Multiple Temptations 1

Transcription

1 Multiple Temptations 1 John E. Stovall 2 University of Rochester JOB MARKET PAPER 3 Forthcoming in Econometrica November 9, I would like to thank Val Lambson, Eddie Dekel, Bart Lipman, numerous seminar audiences for their comments. A co-editor three anonymous referees provided very useful comments. I especially thank my advisor, Larry Epstein, for his guidance. The results presented here were originally distributed in a paper titled Temptation Self- Control as Duals. 2 jstovall@mail.rochester.edu 3 NOTE: Those unfamiliar with the decision theory literature on temptation may want to rst read Appendix D, which provides an illustrative example. Appendix D is supplementary, will not be appearing in the nal version in Econometrica.

2 Abstract We use a preference-over-menus framework to model a decision maker who is a ected by multiple temptations. Our two main axioms on preference, Exclusion Inclusion, identify when the agent would want to restrict his choice set when he would want to exp his choice set. An agent who is tempted would want to restrict his choice set by excluding the normatively worst alternative of that choice set. Simultaneously, he would want to exp his choice set by including a normatively superior alternative. Our representation identi es the agent s normative preference temptations, suggests the agent is uncertain which of these temptations will a ect him. We provide examples to illustrate how our model improves on those of Gul Pesendorfer (2001) Dekel, Lipman, Rustichini (2009).

3 1 Introduction We use a preference-over-menus framework to model a decision maker who is a ected by multiple temptations. We regard temptation as a craving or desire that is di erent from the agent s normative preference (i.e. his view of how he should choose between alternatives, absent temptation). When the agent cannot simultaneously satisfy a temptation his normative desire, he is con icted. He wants to avoid such con ict make normatively good choices. Our two main axioms on preference, Exclusion Inclusion, identify when the agent would want to restrict his choice set when he would want to exp his choice set. An agent who is tempted would want to restrict his choice set by excluding the normatively worst alternative of that choice set, thus avoiding a potential con ict between a temptation his normative preference. At the same time, he would want to exp his choice set by including an alternative that is normatively superior, thus potentially helping him make a normatively good choice. Using the axioms Exclusion Inclusion among others, Theorem 1 characterizes the following representation: I V T (x) = q i [u () + v i ()] v i (), (1) i=1 where q i > 0 for all i, P i q i = 1, u each v i are von Neumann-Morgenstern expected-utility functions. For any singleton menu, V T (fg) = u (). Hence u represents the agent s normative preference. We interpret i to be a subjective state to which the agent assigns probability q i. In state i, v i is the temptation that a ects the agent. He compromises between his normative preference temptation preference chooses the alternative that imizes u + v i. However he experiences the disutility v i (), which is the foregone utility from the most tempting alternative. Thus V T takes an expected-utility form, where [u () + v i ()] v i () is the utility attained in state i, suggesting the agent is uncertain which of his temptations will a ect him. This work is most closely related to the seminal paper by Gul Pesendorfer (2001) a recent paper by Dekel, Lipman, Rustichini (2009) (henceforth GP DLR respectively). Conceptually, our model can be viewed as a compromise between these two as the set of preferences we consider is a special case of DLR s a generalization of GP s. In Section 3, we discuss the relationship between these models. Through examples, we argue for the relaxation of GP s model strengthening of DLR s. 2 The Model The agent in our model chooses between menus, which are sets of lotteries. It is understood, though unmodeled, that he will later choose an alternative from the 1

4 menu he chooses now. Thus we think of a menu as being the agent s future choice set. We assume that when choosing a menu, the agent is in a cool state (meaning he is not tempted by any alternatives when considering a menu), but that he anticipates being in a hot state at the time of choosing an alternative. 1 The agent s normative preference is identi ed with his preference over singleton menus. Formally, let denote the set of probability distributions over a nite set of prizes, call 2 a lottery. Let denote the set of closed nonempty subsets of, call x 2 a menu. We endow with the Hausdor topology de ne the mixture operation x + (1 ) y f + (1 ) 0 : 2 x, 0 2 yg. for 2 [0; 1]. Our primitive is a binary relation over which represents the agent s preference. Consider the following axioms. Axiom 1 (Weak Order) is complete transitive. Axiom 2 (Continuity) The sets fx : x yg fx : y xg are closed. Axiom 3 (Independence) If x y, then for every z 2 2 (0; 1], x + (1 ) z y + (1 ) z. These are straightforward extensions of the stard expected-utility axioms. See Dekel, Lipman, Rustichini (2001) GP for discussion of these axioms. The next axiom was introduced by DLR requires the following de nition. First, for any menu x, let conv (x) denote its convex hull. De nition 1 x 0 conv (x) is critical for x if for all y such that x 0 conv (y) conv (x), we have y x. Observe that if x 0 is critical for x, then x 0 x. We think of a critical set as stripping away the irrelevant alternatives of a menu. That is, suppose x 0 is critical for x, let 2 x n x 0. Then since x 0 is critical, we have x 0 [ fg x. That is, adding to x 0 does not a ect the agents ranking of x 0. Since 2 x, we conclude then that is not important to the decision maker when evaluating x. Axiom 4 (Finiteness) Every menu has a nite critical subset. Following DLR, we assume Finiteness to simplify the analysis. See DLR for more discussion of this axiom. The next two axioms are meant to capture the e ects of temptation on preference. 2 1 See Noor (2007) for a model that relaxes this assumption allows an agent to be tempted at the time of choosing menus. 2 Inclusion has been proposed independently by Nehring (2006) under the name Singleton Monotonicity. Chrasekher (2009) considers an axiom labeled A1 that is similar to, though distinct from, Exclusion. 2

