Intransitive Indifference Under Uncertainty: Expected Scott-Suppes Utility Representation

Transcription

1 Intransitive Indifference Under Uncertainty: Expected Scott-Suppes Utility Representation Nuh Aygün Dalkıran Oral Ersoy Dokumacı : Tarık Kara ; October 23, 2017 Abstract We study preferences with intransitive indifference under uncertainty. Our primitives are semiorders and we are interested in their Scott-Suppes representations. We obtain a Scott-Suppes representation theorem in the spirit of the expected utility theorem of von Neumann and Morgenstern (1944). Our representation offers a decision theoretical interpretation for epsilon equilibrium. Keywords: Semiorder, Instransitive Indifference, Uncertainty, Scott-Suppes Representation, Expected Utility. Bilkent University, Department of Economics; dalkiran@bilkent.edu.tr : University of Rochester, Department of Economics; oralersoy.dokumaci@rochester.edu ; Bilkent University, Department of Economics; ktarik@bilkent.edu.tr 1

2 Table of Contents 1 Introduction Intransitive Indifference and Semiorders Expected Utility Theory Related Literature 5 3 Preliminaries Semiorders Continuity Independence Utility Representations Expected Scott-Suppes Utility Representation The Main Result Uniqueness Independence of the Axioms On Epsilon Equilibrium 32 6 Conclusion 34 7 Appendix 36 References 38 List of Figures 1 Example Example Example Example Example

3 1 Introduction 1.1 Intransitive Indifference and Semiorders The standard rationality assumption in economic theory states that individuals have or should have transitive preferences. 1 A common argument to support the transitivity requirement is that, if individuals do not have transitive preferences, then they are subject to money pumps (Fishburn, 1991). Yet, intransitivity of preferences is frequently observed through choices individuals make in real life and in experiments (May, 1954; Tversky, 1969). Intransitive indifference is a certain type of intransitivity of preferences: an individual can be indifferent between x and y and also y and z, but not necessarily between x and z. Formal studies of the idea of intransitive indifference go back to as early as the 19 th century (Weber, 1834; Fechner, 1860). The Weber-Fechner law states that a small increase in the physical stimulus may not result in a change in perception, which suggests intransitivity of perceptional abilities. 2 A notable example was given by Jules Henri Poincaré (1905): 3 Sometimes we are able to make the distinction between two sensations while we cannot distinguish them from a third sensation. For example, we can easily make the distinction between a weight of 12 grams and a weight of 10 grams, but we are not able to distinguish each of them from a weight of 11 grams. This fact can symbolically be written: A B, B C, A C. Armstrong (1939, 1948, 1950) has repeatedly questioned the assumption 1 An individual has transitive preferences if whenever the individual thinks that x is at least as good as y and y is at least as good as z, then x is at least as good as z. 2 Although Weber s law and Fechner s law are both on human perception, there are some differences among them. We do not go into details here. 3 This quotation appears in Pirlot and Vincke (2013, p19). We also present the rest of the translated passage in the Appendix. 3

4 of transitivity of preferences and concluded: 4 That indifference is not transitive is indisputable, and a world in which it were transitive is indeed unthinkable. A well-known example of intransitive indifference is due to Luce (1956). Suppose an individual prefers a cup of coffee with one cube of sugar to a cup of coffee with five cubes of sugar. We can make four hundred and one cups of coffee, label each cup with i 0, 1,..., 400, and add p1 i{100q cubes of sugar to the i th cup. Since the increase in the amount of sugar from one cup to next is too small to be noticed, the individual would be indifferent between cups i and i 1. However, he is not indifferent between cups 0 and 400. Luce (1956) introduced a way to capture the idea of intransitive indifference. He coined the term semiorder to designate a binary relation satisfying some conditions that allow for intransitive indifference. Since then, semiorders have been studied extensively in preference, choice, and utility theory (Fishburn, 1970a; Aleskerov, Bouyssou, & Monjardet, 2007; Pirlot & Vincke, 2013). 1.2 Expected Utility Theory One of the most fruitful branches of modern economic theory, which has emerged from the seminal work of von Neumann and Morgenstern (1944), is decision making under uncertainty. In many fields, such as decision theory, game theory, and financial economics, the expected utility theorem of von Neumann and Morgenstern has helped in explaining how individuals behave when they face uncertainty. The properties that a decision maker s preferences have to satisfy in order for the decision maker to act as if having an expected utility function have 4 This quotation appears in Armstrong (1948, p3). 4

5 been challenged by many. Some of these properties are modified or removed in order to explain other types of behavior that are frequently observed in different economic settings. With a similar purpose, in this paper, we relax the transitivity axiom and try to understand and characterize the behavior of individuals, for whom indifference is not transitive, under uncertainty. As such, we obtain a representation theorem in the spirit of the expected utility theorem of von Neumann and Morgenstern (1944). 2 Related Literature The behavior we are interested in is often discussed in various contexts when modeling bounded rationality. A decision maker may deviate from rationality by choosing an alternative which is not the optimum but that is rather satisficing (Simon, 1955). Similarly, a player (a decision maker in a game) may deviate slightly from rationality by playing so as to almost, but not quite, maximize utility; i.e., by playing to obtain a payoff that is within epsilon of the maximal payoff, as is the case for epsilon equilibrium (Aumann, 1997; Radner, 1980). What unifies such models is that the decision maker s preferences exhibit thick indifference curves, demonstrating a weaker form of transitivity, which can be captured by intransitive indifference. In this paper, we identify a natural way of representing preferences with intransitive indifference over the set of lotteries. We focus on a particular representation of semiorders that provides utility representation with a positive constant threshold as in Scott and Suppes (1958). Such representations are usually referred to as Scott-Suppes representations. Our representation theorem presents a Scott-Suppes representation that is the natural analog of the expected utility theorem of von Neumann and Morgenstern (1944). A Scott-Suppes representation is initially obtained for semiorders defined on finite sets (Scott & Suppes, 1958). It is also known that when semiorders are defined on a countably infinite set, they admit a Scott-Suppes representa- 5

6 tion (Manders, 1981). Beja and Gilboa (1992) comments on the difficulties for obtaining a Scott-Suppes representation for semiorders defined on uncountable sets. Recently, both necessary and sufficient conditions for semiorders on uncountable sets to have a Scott-Suppes representation have also been obtained (Candeal & Induráin, 2010). Two papers that contain similar analysis to our work are by Vincke (1980) and Nakamura (1988). Vincke (1980) obtains a linear utility function with a non-negative variable threshold representation for semiordered mixture spaces, whereas Nakamura (1988) provides an analogous analysis for interval orders a binary relation that also allows for intransitivity of indifference. Our paper sharpens the result of Vincke (1980) to linear utility functions with positive constant threshold representations, as semiorders are generically associated with such representations (Scott-Suppes representations). Our main result is a counterpart of Scott-Suppes representations in the risky-choice setting. Hence, our representation constitutes an answer to the open problem noted by Fishburn (1968). 3 Preliminaries We begin this section with some observations from the literature and of our own. 5 We use these results in the next section. 3.1 Semiorders Throughout this paper, X denotes a non-empty set. We say that R is a binary relation on X if R X X. Whenever for some x, y P X, we have px, yq P R, we write x R y. Also, if px, yq R R, we write px R yq. Below, we define some common properties of binary relations. 5 We refer interested readers for further details to the following: Fishburn (1970a), Beja and Gilboa (1992), Candeal and Induráin (2010), Fishburn (1970c), Kreps (1988), Ok (2007), Aleskerov et al. (2007), Pirlot and Vincke (2013). 6

