈ P and x in the support of P , i.e., P ( x ) >
0, we denote theconditional lottery P | x as the conditional distribution of outcomes in source 2 given x insource 1, which represents the conditional risk in source 2. Formally, for each y ∈ X , P | x ( y ) = P ( x, y ) /P ( x ). Then we say a preference admits a backward induction bracketingrepresentation if the DM first reduces the conditional risks in source 2 to their certaintyequivalents and then evaluates the risk in source 1. Definition 3 (BIB) . Let % be a binary relation on P and let w : X × X → R , v : X → R be regular functions. The tuple ( w, v ) is a backward induction bracketing (BIB) representation of % if % is represented by V BIB : P → R , which is defined by V BIB ( P ) = X x w ( x, CE v ( P | x )) P ( x ) . In a BIB representation ( w, v ), the DM adopts the following backward inductionevaluation process: i) conditional on each possible outcome x in source 1, the DM firstevaluates the conditional risk in source 2 by replacing the conditional lottery P | x with itscertainty equivalent under EU index v ; ii) Then the DM evaluates the risk in source 1 usingEU index w . It is important to notice that the evaluation of conditional risks is independentof the outcome in source 1, which turns out to be the key behavioral deviation of BIB from11he EU benchmark. Actually, conditional on outcome x in source 1, if we replace v in theBIB representation with w ( · , x ), then we will exactly get the EU representation with index w . Hence, BIB captures the idea of narrowly bracketing risks in source 2. In Section 6, wewill discuss the implications of BIB in time preferences and compare it with the well-knownKreps-Porteus preferences, which also admits a backward induction interpretation.Symmetrically, when the DM only narrowly brackets risks in source 1, then we derive theforward induction bracketing representation. For each P ∈ P and y in the support of P ,we denote P | y as the conditional distribution of outcomes in source 1 given outcome y insource 2. Definition 4 (FIB) . Let % be a binary relation on P and let w : X × X → R , v : X → R be regular functions. The tuple ( w, v ) is a forward induction bracketing (FIB) representation of % if % is represented by V F IB : P → R , which is defined by V F IB ( P ) = X x w ( CE v ( P | y ) , y ) P ( x ) . In applications where there is a natural order on the two sources, FIB might be moreappropriate than BIB. For instance, risk in source 1 can be interpreted as the backgroundrisk or endowment risk, while risk in source 2 can be interpreted as the risk in the currentdecision problem such as the portfolio choice. In an experiment on choices under risk,background risk includes show-up fees, payoffs from other rounds in the experiment andwealth outside the laboratory.
In this section, we consider the second simplifying heuristic: correlation neglect, where theDM finds it difficult to deal with the correlation structure of risks in different sources andhence treats the lottery as if its marginal lotteries are independent from each other. We willintroduce the counterparts of previous representations by imposing correlation neglect.We start with a DM who is only subject to correlation neglect compared to the EUbenchmark.
Definition 5 (EU-CN) . Let % be a binary relation on P and let w : X × X → R be aregular function. The utility index w is an expected utility with correlation neglect EU-CN) representation of % if % is represented by V EU − CN : P → R , which is defined by V EU − CN ( P ) = X x,y w ( x, y ) P ( x ) P ( y ) . The behavior of a DM with an EU-CN representation agrees with the EU benchmark onthe set of product lotteries ˆ P , but she ignores the interdependence of risks from differentsources even if they are not independent.Now we study the interplay of choice bracketing and correlation neglect. First, it is easyto see that NB satisfies correlation neglect as the DM takes certainty equivalents of themarginal lotteries directly. Second, suppose that the DM narrowly brackets marginal risksin source 2 after ignoring the correlation structure, then we get the following representation. Definition 6 (BIB-CN) . Let % be a binary relation on P and let w : X × X → R , v : X → R be regular functions. The tuple ( w, v ) is a backward induction bracketing withcorrelation neglect (BIB-CN) representation of % if % is represented by V BIB − CN : P → R , which is defined by V BIB − CN ( P ) = X x w ( x, CE v ( P )) P ( x ) . Our characterization results in Section 4 allow for a general representation that incorpo-rates both NB and BIB-CN as special cases. The generalization is based on the idea thatwhether the DM narrow brackets the marginal risk in source 1 might depend on the marginalrisk in source 2. For example, suppose the two sources represent today and tomorrowrespectively and the DM’s preference is represented by the following function for some fixedoutcome a ∈ X tomorrow: U ( p, q ) = w ( CE v ( p ) , CE v ( q )) , if CE v ( q ) ≤ a, P x w ( x, CE v ( q )) p ( x ) , if CE v ( q ) > a. That is, the utility representation adopts a threshold structure and the DM might beeither NB or BIB-CN depending on the certainty equivalent of tomorrow’s marginallottery. Intuitively, if tomorrow’s stakes are low, then the DM might make today’s choicesindependent of tomorrow’s outcomes to simplify the decision process. If instead tomorrow’sstakes are high enough, she would be more careful about evaluating today’s risk by takinginto account the income effect of tomorrow’s lottery. To some extent, this example can be13nterpreted as a version of endogenous choice bracketing. In order to keep continuity on theboundary (i.e., when CE v ( q ) = a ), w ( · , a ) must be a positive affine transformation of v .The next definition extends the above idea by generalizing the threshold structure thatdetermines when the DM switches between NB and BIB-CN. Definition 7 (GBIB-CN) . Let % be a binary relation on P , let w : X × X → R , v i : X i → R , i = 1 , , be regular functions and let H be an open subset of X with H . Thetuple ( w, v , v , H ) is a generalized backward induction bracketing with correlationneglect (GBIB-CN) representation of % if % is represented by V GBIB − CN : P → R , whichis defined by V GBIB − CN ( P ) = w ( CE v ( P ) , CE v ( P )) , if CE v ( P ) ∈ X \ H , P x w ( x, CE v ( P )) P ( x ) , if CE v ( P ) ∈ H , where for any y ∈ ∂H , i.e., the boundary of set H , w ( · , y ) is a positive affine transformationof v . Notice that an open subset of the real line can be represented by a countable union ofdisjoint open intervals. Hence, the GBIB-CN representation captures the idea that locallythe DM exhibits narrow bracketing either in both sources, or just in source 2. Specifically,when H is empty, GBIB-CN reduces to NB; when the closure of H is X , GBIB-CN reducesto BIB-CN.Symmetrically, we can modify the definitions of BIB-CN and GBIB-CN to accommodatethe case where the DM narrowly brackets risks in source 1. Definition 8 (FIB-CN) . Let % be a binary relation on P and let w : X × X → R , v : X → R be regular functions. The tuple ( w, v ) is a forward induction bracketing withcorrelation neglect (FIB-CN) representation of % if % is represented by V F IB − CN : P → R , which is defined by V BIB − CN ( P ) = X y w ( CE v ( P ) , y ) P ( y ) . Definition 9 (GFIB-CN) . Let % be a binary relation on P , let w : X × X → R , v i : X i → R , i = 1 , , be regular functions and let H be an open subset of X with H . The The proof is given by Lemma 18 in the appendix. uple ( w, v , v , H ) is a generalized forward induction bracketing with correlationneglect (GFIB-CN) representation of % if % is represented by V GF IB − CN : P → R , whichis defined by V GF IB − CN ( P ) = w ( CE v ( P ) , CE v ( P )) , if CE v ( P ∈ X \ H , P y w ( CE v ( P ) , y ) P ( y ) , if CE v ( P ) ∈ H , where for any x ∈ ∂H , w ( x, · ) is a positive affine transformation of v . We end this section with some remarks on the above representations: EU, EU-CN, BIB,GBIB-CN, FIB and GFIB-CN. It is worthwhile to mention that each of those functionalforms has intuitive and clear implications on the extent to which the DM adopts choicebracketing and correlation neglect. Moreover, each deviation from the EU benchmark dealswith the multi-source risk in a relatively simpler way in terms of computation. Our resultsin the next section characterize those seemingly extreme representations by relaxing thestandard vNM independence axiom in a reasonable manner. This approach is differentfrom a typical decision theory paper which would involve a universal representation. Weinterpret the distinction as an important and necessary feature of our framework instead ofa drawback, since our goal is to model choice bracketing and correlation neglect as simplifyingand intuitive heuristics, while many “intermediate” functional forms in our setup would beeither complicated or hard to interpret.One natural way to unify two representations is to consider their weighted averages.For instance, one popular representation in the literature of choice bracketing (e.g.,Barberis, Huang, and Thaler, 2006, Rabin and Weizsäcker, 2009, Ellis and Freeman, 2020)is “partial narrow bracketing”, which features an α -mixture of EU and NB . Besidesaxiomatic reasons, we justify our exclusion of such “intermediate” representations in twoways. First, the computation of the weighted average utility of EU and NB is arguably morecomplex and involving than the computation of either representation. This contradicts withour interpretation that choice bracketing should simplify the evaluation process comparedto the EU benchmark. Second, using three well-designed experiments, Ellis and Freeman(2020) show that very few (around 5%) of their subjects are best classified to partial narrowbracketing. This suggests that incorporating such intermediate cases might not necessarily Actually the NB representation used in the literature differs from our Definition 2. We will discuss theirdistinction in Section 6 and argue why our version might be more appropriate.
In this section, we present our axioms and characterization theorems. Theorem 1 focuses onthe representations that exhibit correlation neglect, that is, EU-CN, BIB-CN and FIB-CN.Then we extend the result to Theorem 2 and Corollary 1 to incorporate models withoutcorrelation neglect.We start with the axioms shared by the two characterization results. The first axiomassumes rationality of the DM.
Axiom Weak Order : % is complete and transitive.The next axiom is about monotonicity of the preference % with respect to some notionof dominance. In the case with single-source risk, there is an agreed definition of first orderstochastic dominance. However, its extension to multiple sources is not self-obvious. Luckily,we only need a weak notion of dominance, which only involves the comparison of a lotterywith a degenerate lottery. For any lottery P ∈ P and degenerate lottery ( x , x ) ∈ X × X ,we say P dominates ( x , x ) if P = ( x , x ) and y ≥ x , y ≥ x for all y ∈ supp( P ) , y ∈ supp( P ). Symmetrically, we say ( x , x ) dominates P if P = ( x , x ) and y ≤ x , y ≤ x for all y ∈ supp( P ) , y ∈ supp( P ). Then Axiom 2 states that the preference % is monotonicwith respect to dominance. Axiom Monotonicity : For each P ∈ P and ( x , x ) ∈ X × X , P ≻ ( x , x ) if P dominates( x , x ) and ( x , x ) ≻ P if ( x , x ) dominates P .Now we will introduce the continuity axiom. One reasonable candidate is the standardtopological continuity axiom, which guarantees that % has a continuous representation. Axiom Continuity : For each Q ∈ P , the sets { P ∈ P : P ≻ Q } and { P ∈ P : Q ≻ P } areopen subsets of P .However, an important observation is that BIB violates Axiom Continuity generically. Tosee why, recall that a DM with BIB evaluates each lottery P by first replacing the conditional16otteries in source 2 with its certainty equivalent and then taking expected utility for theconstructed new lottery. This would result in discontinuity when a small change in thelottery leads to a drastic change in the conditional lotteries. For instance, suppose that % admits a BIB representation ( w, v ). For each positive integer n , define P n = 1 / δ , δ ) +1 / δ − /n , δ ). Easy to see that P n weakly converges to P = 1 / δ , δ ) + 1 / δ , δ ). ThenAxiom Continuity requires12 w (1 ,
2) + 12 w (1 ,
3) = w (1 , CE v ( 12 δ + 12 δ ))which implies that v should be related to w (1 , · ). Actually, we can show that under AxiomContinuity, a preference that admits a BIB representation also admits an EU representation.In the above example of P n and P , the drastic change in conditional lotteries results fromthe fact that P n are not product lotteries and not all outcomes in the support of P n changeas n increases. This captures the key insights of how BIB violates Axiom Continuity. Similararguments hold for FIB. In order to maintain continuity as strong as possible while allowingfor BIB and FIB, we weaken Axiom Continuity into three parts.The first part guarantees that topological continuity holds on the set of product lotteries. Axiom Topological Continuity over Product Lotteries : For each Q ∈ P , the sets { P ∈ ˆ P : P ≻ Q } and { P ∈ ˆ P : Q ≻ P } are open subsets of ˆ P .The second part states that continuity holds if we only change the probability weightswithout changing the outcomes in the support. This is exactly the notion of mixturecontinuity. Axiom Mixture Continuity : For each
P, R, Q ∈ P , the sets { α ∈ [0 ,
1] : αP + (1 − α ) Q ≻ R } and { α ∈ [0 ,
1] : R ≻ αP + (1 − α ) Q } are open subsets of [0 ,
1] in the relative topology.By comparison, the last part deals with continuity concerning changes of outcomes inthe support instead of the probability weights. To avoid drastic variation in the conditionallotteries, we need to make sure that all outcomes in the same source change by the sameamount unless they have reached the bounds of the outcome space. This can be achievableby a modified notion of convolution with tight upper bounds. For each P ∈ P and a , a > P ∗ ( a , a ) ∈ P such that the probability of ( x, y ) ∈ X × X in P is transferred17o (min { x + a , c } , min { y + a , c } ). Intuitively, lottery P ∗ ( a , a ) is lottery P plus a suregain of a i in source i for i = 1 ,
2, up to the upper bounds imposed by the outcome space.Similarly, we can define p ∗ δ a for p ∈ L ( X ) ∪ L ( X ). The third part of the continuityaxiom guarantees that % is continuous as sure gains converge to 0. Axiom Continuity over Sure Gains : For each
P, Q ∈ P and any two sequences ǫ n , ǫ ′ n such that for each n , ǫ n , ǫ ′ n >
0, and ǫ n , ǫ ′ n → n → ∞ , P ∗ ( δ ǫ n , δ ǫ ′ n ) % Q, ∀ n = ⇒ P % Q and Q % P ∗ ( δ ǫ n , δ ǫ ′ n ) , ∀ n = ⇒ Q % P. Our Axiom Weak Continuity summarizes the above three relaxations of Axiom Continuity.
Axiom Weak Continuity : % satisfies Axiom Topological Continuity over ProductLotteries, Axiom Mixture Continuity and Axiom Continuity over Sure Gains.Now consider the standard vNM independence axiom, which characterizes EU. Axiom Independence : For each
P, Q, R ∈ ˆ P and α ∈ (0 , P ≻ Q = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) R. Under Axiom Weak Continuity, easy to see that Axiom Independence is equivalent to thefollowing stronger axiom.
Axiom Bi-independence : For each
P, Q, R, S ∈ ˆ P and α ∈ (0 , P ≻ Q, R ∼ S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S. In order to introduce our relaxation of Axiom Bi-independence, we first assume correlationneglect and focus on the set of product lotteries. Then we will relax correlation neglect toan independence axiom over lotteries which differ in the correlation structure. The formal definition of P ∗ ( a , a ) is as follows. Recall that X oi is the interior of X i , i = 1 ,
2. Foreach ( x, y ) ∈ X × X , if x + a ∈ X o , y + a ∈ X o , P ∗ ( δ a , δ a )( x + a , y + a ) = P ( x, y ); if y + a ∈ X o , P ∗ ( δ a , δ a )( c , y + a ) = P x + a >c P ( x, y ); if x + a ∈ X o , P ∗ ( δ a , δ a )( x + a , c ) = P y + a >c P ( x, y ).In addition, P ∗ ( δ a , δ a )( c , c ) = P x + a >c ,y + a >c P ( x, y ). .1 Correlation Neglect Axiom Correlation Neglect : For each P ∈ P , P ∼ ( P , P ).Axiom Correlation Neglect states that the DM is indifferent between each lottery and theproduct lottery with the same marginals. Then it suffices to study the preference % restrictedon the set of product lotteries ˆ P . This suits the applications where risks from different sourcesare independent such as experiments on choice bracketing (Barberis, Huang, and Thaler,2006). Another interesting example is the Nash equilibrium in a two-player game. Fishburn(1982) characterizes multilinear utility, which is exactly EU-CN restricted to ˆ P , as afoundation for expected utility in the 2-player game involving mixed strategies. The nextaxiom is key to Fishburn (1982)’s results. Axiom Multilinear Independence : For each
P, Q, R, S ∈ ˆ P , α ∈ (0 ,
1) and i, j ∈{ , } , if P i = R i , Q j = S j , then P ≻ Q, R ∼ S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S. In contrast to Axiom Bi-independence, Axiom Multilinear Independence imposes tworestrictions on the independence property. First, we only consider product lotteries. Second,whenever we want to mix two product lotteries, their marginal lotteries should be the samein at least one source. Technically, this is required to guarantee that the mixed lottery alsohas independent marginals . The following lemma directly follows from Fishburn (1982) andcharacterizes EU-CN. Lemma 1. (Fishburn, 1982)
Let % be a binary relation on P . The following statements areequivalent:i). The relation % satisfies Weak Order, Monotonicity, Weak Continuity, MultilinearIndependence and Correlation Neglect;ii). There exists an EU-CN representation of % . However, when we incorporate choice bracketing, Axiom Multilinear Independence willalso be violated. Notice that the set of product lotteries ˆ P is not a mixture space under the mixture operation definedon P . For instance, ( δ , δ ) and ( δ , δ ) are product lotteries, but their mixture 1 / δ , δ ) + 1 / δ , δ ) isnot. xample 1. Suppose that % admits a NB representation ( w, v , v ) with w ( x, y ) = x + y and v ( x ) = v ( x ) = √ x for all x, y ≥ . Let p = δ , q = δ , r = δ , s = δ , q = δ and p = δ (4+ ǫ ) for some ǫ > . Then V NB ( p , p ) = 25 + (4 + ǫ ) >
16 + 25 = V NB ( q , q ) = ⇒ ( p , p ) ≻ ( q , q ) ,V NB ( p , r ) = 25 + 0 = 16 + 9 = V NB ( q , s ) = ⇒ ( p , r ) ∼ ( q , s ) . However, for α = 1 / , the utilities of the mixed lotteries are V NB ( p , αp + (1 − α ) r ) = 25 + (4 + ǫ ) , V NB ( q , αq + (1 − α ) s ) = 16 + 16 . If < ǫ < √ − , then V NB ( p , αp +(1 − α ) r ) < V NB ( q , αq +(1 − α ) s ) = ⇒ ( p , αp +(1 − α ) r ) ≺ ( q , αq +(1 − α ) s ) . Hence Axiom Multilinear Independence fails.
Now we introduce our main independence axiom in the set of product lotteries. For each i ∈ { , } , we denote − i ∈ { , } with i = − i . Axiom Weak Independence :(i) Axiom Conditional Independence: For p, q, r, s ∈ L ( X ) , p ′ , q ′ , r ′ , s ′ ∈ L ( X ) and α ∈ (0 , s, p ′ ) ≻ ( s, q ′ ) = ⇒ ( s, αp ′ + (1 − α ) r ′ ) ≻ ( s, αq ′ + (1 − α ) r ′ ) . ( p, s ′ ) ≻ ( q, s ′ ) = ⇒ ( αp + (1 − α ) r, s ′ ) ≻ ( αq + (1 − α ) r, s ′ ) . (ii) Axiom Weak Multilinear Independence: For each P, Q, R, S ∈ ˆ P , α ∈ (0 ,
1) and i, j ∈ { , } , if P i = R i , Q j = S j , P − i ∼ − i R − i and Q − j ∼ − j S − j , then P ≻ Q, R ∼ S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S. The first part of the axiom states that if we fix the marginal lottery in one source, thenthe vNM independence axiom holds for marginal lotteries in the other source. For each p ∈ L ( X ), we denote % | p as the restriction of % on { p } × L ( X ). % | p can be interpreted20s the conditional preference in source 2 given lottery p in source 1. When p = δ , % | p agreeswith % , the narrow preference in source 2. Similarly, we can define % | q as the conditionalpreference in source 1 given q ∈ L ( X ). Along with Axiom Weak Order and Axiom WeakContinuity, Conditional Independence guarantees that each conditional preference admitsan EU representation. Hence, choice bracketing differs from the EU benchmark in terms ofhow the evaluations of the two marginal lotteries are aggregated.The second part is a local version of Axiom Multilinear Independence. It requires that theindependence property holds only if the two product lotteries that are mixed are “similar”enough, in the sense that they should agree on the marginal lottery in one source, and theirmarginal lotteries in the other source should be indifferent according to the narrow preference.To see why this reflects choice bracketing, suppose that the DM narrowly brackets risks insource − i , then she will evaluate the marginal lottery in source − i using the narrow preference % − i , regardless of the marginal lottery in source i . P i = R i and P − i ∼ − i R − i suggests that P should be indifferent to R , which implies αP + (1 − α ) R ∼ P by Conditional Independence.Also, we know Q ≻ αQ + (1 − α ) S . Hence Weak Multilinear Independence holds triviallyand the axiom is redundant. Similarly arguments hold if DM narrowly brackets risks insource − j . As a result, Weak Multilinear Independence is not redundant only if either theDM broadly brackets risks or she only narrowly brackets risks in source i = j . In the lattercase, mixture of lotteries only occurs in source − i . To conclude, the second part of AxiomWeak Independence states that the independence property holds locally for a source if theDM does not narrowly brackets risks in that source. This explains why we interpret choicebracketing as violations of the independence property.Now we are ready to state our first representation theorem under correlation neglect. Theorem 1.
Let % be a binary relation on P . The following statements are equivalent:i). The relation % satisfies Weak Order, Monotonicity, Weak Continuity, WeakIndependence and Correlation Neglect;ii). The relation % admits one of the following representations: EU-CN, GBIB-CN andGFIB-CN.Moreover, in all representations H , H are unique, v , v are unique up to a positive affinetransformation and in EU-CN, w is unique up to a positive affine transformation. It is worthwhile to mention that Theorem 1 characterizes three seemingly distant21epresentations with correlation neglect and choice bracketing, while the axioms do notseem to predict such a feature ex ante. Moreover, EU-CN, GBIB-CN and GFIB-CN allsatisfy Axiom Continuity and we keep the Axiom Weak Continuity in Theorem 1 just forconsistency with Theorem 2 below.
In this section we discard Axiom Correlation Neglect to incorporate representations thatare sensitive to the correlation structure of risks in different sources. Since Axiom WeakIndependence only involves product lotteries, we need another independence axiom forlotteries whose marginals are not independent.
Axiom Correlation Consistency : Suppose P ≻ Q with P i = Q i , i = 1 , R ∼ S and supp ( P ) ∩ supp ( R ) = supp ( Q ) ∩ supp ( S ) = ∅ , then for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S. Axiom Correlation Consistency relaxes Axiom Bi-independence as it focuses on theindependence property of the correlation structure. Since P and Q share the same marginallotteries in both sources, P ≻ Q means that the DM prefers the correlation structure of P to that of Q . As she is indifferent between R and S , if the mixture of P, Q and
R, S does not “infect” the original correlation structures of P and Q , then after mixture, theimpacts of R and S will cancel out and the preference ranking between P and Q shouldremain unchanged. How do judge whether or not the correlation structure is “infected”after mixture? Notice that each lottery can be decomposed into a marginal lottery in source1 and a profile of conditional lotteries in source 2. Hence one candidate measure of thecorrelation structure is the profile of conditional lotteries. That is why we need the additionalqualification that supp ( P ) ∩ supp ( R ) = supp ( Q ) ∩ supp ( S ) = ∅ . It says that in source 1,the marginals of two lotteries that are mixed have disjoint supports, which guarantees thatthe profile of conditional lotteries in source 2 are not affected by the mixture. As a result,the preference over the correlation structures of P and Q persist after the mixture and hencethe independence property holds. 22oreover, notice that under Axiom Correlation Neglect, lotteries with the same marginalsshould always be indifferent and hence Axiom Correlation Consistency trivially holds. Thisimplies that Axiom Correlation Consistency relaxes Axiom Bi-independence and AxiomCorrelation Neglect. The next representation theorem generalizes Theorem 1 by simplyreplacing Axiom Correlation Neglect with Axiom Correlation Consistency. Theorem 2.
Let % be a binary relation on P . The following statements are equivalent:i). The relation % satisfies Weak Order, Monotonicity, Weak Continuity, WeakIndependence and Correlation Consistency;ii). The relation % admits one of the following representations: EU, BIB, EU-CN, GBIB-CN and GFIB-CN.Moreover, in all representations H , H are unique, v , v are unique up to a positive affinetransformation and in EU, EU-CN, BIB, w is unique up to a positive affine transformation. By symmetry, an alternative measure of the correlation structure is to decompose eachlottery into a marginal lottery in source 2 and a profile of conditional lotteries in source 1.Then the qualification naturally changes to disjoint supports of marginal lotteries in source2. This observation leads to the following axiom and corollary.
Axiom Forward Correlation Consistency : Suppose P ≻ Q with P i = Q i , i = 1 , R ∼ S and supp ( P ) ∩ supp ( R ) = supp ( Q ) ∩ supp ( S ) = ∅ , then for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S. Corollary 1.