5 Axiom 5 (Exclusion) If fg fg for every 2 x, then x x [ fg. If every alternative in x normatively dominates, then Exclusion states that the agent would prefer to exclude from his menu. We can think of as a bad alternative being added to a good menu x. If tempts the agent, then this temptation will con ict with his normative preference. If the agent thinks he might choose, then he would be choosing over an alternative that is normatively superior. Thus adding can only make the menu less desirable. Axiom 6 (Inclusion) If fg fg for every 2 x, then x [ fg x. If normatively dominates every alternative in x, then Inclusion states that the agent would prefer to include in his menu. We can think of as a good alternative being added to a bad menu x. If tempts the agent, there is no con ict with his normative preference. If the agent thinks he might choose, then he would be choosing over an alternative that is normatively inferior. Thus adding can only make the menu more desirable. Our utility representation takes the form of equation (1), which we call a temptation representation. Theorem 1 The preference satis es Weak Order, Continuity, Independence, Finiteness, Exclusion Inclusion if only if has a temptation representation. The proof is in Appendix B. Remark 1 Finiteness is independent of our other axioms. As an example, preferences represented by: Z V (x) = [u () + v ()] v () (dv), where is a measure (with possibly in nite support) over the set of von Neumann- Morgenstern expected-utility functions, would satisfy all our axioms but Finiteness. 3 This is in contrast to GP s model. Though GP do not assume Finiteness, it is implied by their main axiom, Set Betweenness. 3 In recent work, Dekel Lipman (2007) adapt our proof of Theorem 1 to characterize such preferences. Besides dropping Finiteness, their set of axioms di ers from ours in two ways. First, they add a continuity axiom called Lipschitz Continuity. See Dekel, Lipman, Rustichini, Sarver (2007) for a discussion of this axiom. Second, they replace Exclusion Inclusion with an axiom they call Weak Set Betweenness. Lemma 1 shows that Weak Set Betweenness is equivalent to Exclusion Inclusion, given Weak Order Continuity. 3

6 Remark 2 Though Exclusion Inclusion are necessary, the proof of su ciency does not actually use their full force. Speci cally, we could replace Exclusion with DLR s axiom, Desire for Commitment. Alternatively, because of the symmetry of the representation the axioms, we could replace Inclusion with an axiom symmetric to Desire for Commitment, one which we call Desire for Better Alternatives. However, it is not possible to weaken both Exclusion Inclusion, as shown by a counterexample in Appendix C. 3 Related Literature 3.1 Gul Pesendorfer, 2001 GP were the rst to use a preference-over-menus framework to model temptation. Their key axiom is Set Betweenness. Axiom 7 (Set Betweenness) If x y, then x x [ y y. GP s main theorem states that satis es Weak Order, Continuity, Independence, Set Betweenness if only if has the following representation. De nition 2 A self-control representation is a function V GP such that V GP (x) = [u () + v ()] v (), (2) where u v are von Neumann-Morgenstern expected-utility functions. Observe that a self-control representation is a temptation representation where I = 1. Thus GP s model seems to not allow uncertainty about temptation. The following example, borrowed from DLR, uses uncertainty about temptation to explain a violation of Set Betweenness. This example, however, is consistent with Exclusion Inclusion. Example 1 Suppose an agent is on a diet has three snacks he could eat: broccoli (b), chocolate cake (c), potato chips (p). Broccoli is the healthiest snack while chocolate cake potato chips are equally unhealthy. Hence fbg fcg fpg. The agent thinks he will experience either a salt craving or a sugar craving. If he has a salt craving, then he is better o not having potato chips as an option. Hence fbg fb; pg fb; cg fb; c; pg. 4

7 If he has a sugar craving, then he is better o not having chocolate cake as an option. Hence fbg fb; cg fb; pg fb; c; pg. This violates Set Betweenness since fb; cg [ fb; pg 6 fb; cg ; fb; pg. consistent with Exclusion Inclusion. However it is GP argue that, for preferences with a self-control representation, certain behavior (described below) reveals that an agent anticipates exerting self-control. However this interpretation is not justi ed for preferences with a temptation representation. Uncertainty about temptation is key to this di erence. 4 For preferences with a self-control representation, GP propose that the following reveals that an agent anticipates exerting self-control. De nition 3 has self-control at z if there exists x; y such that z = x [ y x x [ y y. As GP explain, x x [ y captures the fact that y entails greater temptation than x while x [ y y captures the fact that the agent resists this temptation. 5 The intuition motivating this de nition is that there are two requirements for an agent to exert self-control: First, he must be tempted, second, he must resist that temptation. GP support their interpretation of this behavior with the following theorem. 6 First, for any continuous f :! R, de ne c (x; f) arg f (). Theorem 2 (Gul Pesendorfer, 2001, Theorem 2) Suppose has a selfcontrol representation given by (2). The following are equivalent: (i) has self-control at x. (ii) c (x; u + v) \ c (x; v) =?. Property (ii) implies that the agent resists temptation since his anticipated choice is not a most tempting alternative. Consider the following generalization of property (ii) for a temptation representation: There exists i such that c (x; u + v i ) \ c (x; v i ) =?. () 4 Dekel Lipman (2007) argue that GP s de nition can have interpretations other than selfcontrol, even for preferences with a self-control representation. Roughly, they show that a self-control representation can be interpreted alternatively as allowing uncertainty in temptation. 5 pp Notation has been changed in this quote to be consistent with ours. 6 This is not the full statement of GP s theorem. We have omitted a portion as it is not related to the present discussion. 5