7 Definition. A binary relation R on X is reflexive if for each x P X, x R x, irreflexive if for each x P X, px R xq, complete if for each x, y P X, x R y or y R x, symmetric if for each x, y P X, x R y implies y R x, asymmetric if for each x, y P X, x R y implies py R xq, transitive if for each x, y, z P X, x R y and y R z imply x R z. Let R be a reflexive binary relation on X and x, y P X. We define the asymmetric part of R, denoted P, as x P y if x R y and py R xq and symmetric part of R, denoted I, as x I y if x R y and y R x. A binary relation R on X is a weak order if it is complete and transitive. It is an equivalence relation if it is reflexive, symmetric, and transitive. We state without proof that if R is a weak order on X, then I is an equivalence relation on X. Definition. Let P and I be two binary relations on X. The pair pp, Iq is a semiorder on X if I is reflexive (reflexivity), for each x, y P X, exactly one of xp y, yp x, or xi y holds (trichotomy), for each x, y, z, t P X, x P y, y I z, z P t imply x P t (strong intervality), for each x, y, z, t P X, x P y, y P z, z I t imply x P t (semitransitivity). The definition above is slightly different from the definition of a semiorder introduced by Luce (1956). Both definitions are equivalent however, so our analysis remains unaffected. 6 Next, we provide some immediate observations on semiorders. 6 We show the equivalence of both definitions in the Appendix. 7

8 P is irreflexive. Since I is reflexive, by trichotomy, for each x P X, px P xq. P is asymmetric. Suppose for some x, y P X, x P y. By trichotomy, we cannot have y P x. Hence, py P xq. P is transitive. Suppose for some x, y, z P X, x P y P z. 7 Since I is reflexive, x P y I y P z. By strong intervality, we have x P z. I is symmetric. Suppose for some x, y P X, x I y. By trichotomy, we cannot have y P x or x P y. Hence, y I x. x I y if and only if px P yq and py P xq. Directly follows from trichotomy. Every weak order is a semiorder. Directly follows from the fact that for every weak order R, both P and I are transitive. It is useful to keep in mind that every result that applies to semiorders applies to weak orders as well. Similarly, any (counter) example of a weak order showing that a certain statement about weak orders is not true, also shows that this statement is not true for semiorders. Example 1. We give an example of a canonical semiorder. Let k P R. Define pp, Iq on R as: For each x, y P R x P y if x y k, x I y if x y k. y x k x z x k Figure 1: Example 1 We have x P y and x I z. 7 x P y P z is a shorthand notation for x P y and y P z. 8

9 It is straightforward to show that I is reflexive and pp, Iq satisfies trichotomy. Let x, y, z, t P R. Suppose x P y I z P t. This implies x y and y z k. Hence, x z. Moreover, z P t implies z t x t k k. Thus, k, which implies x P t. So, pp, Iq satisfies strong intervality. Now, suppose x P y P z I t. This implies y z k and z t k. Hence, y t. Moreover, x P y implies x y k. Thus, x t k, which implies x P t. So, pp, Iq satisfies semitransitivity. Therefore, pp, Iq is a semiorder on R. If pp, Iq defined in Example 1 were a weak order, then I would be transitive. Yet, we have intransitive indifference: 0 I k and k I 2k but 2k P 0. Therefore, not every semiorder is a weak order. 8 Looking at the definition of a semiorder, one might wonder why we do not impose the following axiom: For each x, y, z, t P X, x I y, y P z, z P t imply x P t (reverse semitransitivity). 9 It turns out that for any pair of binary relations pp, Iq on X, if I is reflexive and pp, Iq satisfies trichotomy and strong intervality, 10 then pp, Iq is semitransitive if and only if it is reverse semitransitive. In order to see this, let pp, Iq on X be a semiorder. Suppose there are x, y, z, t P X such that x I y P z P t but px P tq. By trichotomy, x I t or t P x. If x I t, then y P z P t I x, which, by semitransitivity, implies y P x. This contradicts x I y. If, on the other hand, t P x, then y P z P t P x, which implies y P x. But, we also have x I y, again a contradiction. In order to see the reverse, suppose pp, Iq on X is reflexive, satisfies trichotomy and 8 In the literature, a semiorder that is not a weak order is also referred to as a typical semiorder. 9 As far as we know, there is not a common name for this axiom. Strong intervality is also referred to as pseudotransitivity (Bridges, 1983). Strong intervality, semitransitivity, and reverse semitransitivity are together referred to as generalized pseudotransitivity in Gensemer (1987). It is also worth noting that strong intervality and semitransitivity axioms are usually given in terms of an irreflexive binary relation P. We again show the equivalence of these axioms stated for P or a pair pp, Iq in the Appendix. 10 Such a relation is called an interval order, which is introduced by Fishburn (1970b). 9

10 strong intervality, and is reverse semitransitive. Moreover, suppose there are x, y, z, t P X such that x P y P z I t but px P tq. By trichotomy, x I t or t P x. If x I t, then t I x P y P z, which, by reverse semitransitivity, implies t P z. This contradicts z I t. If, on the other hand, t P x, then t P x P y P z, which implies t P z. Again, this contradicts z I t, which completes the proof. Definition. Let pp, Iq be a pair of binary relations on X that satisfies trichotomy. We define the following binary relations on X: For each x, y P X, x R y if py P xq (i.e., x P y or x I y), x P 0 y if there exists z P X such that x P z R y or x R z P y, x R 0 y if py P 0 xq, x I 0 y if x R 0 y and y R 0 x. It follows from the definition of R 0 that: x R 0 y if and only if for each z P X, ry R z ñ x R zs and rz R x ñ z R ys. Since the contrapositive of ry R z ñ x R zs is rz P x ñ z P ys and the contrapositive of rz R x ñ z R ys is ry P z ñ x P zs, we also have: x R 0 y if and only if for each z P X, ry P z ñ xp zs and rz P x ñ z P ys. Moreover, for each x, y P X, x P y implies x P 0 y since x P y I y. Notation. In the rest of the paper, we refer to a semiorder pp, Iq on X simply as R P Y I. Now, we present a well-known observation: Lemma 1. Let R be a semiorder on X. For each x, y, z P X, if x R 0 y P z or x P y R 0 z, then x P z. 10