Let % be a binary relation on P . The following statements are equivalent:i). The relation % satisfies Weak Order, Monotonicity, Weak Continuity, WeakIndependence and Forward Correlation Consistency;ii). The relation % admits one of the following representations: EU, FIB, EU-CN, GBIB-CN and GFIB-CN.Moreover, in all representations H , H are unique, v , v are unique up to a positive affinetransformation and in EU, EU-CN, FIB, w is unique up to a positive affine transformation. In this section, we briefly discuss the proof sketch of the two representation theorems inSection 4. We will focus on sufficiency of the axioms. It is worthwhile to note that Theorem 1serves as an intermediate result in our proof of Theorem 2. As a result, although Theorem 1seems like a corollary of Theorem 2 , it needs to be proved first.For Theorem 1, by Axiom Correlation Neglect, it suffices to consider the preference overproduct lotteries ˆ P . For any q ∈ L ( X ) and q ′ ∈ L ( X ), we denote the restriction of % on L ( X ) × { q ′ } as % | q ′ and the restriction of % on { q } × L ( X ) as % | q . We first show that % i | q admits an EU representation for each i ∈ { , } and q ∈ L ( X − i ).We define that the independence property holds for tuple ( P, Q, R, S ) ∈ ˆ P with P i = R i , Q j = S j for some i, j ∈ { , } and P % R, Q % S if one of the following conditions hold:• P ≻ Q, R ∼ S and for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S ;• P ∼ Q, R ≻ S and for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S ;• P ∼ Q, R ∼ S and for all α ∈ (0 , αP + (1 − α ) R ∼ αQ + (1 − α ) S ;• P ≻ Q, R ≻ S and for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S .We argue that it suffices to consider the case with P ∼ R and Q ∼ S . Along with AxiomWeak Continuity, Axiom Weak Multilinear Independence states that the independenceproperty holds for any such tuple ( P, Q, R, S ) with P − i ∼ − i R − i and Q − j ∼ − j S − j . The restof the proof proceeds as we discuss to what extent this “local” property can be generalizedin different cases.If the DM narrowly brackets risks in both sources, then easy it is to show that thepreference % admits an NB representation; 24f the DM only narrowly brackets risks in one source, say, in source 2, then it is sufficientto focus on the subset of lotteries L ( X ) × X where the marginal lottery in source 2 isdegenerate. First, we notice that by assuming broad bracketing in source 1, the independenceproperty holds on a nontrivial set of product lotteries. Then we show that if the independenceproperty holds on two sets of product lotteries respectively, then it also holds on their union.Finally, we apply the standard open cover arguments to extend the independence propertyand show that it would lead to a GBIB-CN representation.If the DM does not narrowly bracket risks in either source, then by a similar but morecomplex proof, we can show that % must admit an EU-CN representation. Actually, theinteresting part is to exclude intermediate cases between EU-CN and GBIB-CN/GFIB-CN.The proof of Theorem 2 can be decomposed into four steps. First, we restrict thepreference % to product lotteries ˆ P and derive the corresponding partial representationson ˆ P by Theorem 1. If further Axiom Correlation Neglect holds, then we are done. Fromnow on, suppose that this axiom fails. Second, we show that Axiom Correlation Sensitivitycan be strengthened to a natural relaxation of Axiom Independence, where the marginalsof lotteries that are mixed have disjoint supports in source 1. Third, by embedding the setof lotteries as a subspace of temporal lotteries in Kreps and Porteus (1978), we can extend % to the set of temporal lotteries while satisfying the axioms in Kreps and Porteus (1978).Hence, % admits a KP-style representation on P . Finally, by making use of consistency ofthe two representations in the previous steps on product lotteries ˆ P , we conclude that onlyrepresentations stated in Theorem 2 are feasible. In most economic applications like portfolio choices and labor supply decisions, the outcomein both sources is money or the numeraire. Also, the payoffs in experiments typically takesthe form of tokens, which can be exchanged to money at a fixed rate. We assume that X = X = R in this section.When there is no risk and the DM receives money form both sources simultaneously, weargue that she will evaluate each outcome profile by adding up the monetary prizes. Consider25 trivial example where a worker can choose between two payment schemes after finishingtwo identical tasks. In scheme 1, she will receive $200 from the first task and $220 from thesecond one. In scheme 2, she will get $210 from the first task and $200 from the second one.All payments are made at the same time by cash and tasks have already been done. Thenarguably the worker should choose scheme 1, from which she can get $10 more. The idea issummarized in the following axiom. Axiom Broad Bracketing without Risk : For each x, y ∈ R , ( x, y ) ∼ ( x + y, ∼ (0 , x + y ).For any of the previous representations, this additional axiom just requires that the utilityover degenerate lotteries w : R → R can be replaced by u : R → R such that w ( x, y ) = u ( x + y ) for any x, y ≥
0. Suppose we further assume symmetry of the narrow preferences,that is,
Axiom Symmetry : For each p, q ∈ L ( R ), (0 , p ) % (0 , q ) if and only if ( p, % ( q, V NB ( P ) = u ( CE v ( P ) + CE v ( P )) , ∀ P ∈ P . where u and v are regular functions. Notice that u is strictly monotone, the preference witha NB representation is also represented byˆ V NB ( P ) = CE v ( P ) + CE v ( P ) , ∀ P ∈ P . In other words, a narrow bracketer evaluates a monetary lottery by the summing up thecertainty equivalents of its marginal lotteries.We contrast our representation with the commonly used functional form in the literatureof narrow bracketing, which focuses on product lotteries. We adapt the utility function inBarberis, Huang, and Thaler (2006) and Rabin and Weizsäcker (2009) to our framework asfollows: U ( p, q ) = λ X x,y u ( x + y ) p ( x ) q ( y ) + (1 − λ )[ X x u ( x ) p ( x ) + X y u ( y ) q ( y )] , ∀ ( p, q ) ∈ ˆ P , where λ ∈ [0 ,
1] and 1 − λ determines the degree of narrow bracketing. At the end of Section3.2, we already argued why such λ -mixture models are excluded in our framework. Here we26ocus on the case with λ = 0, that is, the DM admits (fully) narrow bracketing. Insteadof summing up the certainty equivalents of marginal lotteries like in our NB, the abovefunctional form sums up the expected utility of marginal lotteries. This implies that theDM is subject to narrow bracketing even when there is no risk and hence she might prefer( x, y ) over ( x ′ , y ′ ) even if x + y < x ′ + y ′ .One should notice that Barberis, Huang, and Thaler (2006) and Rabin and Weizsäcker(2009) focus on the case where the choice problems in different sources are “independent”.That is, besides assuming risks in two sources are resolved independently, they also assumethat the availability of gambles in one decision problem does not depend on available gamblesin the other one. For instance, if ( x, y ) and ( x ′ , y ′ ) are available options, then ( x, y ′ ) and( x ′ , y ) should also be available. In this restricted choice setup, their narrow bracketingrepresentation will predict the same behavior as our NB model. However, if we consider themore general choice domain, the two models differ and theirs will predict narrow bracketingeven without risk. Now we show how our model can accommodate the experimental evidence of narrowbracketing. Consider the following classic experiment introduced by Tversky and Kahneman(1981) and developed by Rabin and Weizsäcker (2009).
Example 2.
Suppose you face the following pair of concurrent decisions. All lotteries areindependent. First examine both decisions, then indicate your choices. Both choices will bepayoff relevant, i.e., the gains and losses will be added to your overall payment.Decision (1): Choose between:A. A sure gain of $2.40.B. A 25 percent chance to gain $10.00, and a 75 percent chance to gain $0.00.Decision (2): Choose between: If we do not assume Axiom Broad Bracketing without Risk, then U ( p, q ) = P x u ( x ) p ( x ) + P y u ( y ) q ( y )is actually a special case of the intersection of NB and EU. This is the expected discounted utility modelin in time preference with discount factor 1 (see Section 6.2.1). However, as is argued above, Axiom BroadBracketing without Risk is more reasonable in the setting with simultaneous monetary prizes. This againreflects our point that if we want to compare models of choice bracketing under risk, we need to maintainthe same preference over degenerate lotteries. . A sure loss of $7.50.D. A 75 percent chance to lose $10.00, and a 25 percent chance to lose $0.00. Since gains and losses from the two decision problems are aggregated, the DM shouldfocus only on the distribution of overall monetary prizes. For example, the combinationof B and C produces a lottery with a 1/4 chance of gaining $2.50 and a 3/4 chance oflosing $7.50. By comparison, the lottery induced by the combination of A and D is a 1/4chance of gaining $2.40 and a 3/4 chance of losing $7.60. The BC combination is equal tothe AD combination plus a sure gain of $0.10 and hence the AD combination is first-orderstochastically dominated. However, across different treatments in Tversky and Kahneman(1981) and Rabin and Weizsäcker (2009), a reasonably large fraction (above 28%) of subjectschose A in decision (1) and D in decision (2). Notice that BC dominates AD by addinga sure gain and they are reasonably similar in terms of complexity, hence previous modelsthat are monotone or incorporate complexity aversion cannot explain the common choicesof dominated options without choice bracketing.Suppose the DM is narrowly bracketing with the following representation over productlotteries ˆ P : ˆ V NB ( p, q ) = CE v ( p ) + CE v ( q ) . where the utility index u satisfies v ( x ) = √ x, if x ≥ , − √− x, if x < . (1)This is a standard reference-dependent model with the reference point fixed at 0. Easyto show that this model can accommodate the choice of A and D over B and C as CE v ( p A ) = 2 . > .
625 = CE v ( p B ) , CE v ( p D ) = − . > − . CE v ( p C ) . In this section, we interpret the risk in source 1 as the background or endowment risk,and the risk in source 2 as the risky decision at hand. Rabin (2000) formally identifies atension between expected utility and risk aversion regarding small gambles when choicesonly depend on the (distribution of) final wealth. His calibration theorem shows that28 low level of risk aversion with respect to small gambles leads to an absurdly highlevel of risk aversion with respect to large gambles. Safra and Segal (2008) then extendRabin’s calibration results to non-expected utility models satisfying certain differentiabilityconditions. Mu, Pomatto, Strack, and Tamuz (2020) further suggest that “theories that donot account for narrow framing–whereby independent sources of risk are evaluated separatelyby the decision maker–cannot explain commonly observed choices among risky alternatives.”One prominent thought experiment of Rabin’s critique is as follows: if an EU maximizerturns down 50-50 gambles of losing $1000 or gaining $1050 for all initial wealth levels, thenshe would always turn down 50-50 gambles of losing $20,000 or gaining any sum.To avoid the unrealistic behavior by an EU maximizer, we consider a DM who hasdifficulty integrating risks in difference sources and will use narrow bracketing as a simplifyingheuristic. Intuitively, outcomes in the gamble at hand are more “accessible” than backgroundwealth levels (Kahneman, 2011) and the DM might not take into account the backgroundrisk when deciding whether to accept the gamble at hand. For example, suppose that theDM’s NB utility function is given by Equation (1). One can easily show that the DM willalways reject 50-50 of losing $1000 or gaining $1050, and accept 50-50 gamble losing $20,000and gaining $80,050. From now on, we interpret the outcomes in two sources as consumptions or monetary prizesin two different periods. Concretely, source 1 is labeled as period 1 or present, and source 2is labeled as period 2 or future. Sometimes we name an outcome profile as a consumptionprofile or consumption path.The rest of this section consists of two parts. The first part connects our representationswith different seemingly distinct time preference models in the literature and providesa unified framework for them. It is worthwhile to emphasize that our characterizationtheorem originates from simplifying heuristics for multi-source risks, without any ex-antenormative assumptions for intertemporal choices. In the second part, we study implications An alternative way is to assume the DM does not fully ignore the background risk. Instead, she mightonly consider the certainty equivalent of the background risk and admits a FIB-CN representation. However,in order to accommodate the Rabin’s critique, we need to extend the current model to allow for non-expectedutility representations with first-order risk aversion over marginal lotteries like in Gul (1991).
29f some commonly studied axioms in time preferences and propose a new model that canaccommodate many desirable properties, including separation of time and risk preferences,indifference to temporal resolution of uncertainty and dynamic consistency.
In this section, we show that EU, BIB and NB have nice counterparts in time preferencesthat have been well studied in the literature.First, most commonly used time preferences are special cases of EU. The most prominentexample is the Expected Discounted Utility (EDU) model: V EDU ( P ) = E P [ u ( x )] + β E P [ u ( y )] . where u is the EU index in each period and β ∈ [0 ,
1] is the discount factor. The DM evaluateseach lottery by the summation of expected utility of each marginal lottery weighted by thediscount factor. One can easily show that this functional form is also a special case of NB.One natural extension of EDU is the Kihlstrom-Mirman (KM) model (Kihlstrom and Mirman,1974, Dillenberger, Gottlieb, and Ortoleva, 2020) given by V KM ( P ) = E P h φ (cid:16) P t D ( t ) X t =1 D ( t ) u ( x t ) (cid:17)i . where the DM first evaluates each consumption path ( x , x ) using discounted utility andthen takes expected value of the utility profiles by applying additional curvature φ .Second, Selden (1978) and Selden and Stux (1978) introduce a alternative time preferencemodel to EU called Dynamic Ordinal Certainty Equivalent (DOCE), where the DM first takesthe certainty equivalents of marginal lotteries in each period and then evaluate the profileof certainty equivalents. This exactly agrees the time preference interpretation of our NBmodel. V DOCE ( P ) = V NB ( P ) = w ( CE v ( P ) , CE v ( P )) . One nice feature of DOCE/NB is that it can accommodate the tension between stochastic im-patience and risk aversion over time lotteries introduced by DeJarnette, Dillenberger, Gottlieb, and Ortoleva(2020), which is violated by most existing models of time preferences including Epstein-Zin preferences (Epstein and Zin, 1989) and risk-sensitive preferences (Hansen and Sargent,30995) as is shown in Dillenberger, Gottlieb, and Ortoleva (2020).Finally, we connect BIB with the the preferences in Kreps and Porteus (1978) and showthat they have similar functional forms involving backward induction and offer similarpredictions, despite their distinct behavioral motivations and characterizations.Kreps and Porteus (1978) extend lotteries over consumption paths to temporal lotteries in order to model temporal resolution of uncertainty. In the two-period setup, the set oftemporal lotteries is D ∗ := L ( X × L ( X )). To see why the set of temporal lotteries D ∗ isstrictly larger than the set of lotteries P , take the lottery P = 1 / δ , δ ) + 1 / δ , δ ) ∈ P for an instance. There are two temporal lotteries that correspond to P : d = 1 / δ (1 ,δ ) +1 / δ (1 ,δ ) and d = δ (1 , / δ +1 / δ ) . Notice that the marginal lottery of P in the first periodis deterministic and P only involves risk in the second period. It remains unspecified aboutthe timing of resolution of such risk. By comparison, temporal lotteries d and d containexactly the same uncertainty in the second period as P , which resolves in the first periodfor temporal lottery d and in the second period for d . In other words, there are twodated types of mixtures for deterministic consumption paths ( δ , δ ) and ( δ , δ ) in temporallotteries, but only one type of mixture in lotteries. Generically, the DM can have strictranking between temporal lotteries d and d , which reflects her preference for early or lateresolution of uncertainty. We define that a lottery P ∈ P is induced by the temporal lottery d ∈ D ∗ if for any ( x, y ) ∈ X , P ( x, y ) = X ( x,q ) d ( x, q ) q ( y )Following the above example of P and d , d , we can show that every lottery is induced bysome temporal lottery and there exist lotteries induced by more than one temporal lotteries.Thus the domain of temporal lotteries D ∗ is strictly richer than the set of lotteries P .By applying the vNM axioms to temporal lotteries with mixture in period 1 and totemporal lotteries whose uncertainty only resolves in period 2 with mixture in period2, Kreps and Porteus (1978) axiomatize a general class of Kreps-Porteus preferences(henceforth KP). To get proper comparison with BIB, we focus on the history-independentKP, whose representation V KP : D ∗ → R is characterized by a tuple of regular functions( w, v ) such that w : X → R , v : X → R , and V KP ( d ) = X ( x,p ) w ( x, CE v ( p )) d ( x, p ) .
31y comparison, the BIB representation with the same tuple ( w, v ) is given by V BIB ( P ) = X x w ( x, CE v ( P | x )) P ( x ) . The above two representations share the same backward inductive procedure to evaluatemulti-period risk. The DM first reduces the risk (resolving) in period 2 into its certaintyequivalent under some history-independent expected utility index v . This transforms theoriginal (temporal) lottery into one with only uncertainty (resolving) in period 1. Thenthe DM evaluates the new (temporal) lottery based on its expected utility under someindex w . This recursive structure allows for the adoption of dynamic programming methodsin optimization problems, and partially explains the popularity of the Kreps and Porteus(1978) framework in the past decades. Most recursive models, including the famous Epstein-Zin preferences (Epstein and Zin, 1989) (henceforth EZ) and risk-sensitive preferences(Hansen and Sargent, 1995) (henceforth HS) , are generalizations of Kreps and Porteus(1978) to temporal lotteries in an infinite horizon setting.It is worthwhile to mention three differences between BIB and history-independent KPpreferences. First, the history-independent KP representations are defined on a strictlyricher domain than BIB, which allows for resolution of uncertainty in different periods. V KP reduces to V BIB on the subdomain of temporal lotteries where uncertainty about outcomesin period t resolves in period t for each t . Second, as a direct implication of different domains,BIB exhibits indifference to temporal resolution of risk, while the history-independent KPpreference satisfies this property if and only if it reduces to an expected utility representation.This distinction is essential in our discussion about asset market puzzles in Section 6.3.Finally, the history-independent KP preferences satisfy the vNM independence axiom overtemporal lotteries, while BIB violates it on the domain of lotteries. Actually, BIB satisfiesAxiom Independence if and only if it also admits an EU representation .This suggests that shared recursive procedure in the two models is based on differentrationales. In the history-independent KP preferences, it comes from non-indifference to Hansen and Sargent (1995) originally formulate the risk-sensitive preference as an optimal controlproblem with risk-adjusted costs. Bommier, Kochov, and Le Grand (2017) show how it can be interpretedas a monotone recursive preference over temporal lotteries. Another difference is that the history-independent KP preferences can satisfy topological continuityon its domain, which is not true for the BIB preferences. However, we can show that BIB satisfies AxiomIndependence if and only if it satisfies topological continuity. In other words, backward induction bracketingcan be interpreted as a joint relaxation of the independence and continuity properties.
In this section, we will discuss some desirable normative properties of time preferences inthe literature and then propose a new class of models based on BIB that can simultaneouslysatisfy all these properties except for one. For simplicity, assume that X = X = R + .We start with the separation of time and risk preferences. It is well-known that in any EDUmodel, the inverse of the elasticity of intertemporal substitution (EIS) coincides with thecoefficient of relative risk aversion (RRA). That is, the time preference and the risk preferenceand intertwined together. However, enormous empirical evidence in macroeconomics, financeand behavioral economics has shown the necessity to separate the two coefficients . Actually,one motivation of EZ and DOCE is to achieve such separation of time and risk preferences.Suppose that v and v measure the risk preferences with in each period and w measuresthe time preference over deterministic consumption paths. By separation of time and riskpreferences, we mean that for each w , we can represent arbitrary risk preferences in each Indirect evidence includes the failure to explain the equity premium puzzle with EDU(Mehra and Prescott, 1985). See more discussion in Section 6.3. For direct evidence,Barsky, Juster, Kimball, and Shapiro (1997) find that RRA and EIS are uncorrelated through a cross sectionof American households and Andreoni and Sprenger (2012) show similar results in a lab experiment. v and v , and vice versa. The second property is indifference to temporal resolution of uncertainty, which isimplicitly assumed by our choice domain of lotteries. Although EZ can have separateparameters for time and risk preferences, the separation depends on specific preferencesfor temporal resolution of uncertainty. For instance, if the DM is indifferent to when theuncertainty is resolved, then the two parameters must be the same and EZ agrees with EDU.Hence, despite the fact that many people might prefer early or late resolution of uncertainty,it is still worthwhile to separate such preferences with other desirable properties in timepreferences by assuming indifference to temporal resolution of uncertainty.The third property is stationarity. The following two axioms illustrate the idea ofKoopmans (1960) that “the passage of time does not have an effect on preferences”.
Axiom History Independence : For any x, y ≥ p, q ∈ L ( R + ), ( x, p ) % ( x, q ) ifand only if ( y, p ) % ( y, q ). Axiom Stationarity : For any p, q ∈ L ( R + ), p % q if and only if p % q .Axiom History Independence states that, from the ex-ante perspective, the choice inperiod 2 in independent of the history of consumption in period 1. Axiom Stationarity isan adaption of Axiom 5 in Bommier, Kochov, and Le Grand (2017), which is defined ontemporal lotteries with infinite horizons, to our framework. It reflects the time-invariance ofthe DM’s risk preference. Following Koopmans (1960) and Bommier, Kochov, and Le Grand(2017), when there is no confusion, we use the term “stationarity” to represent theconjunction of both axioms.Notice that our analysis takes an ex-ante approach and we haven’t specified thepreference of the DM given that period 2 has arrived. In the infinite horizon setting likeBommier, Kochov, and Le Grand (2017) and other literature on recursive preferences, thepreference from the perspective of period 2 is implicitly assumed to be the same as thepreference in period 1, since the truncated (temporal) lottery from period 2 onwards is stilla (temporal) lottery and the same binary relation can be applied without any modification. Our notion of separation is weaker than the SEP in Definition 4 of Kubler, Selden, and Wei (2020),which further requires that if we just vary the parameters in time preference w , then v and v shouldremain unchanged. Hence, our KM-BIB to be defined below violates SEP and hence is not a counterexampleto the impossibility theorem in Kubler, Selden, and Wei (2020). % as the ”current” time in period 2 is period 2. This approach cannaturally extends each of our representation to a corresponding dynamic model.The fourth property is dynamic consistency, which requires that choices that are optimalin period t , on the basis of the DM’s preferences in period t , remain optimal when evaluatedfrom the perspective of an earlier period t ′ . It connects ex-ante and ex-post choices andpermits the use of backward induction and dynamic programming methods. Suppose thatstationarity holds, then it is straightforward to show that the above dynamic version of BIBsatisfies dynamic consistency. For simplicity, we will just say that BIB satisfies dynamicconsistency in this case.The fifth property is called discounted utility without risk, which states that the preferenceover deterministic consumption paths agrees with EDU and can be represented by thesummation of discounted utility in each period. Assumption 1 – Discounted Utility without Risk : There exists a regular function u : R + → R and β ∈ [0 ,
1] such that for all x , x , y , y ≥
0, ( x , x ) % ( y , y ) if and only if u ( x ) + βu ( x ) ≥ u ( y ) + βu ( y ).Dillenberger, Gottlieb, and Ortoleva (2020) show that KM is exactly the class of EUthat admits a discounted utility representation when there is no risk. Similarly, a BIBrepresentation V BIB satisfies Discounted Utility without Risk if and only if V BIB ( P ) = X x φ (cid:16) u ( x ) + βu ( CE v ( P | x )) (cid:17) P ( x )where φ, u, v : R + → R are regular and β ∈ [0 ,
1] is the discount factor. If we further assumestationarity, then we derive the following KM-BIB representation: V KM − BIB ( P ) = X x φ (cid:16) u ( x ) + βu ( CE φ ◦ u ( P | x )) (cid:17) P ( x ) . (2)where φ, v : R + → R are regular and β ∈ [0 , u . Therisk preference in each period is represented by EU index φ ◦ u . This means that φ is theadditional curvature used only in the case of risk preference and determines the separationbetween risk aversion and intertermporal substitution. We summarize those insights in thefollowing claim. Claim 1.
If the preference % admits a KM-BIB representation in (2), then it satisfiesdiscounted utility without risk, separation of time and risk preferences, indifference totemporal resolution of uncertainty, dynamic consistency and stationarity. By comparison, it is commonly known that KP either violates indifference to temporalresolution of uncertainty or separation of time and risk preferences, while EU (extended to thedynamic version as above) and DOCE violate dynamic consistency. Kubler, Selden, and Wei(2020) adopts an alternative approach by showing that DOCE can satisfy dynamicconsistency in a restricted domain of consumption trees.Now we introduce two notions of “risk aversion” across different periods for KM-BIB asan application. The first is correlation aversion introduced by Bommier (2007).
Axiom – Correlation Aversion : For any x > x and y > y ,12 ( δ x , δ y ) + 12 ( δ x , δ y ) %
12 ( δ x , δ y ) + 12 ( δ x , δ y )Notice the two lotteries agree on the marginal lotteries in both periods and onlydiffer in the correlation structure. By monotonicity, the DM’s most preferred outcomepath is ( δ x , δ y ) and her least preferred outcome path is ( δ x , δ y ). Axiom CorrelationAversion requires that she prefers the mixture between intermediate outcome paths tothe mixture between extreme outcome paths. We can show that a stationary KM-BIBsatisfies Axiom Correlation Aversion if and only if the additional curvature φ is concave.Dillenberger, Gottlieb, and Ortoleva (2020) introduce a similar notion called residual riskaversion, which is also captured by the concavity of φ . This suggests that correlation aversioncoincides with residual risk aversion in our framework.Another relevant notion is called long-run risk aversion. Consider the following twoconsumption plans. In the first one, a coin is flipped independently in each period, andthe payoff is $1 if it lands on heads and $0 if it lands on tails. This scenario is referred to asshort-run risk. In the second plan, a coin is flipped only once at the beginning of period 1,36nd the payoff is either $1 or $0 in both periods. This scenario is referred to as long-run risk.EDU exhibits indifference between long-run risk and short-run risk, while the sensitivity tolong-run risk has been used to explain many financial puzzles (Bansal and Yaron, 2004). Weadapt (and simplify) the notion of long-run risk aversion in Strzalecki (2013) as follows. Axiom – Long-run Risk Aversion : For any x > x ,( 12 δ x + 12 δ x , δ x + 12 δ x ) %
12 ( δ x , δ x ) + 12 ( δ x , δ x )Easy to see that Axiom Long-run Risk Aversion is implied by Axiom Correlation Aversionfor EU models, which include KM. However, for KM-BIB, the two notions differ and long-runrisk attitude might depend on higher-order curvature of φ .We end this section with a property that is generically violated by KM-BIB and relates itto the experimental evidence on choice bracketing in Rabin and Weizsäcker (2009). Considerthe following modification of the ordinal dominance property in Chew and Epstein (1990)and the monotonicity condition in Bommier, Kochov, and Le Grand (2017). Axiom – Ordinal Dominance : For each n >
0, ( x i , x i ) , ( y i , y i ) ∈ R for i = 1 , ..., n and ( π , ..., π n ) ∈ [0 , n with P i π i = 1, if either x i = x j , y i = y j for all i, j = 1 , ..., n , or x i = x j , y i = y j for all i = j , then( δ x i , δ x i ) % ( δ y i , δ y i ) , ∀ i = 1 , ..., n = ⇒ n X i =1 π i ( δ x i , δ x i ) % n X i =1 π i ( δ y i , δ y i ) . Intuitively, ordinal dominance requires that the DM would never choose an action ifanother available action is preferable in every state of the world. Notice that the originalordinal dominance property is defined over temporal lotteries and it holds for mixture oftemporal lotteries in any period t so long as they have the same deterministic history ofconsumptions before period t . In order to maintain similar interpretations in the space oflotteries, we require that ordinal dominance holds either when there is only uncertainty inthe first period (i.e., the case where x i = x j , y i = y j for all i = j ) or in the second period(i.e., the case where x i = x j , y i = y j for all i, j ).Bommier, Kochov, and Le Grand (2017) show that ordinal dominance is a tight restric-tion for recursive preferences. Specifically, the recursive KP preference satisfies ordinaldominance if and only if it is either a risk-sensitive (HS) preference, where the risk attitude37xhibits constant absolute risk aversion, or belongs to the class of Uzawa (1968), whichis a special case of expected utility with infinite horizon. Similar results also hold in ourframework. Axiom Ordinal Dominance is only generically satisfied by EU and it can besatisfied by other representations only if the DM has constant absolute risk aversion.Suppose that instead of intertemporal choices, we interpret marginal lotteries in twosources as simultaneous monetary gambles and assume that the DM broadly bracketsdegenerate marginal lotteries. Then Axiom Ordinal Dominance exactly reduces to “first-order stochastic dominance” considered in Rabin and Weizsäcker (2009). This impliesa deep connection between the empirical evidence of narrow bracketing in experiments(Rabin and Weizsäcker, 2009, Ellis and Freeman, 2020) and the theoretical difficulty tosatisfy ordinal dominance in recursive preferences (Bommier, Kochov, and Le Grand, 2017). Since Mehra and Prescott (1985) introduced the equity premium puzzle, many puzzling factsof asset markets have been observed and challenged the validity of various models, includingthe standard expected discounted utility (EDU) model . It has been well understood thatthose puzzles are quantitative and explanations with extreme parameter values are usuallyregarded as inadequate.One popular approach to address asset pricing puzzles is to use recursive preferencesthat permit the separation of time and risk preferences (Epstein and Zin, 1991). Forinstance, the long-run risks model of Bansal and Yaron (2004) has provided a unifiedrationalization of several puzzling facts in asset markets by combining the EZ preferencewith an endowment process featuring a persistent predictable component for consumptiongrowth and its volatility. However, Epstein, Farhi, and Strzalecki (2014) point out thatthe quantitative assessment of the preference for early resolution of uncertainty has beenignored in the macro-finance literature. The authors show that the parameter values usedin Bansal and Yaron (2004) imply that the DM is willing to give up 25 or 30 percent of herlifetime consumption in order to have all risk resolved in period 1. Such time premium isarguably too high as the risk is about consumption instead of income or asset returns, andthere is no apparent instrumental value of information by early resolution of uncertainty. Actually, the inflexibility of EDU to explain the equity premium puzzle is one major motivation of theliterature on recursive preferences. See Epstein and Zin (1989) for a detailed discussion.