8 This statement would capture the intuition behind GP s de nition of self-control since the agent would anticipate some state where he is tempted but resists the temptation. However, as the following example shows, GP s de nition of self-control does not characterize () for preferences with a temptation representation. Example 2 Suppose the agent has the following normative temptation utilities for two alternatives,!: u v 1 v ! Suppose also that he ranks menus by the temptation representation: V (x) = 2 i=1 1 2 [u () + v i ()] v i (). Then fg f;!g f!g. However, in no state does the agent anticipate exerting self-control. In state 1, the agent is most tempted by (since v 1 () > v 1 (!)), but he also expects to choose (since u () + v 1 () > u (!) + v 1 (!)). Similarly in state 2, the agent is most tempted by!, he expects to choose!. So even though the agent has self-control (according to GP s de nition) at f;!g, condition () does not hold. Though fg f;!g captures the fact that! is tempting in some state, f;!g f!g captures the fact that the agent expects to choose in some state, those two states are not the same. Hence GP s de nition is not enough to characterize () for a temptation representation. In fact, Example 2 shows a little bit more. If has a temptation representation fg f!g, then there are only three possible orderings of fg, f!g, f;!g: (i) fg f;!g f!g, (ii) fg f;!g f!g, (iii) fg f;!g f!g. One can show that fg f;!g in ranking (i) implies 2 c (f;!g ; u + v i ) \ c (f;!g ; v i ) for every i. 7 Similarly, f;!g f!g in ranking (ii) implies! 2 c (f;!g ; u + v i ) \ c (f;!g ; v i ) for every i. Hence, (iii) is the only possible ranking where () might hold. Therefore, Example 2 shows that for an arbitrary set, there is no behavior that would reveal that an agent anticipates exerting self-control. 7 If not, then there exists i such that v i (!) > v i (), which implies u () > u () + v i () v i (!), which implies V T (fg) > V T (f;!g), a contradiction. 6

9 3.2 Dekel, Lipman, Rustichini, 2009 Using Example 1 as part of their motivation to weaken Set Betweenness, DLR propose their own axiom to capture temptation. Axiom 8 (DFC: Desire for Commitment) For every x, there exists 2 x such that fg x. DLR show that satis es Weak Order, Continuity, Independence, Finiteness, DFC a technical axiom called Approximate Improvements are Chosen if only if has the following representation. De nition 4 A DLR temptation representation is a function V DLR such that " I V DLR (x) = q i ( u () + # ) v j () v j (), i=1 j2j i j2j i where q i > 0 for every i, P i q i = 1, where u each v j Morgenstern expected-utility functions. are von Neumann- Observe that if J i is a singleton for every i, then this is a temptation representation. Hence, this is a generalization of a temptation representation where the agent can be a ected by more than one temptation in each state. We argue that a DLR temptation representation permits preferences which are not explained by temptation. Consider the following example. Example 3 Suppose an agent has two snacks he could eat: chocolate cake (c), potato chips (p). Assume the following ranking of menus: fcg fpg fc; pg. This ranking is consistent with DFC but not with Inclusion. 8 It implies that chocolate cake adds some psychic cost to the agent when coupled with potato chips, vice versa. But the agent considers chocolate cake potato chips to be normatively equivalent. Thus temptation is not an explanation for this since there is no con ict with his normative preference. 9 This example suggests a need for a stronger set of axioms. DLR proposed their own strengthening of DFC. 8 This ranking is also consistent with DLR s other axiom, Approximate Improvements are Chosen. 9 We do not mean to say that preferences like Example 3 are unreasonable, only that temptation alone is not a good explanation for them. Such preferences may be explained by, for example, regret (e.g. Sarver (2008)) or perfectionism (e.g. Kopylov (2009)). 7

10 Axiom 9 (Weak Set Betweenness) If fg fg for all 2 x 2 y, then x x [ y y. In an earlier version of their paper, DLR conjectured that Weak Order, Continuity, Independence, Finiteness, Weak Set Betweenness characterize a temptation representation. How does Weak Set Betweenness relate to Exclusion Inclusion? It should be obvious that Weak Set Betweenness implies Exclusion Inclusion. 10 The following lemma shows that in the presence of our other axioms, the other direction holds as well. Lemma 1 If satis es Weak Order, Continuity, Exclusion, Inclusion, then satis es Weak Set Betweenness. Proof. We show that the conclusion holds for nite menus. The result then follows from Continuity the fact that, in the Hausdor topology, any menu is the limit of a sequence of nite menus (see GP, Lemma 0). Let x = f 1 ; :::; M g y = f 1 ; :::; N g satisfy f 1 g f 2 g ::: f M g f 1 g f 2 g ::: f N g. By repeatedly applying Exclusion, we obtain f 1 ; :::; M g f 1 ; :::; M g [ f 1 ; :::; N g, or x x [ y. By repeatedly applying Inclusion, we obtain or x [ y y. f 1 ; :::; M g [ f 1 ; :::; N g f 1 ; :::; N g, Hence Theorem 1 is a proof of DLR s conjecture. However, we prefer Exclusion Inclusion over Weak Set Betweenness because they are more basic axioms. This is desirable because it makes the assumptions on behavior more transparent. 10 In the statement of Weak Set Betweenness, take y as a singleton to get Exclusion, take x as a singleton to get Inclusion. 8

11 Appendix A Notation Let K 3 denote the number of outcomes or prizes. (Results are simple if K = 2.) Let 0 1 denote K-vectors of zeros ones respectively. We will use u, w i, v j, etc., to denote both von Neumann-Morgenstern expected-utility functions as well as K-vectors of payo s of pure outcomes, so that w i () = w i, etc. For any f 2 R K, de ne c (x; f) arg f H f g 2 R K : g f = 0. In particular, H 1 = g 2 R K : g 1 = 0 is the set of vectors whose coordinates sum to zero. Let denote the relative interior of, i.e. f = ( k ) 2 : 0 < k < 1, k = 1; :::; Kg : B Proof of Theorem 1 We prove the su ciency part of the theorem in Section B.1 followed by the necessity part in Section B.2. B.1 Su ciency of Axioms The proof that the axioms are su cient will proceed as follows. In Section B.1.1, we present some useful results for linear functionals. In Section B.1.2, we de ne a key intermediate representation, the nite additive EU representation. We then prove some results for this representation our axioms. In Section B.1.3, we use these results to nish the proof. B.1.1 Some Results for Linear Functionals We present the rst two lemmas of this section without proof. Lemma 2 Suppose u 2 H 1. Suppose x is a sphere in H u \ f; g 2 H 1 n f0g. Write f = au + ~ f g = bu + ~g, where ~ f; ~g 2 H u \ H 1. Then c (x; f) = c (x; g) if only if ~ f is a positive scalar multiple of ~g. Furthermore, if ~ f 6= 0, then c (x; f) is a singleton. We will use Lemma 2 often, especially the following cases: (i) u = 0, (ii) ~ f = g, (iii) f; g 2 H u. 9