11 Proof. Let R be a semiorder on X and x, y, z P X. Suppose x R 0 y P z but z R x. Since x R 0 y and z R x, we have z R y. This contradicts y P z. Now, suppose x P y R 0 z but z R x. Because y R 0 z and z R x, we have y R x, which contradicts x P y. Next, we give a slightly modified version of an important result of Luce (1956), which shows that R 0 induced by a semiorder R is always a weak order. That is, every semiorder induces a natural weak order. 11 Proposition 1. If R is a semiorder on X, then R 0 is a weak order on X. Proof. Let R be a semiorder on X. Claim 1: R 0 is complete. Suppose on the contrary that there are x, y P X such that pxr 0 yq and pyr 0 xq. This implies that there exist z, z 1 P X such that rx R z P y or x P z R ys and ry R z 1 P x or y P z 1 R xs. Case 1: xrz P y and y Rz 1 P x. Since z P y Rz 1 P x, by strong intervality and the fact that P is transitive, we have z P x. This contradicts x R z. Case 2: x R z P y and y P z 1 R x. Since x R z P y P z 1, by reverse semitransitivity and the fact that P is transitive, we have x P z 1. This contradicts z 1 R x. Case 3: x P z R y and y R z 1 P x. Since z 1 P x P z R y, by semitransitivity and the fact that P is transitive, we have z 1 P y. This contradicts y Rz 1. Case 4: xp z Ry and y P z 1 Rx. Since xp z Ry P z 1, by strong intervality and the fact that P is transitive, we have xp z 1. This contradicts z 1 Rx. Claim 2: R 0 is transitive. Let x, y, z, t P X. Suppose x R 0 y R 0 z. Since x R 0 y and t R x, we have t R y. Similarly, since y R 0 z and t R y, we have t R z. Moreover, if z R t, then, by y R 0 z, we have y R t. Similarly, since x R 0 y and y R t, we have x R t. Hence, x R 0 z. Therefore, R 0 is a weak order on X. 11 See Theorem 1 in Luce (1956). 11

12 In Proposition 1, in order to prove that R 0 is complete, we invoke semitransitivity (and reverse semitransitivity). But can we do better, that is, given a binary relation R on X that is reflexive and that satisfies trichotomy and strong intervality, is the associated R 0 on X complete? The answer is negative, as we show in the following example. 12 Therefore, semitransitivity (and hence reverse semitransitivity) is essential for R 0 to be a weak order. Example 2. Define R on r0, 1s as: For each x, y P p0, 1s, we have x P y if x y and x I y if x y, for each x P r0, 1s, we have 0 I x. Clearly, R is reflexive and satisfies trichotomy. Let x, y, z, t P r0, 1s. We claim that if x P y I z P t, then x P t. Suppose x P y I z P t. Since, for each w P r0, 1s we have 0 I w, x P y and z P t implies that x, y, z, t 0. This with x P y I z P t imply x y z t. Hence, x P t. So, R satisfies strong intervality. Finally, since 0.5 P 0.1 I 0 and 0 I 1 P 0.5, we have 0.5 P 0 0 and 0 P Thus, p0 R 0 0.5q and p0.5 R 0 0q. Therefore, R 0 is not complete. In the next section, we analyze continuity and the relationship between a semiorder and the induced weak order in terms of continuity. 3.2 Continuity From this point on, X tx 1, x 2,..., x n u denotes a set with n P N alternatives. A lottery on X is a list p pp 1, p 2,..., p n q such that p i 1 and for each i P t1, 2,..., nu, we have p i 0, where x i occurs with probability p i. We denote the set of all lotteries on X as L. It is easy to see that for each lottery p, q P L and each α P p0, 1q, we have αp p1 αqq P L. Definition. A reflexive binary relation R on L is continuous if for each q P L, the sets 12 This is an example of an interval order. 12

13 UCpqq : tp P L : p R qu and LCpqq : tp P L : q R pu are closed (with respect to the standard metric on R n ), mixture-continuous if for each p, q, r P L, the sets UMCpq; p, rq : tα P r0, 1s : rαp p1 αqrs R qu and LMCpq; p, rq : tα P r0, 1s : q R rαp p1 αqrsu are closed (with respect to the standard metric on R). The following result presents the relationship between continuity and mixture continuity for a semiorder: Lemma 2. If a semiorder R on L is continuous, then it is mixture-continuous. Proof. Let R be a continuous semiorder on L. Let p, q, r P L, α P R, and let pα n q P UMCpq; p, rq N be a sequence such that pα n q Ñ α. Clearly, since r0, 1s is closed, α P r0, 1s. Furthermore, because for each n P N, rα n p p1 α n qrs P UCpqq and UCpqq is closed, the limit rαp p1 αqrs P UCpqq. Hence, UMCpq; p, rq is closed. Similarly, one can show that LMCpq; p, rq is also closed. Next, we investigate the relationship between R on L and the associated weak order R 0 on L in terms of continuity and mixture-continuity. Example 3. We first provide an example of a continuous semiorder whose associated weak order is not mixture-continuous. Define R on r0, 1s as: for each p P r0, 1s, p I 0.5, for each p, p 1 P p0.5, 1s and q, q 1 P r0, 0.5q, p I p 1, p P q, and q I q 1. 13

14 It is easy to see that R is reflexive and satisfies trichotomy. Moreover, since there are no p, q, r, s P r0, 1s such that pp q I r P s or pp q P r I s, R vacuously satisfies strong intervality and semitransitivity. Hence, R is a semiorder. Furthermore, UCp0.5q LCp0.5q r0, 1s and for each p P p0.5, 1s, q P r0, 0.5q, we have UCppq r0.5, 1s, LCppq r0, 1s, UCpqq r0, 1s, LCpqq r0, 0.5s. Thus, R is continuous p 1 Figure 2: Example 3 Gray segment shows UCppq UCp1q r0.5, 1s. Finally, let p, p 1 P p0.5, 1s, q P r0, 0.5q. Since p P q, p P 0 q. Also p I 0 p 1. Moreover, because p P q I 0.5, we have p P So, UMC 0 p1; 1, 0q : tα P r0, 1s : rα1 p1 αq0s R 0 1u p0.5, 1s, which is not closed. Therefore, R 0 is not mixture-continuous Figure 3: Example 3 Gray segment shows UMC 0 p1; 1, 0q p0.5, 1s. Example 4. We now give an example of a semiorder whose associated weak order is continuous but the semiorder itself is not mixture-continuous. Let L be the set of lotteries on X : tx 1, x 2, x 3 u and ɛ P p0, 0.5s. We define R on L as follows: For each p pp 1, p 2, p 3 q, q pq 1, q 2, q 3 q P L, p P q if p 1 q 1 ɛ, p I q if p 1 q 1 ɛ. 14