38t is worthwhile to notice that in most applications of EZ in macroeconomics and finance,the temporal lotteries are not explicitly studied and the temporal resolution of uncertaintyis not directly involved. Instead, there is an implicit assumption that the uncertainty ofconsumption in period t is only resolved in period t (Epstein, Farhi, and Strzalecki, 2014).Thus, the space of temporal lotteries, on which EZ and many other recursive preferencesare defined and characterized, might be unnecessarily complex and our framework involvingonly lotteries over consumption paths might better suit the needs of those applicationsThis provides another justification for our framework. Recall that BIB shares the samepredictions as history-independent KP, whose natural extension to infinite horizon includesthe class of EZ used in Bansal and Yaron (2004). By incorporating the behavioral factor ofchoice bracketing , we provide a foundation for the use of EZ-style preferences in the spaceof lotteries, without the need to worry about temporal resolution of uncertainty and hightime premium. There does, of course, exist evidence for nonindifference to how uncertaintyresolves over time, and our analysis just suggests that it is not a necessary side effect toaddress asset pricing puzzles. Hence our approach shows that the long-run risks model inBansal and Yaron (2004) can be immune to the critiques of Epstein, Farhi, and Strzalecki(2014) by adjusting the domain of choices and introducing choice bracketing. We end upthis section with an analogue to the CRRA-CES EZ model used in Bansal and Yaron (2004)as the special case of KM-BIB in two periods: U KM − BIB ( P ) = X c h (1 − β ) c ρ + β [ E P | x ( c α )] ρ/α i α/ρ P ( c )which is equivalent to assuming u ( x ) = x ρ , φ ( x ) = x α/ρ in the KM-BIB representation. Inthis case, the time preference parameter EIS is 1 / (1 − ρ ) and the risk preference parameterRRA is 1 − α . All the analysis in Bansal and Yaron (2004) follows once this representationis naturally extended the a dynamic model with infinite horizons. We are not the first to include narrow bracketing to explain financial puzzles. Benartzi and Thaler(1995) provide an explanation of the equity premium puzzle by combining loss aversion and narrowbracketing. They argue that investors dislike stocks because they look at their portfolios frequently andevaluate the nominal changes in their accounts with loss aversion, even though they might save for a distantfuture. However, it seems hard to get a unified explanation for other puzzles with this approach. Conclusion
This paper generalizes the expected utility model for preference over lotteries on multi-source outcome profiles to incorporate two simplifying heuristics commonly used in theaggregation of risks: choice bracketing and correlation neglect. We provide characterizationresults for the generalized models by relaxing the vNM independence axiom. We thenapply our framework and representations to different setups by varying the interpretationsof different sources of outcomes. For example, with the interpretation of simultaneousmonetary gambles, our model can explain experimental findings on narrow bracketingin Rabin and Weizsäcker (2009). With the interpretation of background risk, our modelprovides one way to accommodate risk aversion over small favorable gambles in Rabin (2000).With the interpretation of intertemporal choices, we provide a unified framework to studyseveral seemingly distinct models of time preferences in the literature and introduce a newclass of models that can satisfy many desirable normative properties on time preferences.One main point of the paper is that narrow bracketing and correlation neglect can bemodelled as natural distortions of the independence axiom and should not be viewed asmore “irrational” or more behavioral or more exotic than other commonly accepted non-EUtheories in the literature. Hence, we think it might be worthwhile to incorporate these twoheuristics in various economic applications.In a follow-up work, we extend the current the two-source framework to multiple sourcesand infinite horizons and try to axiomatize a recursive version of our KM-BIB model.This will facilitate the comparison between our framework and other recursive modelsbased on Kreps and Porteus (1978) and provide an exact foundation to apply our modelin macroeconomics and finance. In another ongoing work, we consider general models ofcorrelation misperception by modeling the correlation structure among risks in differencesources using copula theory. Moreover, as is mentioned in footnote 7, one can extend ourframework to consider non-EU models in each single source and incorporate factors like firstorder risk aversion and Allais Paradox. 40 eferences
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For simplicity, we use abbreviations for each axiom. We have Axiom Weak Order (WO),Monotonicity (M), Weak Continuity (WC), Weak Independence (WI), Correlation Neglect(CN) and Correlation Sensitivity (CS). Also, We will denote the first part of Axiom WI asAxiom CI and the second part as Axiom WMI. For any q ∈ L ( X ) and q ′ ∈ L ( X ), wedenote the restriction of % on L ( X ) × { q ′ } as % | q ′ and the restriction of % on { q } × L ( X )as % | q . % | q ′ is called the conditional preference in source 1 given lottery q ′ in source 2 and % | q is called the conditional preference in source 2 given lottery q in source 1.If c = + ∞ , then we denote that ( c , q ) ≻ ( p, q ) for all ( p, q ) ∈ ˆ P . If c = −∞ , then wedenote that ( c , q ) ≺ ( p, q ) for all ( p, q ) ∈ ˆ P . Similar notions can be defined for c = + ∞ and c = −∞ . Proof of Theorem 1. ii ) ⇒ i ). We first prove the necessity of these axioms. Axioms WO and CN trivially hold.With Axiom CN, Axiom WC is equivalent to continuity of % on the subdomain of productlotteries ˆ P , which is implied by the continuity and boundedness of w , v and v .For p, q ∈ L ( R ), we denote p % F OSD q if for any x ∈ R , P y ≤ x p ( y ) ≤ P y ≤ x q ( y ) and p ≻ F OSD q if p % F OSD q and p = q . For Axiom M, if P dominates ( δ x , δ x ), then P i % F OSD δ x i for i = 1 , w , v and v guarantees that P ≻ ( δ x , δ x ). By a similar argument, ( δ x , δ x ) ≻ P if ( δ x , δ x ) dominates P . Therefore Axiom M is satisfied.Now we check Axiom WI. First, if % admits an EU-CN representation, then by Lemma 1, % satisfies Axiom Multilinear Independence, and hence Axiom WI. Second, suppose that % admits a GBIB-CN representation ( w, v , v , H ), that is, V GBIB − CN ( P ) = w ( CE v ( P ) , CE v ( P )) , if CE v ( P ) ∈ X \ H P w ( x, CE v ( P )) P ( x ) , if CE v ( P ) ∈ H For each p ∈ L ( X ), % | p is represented by an EU with index v . For each p ′ ∈ L ( X ),when CE v ( p ′ ) ∈ X \ H , then % | p ′ is represented by an EU with index v . Moreover, as0 H , % is admits an EU representation with index v . When CE v ( p ′ ) ∈ H , then % | p ′ is represented by an EU with index w ( · , CE v ( p ′ )). Hence, Axiom CI is satisfied.45hen we check Axiom WMI. Fix P, Q, R, S ∈ ˆ P , α ∈ (0 ,
1) and i, j ∈ { , } with P i = R i , Q j = S j , P − i ∼ − i R − i , Q − j ∼ − j S − j , P ≻ Q and R ∼ S . First we claim that AxiomWMI holds if P ∼ R or Q ∼ S . Suppose that P ∼ R , then by Axiom CI, for all α ∈ (0 , αP + (1 − α ) R ∼ P ∼ R and either Q % αQ + (1 − α ) S or S % αQ + (1 − α ) S . If Q % S , then αP + (1 − α ) R ∼ P ≻ Q % αQ + (1 − α ) S . If instead S ≻ Q , then αP + (1 − α ) R ∼ R ∼ S ≻ αQ + (1 − α ) S . This proves Axiom WMI. Similar argumentshold for Q ∼ S .Now we consider the following three cases.• Case 1: Suppose that i = 1. Then P = R and P ∼ R , which implies CE v ( P ) = CE v ( R ). We know P ∼ R and hence Axiom WMI holds.• Case 2: Suppose that j = 1. Then Q = S and Q ∼ S , which implies CE v ( Q ) = CE v ( S ). We know Q ∼ S and hence Axiom WMI holds.• Case 3: Suppose that i = j = 2.If CE v ( P ) ∈ X \ H or CE v ( Q ) ∈ X \ H , then either P ∼ R or Q ∼ S and weare done. If CE v ( P ) , CE v ( Q ) ∈ H , then the GBIB-CN representation is linear inmarginal lotteries in source 1. Then for any α ∈ (0 , V GBIB − CN ( αP + (1 − α ) R ) = αV GBIB − CN ( P ) + (1 − α ) V GBIB − CN ( R ) > V GBIB − CN ( Q ) + (1 − α ) V GBIB − CN ( S )= V GBIB − CN ( αQ + (1 − α ) S ) . This verifies Axiom WMI.Thus Axiom WI holds for GBIB-CN. A symmetric proof applies if % admits a FBIB-CNrepresentation. This completes the proof for necessity of axioms. i ) ⇒ ii ). Suppose that all axioms hold. For each x i ∈ X i and i = 1 ,
2, denote Π i ( x i ) as the setof marginal lotteries in source i with certainty equivalent x i . Formally, Π i ( x ) = { p ∈ L ( X i ) : p ∼ i δ x i } . For any two product lotteries P, Q ∈ ˆ P with P % Q , let [ Q, P ] denote the set of allproduct lotteries whose utilities lie between P and Q , that is, [ Q, P ] = { S ∈ ˆ P : P % S % Q } .46lso, for each q i ∈ L ( X i ) and x i ∈ X i , i = 1 ,
2, defineΓ( x , x ) = [ P,Q ∈ Π ( x ) × Π ( x ) ,P % Q [ Q, P ]Γ ,q ( x ) = [ P,Q ∈{ q }× Π ( x ) ,P % Q [ Q, P ] , Γ ,q ( x ) = [ P,Q ∈ Π ( x ) ×{ q } ,P % Q [ Q, P ] . Intuitively, Γ( x, y ) includes all product lotteries whose utilities are bounded by lotteries inΠ ( x ) × Π ( y ). Γ ,q ( y ) and Γ ,q ( x ) admit similar interpretations. We further defineΓ ,q = [ x ∈ X Γ ,q ( x ) , Γ ,q = [ x ∈ X Γ ,q ( x ) . For any set of lotteries
A ⊆ P , denote max % A = { P ∈ A : P % Q, ∀ Q ∈ A} whenever itis well-defined. That is, max % A is the set of most preferred lotteries in A under % .Finally, for any set A , we denote A o as its interior and ∂A as its boundary with respectto the appropriate topology. Step 1: Direct implications of axioms.
First, by Axiom CN, P ∼ ( P , P ) for all P ∈ P and it suffices to consider the restrictionof % on product lotteries ˆ P . Then Axiom WC implies that % satisfies topological continuity.The following lemma shows the EU representation of the conditional preference % i | q foreach i = 1 , q ∈ L ( X − i ). Lemma 2.
For each i = 1 , and q ∈ L ( X − i ) , the conditional preference % i | q admits anEU representation with a utility index v i | q , which is continuous, bounded and unique up toa positive affine transformation. Moreover, if q ∈ X − i , then v i | q can be chosen to be strictlymonotone (and hence regular).Proof of Lemma 2 . Fix i = 1 , q ∈ L ( X − i ). By Axiom WC, the conditional preference % i | q is continuous. By Axiom CI, % i | q admits an EU representation with a continuous utilityindex v i | q defined on X i , which is unique up to a positive affine transformation. Normalizethat v i | q (0) = 0. Suppose by contradiction that v i | q is unbounded, then for any positiveinteger n , there exists x n ∈ R such that | v i | q ( x n ) | > n . There exists a subsequence { x n k } k ≥ such that v i | q ( x n k ) > n k for each k or v i | q ( x n k ) < − n k for each k . Suppose, without loss of47enerality, that the former case holds. Consider the marginal lottery p n k = n k δ x nk + n k − n k δ for each k . By continuity of v i | q , we can find ǫ > v i | q ( ǫ ) ∈ (0 , k , the utilityof p n k is U i | q ( p n k ) = n k v i | q ( x n k ) > > v i | q ( ǫ ) = U i | q ( δ ǫ ), which means p n k % i | q δ ǫ . Meanwhile, p n k w −→ δ ≺ i | q δ ǫ as v i | q ( ǫ ) > v i | q (0) = 0. This contradicts with the continuity of % i | q . Asa result, v i | q is bounded. Moreover, if q ∈ X − i , that is, q = δ y for some y ∈ X − i , then byAxiom M, we know v i | q must be strictly monotone.When q = δ , then the conditional preference in source i agrees with the narrow preferencein source i and its EU index is denoted as v i for simplicity. It is worthwhile to note that foreach i = 1 ,
2, if x ∈ X i \ X oi , then Π i ( x ) = { δ x } .A direct corollary of Lemma 2 guarantees the existence of “certainty equivalents”. Corollary 2.
For each P ∈ ˆ P , there exists x , y ∈ X , x , y ∈ X such that P ∼ ( P , x ) ∼ ( x , P ) ∼ ( y , y ) . Proof of Corollary 2 .
Suppose that P X , P X . The case where P ∈ X or P ∈ X is easier to prove. By Lemma 2, we know there exists a, a ′ ∈ X such that v | P ( a ) > P x v | P ( x ) P ( x ) > v | P ( a ′ ). Since v | P is continuous and X is a closed interval, there exists x ∈ X where v | P ( x ) = P x v | P ( x ) P ( x ), which implies P ∼ ( P , x ). Similarly, we canfind x ∈ X with P ∼ ( x , P ). Now let y = x . Repeat the above arguments for productlottery ( P , x ) and we know there exists y ∈ X such that ( y , x ) ∼ ( P , x ) ∼ P .The next lemma summarizes two implications of Axiom WC and Axiom WI. Lemma 3. (i). For each
P, Q, R ∈ ˆ P with P % R % Q , P ≻ Q and P i = Q i for some i ∈ { , } , then there exists a unique λ ∈ [0 , such that R ∼ λP + (1 − λ ) Q .(ii). For each P, Q, R, S ∈ ˆ P , α ∈ (0 , and i, j ∈ { , } , if P i = R i , Q j = S j , P − i ∼ − i R − i and Q − j ∼ − j S − j , then P ∼ Q, R ∼ S = ⇒ αP + (1 − α ) R ∼ αQ + (1 − α ) SP ≻ Q, R ≻ S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S Proof of Lemma 3. (i). Denote I = { η ∈ [0 ,
1] : R ≻ ηP +(1 − η ) Q } and λ = sup I . λ is well-defined as I is bounded. We claim that λP +(1 − λ ) Q ∼ R . If λP +(1 − λ ) Q ≻ R , then λ > This proof technique will be used for multiple times below. For simplicity, we will call it “the subsequencearguments” and denote the subsequence as the original sequence, which is without loss of generality. % , there exists ǫ > λ − ǫ ) P + (1 − λ + ǫ ) Q ≻ R . Thisimplies λ − ǫ I . Since P ≻ Q and P i = Q i for some i , by Axiom CI, for any α, β ∈ [0 , αP + (1 − α ) Q ≻ βP + (1 − β ) Q if any only if α > β , which implies [ λ − ǫ, λ ] ∩ I = ∅ andleads to a contradiction with λ = sup I . If instead R ≻ λP + (1 − λ ) Q , then there exists ǫ > R ≻ ( λ + ǫ ) P + (1 − λ − ǫ ) Q and hence λ + ǫ ∈ I , which again contradicts withthe definition of λ .(ii). Consider the case where P ∼ Q, R ∼ S . If P ∼ R , then the result trivially holds as αP + (1 − α ) R ∼ R ∼ Q ∼ αQ + (1 − α ) S . Without loss of generality, suppose that P ≻ R .Then Q ≻ S and P ≻ αP + (1 − α ) R ≻ R , Q ≻ αQ + (1 − α ) S ≻ S ∼ R . Suppose bycontradiction that αP + (1 − α ) R ≻ αQ + (1 − α ) S . By part (i), there exists a unique λ ∈ (0 ,
1) with αQ + (1 − α ) S ∼ λ ( αP + (1 − α ) R ) + (1 − λ ) R = αλP + (1 − αλ ) R . Noticethat Q ∼ P ≻ αP + (1 − α ) R , S ∼ R , Q j = S j , Q − j ∼ − j S − j , ( αP + (1 − α ) R ) i = P i = R i and ( αP + (1 − α ) R ) − i ∼ − i R − i . The last one holds as % − i admits an EU representation.Hence, Axiom WMI implies that αQ + (1 − α ) S ≻ λ ( αP + (1 − α ) R ) + (1 − λ ) R = αλP + (1 − αλ ) Q which leads to a contradiction. The case for αP + (1 − α ) R ≺ αQ + (1 − α ) S is symmetric.Now assume P ≻ Q, R ≻ S . If P ∼ R , then the result holds as αP + (1 − α ) R ∼ P ≻ max % { Q, S } % αQ + (1 − α ) S for all α ∈ (0 , P ≻ R .If R % Q , then αP + (1 − α ) R ≻ R % max % { Q, S } % αQ + (1 − α ) S for all α ∈ (0 , Q ≻ R , then P ≻ Q ≻ R ≻ S . By part (i) of this lemma, we can find λ ∈ (0 ,
1) suchthat R ∼ λQ + (1 − λ ) S := S ′ . Then S ′ j = S j = Q j and S ′− j = λQ − j + (1 − λ ) S − j ∼ − j Q − j as Q − j ∼ − j S − j . Then the primitives of Axiom WMI hold for the tuple ( P, Q, R, S ′ ) and forany α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S ′ ≻ αQ + (1 − α ) S. The second strict ranking comes from Axiom CI and S ′ ∼ R ≻ S . This completes theproof.A tuple ( P, Q, R, S ) ∈ ˆ P is called proper if P i = R i , Q j = S j for some i, j ∈ { , } and P % R, Q % S . A proper tuple ( P, Q, R, S ) satisfies the independence property if one of thefollowing conditions holds: 49 P ≻ Q, R ∼ S and for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S ;• P ∼ Q, R ≻ S and for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S ;• P ∼ Q, R ∼ S and for all α ∈ (0 , αP + (1 − α ) R ∼ αQ + (1 − α ) S ;• P ≻ Q, R ≻ S and for all α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S .We end this section by showing that for each x ∈ X , y ∈ X , any product lottery inΓ( x, y ) is indifferent to some lottery in Π ( x ) × Π ( y ). Similar results also hold for Γ ,q ( y )and Γ ,q ( x ) for each q ∈ L ( X ) , q ∈ L ( X ). Lemma 4.
Fix q ∈ L ( X ) , q ∈ L ( X ) and x ∈ X , y ∈ X .(i). For each P ∈ Γ( x, y ) , there exists P ′ ∈ Π ( x ) × Π ( y ) with P ′ ∼ P ;(ii). For each P ∈ Γ ,q ( y ) , there exists P ′ ∈ { q } × Π ( y ) with P ′ ∼ P ;(iii). For each P ∈ Γ ,q ( x ) , there exists P ′ ∈ Π ( x ) × { q } with P ′ ∼ P .Proof of Lemma 4. (i). By definition, there exists Q, Q ′ ∈ Π ( x ) × Π ( y ) with Q % P % Q ′ .Denote Q ′′ = ( Q , Q ′ ) ∈ Π ( x ) × Π ( y ). We have either Q ′′ % P % Q ′ or Q % P % Q ′′ .As Q ′′ = Q , Q ′′ = Q ′ , by part (i) of Lemma 3, we know there exists λ ∈ (0 ,
1) with P ∼ λQ ′′ + (1 − λ ) Q ′ or P ∼ λQ ′′ + (1 − λ ) Q . Lemma 2 guarantees that λQ ′′ + (1 − λ ) Q ′ , λQ ′′ + (1 − λ ) Q ∈ Π ( x ) × Π ( y ). The proofs for (ii) and (iii) are similar. Step 2. Suppose that the DM narrowly brackets marginal lotteries in bothsources.
That is, ( p, q ) ∼ ( δ x , δ y ) for all ( x, y ) ∈ X × X and ( p, q ) ∈ Π ( x ) × Π ( y ). Thefollowing lemma shows that % must admit a NB representation. Lemma 5.
Suppose that ( p, q ) ∼ ( δ x , δ y ) for all ( x, y ) ∈ X × X and ( p, q ) ∈ Π ( x ) × Π ( y ) ,then % admits a NB representation.Proof of Lemma 5. By Lemma 2, for i = 1 ,
2, denote v i as the EU index of % i . Since X i is a closed interval and v i is regular, the certainty equivalent function CE v i is well-defined.Then for any ( p, q ) ∈ ˆ P , we know ( p, q ) ∼ ( δ CE v ( p ) , δ CE v ( q ) ).Denote a binary relation ˆ % over X × X such that for all ( x, y ) , ( x ′ , y ′ ) ∈ X × X ,( δ x , δ y ) % ( δ x ′ , δ y ′ ) if and only if ( x, y ) ˆ % ( x ′ , y ′ ). Axiom WC implies that ˆ % is continuous on X × X , which is a separable metric space. By Debreu’s Theorem, ˆ % admits a continuous50epresentation w . Axiom M guarantees that w is strictly monotone. Without loss ofgenerality, we can assume that w (0 ,
0) = 0 and w is bounded, because the bounded monotonetransformation w ′ ( x, y ) = 1 − exp ( − w ( x, y )) for w ( x, y ) ≥ w ′ ( x, y ) = exp ( w ( x, y )) − w ( x, y ) < % . Therefore we can find regular functions w, v and v suchthat for all P, Q ∈ P , P % Q ⇐⇒ ( δ CE v ( P ) , δ CE v ( P ) ) % ( δ CE v ( Q ) , δ CE v ( Q ) ) ⇐⇒ ( CE v ( P ) , CE v ( P )) ˆ % ( CE v ( Q ) , CE v ( Q ) ⇐⇒ w ( CE v ( P ) , CE v ( P )) ≥ w ( CE v ( Q ) , CE v ( Q ))That is, % admits a NB representation ( w, v , v ).From now on, we maintain the assumption that there exist ( x, y ) ∈ X × X and ( p, q ) ∈ Π ( x ) × Π ( y ) such that ( p, q ) ( δ x , δ y ). Step 3: Suppose that the DM narrowly brackets marginal lotteries in source2.
This is equivalent to assuming ( p, q ) ∼ ( p, q ′ ) for all ( x, y ) ∈ X × X , p ∈ Π ( x ) and q, q ′ ∈ Π ( y ). Denote the condition as Assumption 1 . Then for any ( p, q ) , ( p ′ , q ′ ) ∈ ˆ P with q ∼ q ′ , ( p, q ) % ( p ′ , q ′ ) if and only if ( p, CE v ( q )) % ( p ′ , CE v ( q )). Hence we can focus onthe restriction of % on L ( X ) × X .By the assumption at the end of Step 2, we can find ( x , y ) ∈ X × X , p , p ′ ∈ Π ( x )and q ∈ Π ( y ) such that ( p , q ) ≻ ( p ′ , q ). This implies y = 0. By Axiom WC, it iswithout loss of generality to assume y ∈ X o . By Lemma 2, we know that % | q admits anEU representation with a continuous and bounded utility index v | δ y . Recall that X oi is theinterior of X i with respect to R , i = 1 ,
2. For each x ∈ X oi , there exists y, y ′ ∈ X oi with y > x > y ′ . Suppose that there exists x ∈ X o such that ( p , q ) ∼ ( p ′ , q ) for all p , p ′ ∈ Π ( x ). Clearly, x = x . Denote x, x ∈ X oi with x > x > x . As v and v | δ y are uniqueup to positive affine transformations, we can set v | δ y ( x ) = v ( x ) and v | δ y ( x ) = v ( x ).For any x ∈ X with x > x , we can find α ∈ (0 ,
1) with αδ x + (1 − α ) δ x ∈ Π ( x ). Then αv ( x ) + (1 − α ) v ( x ) = v ( x ) and ( αδ x + (1 − α ) δ x , q ) ∼ ( δ x , q ), which implies αv | δ y ( x ) + (1 − α ) v | δ y ( x ) = v | δ y ( x ) = v ( x ) = αv ( x ) + (1 − α ) v ( x ) . Since v | δ y ( x ) = v ( x ) and α ∈ (0 , v | δ y ( x ) = v ( x ). Specifically, we have v | δ y ( x ) =51 ( x ). Now we consider x ∈ X with x < x . There exists β ∈ (0 ,
1) with βδ x + (1 − β ) δ x ∈ Π ( x ). Then βv ( x ) + (1 − β ) v ( x ) = v ( x ) and ( αδ x + (1 − α ) δ x , q ) ∼ ( δ x , q ),which also implies v | δ y ( x ) = v ( x ). Thus v | δ y ≡ v , contradicting with ( p , q ) ≻ ( p ′ , q )as p , p ′ ∈ Π ( x ). Thus, there exists y ∈ X such that for any x ∈ X o , we can find p x , p ′ x ∈ Π ( x ) and q ∈ Π ( y ) with ( p x , q ) ≻ ( p ′ x , q ).Denote Σ := { y ∈ X o : ∃ x ∈ X and p, p ′ ∈ Π ( x ) , q ∈ Π ( y ) s.t. ( p, q ) ≻ ( p ′ , q ) } . Σ isnonempty as y ∈ Σ and is open in X by Axiom WC. Also 0 Σ and hence Σ ⊆ R \{ } .Denote the closure of Σ in X as cl (Σ ).The following lemma provides a sufficient condition for a proper tuple to satisfy theindependence property. Lemma 6.
Suppose that Assumption 1 holds. Then a proper tuple ( P, Q, R, S ) ∈ ( L ( X ) × X ) satisfies the independence property if P = R = δ y , Q = S = δ y with y , y ∈ cl (Σ ) . The proof of Lemma 6 requires several intermediate results. The first one assures that wecan focus on the case where P ∼ Q , R ∼ S . Lemma 7.
Suppose that Assumption 1 holds. ( P, Q, R, S ) ∈ ( L ( X ) × X ) is a propertuple with P = R = δ y , Q = S = δ y with y , y ∈ cl (Σ ) . If the independence propertyholds for any such ( P, Q, R, S ) with P ∼ Q, R ∼ S , then the independence property holds forany such ( P, Q, R, S ) with P % Q, R % S .Proof of Lemma 7. Following similar arguments in the proof of Lemma 3, it suffices toconsider the case where P % Q ≻ R % S . By Lemma 2, there exist α ∈ (0 ,
1] and β ∈ [0 , P ′ = αP + (1 − α ) R ∼ Q and S ′ = βQ + (1 − β ) S ∼ R . Then the independenceproperty holds for ( P ′ , Q, R, S ′ ), that is, for any λ ∈ (0 , λP ′ + (1 − λ ) R ∼ λQ + (1 − λ ) S ′ .By Lemma 2 and P % P ′ , S ′ % S , we have λP + (1 − λ ) R % λP ′ + (1 − λ ) R ∼ λQ + (1 − λ ) S ′ % λQ + (1 − λ ) S. At least one of the above weak preference rankings would be strict if P ≻ Q or R ≻ S .Lemma 8 shows the result in Lemma 6 holds locally, that is, when the utilities of P and R are “close enough”. 52 emma 8. Suppose that Assumption 1 holds. Then a proper tuple ( P, Q, R, S ) ∈ ( L ( X ) × X ) satisfies the independence property if P ∼ Q, R ∼ S , P = R = δ y , Q = S = δ y with y , y ∈ Σ and there exist x , x ∈ X such that P, Q, R, S ∈ Γ( x , y ) ∩ Γ( x , y ) .Proof of Lemma 8. Suppose (
P, Q, R, S ) ∈ ( L ( X ) × X ) is a proper tuple that satisfiesthe conditions stated in the lemma. By Lemma 4, there exist P ′ , R ′ ∈ Π ( x ) and Q ′ , S ′ ∈ Π ( x ) such that P ′ := ( P ′ , P ) ∼ P ∼ Q ∼ Q ′ := ( Q ′ , Q ) and R ′ := ( R ′ , R ) ∼ R ∼ S ∼ S ′ := ( S ′ , S ). By part 2 of Lemma 3, for any α ∈ (0 , αP ′ + (1 − α ) R ′ ∼ αQ ′ + (1 − α ) S ′ .Finally, Lemma 2 implies that αP ′ + (1 − α ) R ′ ∼ αP + (1 − α ) R and αQ ′ + (1 − α ) S ′ ∼ αQ + (1 − α ) S . By transitivity of % , αP + (1 − α ) R ∼ αQ + (1 − α ) S .The next lemma shows that if the independence property holds on two sets of productlotteries respectively, then it also holds on their union. Lemma 9.