12 De nition 5 A set of vectors F is not redundant if for every f; g 2 F, f is not a positive scalar multiple of g. Lemma 3 Suppose u 2 H 1, let ff i g I i=1 fg jg J j=1 sets of vectors in H u \ H 1 n f0g. Then be two nite non-redundant I i=1 f i = J j=1 g j, 8 closed x H u \, if only if I = J, without loss of generality, f i = g i for every i. The following lemma generalizes Lemma 3 by allowing for redundancies zero vectors. Lemma 4 Suppose u 2 H 1, let ff i g I i=1 fg jg J j=1 in H u \ H 1. Then be two nite sets of vectors I i=1 if only if there exists f i = J j=1 (i) I 0 f1; :::; Ig J 0 f1; :::; Jg, g j, 8 closed x H u \, (ii) I 1 ; :::; I N, a partition of I 0, J 1 ; :::; J N, a partition of J 0, (iii) positive scalars f g i2i 0 f g j2j 0, where P i2i n = P j2j n = 1 for every n 2 f1; :::; Ng, (iv) fh 1 ; :::; h N g H u \ H 1 n f0g not redundant, such that f i = 0, 8i 62 I 0, such that for every n, g j = 0, 8j 62 J 0, f i = h n, 8i 2 I n, g j = h n, 8j 2 J n. Proof. The if part is straightforward, so we just prove the only if part. De ne I 0 fi : f i 6= 0g J 0 fj : g j 6= 0g. Partition I 0 into I 1 ; :::; I N, where i; i 0 2 I n if only if f i is a positive scalar multiple of f i 0. Similarly, partition J 0 10

13 into J 1 ; :::; J M, where j; j 0 2 J m if only if g j is a positive scalar multiple of g j 0. Then we have!! N M =, 8 closed x H u \, n=1 N i2i n f i m=1 j2j m g j where P n P o M i2i n f i n=1 j2j m g j are nite sets of vectors in H u \ H 1 n f0g, m=1 each P of which is not redundant. Lemma 3 then implies that N = M, that i2i n f i = P j2j n g j for every n. De ne h n P i2i n f i for every n. Observe that h n 2 H 1 \ H u n f0g for every n, that fh 1 ; :::; h N g is not redundant. For every i 2 I 0, let n (i) denote the n such that i 2 I n. Observe that for every i 2 I 0, f i is a positive scalar multiple of h n(i). So for every i 2 I 0, de ne > 0 by the equation f i = h n(i). Similarly, for every j 2 J 0, let n (j) denote the n such that j 2 J n. For every j 2 J 0, de ne > 0 by the equation g j = h n(j). Hence P i2i n = P j2j n = 1 for every n. B.1.2 Finite Additive EU Representation Some Preliminary Results In their appendix, DLR prove the following. Theorem 3 (Dekel, Lipman, Rustichini, 2009, Theorem 6) The preference satis es Weak Order, Continuity, Independence, Finiteness if only if has the representation V (x) = I i=1 w i () J j=1 v j (), (3) where each w i each v j is a von Neumann-Morgenstern expected-utility function. A representation of the form given in (3) is called a nite additive EU representation. It is a modi ed version of the set of preferences studied by Dekel, Lipman, Rustichini (2001). For such a representation, de ne u w i v j, (4) i j which represents preference over singleton menus. De nition 6 A nite additive EU representation given by (3) is in reduced form if fw 1 ; :::; w I ; v 1 ; :::; v J g H 1 n f0g is not redundant. Lemma 5 If has a nite additive EU representation, then it has a reduced form nite additive EU representation. 11

14 The proof of Lemma 5 is straightforward, so we omit it. The following lemma shows some implications of Exclusion Inclusion. 11 Lemma 6 Suppose has a reduced form nite additive EU representation satis es Exclusion Inclusion. Then for every i, w i is not a positive scalar multiple of u, for every j, v j is not a positive scalar multiple of u. Proof. We prove only the rst part, which uses Exclusion. The proof of the second part is similar uses Inclusion. By way of contradiction, suppose i is such that w i = au for some a > 0. Let x be a sphere in. Since w i 2 H 1 n f0g, Lemma 2 implies c (x; w i ) = f i g is a singleton that f i g = c (x; u), or f i g = arg min u. Since w i v j are not redundant non-zero for every j, Lemma 2 also implies v j > i v j, 8j. Since i 2 u 2 H 1, there exists > 0 such that i u 2 satis es v j > v j, 8j. (5) Observe that w i > i w i since ( u) w i = au u > 0. Also observe that u < i u, which implies that fg f g for every 2 x. Now consider the menu x [ f g. Inequality (5) implies We also have v j = v j, 8j. [f g w i w i, 8i, [f g with a strict inequality for i. Therefore, V (x [ f g) > V (x). But this contradicts Exclusion since fg f g for every 2 x. Next we introduce a key axiom that will be useful for proving an intermediate result later (Lemma 9). De nition 7 A menu x is constant if for every ; 0 2 x, fg f 0 g. Axiom 10 (CMNT: Constant Menus Are Not Tempted) For every constant menu x, for every 2 x, fg x. 11 Lemmas 6 7 can each be proven with the weaker axioms DFC Desire for Better Alternatives replacing Exclusion Inclusion. See Appendix C for a formal statement of Desire for Better Alternatives. Thus Lemma 11 is the only result in the proof where Exclusion or Inclusion is needed, only one of these is needed, not both. See Remark 2. 12