15 1 ɛ ɛ p 2 3 Figure 4: Example 4 We have indifference between p and every lottery in the gray area. Since ɛ 0, for each p, q P L, we cannot have both p P q and q P p. This together with the definition of I give reflexivity and trichotomy. Let p, q, r, s P L be such that pp qi rp s. This implies p 1 q 1 ɛ and q 1 r 1 ɛ. Hence, p 1 r 1. Furthermore, r P s implies r 1 s 1 ɛ. Thus, p 1 s 1 ɛ, which implies p P s. So, R satisfies strong intervality. Similarly, one can show that R is also semitransitive. Therefore, R is a semiorder on L. Moreover, for each p, q P L, p R 0 q if and only if p 1 q 1. To see this, let p, q, r P L. Suppose p 1 q 1, if q 1 r 1 ɛ, then p 1 r 1 ɛ. On the other hand, if r 1 p 1 ɛ, then r 1 q 1 ɛ. That is, for each r P L, [q P r ñ p P r] and [r P p ñ r P q]. Hence, p 1 q 1 implies p R 0 q. Next, suppose p 1 q 1 and r 1 p 1 q 1 ɛ. Then, since q 2 1 r 1 ɛ, q P r and since r 1 p 1 ɛ, r I p. Thus, we have q P r R p, and hence q P 0 p. So, for each p, q P L, if p 1 q 1, then pp R 0 qq, equivalently, p R 0 q implies p 1 q 1. Therefore, p R 0 q if and only if p 1 q 1. An immediate corollary is that R 0 is continuous. Finally, consider UMCpp1, 0, 0q; p1 ɛ, ɛ{2, ɛ{2q, p1, 0, 0qq. It is easy to 15

16 show that UMCpp1, 0, 0q; p1 ɛ, ɛ{2, ɛ{2q, p1, 0, 0qq r0, 1q, which is not closed. Therefore, R is not mixture-continuous. 3.3 Independence Now, we analyze whether the independence axiom is compatible with semiorders representing intransitive indifference. Definition. A reflexive binary relation R on L satisfies independence if for each p, q, r P L and each α P p0, 1q, p P q if and only if rαp p1 αqrs P rαq p1 αqrs, midpoint indifference 13 if for each p, q, r P L, p I q implies r1{2p 1{2rs I r1{2q 1{2rs. It is easy to see that, if a semiorder R on L satisfies independence, then it also satisfies midpoint indifference. Below, we show that a semiorder satisfying the independence axiom cannot have intransitive indifference. 14 that of weak orders are equivalent under independence. Therefore, the study of semiorders and Proposition 2. Let R be a semiorder on L. If R satisfies independence, then I is transitive. Proof. Let R be a semiorder on L that satisfies independence. Suppose there are p, q, r P L such that p I q I r but p P r. Independence and p P r together imply that for each α P p0, 1q, p P rαp p1 αqrs and rαp p1 αqrs P r. Since p P rαp p1 αqrs P r I q, by semitransitivity, p P q. This contradicts p I q. 13 This property is introduced by Herstein and Milnor (1953). 14 A similar observation is made by Fishburn (1968). 16

17 Our main result shows that this is not the case for midpoint indifference. That is, midpoint indifference is compatible with semiorders representing intransitive indifference. 3.4 Utility Representations Let R be a binary relation on X. We say u : X ÝÑ R is a utility representation of R if for each x, y P X, x P y if and only if upxq upyq. A standard utility representation that allows for intransitive indifference is: Definition. Let R be a binary relation on X, u : X ÝÑ R be a function, and k P R. The pair pu, kq is a Scott-Suppes representation of R if for each x, y P X, x P y if and only if upxq upyq k. Here k acts as a threshold of utility discrimination, that is, if the absolute value of the utility difference between two alternatives is less than or equal to k, then it is as if the decision maker cannot consider these two alternatives to be significantly different from each other. Equivalently, one can think that for the decision maker to prefer one alternative over the other, there is a certain utility threshold to be exceeded. If a decision maker s preferences can be represented by such a utility function, then the decision maker acts as if his choice is satisficing when it gives him a utility within k neighborhood of the alternative(s) that maximize(s) the utility function u : X ÝÑ R. A reflexive binary relation R on X is non-trivial if there exist x, y P X such that x P y. We say x P X is maximal with respect to R if for each y P X, x R y. Similarly, x P X is minimal with respect to R if for each y P X, y R x. We denote the set of all maximal and minimal elements of X with respect to R as M R and m R, respectively. We state two more properties that we employ in our main result: Definition. Let R be a semiorder on X (of arbitrary cardinality) and S X. We say S has maximal indifference elements in X with respect to R if 17

18 for each s P S, there exists x P X such that s I x and for each y P X, y P 0 x implies y P s. Definition. Let u : L ÝÑ R be a function. We say that u is linear if for each p, q P L and each α P r0, 1s, upαp p1 αqqq αuppq p1 αqupqq. An important theorem that we use in proving our main result is due to Vincke (1980): Theorem 1. Let pp, Iq be a pair of binary relations on L. Then, pp, Iq is a semiorder, R 0 is mixture-continuous and satisfies midpoint indifference, LzM R has maximal indifference elements in L with respect to R if and only if there exist a linear function u : L ÝÑ R and a non-negative function σ : L ÝÑ R such that for each p, q P L, we have p P q if and only if uppq upqq σpqq, p I q if and only if uppq σppq upqq and upqq σpqq uppq, p I 0 q if and only if uppq upqq, uppq upqq implies uppq σppq upqq σpqq, uppq upqq implies σppq σpqq. Proof. See Vincke (1980). 4 Expected Scott-Suppes Utility Representation Before moving on with our main result, we introduce two more properties that we employ in our main theorem: 18

19 Definition. A reflexive binary relation R on L is regular if there are no p, q P L and no sequences pp n q, pq n q P L N such that for each n P N, we have p P p n and p n 1 P p n or for each n P N, we have q n P q and q n P q n 1. This condition also appears in Manders (1981), Beja and Gilboa (1992), and Candeal and Induráin (2010) in connection with Scott-Suppes representations. In words, a binary relation is regular if its asymmetric part has no infinite up or down chains with an upper or lower bound, respectively. Definition. A reflexive binary relation R on L is mixture-symmetric if for each p, q P L and each α P r0, 1s, pi rαp p1 αqqs implies q I rαq p1 αqps. This axiom is introduced in Nakamura (1988) to obtain a utility representation with a constant threshold for interval orders. We use this axiom for semiorders with similar purposes. 4.1 The Main Result We are now ready to state and prove our main theorem. Theorem 2 (Expected Scott-Suppes Utility Representation). Let R be a non-trivial semiorder on L. Then, R is regular and mixture-symmetric, R 0 is mixture-continuous and satisfies midpoint indifference, and LzM R has maximal indifference elements in L with respect to R if and only if there exists a linear function u : L ÝÑ R and k P R that pu, kq is a Scott-Suppes representation of R. such Proof. ( ùñ ) We first show that the axioms imply the existence of an expected Scott-Suppes utility representation. Since all of the hypotheses of Theorem are satisfied, there is a linear function u : L ÝÑ R and a non-negative function σ : L ÝÑ R such that for each p, q P L, we have: 19