Suppose that Assumption 1 holds. ( P, Q, R, S ) ∈ ( L ( X ) × X ) is a propertuple where P ∼ Q, R ∼ S , P = R = δ y , Q = S = δ y with y , y ∈ Σ . Fix any T i ∈ ˆ P for i = 1 , ..., with T ≻ T ≻ T ≻ T . If the independence property holds for any such ( P, Q, R, S ) with { P, Q, R, S } ⊆ [ T , T ] or { P, Q, R, S } ⊆ [ T , T ] , then it also holds forany such ( P, Q, R, S ) with { P, Q, R, S } ⊆ [ T , T ] .Proof of Lemma 9. Without loss of generality, we can assume P ≻ R and T % (0 , y i ) , i =1 ,
2, otherwise either the lemma is trivial or we can modify T without changing the lemma.Moreover, it suffices to focus on the case where P ∼ Q ≻ T and R ∼ S ≺ T . Fix any W , W with T ≻ W ≻ W ≻ T . Then we have T % P ∼ Q ≻ T ≻ W ≻ W ≻ T ≻ R ∼ S % T . By Lemma 2, we can find ˆ
P , ˆ Q, ˆ R, ˆ S such that ˆ P ∼ ˆ Q ∼ W , ˆ R ∼ ˆ S ∼ W and ˆ P = ˆ R = P = δ y , ˆ Q = ˆ S = Q = δ y . Notice that P, Q, ˆ R, ˆ S ∈ [ T , T ], where the independenceproperty holds. Then there exists λ ∈ (0 ,
1) such that λP + (1 − λ ) ˆ R ∼ ˆ P ∼ λQ + (1 − λ ) ˆ S .Similarly, we can find λ ′ ∈ (0 ,
1) with λ ′ ˆ P + (1 − λ ′ ) R ∼ ˆ R ∼ λ ′ ˆ Q + (1 − λ ′ ) S .Actually in the construction of ˆ R and ˆ S , there exist η , η ∈ (0 ,
1) with η P + (1 − η ) R ∼ ˆ R ∼ ˆ S ∼ η Q + (1 − η ) S.
53e claim that η = η . To see this, as P = R = ˆ P = ˆ R , we know λ ′ ˆ P + (1 − λ ′ ) R ∼ λλ ′ P + (1 − λ ) λ ′ ˆ R + (1 − λ ′ ) R ∼ ˆ R which implies ˆ R ∼ λλ ′ λλ ′ + (1 − λ ′ ) P + 1 − λ ′ λλ ′ + (1 − λ ′ ) R and hence η = λλ ′ λλ ′ +(1 − λ ′ ) . Similarly we can show that η = λλ ′ λλ ′ +(1 − λ ′ ) = η := η w .A symmetric argument shows that there exists η w with η w < η w < η w P + (1 − η w ) R ∼ ˆ P ∼ ˆ Q ∼ η w Q + (1 − η w ) S. Now we consider η with η w < η < η w . Notice that ηP + (1 − η ) R = η − η w η w − η w [ η w P + (1 − η w ) R ] + η w − ηη w − η w [ η w P + (1 − η w ) R ] ∼ η − η w η w − η w ˆ P + η w − ηη w − η w ˆ R. Similarly, ηQ + (1 − η ) S ∼ η − η w η w − η w ˆ Q + η w − ηη w − η w ˆ S. As ˆ P ∼ ˆ Q, ˆ R ∼ ˆ S ∈ [ T , T ] and ˆ P = ˆ R , ˆ Q = ˆ S , the independence property holds for( ˆ P , ˆ Q, ˆ R, ˆ S ) and hence ηP + (1 − η ) R ∼ η − η w η w − η w ˆ P + η w − ηη w − η w ˆ R ∼ η − η w η w − η w ˆ Q + η w − ηη w − η w ˆ S ∼ ηQ + (1 − η ) S. Then we check the independence property for η w < η < P ∼ η w P + (1 − η w ) R = η w − η w η − η w [ ηP + (1 − η ) R ] + η − η w η − η w [ η w P + (1 − η w ) R ] ∼ η w − η w η − η w [ ηP + (1 − η ) R ] + η − η w η − η w ˆ R. Similarly, ˆ Q ∼ η w Q + (1 − η w ) S ∼ η w − η w η − η w [ ηQ + (1 − η ) S ] + η − η w η − η w ˆ S. Note that ηP + (1 − η ) R, ηQ + (1 − η ) S, ˆ R, ˆ S ∈ [ T , T ], and ( ηP + (1 − η ) R ) = ˆ R , ( ηQ +(1 − η ) S ) = ˆ S . By the condition stated in the lemma and the proof of Lemma 7, the54ndependence property holds for ( ηP + (1 − η ) R, ηQ + (1 − η ) S, ˆ R, ˆ S ). Whenever ηP + (1 − η ) R ηQ + (1 − η ) S , we know ˆ P ˆ Q , a contradiction. Thus, ηP + (1 − η ) R ∼ ηQ + (1 − η ) S .The proof for the case with η w > η > η ∈ (0 , ηP + (1 − η ) R ∼ ηQ + (1 − η ) S .Now we extend the local result in Lemma 8 to a bounded set. Recall that ( P, Q, R, S ) ∈ ( L ( X ) × X ) is a proper tuple where P ∼ Q, R ∼ S , P = R = δ y , Q = S = δ y with y , y ∈ Σ . The independence property holds for ( P, Q, R, S ) trivially if y = y . Withoutloss of generality, we assume y > y and P ≻ R .Take any ˆ T ≻ T ≻ T ≻ ( δ a , δ y ) ≻ ( δ a , δ y ) with a ∈ X , ˆ T = T = T = δ y , T = δ z , T = δ z and ˆ T , T , T ∈ ˆ P . As y , y ∈ Σ , by Lemma 2, we can find ˆ x , ˆ x ∈ X o and p , q ∈ Π (ˆ x ), p , q ∈ Π (ˆ x ) such that( q , δ y ) ∼ T ≺ ( p , δ y ) and ( q , δ y ) ∼ T ≺ ( p , δ y ) . By Lemma 4, we know that[ T , ( p , δ y )] ∩ [ T , ( p , δ y )] ⊆ Γ(ˆ x , y ) ∩ Γ(ˆ x , y ) . For any z ∈ [ z , z ], we can choose λ z and η z ∈ [0 ,
1] such that λ z ( q , δ y ) + (1 − λ z )( δ a , δ y ) ∼ ( δ z , δ y ) ∼ η z ( q , δ y ) + (1 − η z )( δ a , δ y ) . By Lemma 2, λ z ( p , δ y ) + (1 − λ z )( δ a , δ y ) ≻ λ z ( q , δ y ) + (1 − λ z )( δ a , δ y ) ∼ ( δ z , δ y ) and η z ( p , δ y ) + (1 − η z )( δ a , δ y ) ≻ η z ( q , δ y ) + (1 − η z )( δ a , δ y ) ∼ ( δ z , δ y ).Denote ˆ x z , ˆ x z for each z ∈ [ z , z ] with λ z q + (1 − λ z ) δ a ∈ Π (ˆ x z ) , and η z q + (1 − η z ) δ a ∈ Π (ˆ x z ) . This leads to[( δ z , δ y ) , ( λ z p + (1 − λ z ) δ a , δ y )] ∩ [( δ z , δ y ) , ( η z p + (1 − η z ) δ a , δ y )] ⊆ Γ(ˆ x z , y ) ∩ Γ(ˆ x z , y ) . z between z and z , and by Axiom WC, we have[ T , T ] ⊆ [ z ≤ z ≤ z (cid:16) Γ(ˆ x z , y ) ∩ Γ(ˆ x z , y ) (cid:17) . (3)In order to get an open cover of [ T , T ], notice that for ǫ > T ≻ ( δ z + ǫ , δ y ) ≻ ( δ z − ǫ , δ y ) ≻ ( δ a , δ y ), we have[ T , T ] ⊆ [ z − ǫ ≤ z ≤ z + ǫ (cid:16) Γ(ˆ x z , y ) ∩ Γ(ˆ x z , y ) (cid:17) . For each z − ǫ ≤ z ≤ z + ǫ , Γ(ˆ x z , y ) ∩ Γ(ˆ x z , y ) has a non-empty interior. Hencewe can find an open cover of [ T , T ] = [( δ z , δ y ) , ( δ z , δ y )] as { C z } z − ǫ ≤ z ≤ z + ǫ with C z ⊂ Γ(ˆ x z , y ) ∩ Γ(ˆ x z , y ). Notice that X ×{ δ y } is isomorphic to X ⊆ R and in the correspondingtopology [ T , T ] = [( δ z , δ y ) , ( δ z , δ y )] is isomorphic to [ z , z ], which is closed and bounded.By Heine–Borel theorem, we can find a finite subcover of { C z } z − ǫ ≤ z ≤ z + ǫ for [ T , T ]. Denotethe subcover as { C z k } Kk =1 .Take any proper tuple ( P, Q, R, S ) with P ∼ Q, R ∼ S , P = R = δ y , Q = S = δ y with y > y ∈ Σ . By Lemma 8, the independence property holds for ( P, Q, R, S ) if
P, Q, R, S ∈ C z k for any k = 1 , ..., K . Then Lemma 9 implies that the independence propertyholds for ( P, Q, R, S ) if
P, Q, R, S ∈ [ T , T ] ⊆ S Kk =1 C z k . By arbitrariness of T , T and a ∈ X , fix any ˆ z, ˆ z ′ ∈ X with ( δ ˆ z , δ y ) ≻ ( δ ˆ z ′ , δ y ), then the given tuple ( P, Q, R, S ) alwayssatisfies the independence property so long as ( δ ˆ z , δ y ) ≻ P, Q, R, S ≻ ( δ ˆ z ′ , δ y ).There are two gaps between the current argument and a complete proof of Lemma 6.First, we have ruled out the possibility that some lottery in the tuple might be indifferent to( δ c , δ y ) or ( δ c , δ y ). Second, we have assumed that y , y ∈ Σ , instead of its closure. Wewill bridge the gap by utilizing Axiom WC. Proof of Lemma 6.
Following the above arguments, it suffices to consider a tuple (
P, Q, R, S )with P ∼ Q ∼ ( δ z , δ y ) ≻ R ∼ S ∼ ( δ z ′ , δ y ), P = R = δ y , Q = S = δ y where y > y ∈ Σ and z > z ′ . We have already shown the case with ( δ c , δ y ) ≻ P ≻ R ≻ ( δ c , δ y ).Now suppose R ∼ ( δ c , δ y ), where c > −∞ . By Axiom M, it must be the case that R = ( δ c , δ y ). Take a sequence of { λ n } n ≥ ⊂ (0 ,
1) with λ n →
0. For each n , denote S n = ( λ n Q + (1 − λ n ) S , S ) and by Lemma 2, we can find β n with R n = ( β n P + (1 − β n ) R , R ) ∼ S n . Clearly, R n ≻ R = ( δ c , δ y ) for each n and hence the independence56roperty holds for ( P, Q, R n , S n ), that is, for each α ∈ (0 , αP + (1 − α ) R n ∼ αQ + (1 − α ) S n . Easy to see that as n goes to infinity, β n converges to 0 and hence S n w −→ S, R n w −→ R . Bycontinuity of % on ˆ P , we have for each α ∈ (0 , αP + (1 − α ) R ∼ αQ + (1 − α ) S. A similar proof works for the case with P ∼ ( δ c , δ y ) and/or R ∼ ( δ c , δ y ). Hence, theindependence property holds for all ( P, Q, R, S ) with P ∼ Q, R ∼ S , P = R = δ y , Q = S = δ y with y , y ∈ Σ .Now we consider y ∈ cl (Σ ) and y ∈ Σ . By definition, we can find a sequence { y n } n ≥ ⊆ Σ such that y n → y as n → ∞ . Using the standard subsequence arguments, we furtherassume that y n ≥ y for all n . The case where y n ≤ y for all n is symmetric. Denote P n = ( P , δ y n ) % P ∼ Q and R n = ( R , δ y n ) % R ∼ S . For each n , we increase y graduallyto y ′ until either ( Q , δ y ′ ) ∼ P n or ( S , δ y ′ ) ∼ R n . Denote such y ′ as y n . Without loss ofgenerality, suppose Q n := ( Q , δ y n ) ∼ P n . Then we can find S n such that S n = ( S n , δ y n ) ∼ R n . This is guaranteed by Axiom M and P ∼ Q ≻ R ∼ S . Hence the independence propertyapplies for ( P n , Q n , R n , S n ) and for each λ ∈ (0 , λP n + (1 − λ ) R n ∼ λQ n + (1 − λ ) S n .Easy to see that P n w −→ P , Q n w −→ Q , R n w −→ R , S n w −→ S . By continuity of % on ˆ P , theindependence property holds for ( P, Q, R, S ). Similarly, the result holds for y ∈ cl (Σ ) and y ∈ Σ .Finally, assume y > y with y , y ∈ cl (Σ ) \ Σ . Suppose that there exists y ∈ Σ with y < y < y . Since P, R ∈ Γ ,δ y ∩ Γ ,δ y , by Axiom M, P, R ∈ Γ ,δ y . Then there exist p ′ , r ′ with P ′ = ( p ′ , δ y ) ∼ Q ∼ P and R ′ = ( r ′ , δ y ) ∼ R ∼ S . By applying the previous result for( P, P ′ , R, R ′ ) and ( Q, P ′ , S, R ′ ) respectively, we know that the independence property holdsfor any ( P, Q, R, S ). Otherwise, we can find a sequence { y n } n ≥ ⊆ Σ such that y n → y as n → ∞ and y n > y for all n . Then the argument in the previous paragraph follows.By Lemma 7, the independence property holds for any proper tuple ( P, Q, R, S ) with P = R = δ y , Q = S = δ y with y , y ∈ cl (Σ ). This completes the proof.We are now ready to show that % must admit a GBIB-CN representation.57 emma 10. Suppose that Assumption 1 holds. Then % admits a GBIB-CN representation.Proof of Lemma 10. The proof idea is analogue to the proof of Lemma 1 in Fishburn (1982).Recall that we can focus on % restricted to L ( X ) × X . For any ( p , δ x ) , ( p , δ x ) ∈ L ( X ) × cl (Σ ) with ( p , δ x ) ≻ ( p , δ x ), we claim that there exists some function f representing % on ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )] such that f is continuous and linear in the firstsource, that is, for any ( q , δ y ) , ( q , δ y ) ∈ ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )] and α ∈ [0 , f ( αq + (1 − α ) q , δ y ) = αf ( q , δ y ) + (1 − α ) f ( q , δ y ). Also, such f is unique up to a positiveaffine transformation. For simplicity, we call f a M AP function.To prove the claim, notice that by Lemma 3, for any Q ∈ [( p , δ x ) , ( p , δ x )], there exists aunique λ Q ∈ [0 ,
1] such that Q ∼ λ Q ( p , δ x ) + (1 − λ Q )( p , δ x ). Define f : [( p , δ x ) , ( p , δ x )] → [0 ,
1] such that f ( Q ) = λ Q . As ( p , δ x ) ≻ ( p , δ x ), we have f ( Q ) ≥ f ( Q ′ ) ⇐⇒ Q ∼ λ Q ( p , δ x ) + (1 − λ Q )( p , δ x ) % λ Q ′ ( p , δ x ) + (1 − λ Q ′ )( p , δ x ) ∼ Q ′ Hence f represents % on [( p , δ x ) , ( p , δ x )]. Continuity of % on ˆ P assures that f is continuous.Then we show the linearity of f in source 1 on ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )]. Takeany ( q , δ y ) , ( q , δ y ) ∈ ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )]. By definition of f , we have( q , δ y ) ∼ f ( q , δ y )( p , δ x ) + (1 − f ( q , δ y ))( p , δ x ) , ( q , δ y ) ∼ f ( q , δ y )( p , δ x ) + (1 − f ( q , δ y ))( p , δ x ) . Clearly, for any α ∈ (0 , α ( q , δ y ) + (1 − α )( q , δ y ) ∈ ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )].By definition of f and Lemma 6, α ( q , δ y ) + (1 − α )( q , δ y ) ∼ [ αf ( q , δ y ) + (1 − α ) f ( q , δ y )]( p , δ x )+ [1 − αf ( q , δ y ) − (1 − α ) f ( q , δ y )]( p , δ x ) ∼ f ( αq + (1 − α ) q , δ y )( p , δ x )+ (1 − f ( αq + (1 − α ) q , δ y ))( p , δ x )By Lemma 2 given δ x in source 2, we know f ( αq +(1 − α ) q , δ y ) = αf ( q , δ y )+(1 − α ) f ( q , δ y ).Easy to see that a positive affine transformation of f also represents % on ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )].Now suppose that f, g represent % on ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )] and they58re continuous and linear in source 1. Without loss of generality, let f ( p , δ x ) = g ( p , δ x ) , f ( p , δ x ) = g ( p , δ x ). Recall that for any Q ∈ ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )],there is a unique λ Q with Q ∼ λ Q ( p , δ x ) + (1 − λ Q )( p , δ x ). By linearity of f and g in source1 on ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )], we have f ( Q ) = λ Q f ( p , δ x ) + (1 − λ Q ) f ( p , δ x )= λ Q f ( p , δ x ) + (1 − λ Q ) f ( p , δ x )= g ( Q )Hence f ≡ g on ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )] and f is unique up to a positive affinetransformation.As p , p are arbitrary and the M AP function is unique up to a positive affinetransformation, for each x ∈ cl (Σ ), we can find a M AP function f that represents % on ( L ( X ) × cl (Σ )) ∩ Γ ,δ x . Also, as f is unique up to a positive affine transformation onany ( L ( X ) × cl (Σ )) ∩ [( p , δ x ) , ( p , δ x )], f is unique up to a positive affine transformationon ( L ( X ) × cl (Σ )) ∩ Γ ,δ x .Now choose y ∈ cl (Σ ) and y = x . Denote the M AP function on ( L ( X ) × cl (Σ )) ∩ Γ ,δ x as f x and the M AP function on ( L ( X ) × cl (Σ )) ∩ Γ ,δ y as f y . If there exist T ≻ T with T , T ∈ Γ ,δ y ∩ Γ ,δ x , then by Lemma 4, we can find p x , p x , p y , p y such that [ T , T ] =[( p x , δ x ) , ( p x , δ x )] = [( p y , δ y ) , ( p y , δ y )]. Since both f x , f y are M AP functions on ( L ( X ) × cl (Σ )) ∩ [ T , T ], they must be positive affine transformations of each other on ( L ( X ) × cl (Σ )) ∩ [ T , T ]. Fix f x and let f y ( T i ) = f x ( T i ), i = 1 ,
2, then we have f x = f y on( L ( X ) × cl (Σ )) ∩ [ T , T ]. Define ˆ f = f x on ( L ( X ) × cl (Σ )) ∩ Γ ,δ x and ˆ f = f y on( L ( X ) × cl (Σ )) ∩ Γ ,δ y . Then easy to show that ˆ f is a M AP function on ( L ( X ) × cl (Σ )) ∩ (Γ ,δ x ∪ Γ ,δ y ) and is unique up to a positive affine transformation. If instead no such T and T exist, then the construction of a M AP function on ( L ( X ) × cl (Σ )) ∩ (Γ ,δ x ∪ Γ ,δ y ) istrivial.By induction, the above arguments can be applied to show the existence of a M AP function on ( L ( X ) × cl (Σ )) ∩ ( S x ∈ A Γ ,δ x ) where A is a finite subset of cl (Σ ), and the M AP function is unique up to a positive affine transformation. As A is an arbitrary finitesubset of cl (Σ ) and S x ∈ X Γ ,δ x = ˆ P , we can find a M AP function V that represents % on L ( X ) × cl (Σ ), which is unique up to a positive affine transformation.59efine w ( x, y ) = V ( δ x , δ y ) for all ( x, y ) ∈ X × cl (Σ ). Easy to show that w is regular on X × cl (Σ ). This implies that for any ( p, δ z ) , ( q, δ z ′ ) ∈ L ( X ) × cl (Σ ),( p, δ z ) % ( q, δ z ′ ) ⇐⇒ X x w ( x, z ) p ( x ) ≥ X x w ( x, z ′ ) q ( x ) . Recall that in the proof of Lemma 5, % restricted to X admits a regular utilityrepresentation. That implies we can extend w to X such that w is regular and represents % on X .Take any ( p, z ) ∈ L ( X ) × X . If z ∈ cl (Σ ), then by continuity of w , we can find a ( p,z ) ∈ X such that ( p, z ) ∼ ( a ( p,z ) , z ), that is, w ( a ( p,z ) , z ) = P x w ( x, z ) p ( x ). If z cl (Σ ),then by definition of Σ , ( p, z ) ∼ ( CE v ( p ) , z ). By Axiom CN and ( p, q ) ∼ ( p, CE v ( q )) for all p, q ∈ L ( X ), we know that for any P ∈ P , P ∼ ( CE v ( P ) , CE v ( P )) if CE v ( P ) cl (Σ )and P ∼ ( a ( P ,δ CEv P ) , CE v ( P )) if CE v ( P ) ∈ cl (Σ ). Hence % is represented by V ( P ) = w ( CE v ( P ) , CE v ( P )) , if CE v ( P ) cl (Σ ) P w ( x, CE v ( P )) P ( x ) , if CE v ( P ) ∈ cl (Σ )Finally, if CE v ( P ) ∈ ∂ Σ = cl (Σ ) \ Σ , then by definition of Σ , P ∼ ( CE v ( P ) , CE v ( P )).This implies w ( · , CE v ( P )) must be a positive affine transformation of v . Thus, we canrewrite the representation as V ( P ) = w ( CE v ( P ) , CE v ( P )) , if CE v ( P ) Σ P w ( x, CE v ( P )) P ( x ) , if CE v ( P ) ∈ Σ and hence % is represented by a GBIB-CN representation ( w, v , v , Σ ), where w, v , v areregular and Σ is open in X with 0 Σ . Step 4: Suppose that the DM narrowly brackets marginal lotteries in source1.
It is equivalent to the assumption that ( p, q ) ∼ ( p ′ , q ) for all ( x, y ) ∈ X , p, p ′ ∈ Π ( x ) and q ∈ Π ( y ). By a symmetric argument to Step 3, we can show that the DM must admit aFBIB-CN representation. Step 5: Suppose that the DM does not narrowly bracket marginal lotteries inboth sources.
That is, we can find x , x ∈ X , y , y ∈ X and ( p , q ) , ( p , q ′ ) ∈ Π ( x ) × Π ( y ), ( p , q ) , ( p ′ , q ) ∈ Π ( x ) × Π ( y ) such that ( p , q ) ≻ ( p , q ′ ) and ( p , q ) ≻ ( p ′ , q ).60e denote this condition as Assumption 2 .The next lemma shows that we can make x = y and x = y in Assumption 2. Lemma 11.
Suppose that Assumption 2 holds. Then there exist ( x o , y o ) ∈ X o × X o and P o , Q o , R o , S o ∈ Π ( x ) × Π ( y ) such that P o = Q o , R o = S o , P o ≻ Q o and R o ≻ S o .Proof of Lemma 11. By Assumption 2, there exist x , x ∈ X , y , y ∈ X and ( p , q ) , ( p , q ′ ) ∈ Π ( x ) × Π ( y ), ( p , q ) , ( p ′ , q ) ∈ Π ( x ) × Π ( y ) such that ( p , q ) ≻ ( p , q ′ ) and( p , q ) ≻ ( p ′ , q ). Clearly, y ∈ X o , x ∈ X o . By Axiom WC, we can also assumethat y ∈ X o and x ∈ X o . By Lemma 2, given p in source 1, as y ∈ X o , wecan find ˆ q , ˆ q ′ ∈ Π ( y ) with ( p , ˆ q ) ≻ ( p , ˆ q ′ ). By Lemma 2 given q in source 2, as x ∈ X o , we can find ˆ p , ˆ p ′ ∈ Π ( x ) with (ˆ p , q ) ≻ (ˆ p ′ , q ). Let ( x o , y o ) = ( x , y ) and P o = ( p , ˆ q ) ≻ Q o = ( p , ˆ q ′ ) , R o = (ˆ p , q ) ≻ S o = (ˆ p ′ , q ). This completes the proof.From now on, take x o , y o and ( P o , Q o , R o , S o ) as given in Lemma 11. Denote the set ofall such pairs of ( x o , y o ) as O . Similar to Σ , we define Σ asΣ := { x ∈ X o : ∃ y ∈ X and p ∈ Π ( x ) , q, q ′ ∈ Π ( y ) s.t. ( p, q ) ≻ ( p, q ′ ) } . One should notice that Σ i ⊆ X oi , i = 1 ,
2. It is possible, for example, that for x = c ,there exists y ∈ X with q, q ′ ∈ Π ( y ) and ( c , q ) ≻ ( c , q ′ ). However, by Axiom WC, thisimplies that ( c − ǫ, c ) ⊆ Σ for some ǫ >
0, which implies c ∈ cl (Σ ). This suggests thatfor each i = 1 , x ∈ X i \ cl (Σ i ), y ∈ X − i , p ∈ Π i ( x ) and q, q ′ ∈ Π − i ( y ), we must have P ∼ Q where P i = Q i = p and P − j = q, Q − j = q ′ .By the proof of Lemma 11, we can show O = Σ × Σ . Also, by continuity of % on ˆ P , Σ is also an open subset of X and 0 Σ . For any x ∈ Σ and x ∈ Σ , we denoteˆΠ ( x ) = n p ∈ Π ( x ) : ∃ x ′ ∈ X , q ∈ Π ( x ′ ) s.t. ( p, q ) ( p, x ′ ) o , ˆΠ ( x ) = n q ∈ Π ( x ) : ∃ x ′ ∈ X , p ∈ Π ( x ′ ) s.t. ( p, q ) ( x ′ , q ) o . Clearly, ˆΠ i ( x i ) ⊆ Π i ( x i ) for each i by definition. Moreover, with the same argument in Step3, for any i = 1 , x i ∈ Σ i , p i ∈ ˆΠ i ( x i ) and x − i ∈ X o − i , we can find p − i ∈ Π − i ( x − i ) such that T T ′ where T i = T ′ i = p i and T − i = p − i , T ′− i = x − i . Lemma 12.
For each i = 1 , and x i ∈ Σ i , cl ( ˆΠ i ( x i )) = Π i ( x i ) . roof of Lemma 12. We will prove the result for i = 1. The proof for i = 2 is symmetric andomitted. By definition of x ∈ Σ , we can find p ∈ ˆΠ ( x ), x ′ ∈ X and q, q ′ ∈ Π ( x ′ ) suchthat ( p, q ) ≻ ( p, q ′ ). For any p o ∈ Π ( x ) \ ˆΠ ( x ), we have ( p o , q ) ∼ ( p o , q ′ ). By Lemma 3,the independence property holds for (( p, q ) , ( p, q ′ ) , ( p o , q ) , ( p o , q ′ )) as p, p o ∈ Π ( x ). Then forany λ ∈ (0 , λp + (1 − λ ) p o , q ) ≻ ( λp + (1 − λ ) p o , q ′ ), which implies λp + (1 − λ ) p o ∈ ˆΠ ( x ).Let λ → p o ∈ cl ( ˆΠ i ( x i )).For any A ⊆ X and A ⊆ X , we denote Y ( A ) = ∪ x ∈ A Π ( x ), Y ( A ) = ∪ x ∈ A Π ( x )and Y ( A , A ) = Y ( A ) × Y ( A ) ⊆ ˆ P . Our goal is to show that the independence propertyholds for certain proper tuple ( P, Q, R, S ) like Lemma 6.
Lemma 13.
Suppose that Assumption 2 holds. Then a proper tuple ( P, Q, R, S ) satisfies theindependence property if P i = R i ∈ Y i ( cl (Σ i )) , Q j = S j ∈ Y j ( cl (Σ j )) for i, j ∈ { , } . Again, the proof will rely on several intermediate lemmas. Similar to Lemma 7, we canfocus on the case where P ∼ Q, R ∼ S without loss of generality.For any ( x , x ) ∈ Σ × Σ , by definition and Lemma 11, there exist ( p, q ) , ( p, q ′ ) , (ˆ p, ˆ q ) , (ˆ p ′ , ˆ q ) ∈ Π ( x ) × Π ( x ) with ( p, q ) ≻ ( p, q ′ ) and (ˆ p, ˆ q ) ≻ (ˆ p ′ , ˆ q ). By Lemma 3, we know that a propertuple ( P, Q, R, S ) satisfies the independence property if
P, Q, R, S ∈ Π ( x ) × Π ( x ) and P ∼ Q, R ∼ S . The next lemma is analogous to Lemma 8. Lemma 14.