15 Intuitively, CMNT states that a constant menu cannot tempt since there can be no con ict between an agent s normative preference a temptation. The following lemma shows that CMNT is implied by our axioms. Lemma 7 Suppose satis es Weak Order, Continuity, Exclusion, Inclusion. Then satis es CMNT. The proof is similar to that of Lemma 1, so we omit it. The next lemma takes care of a trivial case. Lemma 8 Suppose has a reduced form nite additive EU representation satis es CMNT. If u = 0, then I = J = 0. Proof. If u = 0, then fg fg for every ; 2. Hence, for every x 2, x is a constant menu. CMNT then implies V (x) = 0 for every x 2, or i w i = j v j, 8x 2. But since fw 1 ; :::; w I g H 1 n f0g is not redundant fv 1 ; :::; v J g H 1 n f0g is not redundant, Lemma 3 implies I = J w i = v i for every i. But since fw 1 ; :::; w I ; v 1 ; :::; v J g is not redundant, it must be that I = J = 0. The following lemma shows the implications of CMNT. Lemma 9 Suppose has a reduced form nite additive EU representation satis es CMNT. Then there are (i) scalars a 1 ; :::; a I b 1 ; :::; b J, where P I i=1 a P J i j=1 b j = 1, (ii) I 0 f1; :::; Ig J 0 f1; :::; Jg, (iii) I 1 ; :::; I N, a partition of I 0, J 1 ; :::; J N, a partition of J 0, (iv) positive scalars f g i2i 0 f g j2j 0, where P i2i n = P j2j n = 1 for every n 2 f1; :::; Ng, (v) ff 1 ; :::; f N g H u \ H 1 n f0g not redundant, such that w i = a i u, 8i 62 I 0, such that for every n 2 f1; :::; Ng, v j = b j u, 8j 62 J 0, w i = a i u + f n, 8i 2 I n, v j = b j u + f n, 8j 2 J n. 13

16 Proof. If u = 0, then Lemma 8 implies that the result is trivial. So assume u 6= 0. Observe that u 2 H 1. Observe also that for every i, there is ~w i 2 H u \ H 1 scalar a i such that w i = a i u + ~w i, (6) for every j, there is ~v j 2 H u \ H 1 scalar b j such that P Then u = i a P i j b j u + P i ~w i i j, this means P i a P i j b j = 1. i=1 v j = b j u + ~v j. (7) P j ~v j. Since u 6= 0 ~w i ; ~v j 2 H u for every Let x be any constant menu. Set u = u for any 2 x. Then CMNT implies that I J u = w i v j. Hence, using (6) (7), u = I i=1 = u + [a iu + ~w i ] I i=1 ~w i j=1 J j=1 J j=1 [b j u + ~v j ] ~v j, which implies P I ~w i = P J ~v j for any constant menu x. By Lemma 4, there are (i) I 0 f1; :::; Ig J 0 f1; :::; Jg, (ii) I 1 ; :::; I N, a partition of I 0, J 1 ; :::; J N, a partition of J 0, (iii) positive scalars f g i2i 0 f g j2j 0, where P i2i n = P j2j n = 1 for every n 2 f1; :::; Ng, (iv) f 1 ; :::; f N 2 H 1 \ H u n f0g, such that ~w i = 0, 8i 62 I 0, such that for every n, ~v j = 0, 8j 62 J 0, ~w i = f n, 8i 2 I n, ~v j = f n, 8j 2 J n. Inserting these into (6) (7) gives us our result. B.1.3 Finishing the Proof of Su ciency We now show that our axioms are su cient for a temptation representation. 14

17 Let satisfy Weak Order, Continuity, Independence, Finiteness, Exclusion Inclusion. By Theorem 3 Lemma 5, has a reduced form nite additive EU representation V of the form given in (3). De ne u by equation (4). By Lemma 7, P satis es CMNT. We apply Lemma 9 to get (i) scalars a 1 ; :::; a I b 1 ; :::; b J, where I i=1 a P J i j=1 b j = 1, (ii) I 0 f1; :::; Ig J 0 f1; :::; Jg, (iii) I 1 ; :::; I N, a partition of I 0, J 1 ; :::; J N, a partition of J 0, (iv) positive scalars f g i2i 0 f g j2j 0, where P i2i n = P j2j n = 1 for every n 2 f1; :::; Ng, (v) ff 1 ; :::; f N g H u \H 1 nf0g not redundant, such that w i = a i u, 8i 62 I 0, such that for every n 2 f1; :::; Ng, v j = b j u, 8j 62 J 0, w i = a i u + f n, 8i 2 I n, (8) v j = b j u + f n, 8j 2 J n. (9) To interpret this geometrically, for every n, u f n de ne a half-plane, we can think of I n J n as the subsets of the w i s v j s that lie in that half-plane. Since u f n are orthogonal, we can think of the ratio a i as describing the angle that w i makes with u f n. Similarly, we can think of the ratio b j as describing the angle that v j makes with u f n. Figure 1 illustrates. The following notation will be useful. For i 2 I 0, let n (i) denote the n such that i 2 I n. Similarly, for j 2 J 0, let n (j) denote the n such that j 2 J n. For any n 2 f1; :::; Ng for any 2 R, de ne the sets I n () i 2 I n : a i J n () j 2 J n : b j <. Thus I n () is the set of the w i s that make an angle with f n weakly less than arctan ( kuk = kf n k). Similarly, J n () is the set of v j s that make an angle with f n strictly less than arctan ( kuk = kf n k). Before proceeding, we outline the rest of the proof to help the reader follow the argument. Using the fact that fw 1 ; ::; w I ; v 1 ; :::; v J g is not redundant, Lemma 10 shows that we can slightly perturb without changing the sets I n () J n (). Lemma 11 is the key sten the proof. It shows the relationship between the s s for the expected-utility functions in a given half-plane. This is proved by contradiction using Lemma 10 to construct a lottery menu that would otherwise violate Exclusion Alternatively, one could prove Lemma 11 by constructing a lottery menu that would otherwise violate Inclusion. 15