20 (i). p P q if and only if uppq upqq σpqq, (ii). p I q if and only if uppq σppq upqq and upqq σpqq uppq, (iii). p I 0 q if and only if uppq upqq, (iv). uppq upqq implies uppq σppq upqq σpqq, (v). uppq upqq implies σppq σpqq. Moreover, it is straightforward to show that: 15 (vi). p R 0 q if and only if uppq upqq, (vii). p P 0 q if and only if uppq upqq. Our initial aim is to show that for each p, q P LzM R, σppq σpqq 0. Since R is non-trivial, the set of all non-maximal elements of X with respect to R is non-empty. Claim 1: For each p P LzM R, σppq 0. Suppose, on the contrary, that there is a p P LzM R such that σppq 0. Since p is non-maximal, there exists q P L such that q P p. Therefore, upqq uppq. Because u is linear, this implies for each α P p0, 1q we have upqq upαp p1 αqqq uppq. Furthermore, since σppq 0, we have uppq σppq uppq upαp p1 αqqq, which implies, by (i), for each α P p0, 1q, rαp α P p0, 1q, we have σpαp an α P p0, 1q such that σp αp p1 αqqs P p rpqs. This implies that for each p1 αqqq 0. To see why, suppose there is p1 αqqq 0. We have two cases: Case 1: up αp p1 αqqq σp αp p1 αqqq upqq. Since upqq uppq and σpqq 0, we have upqq σpqq uppq. This together with up αp p1 αqqq σp αp p1 αqqq upqq imply, by (ii), 15 Vincke (1980) applies Herstein and Milnor (1953) s utility representation theorem to R 0 and obtains the linear function u : L ÝÑ R. Since R 0 is a weak order and satisfies mixture continuity and midpoint indifference, it follows directly from Herstein and Milnor (1953) s representation theorem that (vi) and (vii) hold. 20

21 q I r αp p1 αqqs. Therefore, mixture symmetry implies p I r αq p1 αqps. By pq, this contradicts tricothomy. Case 2: up αp p1 αqqq σp αp p1 αqqq upqq. Since u is linear, αuppq p1 αqupqq σp αp p1 αqqq α σp αp p1 αqqq 0. Define β P p0, αq as follows: upqq uppq upqq. Hence, β : α σp αp p1 αqqq. upqq uppq By construction, the linearity of u implies upβp p1 βqqq up αp p1 αqqq σp αp p1 αqqq. Since σpβp p1 βqqq 0, we have both upβp p1 βqqq σpβp p1 βqqq up αp p1 αqqq and up αp p1 αqqq σp αp p1 αqqq upβp p1 βqqq. Thus, by (ii), r αp p1 αqqs I rβp p1 βqqs. Moreover, since rβp p1 βqqs α β rp qp αp p1 αqqq p qqs, we get r αp p1 αqqs I rp qp αp p1 β α α β α αqqq p α β α β qqs. So, mixture symmetry implies q I rp qq p α β α qp αp α p1 αqqqqs rp1 β αqq p α βqps. Mixture symmetry once again implies pi rp1 β αqp p α βqqs. By pq, this contradicts trichotomy. 6 If p P LzM R, q P L are such that σppq 0 and q P p, then for each α P p0, 1q we have σpαp p1 αqqq 0. Next, for each n P N, let α n 1{pn 2q. Because, for each α P p0, 1q, σpαp p1 αqqq 0, we have q P P rα n 1 p p1 α n 1 qqs P rα n p p1 α n qqs P P rα 1 p p1 α 1 qqs. This contradicts regularity. Therefore, for each p P LzM R, we have σppq 0. Next, we provide three results that we use in proving our next claim. Lemma 3. For each p, r, s P L, if r P p and upsq uprq σppq, then r I s. Proof. If r P p, by (i), uprq uppq σppq. Therefore, 0 σppq 1. ruprq uppqs 21

22 Define γ P p0, 1q as follows: γ : 1 σppq uprq uppq. Then, by construction, and since u is linear, upγp p1 γqrq uppq σppq. Hence, p I γp p1 γqr. By mixture symmetry, r I γr p1 γqp. Moreover, by definition, σppq p1 γqruprq uppqs. Let s P L be such that upsq uprq σppq. Then, by linearity of u, upsq upγr we also have σpsq σpγr p1 γqpq. Therefore, r I s. p1 γqpq. Thus, by (v), Lemma 4. For each p, q, r, t P L, if r P q P 0 p, upqq uprq σppq, and uptq upqq σppq, then q I t. Proof. If r P q P 0 p, by Lemma 1, r P p. This implies uprq uppq σppq. Hence, by convexity of L and linearity of u, there exists s P L such that upsq uprq σppq. Thus, by Lemma 3, r I s. Since uprq upsq upqq, there is a δ P r0, 1s such that s δr p1 δqq. This means r I δr p1 δqq. By mixture symmetry, q I δq p1 δqr. Furthermore, since upsq uprq σppq, we have σppq p1 δqruprq upqqs. Let t P L such that uptq upqq σppq. This implies, by linearity, uptq δupqq p1 δquprq upδq p1 δqrq. Hence, by (v), we also have σptq σpδq p1 δqrq. Therefore, q I t. Lemma 5. For each p, q, r P L, if r P q P 0 p and upqq uprq σppq, then σpqq σppq. Proof. By convexity of L and linearity of u, there is a t P L such that uptq upqq σppq. Hence, by Lemma 4, qi t. Thus, by (ii), upqq σpqq uptq. Therefore, σpqq σppq. Claim 2: For each p, q P LzM R, σppq σpqq. Suppose, on the contrary, that there are p, q P LzM R such that σppq σpqq. By (v), uppq upqq implies σppq σpqq and, by (iii), p I 0 q if and only if uppq upqq. Hence, pp I 0 qq. Thus, p P 0 q or q P 0 p. Without 22