Suppose that Assumption 2 holds. Then a proper tuple ( P, Q, R, S ) satisfiesthe independence property if there exist some i, j ∈ { , } , x i ∈ Σ i , x j ∈ Σ j and y ∈ X − i , y ′ ∈ X − j such that P i = R i = p ∈ Π i ( x i ) , Q j = S j = q ∈ Π j ( x j ) , P ∼ Q , R ∼ S and P, R ∈ Γ i,p ( y ) ∩ Γ j,q ( y ′ ) .Proof of Lemma 14. By Lemma 4, we can find P ′ , R ′ with P ′ i = R ′ i = p and P ′− i , R ′− i ∈ Π − i ( y ) such that P ∼ P ′ and R ∼ R ′ . By Lemma 2, for any λ ∈ (0 , λP + (1 − λ ) R ∼ λP ′ + (1 − λ ) R ′ . Similarly, we can find Q ′ , S ′ with Q ′ j = S ′ j = q , Q ′− j , R ′− j ∈ Π − j ( y ′ ) and λQ + (1 − λ ) S ∼ λQ ′ + (1 − λ ) S ′ for any λ ∈ (0 , P ′ , Q ′ , R ′ , S ′ ) and hence for any λ ∈ (0 , λP + (1 − λ ) R ∼ λP ′ + (1 − λ ) R ′ ∼ λQ ′ + (1 − λ ) S ′ ∼ λQ + (1 − λ ) S. i,p ( y ) ∩ Γ j,q ( y ′ ) to be nonemptyand hence Lemma 14 can be applied. Lemma 15.
Fix i, j ∈ { , } , p ∈ L ( X i ) , q ∈ L ( X j ) and y ∈ X − i . For any P ∈ Γ i,p ( y ) ∩ Γ j,q , then there exists y ′ ∈ X − j such that P ∈ Γ j,q ( y ′ ) .Proof of Lemma 15. This is by definition of Γ j,q .The following lemma is the counterpart of Lemma 9, which says that if the independenceproperty holds on two sets of product lotteries respectively, then it also holds on their union.
Lemma 16.
Suppose that Assumption 2 holds and ( P, Q, R, S ) ∈ ˆ P is a proper tuple with P ∼ Q and R ∼ S . Fix i ∈ { , } , p ∈ L ( X i ) , y ∈ X − i and T j ∈ ˆ P for j = 1 , ..., with T ≻ T ≻ T ≻ T . If the independence property holds for any such ( P, Q, R, S ) with { P, Q, R, S } ⊆ Γ i,p ( y ) ∩ [ T , T ] or { P, Q, R, S } ⊆ Γ i,p ( y ) ∩ [ T , T ] , then it also holds forany such ( P, Q, R, S ) with { P, Q, R, S } ⊆ Γ i,p ( y ) ∩ [ T , T ] .Proof of Lemma 16. The proof can be directly adapted from the proof of Lemma 9 by notingthat for any W ≻ W ′ , W, W ′ ∈ Γ i,p ( y ) implies that [ W ′ , W ] ⊆ Γ i,p ( y ).Now we extend the local result in Lemma 14 to a bounded set. Fix i, j ∈ { , } , x i ∈ Σ i , x j ∈ Σ j and p ∈ ˆΠ i ( x i ), q ∈ ˆΠ j ( x j ). Without loss of generality, we assume i = 1. Denoteˆ Q ∈ ˆ P with ˆ Q j = q, ˆ Q − j = a ′ for some a ′ ∈ X − j . Similarly, fix a ∈ X .Take any ˆ T ≻ T ≻ T with ˆ T , T , T ∈ ˆ P , ˆ T = T = T = p , T = δ z , T = δ z and T ≻ ˆ Q , T ≻ ( p, a ). Then we know z > z >
0. By definition of p ∈ ˆΠ i ( x ), AxiomM and Axiom WC , we can find z > z , r , r ∈ Π ( z ) such that ( p, r ) ≻ ( p, r ) and( p, δ z ) ∈ [( p, r ) , ( p, r )]. For any 0 < z < z with ( p, δ z ) ≻ ˆ Q , we can find η z ∈ (0 ,
1) with( p, η z δ ¯ z + (1 − η z ) δ a ) ∼ ( p, δ z ).Denote x z such that η z δ ¯ z + (1 − η z ) δ a ∈ Π ( x z ). By Lemma 2, ( p, η z r + (1 − η z ) δ a ) ≻ ( p, η z r + (1 − η z ) δ a ) and( p, δ z ) ∈ [( p, η z r + (1 − η z ) δ a ) , ( p, η z r + (1 − η z ) δ a )] ⊆ Γ ,p ( x z ) .
63y continuity of % on ˆ P , as ¯ z > z and T = ( p, δ z ) ≻ ˆ Q , T = ( p, δ z ) ≻ ( p, δ a ), we canfind ǫ > T = ( p, δ z ) ≻ ( p, η z − ǫ r + (1 − η z − ǫ ) δ a ) , T = ( p, δ z ) ≺ ( p, η z + ǫ r + (1 − η z + ǫ ) δ a ) . Hence we know that { (( p, η z r + (1 − η z ) δ a ) , ( p, η z r + (1 − η z ) δ a )) } z − ǫ ≤ z ≤ z + ǫ is an opencover of [ T , T ] = [( p, δ z ) , ( p, δ z )]. Again, by compactness, it admits a finite subcoverindexed by { z , ..., z n } ⊂ ( z − ǫ, z + ǫ ).Consider a proper tuple ( P, Q, R, S ) with P = R = p ∈ ˆΠ i ( x ), Q j = S j = q ∈ ˆΠ j ( x j ), j ∈ { , } and P ∼ Q, R ∼ S . Fix any y ∈ X − i . For each k = 1 , ..., n , by Lemma 14, theindependence property holds for ( P, Q, R, S ) if
P, R ∈ Γ j,q ( y ) ∩ [( p, η z k r + (1 − η z k ) δ a ) , ( p, η z k r + (1 − η z k ) δ a )] ⊆ Γ j,q ( y ) ∩ Γ ,p ( x z k ) . Lemma 16 implies that the independence property holds for such (
P, Q, R, S ) if
P, R ∈ Γ j,q ( y ) ∩ [ T , T ].Then we show that we can get rid of the constraint that P, R ∈ [ T , T ], where there exist T , T , ˆ T with ˆ T = T = T = p and T ≻ ˆ Q , T ≻ ( p, δ a ). The proof is similar to Step 3by utilizing the arbitrariness of T , T , a, a ′ and the continuity of % on ˆ P . Hence we knowthat the independence property holds for ( P, Q, R, S ) if
P, R ∈ Γ j,q ( y ) for any y ∈ X .Repeat the previous proof technique by varying y , and we can extend the aboveindependence property to the following global property. (Recall that our focus on P ∼ Q, R ∼ S and i = 1 in the previous analysis is without loss of generality.) Lemma 17.
Suppose that Assumption 2 holds. Then a proper tuple ( P, Q, R, S ) satisfies theindependence property if P i = R i = p ∈ ˆΠ i ( x i ) , Q j = S j = q ∈ ˆΠ j ( x j ) for x i ∈ Σ i , x j ∈ Σ j and i, j ∈ { , } . We are now ready to prove Lemma 13.
Proof of Lemma 13.
Lemma 17 is weaker than Lemma 13 as we have assumed that p ∈ ˆΠ i ( x i ), q ∈ ˆΠ j ( x j ), x i ∈ Σ i and x j ∈ Σ j , instead of their closures Π i ( x i ) , Π j ( x j ) , cl (Σ i )and cl (Σ j ) respectively. However, the proof can be completed using the same continuityarguments in the proof of Lemma 6. 64ecall that Σ and Σ are open subsets of R . The following lemma provides acharacterization for a nonempty open set on the real line. The proof is standard and weinclude it for completeness. Lemma 18.
Every non-empty open set I ⊆ R can be expressed as a countable union ofpairwise disjoint open intervals.Proof of Lemma 18. As R is a complete metric space and I is an open set in R , for any x ∈ I ,there exists an open interval I x ⊆ I that contains x . Denote a ( x ) = inf { y : ( y, x ) ⊆ I } and b ( x ) = sup { z : ( x, z ) ⊆ I } . Clearly, a ( x ) < x < b ( x ).Denote J ( x ) = ( a ( x ) , b ( x )) for each x ∈ I . We claim that J ( x ) ⊆ I . To see this, forarbitrary ǫ >
0, as a ( x ) is an infimum, there exists z < a ( x ) + ǫ such that ( z, x ) ⊆ I . Thisimplies ( a ( x ) + ǫ, x ) ⊆ I and hence ( a ( x ) , x ) = S + ∞ n =1 (cid:16) a ( x ) + 1 /n, x (cid:17) ⊆ I . By a similarargument, ( x, b ( x )) ⊆ I . Hence we have J ( x ) = ( a ( x ) , x ) ∪ { x } ∪ ( x, b ( x )) ⊆ I. Suppose now that a ( x ) = −∞ or b ( x ) = + ∞ . If both are the case, it must be I = R and we are done. In other cases, we observe intervals of the type K − ( a ) := ( −∞ , a ) or K + ( a ) = ( a, + ∞ ), both of which are open. Assume that I contains such an interval K − ( a )with a I . By definition, we can always find such a number a , and there can be at most twosuch numbers. For instance, suppose K − ( a ) ⊆ I . Then I = ( I ∩ K − ( a )) ∪ ( I ∩ K + ( a )). Thisimplies I is open if and only if I \ K − ( a ) is open. Then it suffices to show that I \ K − ( a ) canbe decomposed as a countable union of pairwise disjoint open intervals. Therefore, withoutloss of generality, we assume that there is no x ∈ I with a ( x ) = −∞ or b ( x ) = + ∞ .Define a binary relation ˆ ∼ on I by x ˆ ∼ y if and only if J ( x ) = J ( y ). Easy to prove thatˆ ∼ is an equivalent relation and ˆ ∼ partitions I . We claim that the equivalent classes areopen. To see this, let x < y with x, y ∈ I . When x ˆ ∼ y , we have x ∈ J ( x ) = J ( y ) ∋ y .Inversely, when x ∈ J ( y ), then ( x, y ) ⊆ I and hence a ( x ) = a ( y ) , b ( x ) = b ( y ). This implies J ( x ) = J ( y ). Thus, the equivalent class of x is exactly J ( x ).Finally, as J ( x ) is open and nonempty and the set of rational numbers is dense in thereal line, each set in the partition of I can be labelled by a rational number and hence thepartition is countable. This implies I = S n ∈ N ∗ J n and completes the proof.For i = 1 ,
2, since Σ i ⊆ X oi \{ } is open and nonempty, by Lemma 18, we can writeΣ i = S N i n =1 J in where N i ∈ N ∪ { + ∞} and J in = ( b n , b n ) with b n , b n ∈ R ∪ {−∞ , + ∞} and65 n ≥ b n − for each n ≤ N i . Lemma 19.
For each i = 1 , and n ≤ N i , Y i ( cl ( J in )) is a mixture set.Proof of Lemma 19. For any p, q ∈ Y i ( cl ( J in )), there exist x p , x q ∈ cl ( J in ) such that p ∈ Π i ( x p )and q ∈ Π i ( x q ). Without loss of generality, let x p ≥ x q . By Lemma 2, for any λ ∈ (0 , λp + (1 − λ ) q ∈ Π i ( x ′ ) with x ′ ∈ [ x q , x p ]. As cl ( J in i ) is a closed interval and x p , x q ∈ cl ( J in i ), x ′ ∈ cl ( J in i ). This implies λp + (1 − λ ) q ∈ Y i ( cl ( J in )) and hence Y i ( cl ( J in )) is a mixture set.Then for each n ≤ N , n ≤ N , Y ( cl ( J n ) , cl ( J n )) = Y ( cl ( J n )) × Y ( cl ( J n )) is theproduct of two mixture sets. Also, % restricted to Y ( cl ( J n ) , cl ( J n )) is continuous andsatisfies Axiom MI by Lemma 13. By Theorem 1 in Chapter 7.2 (Page 88) of Fishburn(1982), we know that there exists a continuous and multilinear representation V EUn ,n of % on Y ( cl ( J n ) , cl ( J n )), which is unique up to a positive affine transformation.We claim that cl (Σ i ) = X i for i = 1 ,
2. Suppose by contradiction that cl (Σ i ) = X i forsome i . Then we can find a < a < a such that either ( a , a ) ⊆ X i \ cl (Σ i ), ( a , a ) ⊆ cl (Σ i )or ( a , a ) ⊆ X i \ cl (Σ i ), ( a , a ) ⊆ cl (Σ i ). By symmetry, we will focus on the case where i = 2 and ( a , a ) ⊆ X \ cl (Σ ), ( a , a ) ⊆ cl (Σ ). We can further assume ( a , a ) ⊆ cl ( J n )for some n . Choose some n ≤ N and denote J n = ( b , b ) with b < b . We know thatthere exists a multilinear representation V EU for % on Y ([ b , b ] , [ a , a ]).We first focus on the preference % restricted to Y ([ b , b ] , [ a , a ]). By definition, forany P ∈ Y ( cl (Σ ) , X \ cl (Σ )), P = ( P , P ) ∼ ( CE v ( P ) , P ). Lemma 13 guarantees thatindependence property holds for a proper tuple ( P, Q, R, S ) where P = R ∈ Π ( x ), Q = S ∈ Π ( x ′ ) with x, x ′ ∈ [ b , b ] ⊆ cl (Σ ). Then by a similar argument in Step 4 , thereexists a continuous representation V F IB of % on Y ([ b , b ] , [ a , a ]) where V F IB ( P , P ) = V F IB ( δ CE v ( P ) , P ) for each ( P , P ) ∈ Y ([ b , b ] , [ a , a ]), V F IB is linear in the second source(i.e., a
M AP function) and unique up to a positive affine transformation.For each b ∈ ( b , b ) and p ∈ Π ( b ), by Lemma 2, % | p on { p } × L ( X ) admits an EUrepresentation with a regular utility index v | p . When there is no confusion, we also denote v | p as the EU function. The next lemma relates v | p with V F IB . Suppose M , M ⊆ L ( R ) are mixture sets. A function V is multilinear on M × M if V ( λp + (1 − λ ) r, q ) = λV ( p, q ) + (1 − λ ) V ( r, q ) and V ( p, λq + (1 − λ ) s ) = λV ( p, q ) + (1 − λ ) V ( p, s ) for all λ ∈ (0 , p, r ∈ M and q, s ∈ M . emma 20. ∀ b ∈ ( b , b ) and p ∈ Π ( b ) , v | p is a positive affine transformation of V F IB ( δ b , · ) on Y ([ a , a ]) .Proof of Lemma 20. By definition, V F IB ( p, q ) = V F IB ( δ b , q ) for all p ∈ Π ( b ), q ∈ Y ([ a , a ]). Then it suffices to show that v | p is a positive affine transformation of V F IB ( p, · )on Y ([ a , a ]). Suppose, without loss of generality, that v | p ( a i ) = V F IB ( p, δ a i ) for i = 1 , q ∈ Y ([ a , a ]), if δ a % | p q % | p δ a , then there exists a unique λ ∈ [0 ,
1] such that λδ a + (1 − λ ) δ a ∼ | p q and hence v | p ( q ) = v | p ( λδ a + (1 − λ ) δ a )= λv | p ( δ a ) + (1 − λ ) v | p ( δ a )= λV F IB ( p, δ a ) + (1 − λ ) V F IB ( p, δ a )= V F IB ( p, λδ a + (1 − λ ) δ a )= V F IB ( p, q ) . Similar arguments hold for q ≻ | p δ a and δ a ≻ | p q . This completes the proof.A direct corollary is that for all b ∈ ( b , b ) and p ∈ Π ( b ), v | p is a positive affinetransformation of v | δ b on Y ([ a , a ]). We further claim that it holds on L ( X ). Lemma 21. ∀ b ∈ ( b , b ) and p ∈ Π ( b ) , v | p is a positive affine transformation of v | δ b .Proof of Lemma 21. By the corollary of Lemma 20, given b ∈ ( b , b ) and p ∈ Π ( b ), thereexist α p > β p such that v | p ( q ) = α p v | δ b ( q ) + β p for all q ∈ Y ([ a , a ]).Now consider q ∈ L ( X ) \ Y ([ a , a ]). If q ≻ δ a , then we can find λ > λq + (1 − λ ) δ a ∈ Y ([ a , a ]). This implies v | p ( λq + (1 − λ ) δ a ) = α p v | δ b ( λq + (1 − λ ) δ a ) + β p .By linearity of v | p and v | δ b , we have λv | p ( q ) + (1 − λ ) λv | p ( δ a ) = λ [ α p v | δ b ( q ) + β p ] + (1 − λ )[ α p v | δ b ( δ a ) + β p ] . As v | p ( δ a ) = α p v | δ b ( δ a ) + β p and λ >
0, we know v | p ( q ) = α p v | δ b ( q ) + β p . Similar resultscan be shown for q ≺ δ a . Thus v | p ( q ) = α p v | δ b ( q ) + β p for all q ∈ L ( X ).Now we turn to Y ([ b , b ] , [ a , a ]), on which V EU represents % . By a similar argument asLemma 20, for any b ∈ ( b , b ) and p ∈ Π ( b ), V EU ( p, · ) is a positive affine transformation of67 | p on Y ([ a , a ]) and v | p is a positive affine transformation of v | δ b on Y ([ a , a ]). Denote V EU ( p, q ) = ˆ α p v | δ b ( q ) + ˆ β p with ˆ α p > β p ∈ R for each q ∈ Y ([ a , a ]). Noticethat a ∈ ( a , a ). As X is a closed interval, a ∈ X o . Also, ( a , a ) ⊆ X \ cl (Σ ) and( a , a ) ⊆ cl (Σ ) imply that a Σ . Then for each q ∈ Π ( a ), we have ( p, q ) ∼ ( δ b , q ) andthus ˆ α p v | δ b ( q ) + ˆ β p = V EU ( p, q ) = V EU ( δ b , q ) = ˆ α δ b v | δ b ( q ) + ˆ β δ b . As b ∈ Σ and a ∈ X o , there exist q , q ∈ Π ( a ) such that ( δ b , q ) ≻ ( δ b , q ). Hence,ˆ α p v | δ b ( q ) + ˆ β p = ˆ α δ b v | δ b ( q ) + ˆ β δ b , ˆ α p v | δ b ( q ) + ˆ β p = ˆ α δ b v | δ b ( q ) + ˆ β δ b . This implies ˆ α p = ˆ α δ b , ˆ β p = ˆ β δ b for all p ∈ Π ( b ). Then for all p ∈ Π ( b ) and q ∈ Y ([ a , a ]) ,V EU ( p, q ) = ˆ α δ b v | δ b ( q ) + ˆ β δ b = V EU ( δ b , q ) . That is, ( p, q ) ∼ ( δ b , q ) for all p ∈ Π ( b ) and q ∈ Y ([ a , a ]), which suggests that ( a + a ) / Σ , a contradiction with ( a , a ) ⊆ Σ . To conclude, cl (Σ i ) = X i for i = 1 , % admits an EU-CN representation.To summarize, as NB is a special case of GBIB-CN (FBIB-CN), we conclude that underthe axioms stated in the theorem, % admits one of the following representations: EU-CN,GBIB-CN and GFIB-CN. This completes the proof for sufficiency. Proof of Theorem 2. ii ) ⇒ i ). We first prove the necessity of these axioms. First, it is easy to verify that EUsatisfies all the axioms. By Theorem 1, we know representations EU-CN, GBIB-CN andGFIB-CN satisfy Axioms WO, M, WC, WI, and they trivially satisfy Axiom CC as itsprimitive will never be satisfied.For BIB, as it reduces to a special case of GBIB-CN on ˆ P , it suffices to show that BIBsatisfies the first two parts of Axiom WC, Axiom CC and Axiom M.Suppose that % admits a BIB representation ( w, v ), that is, for P ∈ P , V BIB ( P ) = X x w ( x, CE v ( P | x )) P ( x ) .
68o verify part (i) of Axiom WC, for any
P, Q ∈ P and λ ∈ [0 , V BIB ( λP + (1 − λ ) Q ) = λ X x : P ( x ) > ,Q ( x )=0 w ( x, CE v ( P | x )) P ( x )+ (1 − λ ) X x : Q ( x ) > ,P ( x )=0 w ( x, CE v ( Q | x )) Q ( x )+ X x : Q ( x ) > ,P ( x ) > w ( x, CE v ( αP | x + (1 − α ) Q | x ))[ λP ( x ) + (1 − λ ) Q ( x )] , where α = λP ( x ) λP ( x ) + (1 − λ ) Q ( x ) . Then V BIB ( λP + (1 − λ ) Q ) is continuous in λ and mixture continuity holds for % on P .To verify the third part of Axiom WC, i.e., Axiom Continuity over Sure Gains, for each P ∈ P and any two sequences ǫ n , ǫ ′ n → n → ∞ such that for each n , ǫ n , ǫ ′ n >
0. Since P isa simple lottery, for n large enough, we can guarantee that P ∗ ( δ ǫ n , δ ǫ ′ n )( x + ǫ n , y + ǫ ′ n ) = P ( x, y )for all x ∈ X o , y ∈ X o . For such n , we have V BIB ( P ∗ ( δ ǫ n , δ ǫ ′ n )) = X x ∈ X oi w ( x + ǫ n , CE v ( P | x ∗ δ ǫ ′ n )) P ( x )+ w ( c , CE v ( P | x ∗ δ ǫ ′ n )) P ( c ) . If c = + ∞ , then the second term is always 0. Notice that P | x ∗ δ ǫ ′ n w −→ P | x as n goes toinfinity. By continuity of w and v , easy to see that V BIB ( P ∗ ( δ ǫ n , δ ǫ ′ n )) is continuous in( ǫ n , ǫ ′ n ) and hence Axiom Continuity over Sure Gains holds for % .Then we will check Axiom CC. For each P, Q, R, S ∈ P and α ∈ (0 , P i = Q i for i = 1 , P ) ∩ supp( R ) = supp( P ) ∩ supp( S ) = ∅ , then V BIB ( αP + (1 − α ) R ) = αV BIB ( P ) + (1 − α ) V BIB ( R ) ,V BIB ( αQ + (1 − α ) S ) = αV BIB ( Q ) + (1 − α ) V BIB ( S ) . Hence P ≻ Q, R ∼ S implies that αP + (1 − α ) R ≻ αQ + (1 − α ) S .Finally for Axiom M, suppose that P dominates ( δ y , δ y ), then for each x ∈ supp( P ), x ≥ y and P | x % F OSD δ y and there exists x ′ ∈ supp( P ) with x ′ > y or P | x ′ ≻ F OSD δ y .69hen by regularity of w and v , we know V BIB ( P ) = w ( x ′ , CE v ( P | x ′ )) P ( x ′ ) + X x = x ′ w ( x, CE v ( P | x )) P ( x ) > w ( y , y ) P ( x ′ ) + X x = x ′ w ( y , y ) P ( x )= w ( y , y ) = V BIB ( δ y , δ y ) . This implies P ≻ ( δ y , δ y ). Similarly, we can show that P ≺ ( δ x , δ x ) if P is dominated by( δ y , δ y ). This completes the proof for necessity of axioms. i ) ⇒ ii ). The proof of sufficiency is decomposed in the following steps. In Step 1 , we restrictour attention to the set of product lotteries ˆ P and apply Theorem 1. Step 2 studies theimplications of Axiom CC and Axiom M. In
Step 3 , we derive a KP-style representation onthe space of lotteries P . In Step 4 , we utilize the consistency of the two representations in
Step 1 and
Step 3 on ˆ P to finish the proof. Step 1: We restrict % to ˆ P . For each preference % that satisfies the axioms statedin Theorem 2, we can define ˆ % which satisfies Axiom CN and agrees with % on the set ofproduct lotteries ˆ P . Easy to verify that ˆ % also satisfies Axiom WO, Axiom M, Axiom WCand Axiom WI. By Theorem 1, we know that ˆ % admits one of the following representations:EU-CN, GBIB-CN and GFIB-CN. This implies that the restriction of % on ˆ P admits one ofthese representations. Furthermore, if Axiom CN holds, then % admits one of the followingrepresentations: EU-CN, GBIB-CN and GFIB-CN. Step 2: We derive implications of Axiom CC and Axiom M.