18 Figure 1: Here we have two half-planes: one for f 1 one for f 2. w i = a i u + f 1. Also, j 2 J 2 v j = b j u + f 2. So i 2 I 1 Lemma 12 then shows that we can take any w i v j in the same half-plane write d ij w i = c ij u + e ij v j, where c ij, d ij, e ij have appropriate properties. Substituting this into a nite additive EU representation gives the result. Lemma 10 For every n 2 f1; :::; Ng for every 2 R, there exists an interval ; 3 where <, such that I n () = I n ( ), 8 2 ;, J n () = J n ( ), 8 2 ;. Figure 2 illustrates Lemma 10. Proof. Fix n. Since fw 1 ; ::; w I ; v 1 ; :::; v J g is not redundant, a i i 2 I n j 2 J n. Hence, it cannot be that a i So we consider two overlapping cases. Case 1. a i 6=, 8i 2 I n. By de nition, I n ( ) = then I n ( ) = n i 2 I n : a i 6= b j for every = b j = for some i 2 I n j 2 J n. o. But since a i 6= for every i 2 I n, n i 2 I n : a i < o. So there exists > 0 such that I n ( ) = I n ( ) J n ( ) = J n ( ). Set. 16

19 Figure 2: Here = min j2jnnjn( ) b j = i2in( ) a i I n () = I n ( ) J n () = J n ( ) for any 2 ;.. It is easy to see that Case 2. b j 6=, 8j 2 J n. Observe that J n nj n ( ) = n j 2 J n : b j o. But since b j 6=, for every j 2 J n, n then J n nj n ( ) = j 2 J n : b j > o. So there exists > 0 such that J n nj n ( + ) = J n n J n ( ) I n n I n ( + ) = I n n I n ( ), which implies J n ( + ) = J n ( ) I n ( + ) = I n ( ). Set +. Our next lemma shows the relationship between the w i s v j s in one of the half-planes. Lemma 11 For every n 2 f1; :::; Ng for every 2 R, we have 0. i2i n() j2j n() Proof. If u = 0, then I = J = ; by Lemma 8, the result is vacuous. So assume u 6= 0: 17

20 Figure 3: This table shows where the imum for the respective expected-utility function is attained. For example, the third row rst column show [f g w i = w i, for i 2 I 0 n I n. Now suppose the result is not true, i.e. there exists n such that > 0. (10) i2i n ( ) j2j n ( ) By Lemma 10, there exists a non-singleton interval ; 3, such that I n () = I n ( ), 8 2 ;, (11) J n () = J n ( ), 8 2 ;. (12) In the remainder of the proof, we construct a menu, x [ f g, lottery,, that violate Exclusion given (10). We show this by determining where each expectedutility function (i.e. w 1 ; :::; w I ; v 1 ; :::; v J ) attains its imum on x [ f g then again on x [ f g [ f g. Figure 3 collects these results. As Figure 3 shows, the only functions whose imum changes when is added to the menu are those associated with I n ( ) J n ( ). This allows us to use (10) to show that V (x [ f g) < V (x [ f g [ f g), violating exclusion. Let x be a sphere in H u \. Observe that x is a constant menu. Since f n 2 H u \ H 1 n f0g, Lemma 2 implies that c (x; f n ) = f n g is a singleton that w i = n w i, 8i 2 I n. (13) 18

21 Also, since ff 1 ; :::; f N g H u \ H 1 n f0g is not redundant, Lemma 2 implies w i > n w i, 8i 2 I 0 n I n. (14) Similarly, we can show v j = n v j, 8j 2 J n (15) v j > n v j ; 8j 2 J 0 n J n : (16) Observe that f 1 f n f n n u 2 uu H1 since u; f n 2 H 1. Hence, n 2 inequalities (14) (16) imply that there exists > 0 such that 1 n + f n f n f n u u u 2, w i > w i, 8i 2 I 0 n I n, (17) v j > v j, 8j 2 J 0 n J n. (18) Observe that u = n u, which implies u = min [f g u since x is constant. Geometrically, is a lottery that is an move from n in the direction of f 1 f n f n n u. uu Consider the menu x [ f g. To ll in the column x [ f g in Figure 3, we must determine the imizers over x [ f g for the expected-utility functions fw 1 ; :::; w I ; v 1 ; :::; v J g. We show this only for the expected-utility functions associated with I n J n leave the rest to the reader. Using (8), observe that w i = n w i + a i for i 2 I n. Since > 0 > 0, this implies w i n w i if only if i 2 I n (). Hence equations (11) (13) imply w [f i = w i, 8i 2 I n ( ) g w [f i = w i, 8i 2 I n n I n ( ). g Similarly, using equations (12) (15), we obtain v [f j = v j, 8j 2 J n ( ) g [f g v j = v j, 8j 2 J n n J n ( ). 19

22 Now we construct the lottery that will lead to a contradiction of Exclusion. The idea is that we will take the lottery move a small distance in the direction of f n. Since f n u are orthogonal, this will not change the commitment utility u. However, we will use inequality (10) to show that adding this lottery to our menu x must increase the utility of the menu. Equations (17) (18) imply w i > w i, 8i 2 I 0 n I n [f g v j > v j, 8j 2 J 0 n J n. [f g Since f n 2 H 1 2, there exists ^ 0 > 0 such that + ^ 0 f n 2 De ne ^ 00 w i > ( + ^ 0 f n ) w i, 8i 2 I 0 n I n (19) [f g v j > ( + ^ 0 f n ) v j, 8j 2 J 0 n J n. (20) [f g f n f n > 0. Observe that by the de nition of, + ^ 00 f n = n + f n f n f n Hence, using (8), ( + ^ 00 f n ) w i = n w i + 1 u u u. a i ( + ^ 00 f n ) w i n w i if i 2 I n n I n. Hence by (11), Similarly, using (12), we obtain for i 2 I n. This implies n w i ( + ^ 00 f n ) w i, 8i 2 I n n I n ( ). (21) n v j > ( + ^ 00 f n ) v j, 8j 2 J n n J n ( ). (22) Set ^ min f^ 0 ;^ 00 g + ^f n. Hence 2, so consider the menu x [ f g [ f g. First, observe that u = u = min [f g u. Hence, fg f g for every 2 x[f g. To ll in the column x[f g[f g in Figure 3, we must determine the imizers over x [ f g [ f g for the expected-utility functions fw 1 ; :::; w I ; v 1 ; :::; v J g. Again, we show this only for the expected-utility functions associated with I n J n leave the rest to the reader. (Inequalities (19) (20) essentially give us the results for I 0 n I n J 0 n J n.) Inequality (21) implies n w i w i, 8i 2 I n n I n ( ), 20