23 loss of generality, suppose q P 0 p. By (vii), upqq uppq. Moreover, since q is non-maximal, there is an r P L such that r P q. This implies r P q P 0 p. We have two cases: Case 1: upqq uprq σppq. Then, by Lemma 5, σpqq σppq. Since σpqq σppq, we have σpqq σppq. Let s P L such that upsq uprq σpqq. Because r P q, by Lemma 3, r I s. Since uprq upsq upqq uppq, there is an η P p0, 1q such that s ηr p1 ηqp. This implies r I ηr p1 ηqp. Hence, by mixture symmetry, p I ηp p1 ηqr. Then, by (ii), we have uppq σppq upηp p1 ηqrq. This means, by linearity, σppq p1 ηqruprq uppqs. But, upηr p1 ηqpq uprq σpqq, which implies p1 ηqruprq uppqs σpqq. Therefore, σppq σpqq, which contradicts σpqq σppq. Case 2: upqq uprq σppq. Now, let s P L be such that upsq uprq σppq. Then, upqq upsq, which implies, by (vi), q P 0 s. Hence, r P q P 0 s. Thus, by Lemma 1, r P s. But, by Lemma 3, we also have r I s. This contradicts trichotomy. 6 For each p, q P LzM R, we have σppq σpqq. Now, for each p P LzM R, let k : σppq 0. Since u is linear (and hence continuous) and L is compact, there are r P L such that for each q P L, p, up rq upqq and upqq up pq. Clearly, σp pq k. If r P M R, then for each q P L, uprq σprq upqq. Therefore, if for each r P M R, up rq uprq k, then for each r P M R, replacing σprq with k yields pu, kq as a Scott-Suppes representation of R. We now complete our proof by showing that this is, indeed, the case. Claim 3: For each r P M R, up rq uprq k. Suppose, on the contrary, that there exists r 1 P M R such that up rq upr 1 q k. Since upr 1 q σpr 1 q up rq, by (ii), r I r 1. Moreover, since up rq upr 1 q up pq, by linearity of u, there is a λ P r0, 1s such that 23

24 upr 1 q λup rq p1 λqup pq upλ r p1 λq pq. Then, by (v), we have r I λ r p1 λq p. Hence, mixture symmetry implies I λ p p p1 λq r. This implies, by (ii), up pq k upλ p p1 λq rq. Thus, by linearity of u, k p1 λqrup rq up pqs. But, since up rq upr 1 q k and upr 1 q upλ r p1 λq pq, by linearity of u, p1 λqrup rq up pqs k, a contradiction. ( ðù ) Next, we show that expected Scott-Suppes utility representation implies our axioms. Suppose there exists a linear function u : L ÝÑ R and k P R such that pu, kq is a Scott-Suppes representation of R. It is straightforward to show that since k P R, I is reflexive and R satisfies trichotomy. Let p, q, r, s P L. Suppose p P q I r P s. This implies uppq upqq k and upqq uprq k. Hence, uppq uprq. Moreover, r P s implies uprq upsq k. Thus, uppq upsq k, which implies p P s. So, R satisfies strong intervality. Now, suppose p P q P r I s. This implies, uppq upqq k, upqq uprq k, and uprq upsq k. Hence, uppq uprq 2k, and hence uppq upsq k. This means p P s. That is, R satisfies semitransitivity as well. Therefore, R is a semiorder on L. Suppose R is not regular, that is, there exist p P L and pp n q P L N such that for each n P N, we have p P p n and p n 1 P p n, or there exist q P L and pq n q P L N such that for each n P N, we have q n P q and q n P q n 1. Suppose that there exist p P L and pp n q P L N such that for each n P N, we have p P p n and p n 1 P p n. That is, suppose p is an upperbound for an infinite chain of P, the asymmetric part of R. Let d uppq upp 1 q. Clearly, d k. Take i rd{ks. Then, upp i 1 q upp 1 q ik upp 1 q d upp 1 q uppq upp 1 q uppq. This contradicts p P p i 1. The existence of a lower bound for an infinite chain of P yields a similar contradiction. Hence, P is regular. Let p, q P L and α P p0, 1q. Suppose p I αp p1 αqq. This implies uppq upαp p1 αqqq k. Since u is linear, uppq rαuppq p1 αqupqqs k. Rearranging the terms gives rαupqq p1 αquppqs upqq k. Hence, 24

25 q I αq p1 αqp. Thus, R is mixture-symmetric. It is easy to show that for each p, q P L, p R 0 q if and only if uppq upqq. Since u is a continuous function, the preimage of a closed set is closed. Hence, R 0 is continuous. This implies that R 0 is mixture-continuous. Let p, q P L with p I 0 q. This implies uppq upqq. Hence, for each r P L, 1{2uppq 1{2uprq 1{2upqq 1{2uprq. The linearity of u implies up1{2p 1{2rq up1{2q 1{2rq. Thus, r1{2p 1{2rs I 0 r1{2q 1{2rs. So, R 0 satisfies midpoint indifference. Finally, suppose p P LzM R. This implies that there is an r P L such that r P p. Hence, uprq uppq k. Thus, by linearity of u, there is a q P L such that upqq uppq k. So, p I q. Moreover, if for some s P L, s P 0 q, then upsq uppq k. This implies s P p. Therefore, LzM R has maximal indifference elements in L with respect to R. 4.2 Uniqueness Next, we note that the expected Scott-Suppes utility representation is unique up to affine transformations: Proposition 3. Let pu, kq be an expected Scott-Suppes utility representation of a semiorder R on L, α P R, and β P R. If v : L ÝÑ R is such that for each p P L, vppq αuppq β, then pv, αkq is also an expected Scott-Suppes utility representation of R. Proof. Let p, q P L. Since α P R and β P R, we have p P q if and only if uppq upqq k if and only if αpuppqq β αpupqq kq β if and only if vppq vpqq αk. Hence, pv, αkq is a Scott-Suppes representation of R. Moreover, as an affine transformation of a linear function, v is also linear. 4.3 Independence of the Axioms Let R be a non-trivial semiorder on L. Consider the following axioms: 25

26 R is regular (reg), R is mixture-symmetric (mix-sym), R 0 is mixture-continuous (mix-cont), R 0 satisfies midpoint indifference (mid indiff), LzM R has maximal indifference elements in L with respect to R (max indiff). First, we present an example that shows that the axioms listed above are compatible, i.e., in the example below, they all hold simultaneously: Example 5. Let L be the set of lotteries on X : tx 1, x 2, x 3 u, p, q P L, and ɛ P p0, 0.5s. We define R on L such that: p P q if p 1 q 1 ɛ, p I q if p 1 q 1 ɛ. 1 ɛ ɛ p 2 3 Figure 5: Example 5 We have indifference between p and every lottery in the gray area. It is straightforward to show that R is a non-trivial semiorder. 26