Suppose, from now on, that Axiom CN does not hold, that is, there exists P ≻ ˜ P with P i = ˜ P i for i = 1 ,
2. For any ( p, q ) ∈ L ( X ) × L ( X ), denote M ( p, q ) as the set of lotterieswhose marginal lotteries are p and q respectively. For any P, R ∈ P , we say P and R are compatible , or P is compatible with R if supp( P ) ∩ supp( R ) = ∅ . Easy to see that if R is compatible with both P and Q , then R is also compatible with λP + (1 − λ ) Q for any λ ∈ (0 , P is compatible with Q , then P is compatible with all Q ′ ∈ M ( Q , Q ).One main difficulty is that betweenness does not hold, that is, for λ ∈ (0 , P ≻ λP + (1 − λ ) ˜ P ≻ ˜ P . However, we have the following weaker and localversion of the betweenness property. 70 emma 22. For any Q ≻ Q ′ , there exists Q ∗ = λ ∗ Q + (1 − λ ∗ ) Q ′ for some λ ∗ ∈ [0 , suchthat for any ǫ > , we can find λ ǫ ∈ ( λ ∗ − ǫ, λ ∗ + ǫ ) ∩ [0 , with Q ∗ λ ǫ Q + (1 − λ ǫ ) Q ′ .Proof of Lemma 22. Suppose the result fails. Then for any λ ∈ [0 , ǫ λ > λ ′ ∈ ( λ − ǫ λ , λ + ǫ λ ) ∩ [0 , λQ + (1 − λ ) Q ′ ∼ λ ′ Q + (1 − λ ′ ) Q ′ . Notice that { ( λ − ǫ λ , λ + ǫ λ ) } λ ∈ [0 , forms an open cover of the compact set [0 , , % , we know that λQ + (1 − λ ) Q ′ ∼ λ ′ Q + (1 − λ ′ ) Q ′ forall λ, λ ′ ∈ [0 , Q ∼ Q ′ and a contradiction.For P ≻ ˜ P with ˜ P ∈ M ( P , P ), denote P ∗ = λ ∗ P + (1 − λ ∗ ) ˜ P as the lottery foundin Lemma 22. Clearly, either P ∗ P or P ∗ ˜ P . Also, P , P are not degenerate. ByLemma 22, for any n >
0, there exists λ n ∈ ( λ ∗ − /n, λ ∗ + 1 /n ) ∩ [0 ,
1] with P ∗ λ n P +(1 − λ n ) ˜ P := P n . By completeness, for each n , either P n ≻ P ∗ or P ∗ ≻ P n . Then we canfind a subsequence of { P n } (still denoted as { P n } when there is no confusion) such thateither P n ≻ P ∗ for all n or P ∗ ≻ P n for all n . Suppose that the former case holds. Take any R ∼ S and R, S compatible with P . Axiom CC implies that for all α ∈ (0 ,
1) and n ≥ αP n + (1 − α ) R ≻ αP ∗ + (1 − α ) S . By mixture continuity of % (the second part of AxiomWC), as n goes to infinity, that is, λ n goes to λ ∗ , we have αP ∗ + (1 − α ) R % αP ∗ + (1 − α ) S .This holds for all R ∼ S with R, S compatible with P . By symmetry, we can just changethe place of R and S , and get αP ∗ + (1 − α ) S % αP ∗ + (1 − α ) R . Thus, for all α ∈ (0 , R ∼ S with R, S compatible with P , αP ∗ + (1 − α ) S ∼ αP ∗ + (1 − α ) R. If instead P ∗ ≻ P n for all n , then the same result holds as the conclusion is an indifferencerelation. Without loss of generality, we assume that P n ≻ P ∗ for all n and P ∗ ≻ ˜ P fromnow on.Fix any Q compatible with P and we know Q is also compatible with ˜ P , P ∗ and P n for each n . By Axiom CC, for any β ∈ (0 , βP ∗ + (1 − β ) Q ≻ β ˜ P + (1 − β ) Q and βP ∗ + (1 − β ) Q, β ˜ P + (1 − β ) Q ∈ M ( βP + (1 − β ) Q , βP + (1 − β ) Q ). Similarly, as P n ≻ P ∗ for all n , for any β ∈ (0 , βP n + (1 − β ) Q ≻ βP ∗ + (1 − β ) Q and βP n + (1 − β ) Q ∈ M ( βP + (1 − β ) Q , βP + (1 − β ) Q ). For any R ∼ S with R, S compatible with both
P, Q ,we know
R, S are also compatible with βP n + (1 − β ) Q and βP ∗ + (1 − β ) Q . With the same71rguments as above, we can show that for any α ∈ (0 ,
1) and β ∈ (0 , α [ βP ∗ + (1 − β ) Q ] + (1 − α ) R ∼ α [ βP ∗ + (1 − β ) Q ] + (1 − α ) S. This can be rearranged as β [ αP ∗ + (1 − α ) R ] + (1 − β )[ αQ + (1 − α ) R ] ∼ β [ αP ∗ + (1 − α ) S ] + (1 − β )[ αQ + (1 − α ) S ]Again by mixture continuity of % , let β → + and we have αQ + (1 − α ) R ∼ αQ + (1 − α ) S, (4)for any α ∈ (0 , R ∼ S , Q compatible with P, R, S and P compatible with Q, R, S .Fix P, ˜ P and Q such that P is compatible with Q , we want to strengthen property (4) bydiscarding the constraint that R, S are compatible with P . By the third part of Axiom WC,as P ≻ ˜ P , we can find ¯ ǫ > ǫ ∈ (0 , ¯ ǫ ) and P ǫ = P ∗ ( δ ǫ , δ ), ˜ P ǫ = ˜ P ∗ ( δ ǫ , δ ),we have P ǫ ≻ ˜ P ǫ . Note hat ˜ P ǫ , P ǫ ∈ M ( P ∗ δ ǫ , P ). Since supp( P ) ∪ supp( Q ) is finite, wecan make ¯ ǫ small enough such that for all ǫ ∈ (0 , ¯ ǫ ), supp( P ∗ δ ǫ ) ∩ supp( Q ) = ∅ and ˜ P ǫ , P ǫ are compatible with Q . Then any Q compatible with P , αQ + (1 − α ) R ∼ αQ + (1 − α ) S, for any ǫ ∈ (0 , ¯ ǫ ), α ∈ (0 , R ∼ S , Q compatible with R, S and P ǫ compatible with R, S .Now we show that by varying ǫ , we can further get rid of the constraint that R, S arecompatible with P ǫ for some ǫ ∈ (0 , ǫ ). This is again guaranteed by the fact that each lotteryin P has a finite support. Concretely, for any R ∼ S with R, S compatible with Q , we canalways find ǫ ∗ such that R, S are compatible with P ǫ ∗ . Thus, αQ + (1 − α ) R ∼ αQ + (1 − α ) S, for any α ∈ (0 , R ∼ S and Q compatible with R, S, P .The same argument can be applied to relax the requirement that Q is compatible with P and hence we end up with the result that for any Q ∈ P , αQ + (1 − α ) R ∼ αQ + (1 − α ) S, (5)for any α ∈ (0 , R ∼ S , Q compatible with R, S .72or each y ∈ X , recall thatΓ ,δ y = [ x ∈ X Γ ,δ y ( x ) = [ x ∈ X [ P,Q ∈ Π ( x ) ×{ δ y } ,P % Q [ Q, P ] ⊆ ˆ P . We define Φ ,δ y ⊆ P such that for each y ∈ X ,Φ ,δ y = n P ∈ P : ∃ T, T ′ ∈ Γ ,δ y s.t. T ≻ P ≻ T ′ o ; Lemma 23. (i). For each
P, Q, R ∈ P with P ≻ Q ≻ R , there exists λ ∈ (0 , such that λP + (1 − λ ) R ∼ Q .(ii). For each P ∈ P , there exists ( x , x ) ∈ X × X such that P ∼ ( δ x , δ x ) . Moreover, if P ∈ Φ ,δ y for some y ∈ X , then we can choose x = y .Proof of Lemma 23. (i). Denote A = { α ∈ (0 ,
1) : αP + (1 − α ) R ≻ Q } and λ = inf A .We claim that λP + (1 − λ ) R ∼ Q . Suppose by contradiction that λP + (1 − λ ) R Q . If λP + (1 − λ ) R ≻ Q , then λ ∈ A , which is open by mixture continuity of % . Hence thereexists λ ′ < λ with λ ′ ∈ A , which contradicts with the definition of λ . If λP + (1 − λ ) R ≺ Q ,then λ ∈ { α ∈ (0 ,
1) : αP + (1 − α ) R ≺ Q } , which is also open. We can find ǫ > λ, λ + ǫ ) ⊆ (0 , \ A . Again a contradiction with the definition of λ . Hence λP +(1 − λ ) R ∼ Q .(ii). For each P ∈ P , denote x i = max supp( P i ) , y i = min supp( P i ) for i = 1 ,
2. By AxiomM, ( δ x , δ x ) % P % ( δ y , δ y ). Then either ( δ x , δ x ) % P % ( δ x , δ y ) or ( δ x , δ y ) % P % ( δ y , δ y ). By symmetry, suppose the former case holds. Using the same argument as theproof of part (i) in Lemma 3, we can find λ ∈ [0 ,
1] such that P ∼ ( δ x , λδ y + (1 − λ ) δ y ).By Lemma 2, there exists x ′ ∈ X where P ∼ ( δ x , λδ y + (1 − λ ) δ y ) ∼ ( δ x , δ x ′ ).If further P ∈ Φ ,δ y for some y ∈ X , then we can find p , p ′ ∈ L ( X ) with ( p , δ y ) ≻ P ≻ ( p ′ , δ y ). By the same argument, we can find x ′ ∈ X such that P ∼ ( δ x ′ , δ y ).The next lemma generalize Axiom CC on each Φ ,δ y by relaxing the requirement that P and Q must agree on the marginal lotteries. Lemma 24.
Suppose that Axiom CN fails. For each y ∈ X and P, Q, R, S ∈ Φ ,δ y , thefollowing properties hold:i). P ∼ Q and P is compatible with Q = ⇒ αP + (1 − α ) Q ∼ P ∼ Q for all α ∈ (0 , ;ii). P ≻ Q and P is compatible with Q = ⇒ P ≻ αP + (1 − α ) Q ≻ Q for all α ∈ (0 , ; ii). P ≻ Q , R ∼ S , P is compatible with R and Q is compatible with S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S for all α ∈ (0 , ;iv). P ∼ Q , R ∼ S , P is compatible with R and Q is compatible with S = ⇒ αP + (1 − α ) R ∼ αQ + (1 − α ) S for all α ∈ (0 , .Proof of Lemma 24. We first prove (i) and (ii). Suppose
P, Q ∈ Φ ,δ y for some y ∈ X and P, Q are compatible. By Lemma 23, there exists x P , x Q ∈ X o such that P ∼ ( δ x P , δ y ) and Q ∼ ( δ x Q , δ y ). By Lemma 2, we can find ǫ > z P ∈ [ x P − ǫ, x P ] , z Q ∈ [ x Q − ǫ, x Q ], there exist z ′ P ≥ x P , z ′ Q ≥ x Q such that P ∼ (1 / δ z P + 1 / δ z ′ P , δ y ) and Q ∼ (1 / δ z Q + 1 / δ z ′ Q , δ y ). Moreover, as z P , z Q increases, z ′ P , z ′ Q will be decreasing continuously.Since P, Q are simple, that is, supp( P ) ∪ sup( Q ) is finite, we can construct z ∗ P = z ∗ Q , z ∗ ′ P = z ∗ ′ Q and z ∗ P , z ∗ Q , z ∗ ′ P , z ∗ ′ Q supp( P ) ∪ sup( Q ). Denote P ′ = (1 / δ z ∗ P + 1 / δ z ∗′ P , δ y ), Q ′ = (1 / δ z ∗ Q + 1 / δ z ∗′ Q , δ y ). Then P ∼ P ′ , Q ∼ Q ′ and P, Q, P ′ , Q ′ are compatible with eachother. Apply indifference relation (5) twice and we get for any α ∈ (0 , αP + (1 − α ) Q ∼ αP + (1 − α ) Q ′ ∼ αP ′ + (1 − α ) Q ′ . Again by Lemma 2 given marginal lottery in source 2 as δ y , we know P ∼ Q = ⇒ P ′ ∼ Q ′ = ⇒ αP + (1 − α ) Q ∼ αP ′ + (1 − α ) Q ′ ∼ Q ′ ∼ Q,P ≻ Q = ⇒ P ′ ≻ Q ′ = ⇒ αP + (1 − α ) Q ≻ αP ′ + (1 − α ) Q ′ ≻ Q ′ ∼ Q. Then we show (iii) and (iii) in a similar way. For
P, Q, R, S ∈ Φ ,δ y , we can construct P ′ ∼ P, Q ′ ∼ Q, R ′ ∼ R and S ′ ∼ S such that P ′ , Q ′ , R ′ , S ′ ∈ ˆ P , P ′ = Q ′ = R ′ = S ′ = δ y , P, R, P ′ , R ′ are compatible with each other and Q, S, Q ′ , S ′ are compatible with each other.Then for any α ∈ (0 , αP + (1 − α ) R ∼ αP + (1 − α ) R ′ ∼ αP ′ + (1 − α ) R ′ ,αQ + (1 − α ) S ∼ αQ + (1 − α ) S ′ ∼ αQ ′ + (1 − α ) S ′ . By Lemma 2 given marginal lottery in source 2 as δ y , we know P ∼ Q, R ∼ S = ⇒ αP + (1 − α ) R ∼ αP ′ + (1 − α ) R ′ ∼ αQ ′ + (1 − α ) S ′ ∼ αQ + (1 − α ) S,P ≻ Q, R ∼ S = ⇒ αP + (1 − α ) R ∼ αP ′ + (1 − α ) R ′ ≻ αQ ′ + (1 − α ) S ′ ∼ αQ + (1 − α ) S. emma 25. Suppose that Axiom CN fails. For each y ∈ X and P, Q, R, S ∈ ∪ y ∈ X Φ ,δ y ,the following properties hold:i). P ∼ Q and P is compatible with Q = ⇒ αP + (1 − α ) Q ∼ P ∼ Q for all α ∈ (0 , ;ii). P ≻ Q and P is compatible with Q = ⇒ P ≻ αP + (1 − α ) Q ≻ Q for all α ∈ (0 , ;iii). P ≻ Q , R ∼ S , P is compatible with R and Q is compatible with S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S for all α ∈ (0 , ;iv). P ∼ Q , R ∼ S , P is compatible with R and Q is compatible with S = ⇒ αP + (1 − α ) R ∼ αQ + (1 − α ) S for all α ∈ (0 , .Proof of Lemma 25. First, for any
P, Q, R, S ∈ ∪ y ∈ X Φ ,δ y , we claim that there exist finitelymany z k ∈ X , k = 1 , ..., K such that z < z < ... < z K and P, Q, R, S ∈ ∪ Kk =1 Φ ,δ zk .Choose P , P ∈ { P, Q, R, S } such that P % P, Q, R, S % P . Suppose that P ∈ Φ ,δ y and P ∈ Φ ,δ y with y ≥ y . If y = y , then P, Q, R, S ∈ Φ ,δ y and we are done.Now suppose that y > y and by Lemma 2, we can find t, t ′ ∈ L ( X ) with ( t, δ y ) ≻ P ≻ ( t ′ , δ y ) , ( t, δ y ) ≻ P ≻ ( t ′ , δ y ). Notice that for each y ∈ [ y , y ], H ( y ) := { P ∈ P : ( t, δ y ) ≻ P ≻ ( t ′ , δ y ) } ⊆ Φ ,δ y . By Axiom WC, { P ∈ P : P % P % P } ⊆ ∪ y ≤ y ≤ y H ( y ) and for all y ∈ [ y , y ], there exists ǫ y > H ( y ) ∩ H ( y ′ ) = ∅ for all y ′ ∈ [ y − ǫ y , y + ǫ y ] ∩ [ y , y ].By Finite Cover Theorem, we can find finitely many z < z < ... < z K ∈ [ y , y ] with[ y , y ] ⊆ ∪ Kk =1 [ z k − ǫ z k , z k + ǫ z k ]. This implies P, Q, R, S ∈ { P ∈ P : P % P % P } ⊆ ∪ y ≤ y ≤ y H ( y ) = ∪ Kk =1 H ( z k ) ⊆ ∪ Kk =1 Φ ,δ zk . Then we use induction to show that the four properties stated in the lemma holdfor
P, Q, R, S ∈ ∪ Kk =1 Φ ,δ zk . The proof idea is similar to the proof of Lemma 9. ByLemma 24, the four properties hold if P, Q, R, S ∈ Φ ,δ z . Suppose by induction thatthey also hold if P, Q, R, S ∈ ∪ tk =1 Φ ,δ zk for some 1 ≤ t < K . By our constructionof { z k } , Φ ,δ zt ∩ Φ ,δ zt +1 has nonempty interior. Choose T , T ∈ Φ ,δ zt ∩ Φ ,δ zt +1 with T ≻ T . By Lemma 23 and Lemma 2, as P , Q , R , S have finite supports, we can find p , p , q , q ∈ L ( X ) such that ( p , δ z t +1 ) ∼ ( q , δ z t ) ∼ T , ( p , δ z t +1 ) ∼ ( q , δ z t ) ∼ T and( p , δ z t +1 ) , ( q , δ z t ) , ( p , δ z t +1 ) , ( q , δ z t ) are compatible with P, Q, R, S .For properties (i) and (ii), suppose P % Q , P is compatible with Q and P, Q ∈ ∪ t +1 k =1 Φ ,δ zk .75f P ∼ Q , then P, Q ∈ Φ ,δ zk for some k = 1 , ..., t + 1 and hence property (i) holds by theinductive hypothesis.If P ≻ Q , then it suffices to consider the case where P ∈ Φ ,δ zt +1 \ ( ∪ tk =1 Φ ,δ zk ) and Q ∈ ( ∪ tk =1 Φ ,δ zk ) \ Φ ,δ zt +1 . This implies P ≻ T ≻ T ≻ Q . By Lemma 23, there exist λ = λ ∈ (0 ,
1) such that T ∼ λ P + (1 − λ ) Q and T ∼ λ P + (1 − λ ) Q . Then property(ii) holds for λ = λ , λ .Notice that at the moment we cannot conclude that λ > λ . Suppose that λ i > λ − i for some i = 1 ,
2. By Lemma 23, we can find P ′ , Q ′ ∈ ˆ P with Q ′ ∼ Q, P ′ ∼ P and P, P ′ , Q, Q ′ , ( p , δ z t +1 ) , ( q , δ z t ) , ( p , δ z t +1 ) , ( q , δ z t ) compatible with each other. Thisguarantees T ∼ λ P + (1 − λ ) Q ∼ λ P ′ + (1 − λ ) Q ∼ λ P + (1 − λ ) Q ′ ∼ λ P ′ + (1 − λ ) Q ′ ,T ∼ λ P + (1 − λ ) Q ∼ λ P ′ + (1 − λ ) Q ∼ λ P + (1 − λ ) Q ′ ∼ λ P ′ + (1 − λ ) Q ′ . By property (i), for all β, β ′ ∈ (0 , βP + (1 − β ) P ′ ∼ P, β ′ Q + (1 − β ′ ) Q ′ ∼ Q . Applyindifference relation (5) twice and we have for each λ, β, β ′ ∈ (0 , λP + (1 − λ ) Q ∼ λ ( βP + (1 − β ) P ′ ) + (1 − λ )( β ′ Q + (1 − β ′ ) Q ′ ) . (6)For any λ ∈ ( λ − i , λ i ), let β = 1, β ′ = λ i − λλ i (1 − λ ) , and (6) becomes λP + (1 − λ ) Q ∼ λλ i ( λ i P + (1 − λ i ) Q ′ ) + (1 − λλ i ) Q ∼ λλ i ( q i , δ z t ) + (1 − λλ i ) Q The second indifference comes from the fact that λ i P + (1 − λ i ) Q ′ ∼ T i ∼ ( q i , δ z t ) and (5).Then by the inductive hypothesis on ∪ tk =1 Φ ,δ zk , we have P ≻ ( q i , δ z t ) ≻ λP + (1 − λ ) Q ∼ λλ i ( q i , δ z t ) + (1 − λλ i ) Q ≻ Q. If λ > λ i , then let β = λ − λ i λ (1 − λ i ) , β ′ = 0 and (6) becomes λP + (1 − λ ) Q ∼ λ − λ i − λ i P + (1 − λ − λ i − λ i )( λ i P ′ + (1 − λ i ) Q ) ∼ λ − λ i − λ i P + (1 − λ − λ i − λ i )( p i , δ z t +1 )76he second indifference comes from the fact that λ i P ′ + (1 − λ i ) Q ∼ T i ∼ ( p i , δ z t +1 ) and (5).Then by Lemma 24 on Φ ,δ zt +1 , we have P ≻ λP + (1 − λ ) Q ∼ λ − λ i − λ i P + (1 − λ − λ i − λ i )( p i , δ z t +1 ) ≻ ( p i , δ z t +1 ) ≻ Q. A symmetric proof applies for the case where λ < λ − i . This completes the proof for property(ii) on ∪ t +1 k =1 Φ ,δ zk .Now consider P, Q, R, S ∈ ∪ t +1 k =1 Φ ,δ zk where P is compatible with R and Q is compatiblewith S . We first suppose P ∼ Q, R ∼ S and prove property (iv). If P ∼ R , thenthe result is trivial by property (i). Without loss of generality, we assume P ≻ R .By the inductive hypothesis, if suffices to prove the case for P, Q ∈ Φ ,δ zt +1 \ ( ∪ tk =1 Φ ,δ zk )and R, S ∈ ( ∪ tk =1 Φ ,δ zk ) \ Φ ,δ zt +1 . Following the proof for property (ii), we construct T , T , ( p , δ z t +1 ) , ( q , δ z t ) , ( p , δ z t +1 ) , ( q , δ z t ) , P ′ , Q ′ , R ′ , S ′ . Concretely, these lotteries aremutually compatible and each of them is compatible with P, Q, R, S such that P ∼ Q ∼ P ′ ∼ Q ′ , R ∼ S ∼ R ′ ∼ S ′ ; T ∼ ( p , δ t t +1 ) ∼ ( q , δ z t ) , T ∼ ( p , δ t t +1 ) ∼ ( q , δ z t ) . By the inductive hypothesis, we can find λ, λ ′ ∈ (0 ,
1) such that λP + (1 − λ )( p , δ z t +1 ) ∼ T ∼ λQ + (1 − λ )( q , δ z t ) ,λ ′ ( p , δ z t +1 ) + (1 − λ ′ ) R ∼ T ∼ λ ′ ( q , δ z t ) + (1 − λ ′ ) S. By Lemma 23, there exist η, η ′ ∈ (0 ,
1) with ηP + (1 − η ) R ∼ T ∼ η ′ Q + (1 − η ′ ) S. We claim that we can choose η = η ′ . To see this, notice that ( p , δ z t +1 ) ∼ T ∼ λP + (1 − λ )( p , δ z t +1 ), all of which are compatible with R , we have λ ′ ( p , δ z t +1 ) + (1 − λ ′ ) R ∼ λλ ′ P + (1 − λ ) λ ′ ( p , δ z t +1 ) + (1 − λ ′ ) R ∼ ( p , δ z t +1 ) . Again, as ( p , δ z t +1 ) is compatible with both P and R , by property (i) and (ii), it must bethe case that T ∼ ( p , δ z t +1 ) ∼ λλ ′ λλ ′ + (1 − λ ′ ) P + 1 − λ ′ λλ ′ + (1 − λ ′ ) R. η = λλ ′ λλ ′ +(1 − λ ′ ) := η . Similarly we can show that η ′ = η guarantees T ∼ η ′ Q + (1 − η ) S .A symmetric argument shows that there exist η = λλ +(1 − λ )(1 − λ ′ ) ∈ ( η ,
1) with η P + (1 − η ) R ∼ T ∼ η Q + (1 − η ) S. Now we consider η with η < η < η . By (6), we can set β = ( η − η ) η ( η − η ) η , β ′ = ( η − η )(1 − η )( η − η )(1 − η ) and then ηP + (1 − η ) R ∼ η − η η − η [ η P + (1 − η ) R ] + η − ηη − η [ η P ′ + (1 − η ) R ′ ] ∼ η − η η − η ( p , δ z t +1 ) + η − ηη − η ( p , δ z t +1 ) . Similarly, ηQ + (1 − η ) S ∼ η − η η − η ( p , δ z t +1 ) + η − ηη − η ( p , δ z t +1 ) . Hence ηP + (1 − η ) R ∼ ηQ + (1 − η ) S for η ∈ ( η , η ).Suppose that η < η <
1. By (6), we can set β = ( η − η ) η ( η − η ) η , β ′ = ( η − η )(1 − η )( η − η )(1 − η ) and then( p , δ z t +1 ) ∼ η P + (1 − η ) R ∼ η − η η − η [ ηP + (1 − η ) R ] + η − η η − η [ η P ′ + (1 − η ) R ′ ] ∼ η − η η − η [ ηP + (1 − η ) R ] + η − η η − η ( p , δ z t +1 ) . Similarly,( p , δ z t +1 ) ∼ η Q + (1 − η ) S ∼ η − η η − η [ ηQ + (1 − η ) S ] + η − η η − η ( q , δ z t ) . We claim that ηP + (1 − η ) R ∈ ∪ t +1 k =1 Φ ,δ zk . To see this, note that we can construct P ′ , R ′ such that P ′ = δ z t +1 , R ′ = δ z k for some k ≤ t and ηP +(1 − η ) R ∼ ηP ′ +(1 − η ) R ′ . By AxiomM and definition of Φ ,δ zk , there exist x, x ′ ∈ X such that ( δ x , δ z t +1 ) ≻ ηP + (1 − η ) R ∼ ηP ′ + (1 − η ) R ′ ≻ ( δ x ′ , δ z k ). This implies ηP + (1 − η ) R ∈ ∪ t +1 k =1 Φ ,δ zk . Similarly we know ηQ + (1 − η ) S ∈ ∪ t +1 k =1 Φ ,δ zk .If ηP + (1 − η ) R ≻ T ≻ ηQ + (1 − η ) S or ηQ + (1 − η ) S ≻ T ≻ ηP + (1 − η ) R ,then ( p , δ z t +1 ) ≻ T ≻ ( p , δ z t +1 ), a contradiction. Hence either ηP + (1 − η ) R, ηQ + (1 − η ) S ∈ Φ ,δ zt +1 or ηP + (1 − η ) R, ηQ + (1 − η ) S ∈ ∪ tk =1 Φ ,δ zk . By the inductive hypothesis,78s ( p , δ z t +1 ) ∼ ( q , δ z t ) ∈ Φ ,δ zt ∩ Φ ,δ zt +1 , independence properties (iii) and (iv) hold for( ηP + (1 − η ) R, ηQ + (1 − η ) S, ( p , δ z t +1 ) , ( q , δ z t )). Thus we must have ηP + (1 − η ) R ∼ ηQ + (1 − η ) S .The proof for the case with η ∈ (0 , η ) is symmetric. Therefore for all η ∈ (0 , ηP +(1 − η ) R ∼ ηQ + (1 − η ) S and property (iv) holds on ∪ t +1 k =1 Φ ,δ zk .Before proving property (iii), we claim that for each P, Q ∈ ∪ t +1 k =1 Φ ,δ zk with P ≻ Q , P compatible with Q and 1 > λ > λ >
0, we have λ P + (1 − λ ) Q ≻ λ P + (1 − λ ) Q . Tosee this, by (6), we can find P ′ ∼ P where P ′ is compatible with both P and Q such that λ P + (1 − λ ) Q ∼ λ − λ − λ P ′ + 1 − λ − λ [ λ P + (1 − λ ) Q ] ≻ λ P + (1 − λ ) Q The second strict preference holds since P ∼ P ′ ≻ λ P + (1 − λ ) Q by property (ii).For property (iii), suppose P ≻ Q , R ∼ S . If P % R % Q , then by properties (i) and(ii), λP + (1 − λ ) R % R ∼ S % λQ + (1 − λ ) S . As P ≻ Q , at least one of the above weakpreference rankings should be strict and we are done. Then either P ≻ Q ≻ R ∼ S or R ∼ S ≻ P ≻ Q . We start with the former case.By Lemma 23, we can find α ∈ (0 ,
1) and P ′ , R ′ where P ′ , R ′ are compatible and both ofthem are compatible with P, Q, R, S such that R ∼ R ′ and αP + (1 − α ) R ′ ∼ P ′ ∼ Q. Then by property (iv), for any λ ∈ (0 , λQ + (1 − λ ) S ∼ λP ′ + (1 − λ ) R ∼ λαP + (1 − λ ) R + (1 − α ) λR ′ . Since R ∼ R ′ , property (i) implies that − λ − λα R + (1 − α ) λ − λα R ′ ∼ R . By indifference relation (5), λQ + (1 − λ ) S ∼ λαP + (1 − λα ) R ≺ λP + (1 − λ ) R by the previous claim and α ∈ (0 , R ∼ S ≻ P ≻ Q . Thiscompletes the proof for property (iii) on ∪ t +1 k =1 Φ ,δ zk .By induction, the four properties hold for P, Q, R, S ∈ ∪ Kk =1 Φ ,δ zk and hence arbitrary P, Q, R, S ∈ ∪ y ∈ X Φ ,δ y .It is worthwhile to notice that ∪ y ∈ X Φ ,δ y might not be the same as P . The next lemmas79hows that they only possibly differ in the worst and the best possible lottery. Concretely, if c , c < + ∞ , then ( c , c ) ∈ P\ ( ∪ y ∈ X Φ ,δ y ); if c , c > −∞ , then ( c , c ) ∈ P\ ( ∪ y ∈ X Φ ,δ y ). Lemma 26.
Suppose that Axiom CN fails. P\ ( ∪ y ∈ X Φ ,δ y ) = { ( c , c ) , ( c , c ) } ∩ R . Proof of Lemma 26.
We will focus on the case with c , c > −∞ and c , c < + ∞ . Theproof for the other case is simpler as it only involves the worst or the best possible lottery.First, for each P ∈ P with ( c , c ) ≻ P ≻ ( c , c ), there exists Q, Q ′ ∈ ∪ y ∈ X Φ ,δ y with Q ≻ P ≻ Q ′ , which implies Q ∈ ∪ y ∈ X Φ ,δ y . Hence P = ( ∪ y ∈ X Φ ,δ y ) ∪ { P ∈ P : P ∼ ( c , c ) or P ∼ ( c , c ) } . It suffices to show that P ∼ ( c , c ) if and only if P = ( c , c ), P ∼ ( c , c ) if and only if P = ( c , c ). This is trivial by Axiom M as for any P = ( c , c ) , ( c , c ), P dominates ( c , c )and is dominated by ( c , c ).Using the same proof as in Lemma 24, we can easily show that the independence propertyholds for P, Q, R, S ∈ Φ ,δ ∪ { ( c , c ) } if c , c > −∞ or P, Q, R, S ∈ Φ ,δ c ∪ { ( c , c ) } if c , c < + ∞ . Then a direct corollary of Lemma 26 follows. Corollary 3.