23 which implies Inequality (22) implies which implies w i = w i, 8i 2 I n n I n ( ). [f g[f g [f g n v j > v j, 8j 2 J n n J n ( ), v j = v j, 8j 2 J n n J n ( ). [f g[f g [f g Using (8), observe that w i = w i +^ f n f n for i 2 I n. But ^ f n f n > 0, so this implies w i > w i for every i 2 I n. Hence, this holds for every i 2 I n ( ), or w i > w i, 8i 2 I n ( ), which implies w i = w i, 8i 2 I n [f g[f ( ), g since [f g w i = w i, for i 2 I n ( ). Similarly, which implies v j > v j, 8j 2 J n ( ), v j = v j, 8j 2 J n [f g[f ( ), g since [f g v j = v j, for j 2 J n ( ). As is evident from viewing Figure 3, the only expected-utility functions that increase by going from x [ f g to x [ f g [ f g are those associated with the sets I n ( ) J n ( ). Hence, V (x [ f g [ f g) V (x [ f g) = ( ) w i ( ) v j = = i2i n ( ) i2i n ( ) i2i n ( ) ^f n w i ^ f n f n 0 = ^ (f n f n i2i n ( ) j2j n ( ) j2j n ( ) j2j n ( ) ^f n v j ^ f n f n j2j n ( ) > 0 since ^ (f n f n ) > 0 P i2i n ( ) p P i j2j n ( ) > 0 by (10). This implies x [ f g [ f g x [ f g, which contradicts Exclusion since fg f g for every 2 x [ f g A

24 Lemma 12 For every n, there exist nonnegative scalars fc ij g i2in;j2j n, fd ij g i2in;j2j n, fe ij g i2in;j2j n, where P j2j n d ij = 1 for all i 2 I n P i2i n e ij = 1 for all j 2 J n, such that d ij w i = c ij u + e ij v j, 8i 2 I n ; 8j 2 J n. Proof. The proof is by construction. Fix n. For every i 2 I n j 2 J n, de ne >< >= >< >= ^L ij min 0; 0; >: i 0 2I n: a i 0 p a i >; >: i i 0 p 0 2I n: a i 0 i p > a i >; i 0 0 j 0 2J b j n: 0 q b j j 0 L ij n0; ^L o ij. 0 j 0 2J b j n: 0 q > b j j 0 Hence, L ij 0 for every i 2 I n j 2 J n. One can show that for every i 2 I n, P j2j n L ij =, that for every j 2 J n, P i2i n L ij = : Now for every i 2 I n j 2 J n, de ne d ij L ij, e ij L ij, c ij a i d ij b j e ij. (These are well-de ned since are strictly positive.) Fix i 2 I n j 2 J n. Observe that d ij 0, e ij 0, P j 0 2J n d ij 0 P = 1, i 0 2I n e i 0 j = 1. Using (8) (9), one can show that d ij w i = c ij u + e ij v j. Now we show that c ij 0. If L ij = 0, then c ij = 0. So suppose L ij > 0. Observe that ai b j c ij = L ij. Hence, c ij 0 if only if a i b j. Since L ij > 0, it must be that 0 > 0: j 0 2J b j n: 0 q b j i 0 2I n: a i 0 j 0 q p > a i j i 0 But Lemma 11 implies Together, these imply i 0 2I n: a i 0 0 > b j 0 i 0 2I n: a i 0 0 > b j 0 > 0: j 0 2J b j n: 0 q b j j 0 0 i 0 2I n: a i 0 p > a i i 0 which can only be true if a i > b j. Now we nish the proof of su ciency. Observe that we can write V as ( ) N V (x) = w i v j + w i v j. n=1 i2i n j2j n i62i 0 j62j 0 22

25 Using Lemma 12, we get ( N V (x) = d ijw i n=1 i2i n j2j n ( N = [ c iju + e ij v j ] n=1 i2i n j2j n + i62i 0 w i j62j 0 v j ) e ijv j + w i i62i 0 ) e ijv j j62j 0 v j If c ij = 0, then d ij = e ij = 0 since w i v j are not redundant. So let M denote the number of (i; j) pairs in I 0 J 0 such that n (i) = n (j) c ij 6= 0. Let m (i; j) denote a distinct element of f1; :::; Mg. De ne ^v m(i;j) e ij c ij v j ^c m(i;j) c ij. Hence we can write M V (x) = m=1 ^c m [ u + ^v m] ^v m + w i i62i 0 j62j 0 v j. For i 62 I 0, recall that Lemma 6 the fact that w i 6= 0 imply that w i = a i u where a i > 0. So for i 62 I 0, de ne ~v i 0. Similarly, for j 62 J 0, Lemma 6 the fact that v j 6= 0 imply that v j = b j u where b j > 0. So for j 62 J 0, de ne v j u. Hence V (x) = M m=1 ^c m [ u + ^v m] ^v m + i62i 0 a i [ u + ~v i] ~v i + j62j 0 b j [ u + v j] Finally, observe that u = w i v j + w i v j i2i 0 j2j 0 i62i 0 j62j ( ) 0 N = (d ij w i e ij v j ) + w i n=1 i2i n j2j n i62i ( ) 0 N = c ij u + a i u + b j u n=1 i2i n j2j n i62i 0 j62j 0 M = ^c m + a i +! b j u, i62i 0 j62j 0 m=1 v j. (23) j62j 0 v j which implies P M m=1 ^c m + P i62i 0 a i + P j62j 0 b j = 1. So f^c 1 ; :::; ^c m g [ fa i g i62i 0 [ fb j g j62j 0 is a set of positive scalars that sum to 1. Hence (23) is a temptation representation. 23