27 Reg. Since ɛ 0, it is easy to show that R is regular. Mix-sym. Let p, q P L and α P p0, 1q. Suppose p I rαp implies p 1 αp 1 q 1 αq 1 ɛ. p1 αqqs. This Rearranging the terms gives αq 1 p1 αqp 1 q 1 ɛ. Hence, q I rαq p1 αqps. Thus, R is mixture-symmetric. Mix-cont. It is easy to see that for each p, q P L, pr 0 q if and only if p 1 q 1. Hence, R 0 is continuous, which implies that it is mixture-continuous. Mid indiff. Let r P L. Suppose for some p, q P L, p I 0 q. Because for each p, q P L, p I 0 q if and only if p 1 q 1, we have p 1 q 1. Hence, 1{2p 1 1{2r 1 1{2q 1 1{2r 1. Thus, r1{2p 1{2rs I 0 r1{2q 1{2rs. Max indiff. Let p P LzM R. This implies 0 p 1 1 ɛ. Define p 1 pp 1 ɛ, 1 pp 1 ɛq, 0q. Since 1 p 1 ɛ 0, p 1 P L. Moreover, p I p 1. Let q P L. Suppose q P 0 p 1. Because for each q P L, q P 0 p 1 if and only if q 1 p 1 1, we have q 1 p 1 ɛ. Hence, q P p. Now, we show that the axioms in our main result (Theorem 2) are mutually independent by providing an example for each axiom. Example 6 (Reg, Mix-sym, Mix-cont, Mid indiff Max indiff). Let L be the set of lotteries on X : tx 1, x 2, x 3 u, p, q P L, and ɛ P p0, 0.5s. We define R on L such that: p P q if p 1 q 1 ɛ, p I q if p 1 q 1 ɛ. Since ɛ P p0, 1q, R is non-trivial. In Example 4 we show that R is a semiorder. Reg. Since ɛ 0, it is straightforward to show that R is regular. 27

28 Mix-sym. Let p, q P L and α P p0, 1q. Suppose p I rαp implies p 1 αp 1 q 1 αq 1 ɛ. p1 αqqs. This Rearranging the terms gives αq 1 p1 αqp 1 q 1 ɛ. Hence, q I rαq p1 αqps. Mix-cont. In Example 4, we also show that R 0 is continuous, and hence R 0 is mixture-continuous. Mid indiff. It is easy to show that for each p, q P L, p I 0 q if and only if p 1 q 1. Let r P L. Suppose for some p, q P L, p I 0 q. This implies p 1 q 1. Hence, 1{2p 1 1{2r 1 1{2q 1 1{2r 1, which implies r1{2p 1{2rs I 0 r1{2q 1{2rs. Max indiff. Let p p0, 0, 1q. Since p1, 0, 0q P p, p P LzM R. Take p 1 I p, then p 1 1 ɛ. Let q prp1 1 ɛs{2, 1 rp1 1 ɛs{2, 0q. It is easy to see that q P 0 p 1 but p I q. Thus, LzM R does not have maximal indifference elements in L with respect to R. Example 7 (Reg, Mix-sym, Mix-cont, Max indiff Mid indiff). Let L be the set of lotteries on X : tx 1, x 2 u and p, q P L. We define R on L such that: p P q if p 1 q 1 0.6, p I q if p 1 q It is similar to Example 5 to show that R is a non-trivial semiorder. Reg, Mix-sym, Max indiff. Showing that these axioms hold is also similar to Example 5. 28

29 Mix-cont, Mid indiff. However, unlike Example 5, for each p, q P L, we neither have pr 0 q if and only if p 1 q 1 nor pi 0 q if and only if p 1 q 1. This is because for each p, q P L, if p 1, q 1 P r0.4, 0.6s, then p I 0 q. On the other hand, for each p, q P L, if p 1 P r0, 0.4q Y p0.6, 1s and q 1 P r0, 1s, then still pr 0 q if and only if p 1 q 1, and p I 0 q if and only if p 1 q 1. Hence, it is straightforward to see that R 0 is mixture-continuous. Moreover, p0.6, 0.4q I 0 p0.4, 0.6q but 1{2p0.6, 0.4q 1{2p1, 0q p0.8, 0.2q P 0 p0.7, 0.3q 1{2p0.4, 0.6q 1{2p1, 0q. This is because if p 1 P r0, 0.4q Y p0.6, 1s, then p P 0 q if and only if p 1 q 1. Thus, R 0 does not satisfy midpoint indifference. Example 8 (Reg, Mix-sym, Mid indiff, Max indiff Mix-cont). Let L be the set of lotteries on X : tx 1, x 2 u and p, q P L. We define R on L such that: p P q if p 1 1 and q 1 0, p I q if pp P qq and pq P pq. Since p1, 0q P p0, 1q, R is non-trivial. Moreover, it is straightforward to show that R is reflexive and satisfies trichotomy. Because the only strict preference is p1, 0q P p0, 1q, there are no p, q, r, s P L such that p P q I r P s or p P q P r I s. Therefore, R vacuously satisfies strong intervality and semitransitivity. Hence, R is a semiorder. Reg. Since the only strict preference is p1, 0qP p0, 1q, R is trivially regular. Mix-sym. Let p P L. Suppose p 1 P p0, 1q. Then, for each q P L, p I q. Moreover, for each α P p0, 1q, p0, 1q I rαp0, 1q p1 αqp1, 0qs I p1, 0q. Hence, R is mixture-symmetric. Mid indiff. For each p P Lztp1, 0q, p0, 1qu, we have p1, 0q P 0 p P 0 p0, 1q. Furthermore, for each p, q P Lztp1, 0q, p0, 1qu, we have p I 0 q. Therefore, R 0 29

30 satisfies midpoint indifference. Max indiff. Since the only strict preference is p1, 0q P p0, 1q, LzM R tp0, 1qu. Moreover, p0.5, 0.5q P 0 r only if r p0, 1q. Hence, p1, 0q P 0 p0.5, 0.5q and p1, 0q P p0, 1q together imply that LzM R has maximal indifference elements in L with respect to R. Mix-cont. UMC 0 pp0.5, 0.5q; p0, 1q, p1, 0qq : tα P r0, 1s : rαp0, 1q p1 αqp1, 0qs R 0 p0.5, 0.5qu r0, 1q, which is not closed. Therefore, R 0 is not mixture-continuous. Example 9 (Reg, Mix-cont, Mid indiff, Max indiff Mix-sym). Let L be the set of lotteries on X : tx 1, x 2 u and p, q P L. We define R on L such that: p P q if 2p 1 3q 1 0.5, p I q if 2p 1 3q Since p1, 0q P p0, 1q, R is non-trivial. Let p P L. Define u : L ÝÑ R as uppq lnpp 1 0.5q. It is straightforward to show that pu, lnp3{2qq is a Scott- Suppes representation of R. Hence, R is a semiorder Reg. Since R has a Scott-Suppes representation, it is regular. Mix-cont. It is straightforward to show that for each p, q P L, p R 0 q if and only if p 1 q 1. Hence, R 0 is continuous, which implies that it is mixture-continuous. Mid indiff. It is also easy to see that for each p, q P L, p I 0 q if and only if p 1 q 1. Let r P L. Suppose that for some p, q P L, p I 0 q. This implies p 1 q 1. Hence, 1{2p 1 1{2r 1 1{2q 1 1{2r 1. Thus, r1{2p 1{2rs I 0 r1{2q 1{2rs. 30