Suppose that Axiom CN fails. Then the following properties hold:i). P ∼ Q and P is compatible with Q = ⇒ αP + (1 − α ) Q ∼ P ∼ Q for all α ∈ (0 , ;ii). P ≻ Q and P is compatible with Q = ⇒ P ≻ αP + (1 − α ) Q ≻ Q for all α ∈ (0 , ;iii). P ≻ Q , R ∼ S , P is compatible with R and Q is compatible with S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S for all α ∈ (0 , ;iv). P ∼ Q , R ∼ S , P is compatible with R and Q is compatible with S = ⇒ αP + (1 − α ) R ∼ αQ + (1 − α ) S for all α ∈ (0 , . We end this section by slightly relaxing the requirement of compatibility. For each
P, Q ∈P , we say P and Q are weakly compatible if the following properties hold:• supp( P ) ∩ supp( Q ) ⊆ { c , c } ;• when c ∈ supp( P ) ∩ supp( Q ), we have P | c = Q | c = δ c ;• when c ∈ supp( P ) ∩ supp( Q ), we have P | c = Q | c = δ c .80n other words, for P weakly compatible with Q , we allow outcome c or c to be containedin the overlapping supports of P and Q only if the conditional lotteries of P and Q givenoutcome x are both δ c or δ c . Lemma 27.
Suppose that Axiom CN fails. Then the following properties hold:i). P ∼ Q and P is weakly compatible with Q = ⇒ αP + (1 − α ) Q ∼ P ∼ Q for all α ∈ (0 , ;ii). P ≻ Q and P is weakly compatible with Q = ⇒ P ≻ αP + (1 − α ) Q ≻ Q for all α ∈ (0 , ;iii). P ≻ Q , R ∼ S , P is weakly compatible with R and Q is weakly compatible with S = ⇒ αP + (1 − α ) R ≻ αQ + (1 − α ) S for all α ∈ (0 , ;iv). P ∼ Q , R ∼ S , P is weakly compatible with R and Q is weakly compatible with S = ⇒ αP + (1 − α ) R ∼ αQ + (1 − α ) S for all α ∈ (0 , .Proof of Lemma 27. Suppose
P, Q are weakly compatible but not compatible, that is, ∅ 6 =supp( P ) ∩ supp( Q ) ⊆ { c , c } . We claim that we can find ˜ P ∼ P and ˜ Q ∼ Q such that˜ P is compatible with ˜ Q and for any α ∈ (0 , αP + (1 − α ) Q ∼ α ˜ P + (1 − α ) ˜ Q , unless P = Q = ( c , c ) or P = Q = ( c , c ). Case 1 : If supp( P ) ∩ supp( Q ) = { c , c } , then P | c = Q | c = δ c and P | c = Q | c = δ c . First, we suppose that P ( c ) + P ( c ) < Q ( c ) + Q ( c ) <
1. By symmetry, itsuffices to focus on the former case. Denote P o = P x = c ,c P ( x )1 − P ( c ) − P ( c ) ( δ x , P | x ). Then P = P ( c )( δ c , δ c ) + P ( c )( δ c , δ c ) + (1 − P ( c ) − P ( c )) P o . We know P o and Q are compatible and c , c supp( P o ). We can similarly define Q o if Q ( c ) + Q ( c ) <
1. Otherwise, just choose an arbitrary Q o so long as c , c supp( Q o ).By Axiom M, ( c , c ) ≻ P o ≻ ( c , c ). Then we can find ǫ P > c − ǫ P , ǫ P supp( P ) ∪ supp( Q ), ǫ P = c − ǫ P and( c − ǫ P , c ) ≻ P o ≻ ( c + ǫ P , c ) . By Lemma 23, we can find λ P ∈ (0 ,
1) such that P o ∼ λ P ( δ c − ǫ P , δ c ) + (1 − λ P )( δ c + ǫ P , δ c ) := P o ′ . P o ′ is compatible with P, Q .By Corollary 3, we know P ′ := P ( c )( δ c , δ c ) + P ( c )( δ c , δ c ) + (1 − P ( c ) − P ( c )) P o ′ ∼ P. Notice that P ′ =[ P ( c )( δ c , δ c ) + (1 − P ( c ) − P ( c ))(1 − λ P )( δ c + ǫ P , δ c )]+ [ P ( c )( δ c , δ c ) + (1 − P ( c ) − P ( c )) λ P ( δ c − ǫ P , δ c )] . By Lemma 2 given δ c or δ c in source two, we can then find p, p ∈ L ( X ) with( p, δ c ) , ( p, δ c ) , P, P ′ , Q, Q ′ are pairwise compatible and( p, δ c ) ∼ P ( c )( δ c , δ c ) + λ P (1 − P ( c ) − P ( c ))( δ c − ǫ P , δ c ) P ( c ) + λ P (1 − P ( c ) − P ( c )) ;( p, δ c ) ∼ P ( c )( δ c , δ c ) + (1 − λ P )(1 − P ( c ) − P ( c ))( δ c + ǫ P , δ c ) P ( c ) + (1 − λ P )(1 − P ( c ) − P ( c )) . It is important to notice that p = δ c and p = δ c .Again by Corollary 3, we have˜ P = λ ∗ P ( p, δ c ) + (1 − λ ∗ P )( p, δ c ) ∼ P ′ ∼ P, where λ ∗ P = P ( c ) + (1 − λ P )(1 − P ( c ) − P ( c )). Easy to see that ˜ P is compatible with P and Q .We then want to show that for each α ∈ (0 , αP + (1 − α ) Q ∼ α ˜ P + (1 − α ) Q . To seethis, notice that for each α ∈ (0 , αP + (1 − α ) Q = ( αP ( c ) + (1 − α ) Q ( c ))( δ c , δ c ) + ( αP ( c ) + (1 − α ) Q ( c ))( δ c , δ c )+ α (1 − P ( c ) − P ( c )) P o + (1 − α )(1 − Q ( c ) − Q ( c )) Q o . Recall that P o ∼ P o ′ , P o is compatible with Q and P . Then Corollary 3 implies that αP + (1 − α ) Q ∼ ( αP ( c ) + (1 − α ) Q ( c ))( δ c , δ c ) + ( αP ( c ) + (1 − α ) Q ( c ))( δ c , δ c )+ α (1 − P ( c ) − P ( c )) P o ′ + (1 − α )(1 − Q ( c ) − Q ( c )) Q o = αP ′ + (1 − α ) Q α [ P ( c )( δ c , δ c ) + (1 − P ( c ) − P ( c ))(1 − λ P )( δ c + ǫ P , δ c )]+ (1 − α ) Q ( c )( δ c , δ c )+ α [ P ( c )( δ c , δ c ) + (1 − P ( c ) − P ( c )) λ P ( δ c − ǫ P , δ c )]+ (1 − α ) Q ( c )( δ c , δ c )+ (1 − α )(1 − Q ( c ) − Q ( c )) Q o . Notice that the first two terms in the last equation have δ c in source two, while the thirdand four term have δ c in the source two. Apply Lemma 2 given δ c or δ c in source two andCorollary 3 sequentially, we know αP + (1 − α ) Q ∼ αλ ∗ P ( p, δ c ) + (1 − α ) Q ( c )( δ c , δ c )+ α (1 − λ ∗ P )( p, δ c ) + (1 − α ) Q ( c )( δ c , δ c )+ (1 − α )(1 − Q ( c ) − Q ( c )) Q o = α ˜ P + (1 − α ) Q. Now we suppose P ( c )+ P ( c ) = Q ( c )+ Q ( c ) = 1. As supp( P ) ∩ supp( Q ) = { , c } ,we have P ( c ) , Q ( c ) ∈ (0 , δ c , δ c ) ≻ P, Q ≻ ( δ c , δ c ). Then we canfind ˜ P ∼ P such that c , c supp( ˜ P ). This implies ˜ P is compatible with P and Q . For any β ∈ (0 , P ∼ βP + (1 − β ) P ′ := P β . Clearly, P β ( c ) + P β ( c ) <
1. We canapply the previous result for P β and Q , that is, for each β , we can find ˜ P β ∼ P β with ˜ P β compatible with Q such that for each α ∈ (0 , αP β + (1 − α ) Q ∼ α ˜ P β + (1 − α ) Q . Againby Corollary 3, we can actually choose ˜ P β to be the same across all β ∈ (0 , P . Hence, for each β, α ∈ (0 , αβP + α (1 − β ) P ′ + (1 − α ) Q ∼ α ˜ P + (1 − α ) Q By mixture continuity of ∼ , let β → α ∈ (0 , αP + (1 − α ) Q ∼ α ˜ P + (1 − α ) Q. Case 2 : If supp( P ) ∩ supp( Q ) = { c } , then P | c = Q | c = δ c . By our assumption,either P ( c ) < P ( c ) <
1. Without loss of generality, assume P ( c ) < P ( c ) < − P ( c ) or P | c = δ c , denote P o = P x = c P ( x )1 − P ( c ) ( δ x , P | x ). Then P = P ( c )( δ c , δ c ) + (1 − P ( c )) P o . We know P o and Q are compatible and 0 supp( P o ). We can similarly define Q o if Q ( c ) <
1. Otherwise, just choose an arbitrary Q o so long as c supp( Q o ).By Lemma 26, ( c , c ) ≻ P o ≻ ( c , c ). Then we can find ǫ P > c − ǫ P , ǫ P supp( P ) ∪ supp( Q ), ǫ P = c − ǫ P and( δ c − ǫ P , δ c ) ≻ P o ≻ ( δ c + ǫ P , δ c ) . By Lemma 23, we can find λ P ∈ (0 ,
1) such that P o ∼ λ P ( δ c − ǫ P , δ c ) + (1 − λ P )( δ c + ǫ P , δ c ) := P o ′ . and P o ′ is compatible with P, Q .By Corollary 3, we know P ′ := P ( c )( δ c , δ c ) + (1 − P ( c )) P o ′ ∼ P. Notice that P ′ =[ P ( c )( δ c , δ c ) + (1 − P ( c ))(1 − λ P )( δ c + ǫ P , δ c )]+ (1 − P ( c )) λ P ( δ c − ǫ P , δ c ) . By Lemma 2 given δ c or δ c in source two, we can then find p, p ∈ L ( X ) with( p, δ c ) , ( p, δ c ) , P, P ′ , Q, Q ′ are pairwise compatible and( p, δ c ) ∼ ( δ c − ǫ P , δ c );( p, δ c ) ∼ P ( c )( δ c , δ c ) + (1 − λ P )(1 − P ( c ))( δ c + ǫ P , δ c ) P ( c ) + (1 − λ P )(1 − P ( c )) . It is important to notice that p = δ c and p = δ c .Again by Corollary 3, we have˜ P = λ ∗ P ( p, δ c ) + (1 − λ ∗ P )( p, δ c ) ∼ P ′ ∼ P, λ ∗ P = P ( c ) + (1 − λ P )(1 − P ( c )). Easy to see that ˜ P is compatible with P and Q .We then want to show that for each α ∈ (0 , αP + (1 − α ) Q ∼ α ˜ P + (1 − α ) Q . To seethis, notice that for each α ∈ (0 , αP + (1 − α ) Q = ( αP ( c ) + (1 − α ) Q ( c ))( δ c , δ c )+ α (1 − P ( c )) P o + (1 − α )(1 − Q ( c )) Q o . Recall that P o ∼ P o ′ , P o is compatible with Q and P . Then Corollary 3 implies that αP + (1 − α ) Q ∼ ( αP ( c ) + (1 − α ) Q ( c ))( δ c , δ c )+ α (1 − P ( c )) P o ′ + (1 − α )(1 − Q ( c )) Q o = αP ′ + (1 − α ) Q = α [ P ( c )( δ c , δ c ) + (1 − P ( c ))(1 − λ P )( δ c + ǫ P , δ c )]+ (1 − α ) Q ( c )( δ c , δ c )+ α (1 − P ( c )) λ P ( δ c − ǫ P , δ c )+ (1 − α )(1 − Q ( c )) Q o . Apply Lemma 2 given δ c in source two and Corollary 3 sequentially, we know αP + (1 − α ) Q ∼ αλ ∗ P ( p, δ c ) + (1 − α ) Q ( c )( δ c , δ c )+ α (1 − λ ∗ P )( p, δ c )+ (1 − α )(1 − Q ( c )) Q o = α ˜ P + (1 − α ) Q. If P ( c ) = 1 − P ( c ) or P | c = δ c , then the result can be proved by the same continuityargument in Case 1. Case 3 : If supp( P ) ∩ supp( Q ) = { c } , then the proof is symmetric to the proof ofCase 2 and hence omitted.As an intermediate summary, for each P, Q weakly compatible, we can find ˜ P ∼ P and ˜ Q ∼ Q such that ˜ P is compatible with ˜ Q and for any α ∈ (0 , αP + (1 − α ) Q ∼ α ˜ P + (1 − α ) ˜ Q , unless P = Q = ( δ c , δ c ) or P = Q = ( δ c , δ c ).Now we are ready to prove the four properties.85or (i) and (ii), if P = Q = ( δ c , δ c ) or P = Q = ( δ c , δ c ), then the result is trivial.Otherwise, there exist ˜ P ∼ P and ˜ Q ∼ Q such that for any α ∈ (0 , P ∼ Q = ⇒ P ′ ∼ Q ′ = ⇒ αP + (1 − α ) Q ∼ αP ′ + (1 − α ) Q ′ ∼ P,P ≻ Q = ⇒ P ′ ≻ Q ′ = ⇒ αP + (1 − α ) Q ≻ αP ′ + (1 − α ) Q ′ ∼ P. For (iii) and (iv), if P = R = ( δ c , δ c ) or Q = S = ( δ c , δ c ) or P = R = ( δ c , δ c ) or Q = S = ( δ c , δ c ), then by Lemma 26, the primitives of (iii) or (iv) hold only if P = Q = R = S ,in which case the result holds trivially. By excluding those cases, we can construct ˜ P ∼ P ,˜ Q ∼ Q , ˜ R ∼ P and ˜ S ∼ S such that ˜ P is compatible with ˜ R , ˜ Q is compatible with ˜ S andfor any α ∈ (0 , αP + (1 − α ) R ∼ α ˜ P + (1 − α ) ˜ R , αQ + (1 − α ) S ∼ α ˜ Q + (1 − α ) ˜ S .By Corollary 3, we know P ∼ Q, R ∼ S = ⇒ αP + (1 − α ) R ∼ α ˜ P + (1 − α ) ˜ R ∼ α ˜ Q + (1 − α ) ˜ S ∼ αQ + (1 − α ) S,P ≻ Q, R ∼ S = ⇒ αP + (1 − α ) R ∼ α ˜ P + (1 − α ) ˜ R ≻ α ˜ Q + (1 − α ) ˜ S ∼ αQ + (1 − α ) S. This completes the proof.
Step 3: Then we show that % admits a KP-style representation. Recall that the general (history-dependent) KP representation in two periods is given by V KP as V KP ( d ) = X ( x,p ) w ( x, CE v x ( p )) d ( x, p )This next lemma introduces a KP-style representation in the space of lotteries P . Notice thatthe difference from the BIB model is that the conditional preference in source 2 is allowedto depend on the outcome in source 1. Lemma 28.
Suppose that Axiom CN fails. Then % on P admits the a representation U where for each P ∈ P , U ( P ) = X x ˆ u D ( x, CE v x ( P | x )) P ( x ) with regular ˆ u D and v x for all x ∈ X . Before proving Lemma 28, we introduce a mapping from the space of lotteries to the space86f temporal lotteries. Denote D ∗ := L ( X × L ( X )) as the set of temporal lotteries andˆ D ∗ as a subset of D ∗ such thatˆ D ∗ := n d ∈ D ∗ : d ( x, p ) d ( x, p ′ ) = 0 , ∀ x ∈ X , p = p ′ ∈ L ( X ) o . Notice that for i = 1 , X i with the standard topology is separable. By Kreps and Porteus(1978), we know that the P and D ∗ with weak topology can be metrizable by the Prokhorovmetric. Endow ˆ P with the relative topology with respect to the weak topology on P and ˆ D ∗ with the relative topology with respect to the weak topology on D ∗ .Define a mapping f : P → ˆ D ∗ as follows: for P ∈ P , denote f [ P ] = d ∈ D ∗ such thatfor any ( x, q ) ∈ X × L ( X ), f [ P ]( x, q ) = P ( x ) if q = P | x and f [ P ]( x, q ) = 0 if q = P | x .Clearly, for all q ′ = P | x , f [ p ]( x, q ) = 0. Hence f [ P ] ∈ ˆ D ∗ and f is well-defined. Inversely, f − : ˆ D ∗ → P such that f − [ d ]( x, y ) = P q ∈L ( X ) d ( x, q ) q ( y ). This is also well-defined as foreach x ∈ X there exists at most one q ∈ L ( X ) with d ( x, q ) > d ∈ ˆ D ∗ . Thus, f isa bijective mapping between P and ˆ D ∗ . It is worth noting that f is not a homeomorphismas f is not continuous, although f − is continuous.Now we define a binary relation % ′ on ˆ D ∗ by d % ′ d ′ if and only if f − [ d ] % f − ( d ′ ). ≻ ′ and ∼ ′ are defined correspondingly. We have the following corollary of Lemma 27. Corollary 4.
Suppose λ i > for all i and P ni =1 λ i = 1 . For d = P ni =1 λ i δ ( x i ,p i ) , d = P ni =1 λ i δ ( y i ,q i ) with x i = x j , y i = y j for all i = j and δ ( x i ,p i ) ∼ ′ δ ( y i ,q i ) for all i , then d ∼ ′ d .Proof of Corollary 4. By Lemma 27, as x i = x j , y i = y j for all i = j and δ ( x i ,p i ) ∼ ′ δ ( y i ,q i ) forall i , we have λ λ + λ ( δ x , p ) + λ λ + λ ( δ x , p ) ∼ λ λ + λ ( δ y , q ) + λ λ + λ ( δ y , q ) . Then by induction, we can get n X i =1 λ i ( δ x i , p i ) ∼ n X i =1 λ i ( δ y i , q i ) . By the definition of % ′ and note that d = f − ( P ni =1 λ i ( δ x i , p i )), d = f − ( P ni =1 λ i ( δ y i , q i )),we conclude that d ∼ ′ d .Then we extend % ′ to % ∗ on the entire space of temporal lotteries D ∗ . For any d ∈ D ∗ \ ˆ D ∗ ,denote d as the marginal lottery in source 1 and d | x as the lottery over marginal lotteries87onditional on outcome x in source 1. Denote supp( d ) = { x , ..., x N } with x < · · · < x N and for each k = 1 , ..., N , supp( d | x k ) = { p k, , ..., p k,t k } ⊆ L ( X ) with t k ≥
1. Since d ˆ D ∗ ,there exists some k ′ with t k ′ >
1. We will construct % ∗ by relating d to some temporallottery in ˆ D ∗ as follows. Stage 1. i = 1. If t = 1, then define d such that for all x = x and q ∈ L ( X ), d ( x, q ) = d ( x, q ). Note that ( δ x , q ) % ( δ x , δ c ) if c > −∞ . Denote z , = x . By Lemma 2,we can find ˆ z , with ( δ x , δ ˆ z , ) ∼ ( δ x , p , ). Denote d ( x , δ ˆ z , ) = d ( x , p , ) and d ( x , q ) =0 for all q = δ ˆ z , .If t > d ( c , δ c ) >
0, that is, x = c and δ c ∈ supp( d | δ c ), then c , c > −∞ andwe can reorder lotteries in supp( d | δ c ) such that ( δ x , p , ) - ( δ x , p , ) - · · · - ( δ x , p ,t ).By Axiom M, we know ( δ x , p , ) = ( δ c , δ c ) ≺ ( δ x , p ,i ) for each i > % on ˆ P , we can find x + x > z > c = x such that ( δ x , p , ) ≻ ( δ z , δ c ).Also, by Axiom M, for each i > x ≤ z ≤ z , we have ( δ x , p ,i ) ∈ Γ ,δ z . Define z , = x = c and z ,i = i − t z for all i = 2 , ..., t . Clearly, z , < z , < · · · < z ,t < z .By Lemma 4 and Lemma 2, we can find ˆ z ,i for i ≥ δ z ,i , δ ˆ z ,i ) ∼ ( δ x , p ,i ). Then wedefine d ∈ D ∗ such that for any x
6∈ { z ,i } t i =1 , q ∈ L ( X ), d ( x, q ) = d ( x, q ), and for each i = 1 , ..., t , d ( z ,i , δ ˆ z ,i ) = d ( x , p ,i ) and d ( z ,i , q ) = 0 for q = δ ˆ z ,i .If t > d ( c , δ c ) = 0, then we can apply a similar construction method by choosing z , > c . Stage 2. i ≥
2. Consider x i > x i − ≥ x ≥ c . If t i = 1, then define d i ∈ D ∗ such thatfor all x = x i and q ∈ L ( X ), d i ( x, q ) = d i − ( x, q ). Denote z i, = x i and by Lemma 2, thereexists ˆ z i, with ( δ x i , δ ˆ z i, ) ∼ ( δ x i , p i, ). Define d i ( x i , δ ˆ z i, ) = d ( x i , p i, ) and d i ( x i , q ) = 0 for q = δ ˆ z i, .If instead t i >
1, again we assume that without loss of generality, ( δ x i , p i, ) - · · · - ( δ x i , p i,t i ). As x i > x ≥ c , ( δ x i , p i, ) ≻ ( δ xi + xi − , δ c ) ≻ ( δ x , δ c ) . If p i, = δ c , then ( δ x i , p i, ) = ( δ x i , p i, ) implies that ( δ x i , p i, ) ≺ ( δ x i , p i, ). Again, we canfind x i + x i +1 > z i > x i such that ( δ x i , p , ) ≻ ( δ z i , δ c ). By Axiom M, for each j > x i ≤ z ≤ z i , we have ( δ x i , p i,j ) ∈ Γ ,δ z .Denote z i,j = x i + z i j − t i for j = 1 , , ..., t i . Clearly, x i = z i, < z i, < · · · < z i,t i < z i . ByLemma 4 and Lemma 2, we can find ˆ z i,j for j ≥ δ z i,j , δ ˆ z i,j ) ∼ ( δ x i , p i,j ). Then wedefine d i ∈ D ∗ such that for any x
6∈ { z i,j } t i j =1 , q ∈ L ( X ), d i ( x, q ) = d i − ( x, q ), and for each88 = 1 , ..., t i , d i ( z i,j , δ ˆ z i,j ) = d ( x i , p i,j ) and d i ( z i,j , q ) = 0 for q = δ ˆ z i,j .The algorithm ends at i = N < + ∞ . We know that supp( d N ) = S Nk =1 { z k, , · · · , z k,t k } .The discussion with i = N is similar to the discussion with i = 1 as we need to considerthe cases where t N >
1, and d ( c , δ c ) > d ( c , δ c ) = 0. For each z k,i ∈ supp( d N ), wehave ˆ z k,i with ( δ z k,i , δ ˆ z k,i ) ∼ ( δ x k , p k,i ) and d k ( z k,i , δ ˆ z k,i ) = d ( x k , p k,i ) for all 1 ≤ i ≤ t k and1 ≤ k ≤ N . Also, { z k,i } admits a lexicographic order, that is, z k,i < z k ′ ,i ′ if i < i ′ or i = i ′ and j < j ′ . This implies d N ∈ ˆ D ∗ . In this way, we have defined a mapping h : D ∗ \ ˆ D ∗ → ˆ D ∗ where h ( d ) = d N . When there is no confusion, we can also use the same technique to derive h ( d ) for d ∈ ˆ D ∗ . Although it might be the case that h ( d ) = d , we must have h ( d ) ∼ ′ d as isshown in the next paragraph. Easy to show that h ( h ( d )) = h ( d ) for all d ∈ D ∗ .Now we can define % ∗ on D ∗ such that % ∗ agrees with % ′ on ˆ D ∗ and d ∼ ∗ h ( d ) for d ∈ D ∗ \ ˆ D ∗ . Then we know d % ∗ d ′ if and only if h ( d ) % ′ h ( d ′ ) for all d, d ′ ∈ D ∗ . To verifythat % ∗ is well-defined, we need to argue that the arbitrary choice of { z k,i } does not affect thedefinition of % ∗ . Consider two constructions h and ˆ h . For each d = P k,i d ( x k , p k,i ) δ ( x k ,p k,i ) , h ( d ) = P k,i d ( x k , p k,i ) δ ( z k,i ,δ ˆ zk,i ) and ˆ h ( d ) = P k,i d ( x k , p k,i ) δ ( z ′ k,i ,δ ˆ z ′ k,i ) such that z k,i = z k ′ ,i ′ , z ′ k,i = z ′ k ′ ,i ′ for all ( k, i ) = ( k ′ , i ′ ) and for all k, i , ( δ z ′ k,i , δ ˆ z ′ k,i ) ∼ ( δ z k,i , δ ˆ z k,i ) ∼ ( x k , p k,i ).By Corollary 4, h ( d ) ∼ ′ ˆ h ( d ). Hence the definition of % ∗ is not affected by the specificconstruction of h .The next lemma extends Corollary 4 to temporal lotteries in D ∗ \ ˆ D ∗ . Lemma 29.
Suppose λ i > for all i and P ni =1 λ i = 1 . For d = P ni =1 λ i δ ( x i ,p i ) , d = P ni =1 λ i δ ( y i ,q i ) with δ ( x i ,p i ) ∼ ′ δ ( y i ,q i ) for all i , then d ∼ ∗ d .Proof of Lemma 29. We first prove the result for the case that ( x i , p i ) = ( x j , p j ) , ( y i , q i ) =( y j , q j ) for all i = j . Notice that h ( d ) = P ni =1 λ i δ ( z i ,δ ˆ z i ) with z i = z j for i = j and( δ z i , δ ˆ z i ) ∼ ( δ x i , p i ) for each i = 1 , ..., n . Similarly, h ( d ) = P ni =1 λ i δ ( z i ,δ ˆ z i ) with z i = z j for i = j and ( δ z i , δ ˆ z i ) ∼ ( δ x i , p i ) for each i = 1 , ..., n . We know that h ( d ) ∼ ∗ d and h ( d ) ∼ ∗ d . By Corollary 4, we have h ( d ) ∼ ∗ h ( d ) and hence d ∼ ∗ d .Now we consider the general case. If i = 1, then the result is trivial. Suppose i ≥ x i , p i ) = ( δ c , δ c ) for i ≤ k for some k ≥ x i , p i ) ≻ ( δ c , δ c ) for i > k . Since ( δ x i , p i ) ∼ ( δ y i , q i ), we know that ( y i , q i ) = ( δ c , δ c )89or i ≤ k and ( y i , q i ) ≻ ( δ c , δ c ) for i > k . Then we can write d = [ k X i =1 λ i ]( δ c , δ c ) + n X i = k +1 λ i δ ( x i ,p i ) , d = [ k X i =1 λ i ]( δ c , δ c ) + n X i = k +1 λ i δ ( y i ,q i ) . This implies that we can assume that ( δ x i , p i ) = ( δ c , δ c ) for all i ≥
2. By a similar argument,we can assume ( δ x i , p i ) = ( δ c , δ c ) for all i < n .Without loss of generality, we can further assume c > x i > c for all 2 ≤ i ≤ n − δ c , p i ) = ( δ c , δ c ) with ( δ a , q ) for some a > c , and ( δ c , p i ) = ( δ c , δ c )with ( δ b , q ) for some c < a, b < c without changing the preference ranking of d .i). Suppose that ( δ x , p ) ≻ ( δ c , δ c ) and ( δ x n , p n ) ≺ ( δ c , δ c ). By reordering, there existsa partition of { , ..., n } as { , ..., t } , · · · , { t k − + 1 , ..., n } such that ( x i , p i ) = ( x j , p j ) for all t l + 1 ≤ i, j ≤ t l +1 with 0 ≤ l ≤ k − t = 0 , t k = n .For l = 0, that is, 1 ≤ i ≤ t , by continuity on ˆ P and Lemma 4, we can construct z i > c ,ˆ z i ≥ c with ( δ z i , δ ˆ z i ) ∼ ( δ x i , p i ) = ( δ x , p ) for all i = 1 , ..., t and z i = z j for all i = j . Byapplying Lemma 27 repeatedly, we derive t X i =1 λ i P t j =1 λ j ( δ z i , δ ˆ z i ) ∼ ( δ x , p ) . The same result holds for l = 1 , ..., k − d = P ni =1 λ i δ ( x i ,p i ) = P k − l =0 ( P t l +1 i = t l +1 λ i ) δ ( x i ,p i ) . By definition of h , wecan find h ( d ) ∼ ∗ d with h ( d ) ∈ ˆ D ∗ . Denote h ( d ) = P k − l =0 ˆ λ l +1 δ ( x ′ l +1 , ˆ x ′ l +1 ) , whereˆ λ l +1 = P t l +1 i = t l +1 λ i and ( δ x ′ l +1 , δ ˆ x ′ l +1 ) ∼ ( δ x tl +1 , p t l +1 ) ∼ P t l +1 i = t l +1 λ i P tl +1 j = tl +1 λ j ( δ z i , δ ˆ z i ) for each l .Denote R l +1 = P t l +1 i = t l +1 λ i P tl +1 j = tl +1 λ j ( δ z i , δ ˆ z i ). Note that R l and R l ′ are compatible, ( δ x ′ l , δ ˆ x ′ l ) and( δ x ′ l ′ , δ ˆ x ′ l ′ ) are compatible for all l = l ′ . By Lemma 27 and the definition of % ′ , d ∼ ∗ h ( d ) = k X l =1 ˆ λ l δ ( x ′ l ,δ ˆ x ′ l ) ∼ ′ k X l =1 h (ˆ λ l R l ) = n X i =1 λ i δ ( z i ,δ ˆ zi ) ∈ ˆ D ∗ . Similarly, we can find z ′ i , ˆ z ′ i for i = 1 , ..., n such that z ′ i = z ′ j for all i = j , ( δ z ′ i , δ ˆ z ′ i ) ∼ ( δ y i , q i )for each i and d ∼ ∗ n X i =1 λ i δ ( z ′ i ,δ ˆ z ′ i ) ∈ ˆ D ∗ .