26 B.2 Theorem 1: Necessity of Axioms We show only the necessity of Exclusion. The proof for Inclusion is similar. The necessity of the other axioms is immediate from DLR Theorem 6. Suppose has a temptation representation of the form in (1). Let x 2 2 be such that fg fg for every 2 x. Then min u () u (). We will show that for every i, [u () + v i ()] v i () [u () + v i ()] v i (), [fg [fg thus proving that V (x) V (x [ fg). Fix i. Case 1. u () + v i () [u () + v i ()]. Since min u () u (), it must be that v i () v i (). Hence [u () + v i ()] v i () = u () [fg [fg min [u () + v i ()] v i (), where the second inequality can easily be veri ed by making the substitution w i u v i. Case 2. [u () + v i ()] > u () + v i (). Then [u () + v i ()] v i () [fg [fg = i ()] i ()] v i () [fg v i (). C Counterexample Here we provide an example of preferences that satisfy DFC Desire for Better Alternatives ( all our other axioms) but not Exclusion or Inclusion. Since Exclusion Inclusion are necessary, this proves that such preferences do not have a temptation representation. We thank an anonymous referee for providing this example. First we state formally the axiom Desire for Better Alternatives. Axiom 11 (Desire for Better Alternatives) For every x, there exists 2 x such that x fg. 24

27 Fix any vectors w 1 ; f 2 H 1 n f0g such that f? w 1. Set v 1 w 1, v 2 f + aw 1, w 2 f aw 1 for 0 < a < 1=2. Let preferences be represented by V (x) = w 1 () + w 2 () v 1 () v 2 (), which is a nite additive EU representation, hence satis es Weak Order, Continuity, Independence, Finiteness. Set u w 1 + w 2 v 1 v 2 = (2 2a) w 1. Then we can rewrite V as V (x) = 1 2 2a u () + 1 2a 2 2a [u () + ^v 1 () + ^v 2 ()] ^v 1 () ^v 2 () where ^v 1 = 2 2av 1 2a 1 ^v 2 = 2 2av 1 2a 2, which is a DLR temptation representation. Hence, this preference satis es DFC. Using the symmetry of the representation, one could similarly show that V can also be written as a DLR temptation representation. Hence the preference represented by V satisfy DFC. But this implies that the preference represented by V satis es Desire for Better Alternatives. (This preference also satis es DLR s other axiom, Approximate Improvements are Chosen, as well as an axiom symmetric to this. So even adding these axioms would be insu cient to get a temptation representation.) Now we show that this preference violates Exclusion. (The following is similar to the method used to prove Lemma 11 in the proof of su ciency for Theorem 1.) Fix any 2. Let > 0 be such that + u 2. Observe then that w 1 > w 1, w 2 > w 2, v 1 > v 1, v 2 > v 2. Let 0 > 0 be such that f 2 v 2 > 0 v 2. Then V (f; ; 0 g) > V (f; g) since w 2 is the only vector whose imum changes when 0 is added to f; g. But this violates Exclusion since 0 u = u < u. A violation of Inclusion could be constructed in a similar manner. D Supplement: An Illustrative Example In this appendix, we use a simple example to illustrate how choice over sets of alternatives (called menus in the paper) can capture the e ects of temptation. Consider an agent who is deciding on a restaurant for lunch. 13 One restaurant serves only salad, s, while another serves only hamburgers, h. The agent is on a diet prefers the salad restaurant to the hamburger restaurant, or fsg fhg. 13 The following example is adapted from Gul Pesendorfer (2001). 25

28 How would he then rank a third restaurant that serves both salad hamburger, fs; hg? One can view the third restaurant as adding hamburgers to the menu fsg, which can only make it worse to the agent since he might be tempted by hamburgers. So fsg fs; hg. Alternatively, one can view the third restaurant as adding salad to the menu fhg, which can only make it better to the agent since he might exert self-control choose salad instead of hamburger. So fs; hg fhg. This example illustrates two (somewhat opposite) e ects temptation has on preference. First, the agent wishes to restrict his choice set (e.g. fsg fs; hg) to avoid temptation. Second, the agent wishes to exp his choice set (e.g. fs; hg fhg) to help make a better choice. Our two main axioms on preference, Exclusion Inclusion, demarcate these two e ects. An agent who is tempted would want to restrict his choice set by excluding the normatively worst alternative of that choice set, thus avoiding a potential con ict between a temptation his normative preference. At the same time, he would want to exp his choice set by including an alternative that is normatively superior, thus potentially helping him make a normatively good choice. References Chrasekher, M. (2009): A theorey of local menu preferences, working paper, Arizona State University. Dekel, E., B. Lipman (2007): Self-control rom Strotz representations, working paper, Boston University. Dekel, E., B. Lipman, A. Rustichini (2001): Representing preferences with a unique subjective state space, Econometrica, 69, (2009): Temptation-driven preferences, Review of Economic Studies, 76, Dekel, E., B. Lipman, A. Rustichini, T. Sarver (2007): Representing preferences with a unique subjective state space: Corrigendum, Econometrica, 75(2), Gul, F., W. Pesendorfer (2001): Temptation self-control, Econometrica, 69, Kopylov, I. (2009): Perfectionism choice, working paper, UC Irvine. 26

29 Nehring, K. (2006): Self-control through second-order preferences, working paper, UC Davis. Noor, J. (2007): Commitment self-control, Journal of Economic Theory, 135(1), Sarver, T. (2008): Anticipating regret: Why fewer options may be better, Econometrica, 76(2),