31 Max indiff. Let p P LzM R. This implies p 1 1{2. Define p 1 pp6p 1 1q{4, 1 rp6p 1 1q{4sq. Because 1{2 p 1 0, p 1 P L. Furthermore, pi p 1. Since for each q P L, q P 0 p 1 if and only if q 1 p 1 1, we have q P p. Hence, LzM R has maximal indifference elements in L with respect to R. Mix-sym. We have p1, 0q I p0.5, 0.5q 1{2p1, 0q 1{2p0, 1q. But, pp0, 1q I r1{2p0, 1q 1{2p1, 0qsq. Hence, R 0 is not mixture-symmetric. Example 10 (Mix-sym, Mix-cont, Mid indiff, Max indiff Reg). Let L be the set of lotteries on X : tx 1, x 2 u and p, q P L. We define R on L such that: p P q if p 1 q 1, p I q if p 1 q 1. Since p1, 0q P p0, 1q, R is non-trivial. Moreover, because R is a weak order, it is a semiorder. Mix-sym. Let p, q P L and α P p0, 1q. If p I rαp p1 αqqs, then p 1 αp 1 p1 αqq 1. This implies q 1 αq 1 p1 αqp 1. Hence, q I rαq p1 αqps. Mix-cont. It is straightforward to show that for each p, q P L, p R q if and only if p R 0 q if and only if p 1 q 1. Hence, R 0 is continuous, which implies that it is mixture-continuous. Mid indiff. It is also easy to show that for each p, q P L, p I q if and only if p I 0 q if and only if p 1 q 1. Let r P L. Suppose for some p, q P L, p I 0 q. This implies p 1 q 1. Hence, 1{2p 1 1{2r 1 1{2q 1 1{2r 1. Thus, r1{2p 1{2rs I 0 r1{2q 1{2rs. Max Indiff. Let p P LzM R. This implies p 1 1. Moreover, p I p 1 only if p p 1. Let q P L. Suppose q P 0 p. Since for each q P L, q P 0 p if and only 31

32 if q P p if and only if q 1 p 1, we have q P p. Hence, LzM R has maximal indifference elements in L with respect to R. Reg. It is easy to see that for each p, q P L, p P q if and only if p 1 q 1. Take p n p n, 1 q. It follows that for each n n 1 n 1 P N, we have p1, 0q P p n and p n 1 P p n. Hence, R is not regular. 5 On Epsilon Equilibrium Next, we consider the relationship between our expected Scott-Suppes representation and the concept of epsilon equilibrium. Let N be a set of players, A i be the set of actions available to player i P N, and R i be the reflexive binary relation that represents the preferences of player i P N over the set of lotteries on the set of action profiles. We denote the set of all (pure) action profiles as A : ipn A i and the set of all lotteries on A as paq. That is, xn, pa i q ipn, pr i q ipn y is a normal form game. A (possibly) mixed action profile is an equilibrium if no player has a unilateral deviation that makes him strictly better off. 16 This translates into the following definition: Definition. A (possibly mixed) action profile σ pσ i, σ i q P paq is an equilibrium of xn, pa i q ipn, pr i q ipn y if for each i P N there does not exist a i P A i such that pa i, σ i q P i σ. Suppose now that the reflexive binary relation representing the preferences of each agent i P N, R i, is a non-trivial semiorder that satisfies the axioms of our representation theorem. Then, by our main result, R i has an expected Scott-Suppes utility representation pu i, k i q. That is, there is a linear 16 We remark that this is one of the standard definitions of Nash equilibrium. But, we refrain from referring it as a Nash equilibrium since an analogous construction can be adopted to other equilibrium concepts as well. 32

33 utility function u i : paq ÝÑ R and k P R we have p P i q if and only if u i ppq u i pqq equilibrium, under our axioms, is equivalent to: such that for each p, q P paq, k i. Hence, the definition of an Definition. A (possibly mixed) action profile σ pσ i, σ i q P paq is an equilibrium of xn, pa i q ipn, pr i q ipn y if for each i P N there does not exist a i P A i such that u i ppa i, σ i qq u ipσ q k i. Furthermore, our representation is unique up to affine transformations as described in Proposition 3. Now, fix an ɛ 0. For each i P N, let γ i ɛ k i and v i : paq ÝÑ R be defined as, for each p P paq, v i ppq : γ i u i ppq. Thus, Proposition 3 implies that pv i, ɛq is another expected Scott-Suppes representation of R i. Therefore, the definition of an equilibrium, under this expected Scott-Suppes representation, becomes: Definition. A (possibly mixed) action profile σ pσ i, σ i q P paq is an equilibrium of xn, pa i q ipn, pr i q ipn y if for each i P N and for each a i P A i we have v i pσ q v i ppa i, σ i qq ɛ. We would like to point out that the epsilon in the above definition is the same for each player. We are able to obtain such a fixed epsilon by rescaling the linear utility functions obtained through our expected Scott- Suppes representation theorem for each player i P N. 17 Moreover, this is the same definition given by Radner (1980) for epsilon equilibrium. Therefore, our representation theorem provides a reinterpretation of the concept of epsilon equilibrium: 17 Argenziano and Gilboa (2017) points out that the utility functions that represent semiorders carry a cardinal meaning, and hence just noticeable differences provide a common unit of measure for interpersonal comparisons of utility differences. Whether we are able to make interpersonal utility comparisons between players by rescaling for each player i P N the constant k i in our expected Scott-Suppes utility representation is an interesting question. We are moot on this question since the welfare implications of just noticeable differences are not as clear under uncertainty. 33

34 In most of the applications, economists construct preferences of agents after observing their choice behavior. The reason why preferences are constructed as weak orders is mainly due to tractability, i.e., to have measurable utility functions. However, it is possible that the underlying preferences exhibit intransitive indifference and because of missing choice data (and due to the weak order convention), we might observe outcomes that look like an epsilon equilibrium. It might also be the case that the revealed preferences of agents look like a weak order over deterministic outcomes. But, this does not have to be the case for lotteries over these outcomes especially when respective probabilities are close to each other. 18 Whatever the underlying reason, our representation highlights a decision theoretical foundation for why we might observe outcomes that look like an epsilon equilibrium. 6 Conclusion In this paper, we study intransitive indifference under uncertainty. In particular, we identify necessary and sufficient conditions for a semiorder over the set of lotteries to have a Scott-Suppes representation via a linear utility function. Our main result (Theorem 2) employs all of the axioms in Vincke (1980). On top of these axioms, we employ two more axioms, one introduced by Nakamura (1988), and another well-known axiom in the literature (Manders, 1981; Beja & Gilboa, 1992; Candeal & Induráin, 2010). These two additional axioms help us convert the non-negative threshold function of Vincke (1980) s representation into a positive constant threshold, which leads to our expected Scott-Suppes representation. We also show that all of these axioms are compatible and mutually independent. Our representation theorem is the natural analog of the expected utility theorem of von Neumann and Morgenstern (1944) for semiorders in the sense 18 Another reason why we observe epsilon equilibrium might be due to learning as in Kalai and Lehrer (1993). 34