90y Corollary 4, we have d ∼ ∗ n X i =1 λ i δ ( z ′ i ,δ ˆ z ′ i ) ∼ ∗ n X i =1 λ i δ ( z i ,δ ˆ zi ) ∼ ∗ d . (ii). Now we turn to the case where ( δ x , p ) = ( δ c , δ c ) or ( δ x n ,p n ) = ( δ c , δ c ) or both.This implies ( δ y , q ) = ( δ c , δ c ) or ( δ y n ,q n ) = ( δ c , δ c ) or both. If λ + λ n = 1, then the resultis trivial as n = 2 and we are back to the special case where ( x i , p i ) = ( x j , p j ) , ( y i , q i ) = ( y j , q j )for all i = j .Recall that ( δ c , δ c ) ≻ ( δ x i , p i ) ≻ ( δ c , δ c ) , ( δ c , δ c ) ≻ ( δ y i , q i ) ≻ ( δ c , δ c ) for all 2 ≤ i ≤ n −
2. Define ˆ d = 11 − λ − λ n n X i =2 λ i δ ( x i ,p i ) , ˆ d = 11 − λ − λ n n X i =2 λ i δ ( y i ,q i ) . Then we are back to case (i) and ˆ d ∼ ∗ ˆ d . By Lemma 27 and the definition of % ′ , we know h ( d ) = λ δ (0 ,δ ) + λ n δ ( c ,δ c ) + (1 − λ − λ n ) h ( ˆ d ) ∼ λ δ (0 ,δ ) + λ n δ ( c ,δ c ) + (1 − λ − λ n ) h ( ˆ d ) = h ( d ). Thus, by definition of % ∗ , we conclude that d ∼ ∗ d .As a summary, we have defined a preference relation % ∗ on D ∗ , which is a mixture space.For d, d ′ ∈ D ∗ and α ∈ (0 , α -mixture of d and d ′ as[ αd + (1 − α ) d ′ ]( x, q ) = αd ( x, q ) + (1 − α ) d ′ ( x, q ) , ∀ α ∈ (0 , , ( x, q ) ∈ X × L ( X ) . We claim that % ∗ satisfies the vNM independence property and mixture continuity.For the independence property, fix d, d ′′ ∈ D ∗ and α ∈ (0 , d = P ni =1 λ i δ ( x i ,p i ) and d ′′ = P mj =1 η j δ ( y j ,q j ) . As supp( d ) ∪ supp( d ′′ ) is finite, for any i such that ( x i , p i ) = ( δ c , δ c )and ( δ c , δ c ), we can find ( δ z di , δ ˆ z di ) ∼ ( δ x i , p i ); for any j such that ( y j , q j ) = ( δ c , δ c ) and( δ c , δ c ), we can find ( δ z d ′′ j , δ ˆ z d ′′ j ) ∼ ( δ y j , q j ). Moreover, we require that c > z di , z d ′′ j > c and z di = z d ′′ j for all i, j . For any i, j with ( x i , p i ) = ( δ c , δ c ) or ( y j , q j ) = ( δ c , δ c ), denote z di = z d ′′ j = c and ˆ z di = ˆ z d ′′ j = c . For any i, j with ( x i , p i ) = ( δ c , δ c ) or ( y j , q j ) = ( δ c , δ c ),denote z di = z d ′′ j = c and ˆ z di = ˆ z d ′′ j = c .By Lemma 29, we know αd + (1 − α ) d ′′ ∼ ∗ α ˆ d + (1 − α ) ˆ d ′′ . d = P ni =1 λ i δ ( z di ,δ ˆ zdi ) ∼ ∗ d and ˆ d ′′ = P mj =1 η j δ ( z d ′′ i ,δ ˆ zd ′′ i ) ∼ ∗ d ′′ . As ˆ d, ˆ d ′′ ∈ ˆ D ∗ , we candenote P = f − [ ˆ d ] and R = f − [ ˆ d ′′ ]. Easy to see that P and R are weakly compatible, and αd + (1 − α ) d ′′ ∼ ∗ αf ( P ) + (1 − α ) f ( R ) for any α ∈ (0 , d ′ with d ≻ ∗ d ′ . Using the same argument, we can find Q and S suchthat d ′ ∼ ∗ f ( Q ), d ′′ ∼ ∗ f ( S ), Q and S are weakly compatible and αd ′ + (1 − α ) d ′′ ∼ ∗ αf ( Q ) + (1 − α ) f ( S ) for any α ∈ (0 , d ≻ ∗ d ′ if and only if P ≻ Q and d ′′ = d ′′ implies that R ∼ S . By Lemma 27, we know αP + (1 − α ) R ≻ αQ + (1 − α ) S for any α ∈ (0 , f ( αP + (1 − αR )) = αf ( P ) + (1 − α ) f ( R ) and f ( αQ + (1 − αS )) = αf ( Q ) + (1 − α ) f ( S ) since P and R are weakly compatible and Q and S are weakly compatible. Thus for each α ∈ (0 , αP + (1 − α ) R ≻ αQ + (1 − α ) S ⇐⇒ f ( αP + (1 − α ) R ) ≻ ∗ f ( αQ + (1 − α ) S ) ⇐⇒ αf ( P ) + (1 − α ) f ( R ) ≻ ∗ αf ( Q ) + (1 − α ) f ( S ) ⇐⇒ αd + (1 − α ) d ′′ ≻ ∗ αd ′ + (1 − α ) d ′′ . Hence % ∗ satisfies the vNM independence property on D ∗ .Next we show the mixture continuity of % ∗ on D ∗ . For any d, d ′ , d ′′ ∈ D ∗ , by the aboveproof for independence, we can find P, Q, R ∈ P such that f ( P ) ∼ ∗ d, f ( Q ) ∼ ∗ d ′ , f ( R ) ∼ ∗ d ′′ and for each α ∈ (0 , αP + (1 − α ) Q ∈ P , f ( αP + (1 − α ) Q ) = αf ( P ) + (1 − α ) f ( Q ) and αd + (1 − α ) d ′ ∼ ∗ αf ( P ) + (1 − α ) f ( Q ). Then A = n α ∈ [0 ,
1] : αd + (1 − α ) d ′ ≻ ∗ d ′′ o = n α ∈ [0 ,
1] : αf ( P ) + (1 − α ) f ( Q ) ≻ ∗ f ( R ) o = n α ∈ [0 ,
1] : f ( αP + (1 − α ) Q ) ≻ ∗ f ( R ) o = n α ∈ [0 ,
1] : αP + (1 − α ) Q ≻ R o . By mixture continuity of % on P , we know A is open in [0 , n α ∈ [0 ,
1] : αd + (1 − α ) d ′ ≺ ∗ d ′′ o is also open in [0 , % ∗ satisfies mixture continuity on D ∗ .We are now prepared to finish the proof of Lemma 28. Proof of Lemma 28.
Since D ∗ is a mixture space and the preference relation % ∗ satisfiesmixture continuity and the independence axiom, by the Mixture Space Theorem, % ∗ on D ∗ U with a utility index w D : X × L ( X ) → R . That is, theexpected utility of d ∈ D ∗ is given by U ( d ) = X x,p w D ( x, p ) d ( x, p ) . We also know that w D is unique up to a positive affine transformation.Recall that % ∗ extends % ′ from ˆ D ∗ to D ∗ and d % ′ d ′ if and only if f − ( d ) % f − ( d ′ ) forall d, d ′ ∈ ˆ D ∗ . Then the utility function V ( P ) = X x P ( x ) w D ( x, P | x ) , ∀ P ∈ P represents % on P .Then we derive the implications of Axiom CI. Recall that % restricted to { δ x } × L ( X )for each x ∈ X admits an EU representation with some regular utility index v x . Then byuniqueness up to a positive affine transformation, there exists a continuous and monotonefunction φ x such that for all p ∈ L ( X ), V ( δ x , p ) = w D ( x, p ) = φ x ( CE v x ( p ))Define ˆ u D : X × X → R as ˆ w ( x, y ) = φ x ( y ) for all ( x, y ) ∈ X . Then the representation canbe rewritten as V ( P ) = X x ˆ u D ( x, CE v x ( P | x )) P ( x ) , ∀ P ∈ P where v x is a regular function for each x ∈ X . This is exactly the functional form stated inLemma 28. The final step is to verify that ˆ u D is a regular function.Monotonicity can be guaranteed by Axiom M. WLOG, let ˆ u D (0 ,
0) = 0. To see whyˆ u D is bounded, notice that % satisfies mixture continuity on ˆ P . Suppose by contradictionthat ˆ u D is unbounded from above. Then c >
0. Denote ˆ u D ( c / ,
0) = a >
0. For any n , we can find z n > z n − > z ′ n ≥ u D ( z n , z ′ n ) > n a , which implies V (( n δ z n + n − n δ , n δ z ′ n + n − n δ )) > a and hence ( n δ z n + n − n δ , n δ z ′ n + n − n δ ) ≻ ( δ c / , δ )for each n . However, by continuity of % over product lotteries, as n → ∞ , we must have( δ , δ ) % ( δ c / , δ ), a contradiction.Then we show that ˆ u D is continuous. Again, we normalize ˆ u D (0 ,
0) = 0. Suppose bycontradiction that ˆ u D is not continuous, then we can find ( x, y ) ∈ X × X and a sequence93 x n , y n ) → ( x, y ) such that lim n →∞ ˆ u D ( x n , y n ) = ˆ u D ( x, y ). Then there exists a boundedsubsequence of { ( x n , y n ) } (still denoted as { ( x n , y n ) } given there is no confusion) such thateither ˆ u D ( x n , y n ) ≤ ˆ u D ( x, y ) for all n or ˆ u D ( x n , y n ) ≥ ˆ u D ( x, y ) for all n . By symmetry, wewill focus on the former case. Since ˆ u D is bounded, { ˆ u D ( x n , y n ) } n ≥ admits a convergentsubsequence (again we still denote the subsequence as the sequence itself). Then it must bethe case that lim n →∞ ˆ u D ( x n , y n ) = a < b = ˆ u D ( x, y ).We claim that we can find some P ∈ P with V ( P ) = a + b . To see this, first suppose that x n = x for all x large enough. Without loss of generality, we can assume x n = x for all x . If x = c , then fix any c < x ′ < x , we have ˆ u D ( x ′ , y m ) < a + b for some m large enoughsince ˆ u D is monotone and lim n →∞ ˆ u D ( x n , y n ) = a . Then there exists η ∈ (0 ,
1) such that V ( η ( δ x ′ , δ y m ) + (1 − η )( δ x , δ y )) = η ˆ u D ( x ′ , y m ) + (1 − η ) b = a + b . If x = c , then fix any x < x ′
1) such that V ( η ( δ x , δ y m )+(1 − η )( δ x ′ , δ y )) = η ˆ u D ( x, y m )+(1 − η )ˆ u D ( x ′ , y ) = a + b . Nowsuppose that we can find m large enough such that x m = x and ˆ u D ( x m , y m ) < a + b . Thenthere exists η ∈ (0 ,
1) such that V ( η ( δ x m , δ y m )+(1 − η )( δ x , δ y )) = η ˆ u D ( x m , y m )+(1 − η ) b = a + b .As lim n →∞ ˆ u D ( x n , y n ) = a < a + b , for n large enough, we have ˆ u D ( x n , y n ) < a + b < b ,that is, ( δ x n , δ y n ) ≺ P . Let n goes to infinity and we know ( δ x n , δ y n ) w −→ ( δ x , δ y ). By AxiomTopological Continuity over Product Lotteries (the second part of Axiom WC), ( δ x , δ y ) - P ,that is, b < a + b , a contradiction. Hence ˆ u D is continuous and this completes the proof. Step 4: Finally we check the consistency of the previous representations on ˆ P . Now we have two representations on ˆ P : the EU-CN, GBIB-CN and GFIB-CN represen-tations in Step 1 and the KP-style representations in Lemma 28. Both of them represent % on ˆ P and we will explore the implications of such consistency.By Lemma 28, we know that % on ˆ P can be represented by U ( P , P ) = X x ˆ u D ( x, CE v x ( P )) P ( x ) , ∀ ( P , P ) ∈ ˆ P where ˆ u D and v x for all x ∈ X are regular.First, suppose that % admits an EU-CN representation w on ˆ P . Assume that c >
0. Thecase with c = 0 is symmetric. Fix 0 < a < c . As w and ˆ u D are unique up to a positive affinetransformation, we can normalize w (0 ,
0) = ˆ u D (0 ,
0) = 0 and w ( a,
0) = ˆ u D ( a,
0) = b > φ : w ( X , X ) → R such that for all P ∈ ˆ P , U KP ( P , P ) = φ ◦ U EU − CN ( P , P ) . Now focus on L ( X ) × { δ y } for some y ∈ X . We know for all p ∈ L ( X ), U KP ( p, δ y ) = X x ˆ u D ( x, y ) p ( x ) = φ ◦ U EU − CN ( p, δ y ) = φ [ X x w ( x, y ) p ( x )] . Then we know φ must be linear on w ( X , y ) for each y. By continuity of w , by ranging over y ∈ X , we know φ must be linear on its domain w ( X , X ). That is, for all t , t ∈ w ( X , X )and α ∈ (0 , φ ( αt +(1 − α ) t ) = αφ ( t )+(1 − α ) φ ( t ). Also, by our normalization, φ (0) = 0and φ ( b ) = b . Then for any t ∈ (0 , b ), φ ( t ) = φ ( tb · b + (1 − tb ) ·
0) = tb φ ( b ) + (1 − tb ) φ (0) = t .We show also that h ( t ) = t for t > b or t <
0. Thus w ≡ ˆ u D .Second, suppose that % admits a GBIB-CN representation ( w, v ′ , v ′ , H ). Then for each x ∈ X , on { δ x } × L ( X ), U GBIB − CN ( δ x , p ) = w ( x, CE v ′ ( p ))Similarly, U KP ( δ x , p ) = ˆ u D ( x, CE v x ( p )). Since v x and v ′ are regular, consistency of BIB-CNand KP on { δ x } × L ( X ) requires that v x is a positive affine transformation of v ′ , that is, CE v x ( p ) = CE v ′ ( p ) for all x ∈ X . Hence for any P ∈ P , U KP ( P ) = X x ˆ u D ( x, CE v x ( P | x )) P ( x )= X x ˆ u D ( x, CE v ′ ( P | x )) P ( x )Thus, % admits an BIB representation (ˆ u D , v ′ ).Finally, suppose that % admits a GFIB-CN representation ( w, v ′ , v ′ , H ). That is, V GF IB − CN ( P , P ) = w ( CE v ′ ( P ) , CE v ′ ( P )) , if CE v ′ ( P ) H P w ( CE v ′ ( P ) , y ) P ( y ) , if CE v ′ ( P ) ∈ H As both utility functions represent % on ˆ P , we can find a monotone and continuousfunction φ : w ( X , X ) → R with U KP ( P , P ) = φ ◦ U GF IB − CN ( P , P ) , ∀ P ∈ ˆ P .
95e first focus on L ( X ) × { p } for some p ∈ L ( X ). Based on the GFIB-CN, thepreference % | p admits an EU representation with the index v ′ . Also, note that U KP islinear in the first source for fixed p . Hence for each p ∈ L ( X ), w ( · , CE v x ( p )) must bea positive affine transformation of v ′ . That is, there exists functions ˆ α and ˆ β defined on L ( X ) such that ˆ a ( p ) >
0, ˆ b ( p ) ∈ R for all p ∈ L ( X ) and w ( x, CE v x ( p )) = ˆ a ( p ) v ′ ( x ) + ˆ b ( p ) , ∀ x ∈ X . Specifically, if p = δ y for some y ∈ X , then we know w ( x, y ) = ˆ a ( δ y ) v ′ ( x ) + ˆ b ( δ y ) , ∀ ( x, y ) ∈ X × X . Define a, b as functions on X by a ( y ) = ˆ a ( δ y ) and b ( y ) = ˆ b ( δ y ). This implies w ( x, CE v x ( p )) = ˆ a ( p ) v ′ ( x ) + ˆ b ( p )= a ( CE v x ( p )) v ′ ( x ) + b ( CE v x ( p )) , ∀ x ∈ X , q ∈ L ( X )If H = ∅ , then GFIB-CN reduces to NB, which is a special case of GBIB-CN and it hasbeen covered in the second case. From now on, assume H = ∅ . For any fixed x , x ∈ X with x < x , we can always normalize v ′ ( x ) = 0 and v ′ ( x ) = 1. Plug the two values intothe previous equation, we get for any p ∈ L ( X ),ˆ b ( p ) = b ( CE v x ( p )) , ˆ a ( p ) = a ( CE v x ( p )) + b ( CE v x ( p )) − b ( CE v x ( p )) . First, suppose that there exists x < x with x , x H . Then ˆ u D ( x i , CE v xi ( p )) = φ ( w ( x i , CE v ′ ( p ))) for i = 1 , x ∈ X , p ∈ L ( X ),ˆ u D ( x, CE v x ( p )) = φ ( w ( x , CE v ′ ( p ))) · v ′ ( x ) + φ ( w ( x , CE v ′ ( p ))) · (1 − v ′ ( x )) . Clearly, the RHS depends on p only through its certainty equivalent under v ′ . Then v x must be a positive affine transformation of v ′ for all x ∈ X and the KP representationreduces to a BIB representation (ˆ u D , v ′ ).Second, suppose that there exists x < x with x , x ∈ H and w ( x , · ) is a positive affinetransformation of w ( x , · ). As x , x ∈ H , we know ˆ u D ( x i , CE v xi ( p )) = φ ( P y w ( x i , y ) p ( y ))96or i = 1 ,
2, which further implies for any x ∈ X , p ∈ L ( X ),ˆ u D ( x, CE v x ( p )) = φ ( X y w ( x , y ) p ( y )) · v ′ ( x ) + φ ( X y w ( x , y ) p ( y )) · (1 − v ′ ( x )) . Since w ( x , · ) is a positive affine transformation of w ( x , · )., the above equation can berewritten as ˆ u D ( x, CE v x ( p )) = g ( x, X y w ( x , y ) p ( y ))for some function g . Notice that g depends on p only through its expected value under w ( x , · ). Hence the v x must be a positive affine transformation of w ( x , · ) for all x ∈ X .Denote w ( x , · ) as w ( · ), then the KP representation reduces to a BIB representation (ˆ u D , w ).Finally, suppose cl ( H ) = X and for all x = x , w ( x , · ) is not a positive affinetransformation of w ( x , · ). Without loss of generality, we suppose c > w (0 ,
0) = 0, x = 0 , x = c /
2. Denote w ( x , · ) = w ( · ) and w ( x , · ) = w ( · ). Then we know w is not apositive affine transformation of w . We denote the preference represented by EU index w i as % w i . Then % w = % w . Since φ is strictly increasing and continuous, φ is almost everywheredifferentiable. Recall our normalization that v ′ ( x ) = 0 and v ′ ( x ) = 1. Then the previousargument applies and we can show that for any x ∈ X , p ∈ L ( X ),ˆ u D ( x, CE v x ( p )) = φ ( w ( p )) v ′ ( x ) + φ ( w ( p ))(1 − v ′ ( x )) . Denote H = n p ∈ L ( X ) : ∃ q ∈ L ( X ) , s.t. (cid:16) w ( p ) − w ( q ) (cid:17) · (cid:16) w ( p ) − w ( q ) (cid:17) < o That is, H is the set of single-source lotteries such that there exists another lottery wherethe two preferences represented by w and w disagree on the ranking of the two lotteries.We claim that H = L ( X ) \{ δ c , δ c } . Clearly δ c , δ c
6∈ H .For any q = δ c , δ c with q ∼ w i δ y for i = 1 ,
2, we know y ∈ ( c , c ). Since % w = % w , wecan find q ′ ∼ w δ y and q ′ w δ y , otherwise % w and % w share the same indifference curvewith certainty equivalent y ∈ ( c , c ) and they should be the same EU preference. Then wecan assume q ′ w q , otherwise, we can take a mixture between q ′ and δ y . Without lossof generality, assume q ′ ≻ w q . Then we can choose q ′′ slightly dominated by q ′ and bycontinuity, we have q ′′ ≻ w q and q ≻ w q ′′ . 97ow for any q ∈ H , take p such that, without loss of generality, p ≻ w q and q ≻ w p .Then we can find x ∈ (0 ,
1) such that1 − v ′ ( x ) v ′ ( x ) = φ ( w ( p )) − φ ( w ( q )) φ ( w ( q ) − φ ( w ( p )) . Such x exists as the RHS is strictly positive and the the range of − v ′ ( x ) v ′ ( x ) for x ∈ (0 ,
1) is(0 , + ∞ ). Rearrange the above equation, we can get φ ( w ( q )) v ′ ( x ) + φ ( w ( q ))(1 − v ′ ( x )) = φ ( w ( p )) v ′ ( x ) + φ ( w ( p ))(1 − v ′ ( x ))that is, ( δ x , p ) ∼ ( δ x , q ). By conditional independence, for any β ∈ [0 , δ x , βp + (1 − β ) q ) ∼ ( δ x , p ) ∼ ( δ x , q ). This implies that for all β = β ′ φ ( βw ( p ) + (1 − β ) w ( q )) − φ ( β ′ w ( p ) + (1 − β ′ ) w ( q )) φ ( β ′ w ( p ) + (1 − β ′ ) w ( q )) − φ ( βw ( p ) + (1 − β ) w ( q )) = 1 − v ′ ( x ) v ′ ( x ) = φ ( w ( p )) − φ ( w ( q )) φ ( w ( q ) − φ ( w ( p )) . Take β ′ = 0 and let β → + , we have ∂ + φ ( w ( q )) /∂x∂ − φ ( w ( q )) /∂x = φ ( w ( p )) − φ ( w ( q )) φ ( w ( q ) − φ ( w ( p )) w ( q ) − w ( p ) w ( p ) − w ( q ) . We argue that the two semi-derivatives are well-defined. The RHS is always well-defined.Notice that by continuity of φ , we can change q slightly to change w ( q ) without changing w ( q ). This is possible as % w = % w . If the semi-derivative ∂ − φ ( w ( q )) /∂x does not exist,then ∂ + φ ( x ) /∂x does not exist for x in a open interval, which contradicts with the factthat φ is almost everywhere differentiable. A similar proof can show that ∂ + φ ( w ( q )) /∂x iswell-defined.Let β = 1 and β ′ → − , we can get ∂ − φ ( w ( p )) /∂x∂ + φ ( w ( p )) /∂x = φ ( w ( p )) − φ ( w ( q )) φ ( w ( q ) − φ ( w ( p )) w ( q ) − w ( p ) w ( p ) − w ( q ) = ∂ + φ ( w ( q )) /∂x∂ − φ ( w ( q )) /∂x . (7)Again, all the semi-derivatives are well-defined.Fix p and choose q ′ ∼ w q . we can find q ′ w q and q ′ ≻ w p . Without loss of generality,let w ( q ′ ) > w ( q ). The other case can be proved symmetrically. Then for any α ∈ (0 , αq ′ + (1 − α ) q ≻ w p and p ≻ w q ′ + (1 − α ) q ∼ w q .98hen the left equality of equation (7) becomes: ∂ − φ ( w ( p )) /∂x∂ + φ ( w ( p )) /∂x = φ ( w ( p )) − φ ( w ( q )) w ( p ) − w ( q ) αw ( q ) + (1 − α ) w ( q ′ ) − w ( p ) φ ( αw ( q ) + (1 − α ) w ( q ′ )) − φ ( w ( p )) . This implies that φ ( αw ( q ) + (1 − α ) w ( q ′ )) − φ ( w ( p )) αw ( q ) + (1 − α ) w ( q ′ ) − w ( p )is a constant as α varies in (0 , φ ′ ( z ) = C for all z ∈ ( w ( q ) , w ( q ′ )), where C > z , z ∈ w ( X , X ) with w ( c , c ) > z > z > w ( c , c ). Given y ∈ ( c , c ), for any β ∈ (0 , βp + (1 − β ) δ y % w βq + (1 − β ) δ y ∼ w βq ′ + (1 − β ) δ y and βq ′ + (1 − β ) δ y % w βq + (1 − β ) δ y ∼ w βp + (1 − β ) δ y . By the above argument, we know that φ ′ ( z ) is a constant for z ∈ ( βw ( q ) + (1 − β ) w ( δ y ) , βw ( q ′ ) + (1 − β ) w ( δ y )). By continuityof w , those open intervals are intersecting with each other. We make y large enough andsmall enough respectively, so that we can get an open cover of [ z , z ]. Then there exists afinite subcover can we have φ ′ ( z ) = φ ′ ( z ) = C for any w ( c , c ) > z > z > w ( c , c ) with z , z ∈ w ( X , X ). This implies φ ( z ) = Cz + b for z ∈ ( w ( c , c ) , w ( c , c )) ∩ w ( X , X ) andby continuity of φ , φ ( z ) = Cz + b holds for all z ∈ w ( X , X ).Since w and ˆ u D are unique up to positive affine transformation, we can set φ ( z ) = z forall z without loss of generality. Then we know that for all x ∈ X , p ∈ L ( X )ˆ u D ( x, CE v x ( p )) = X y w ( y ) p ( y ) v ′ ( x ) + X y w ( y ) p ( y )(1 − v ′ ( x ))Thus the representation on P is given by U KP ( P ) = X x ˆ u D ( x, CE v x ( P | x )) P ( x )= X x,y (cid:16) [ w ( y ) − w ( y )] v ′ ( x ) + w ( y ) (cid:17) P ( x, y ) , for all P ∈ P . This is of course an EU representation.As a summary of Step 4 , in all possible cases, % on P can be represented by either anEU or a BIB representation.Combining the results in Step 1 and
Step 4 , given the axioms stated in Theorem 2, therelation %%