MULTIPLE TEMPTATIONS

Transcription

1 Econometrica, Vol. 78, No. 1 (January, 2010), MULTIPLE TEMPTATIONS JOHN E. STOVALL University of Rochester, Rochester, NY 14627, U.S.A. The copyright to this Article is held by the Econometric Society. It may be downloaded, printed reproduced only for educational or research purposes, including use in course packs. No downloading or copying may be done for any commercial purpose without the explicit permission of the Econometric Society. For such commercial purposes contact the Office of the Econometric Society (contact information may be found at the website or in the back cover of Econometrica). This statement must be included on all copies of this Article that are made available electronically or in any other format.

2 Econometrica, Vol. 78, No. 1 (January, 2010), NOTES AND COMMENTS MULTIPLE TEMPTATIONS BY JOHN E. STOVALL 1 We use a preference-over-menus framework to model a decision maker who is affected 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 identifies the agent s normative preference temptations, suggests the agent is uncertain which of these temptations will affect him. We provide examples to illustrate how our model improves on those of Gul Pesendorfer (2001) Dekel, Lipman, Rustichini (2009). KEYWORDS: Temptation, self-control, exclusion, inclusion. 1. INTRODUCTION WE USE A PREFERENCE-OVER-MENUSFRAMEWORKto model a decision maker who is affected by multiple temptations. We regard temptation as a craving or desire that is different 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 conflicted. He wants to avoid such conflict 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 conflict 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 exclusion inclusion axioms among others, Theorem 1 characterizes the representation (1) V T (x) = I i=1 } q i { [u(β) + v i(β)] v i(β) β x β x 1 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 The Econometric Society DOI: /ECTA8090

3 350 JOHN E. STOVALL where q i > 0foralli, q i i = 1, u each v i are von Neumann Morgenstern expected-utility functions. For any singleton menu, V T ({β}) = u(β). 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 affects the agent. He compromises between his normative preference temptation preference chooses the alternative that imizes u + v i. However, he experiences the disutility β x v i (β), which is the foregone utility from the most tempting alternative. Thus V T takes an expected-utility form, where β x [u(β) + v i (β)] β x v i (β) is the utility attained in state i, suggesting the agent is uncertain which of his temptations will affect him. This work is most closely related to the seminal paper by Gul Pesendorfer (2001) arecent paperbydekel, 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, wediscuss the relationship between these models. Through examples, we argue for the relaxation of GP s model the strengthening of DLR s model. 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 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. 2 The agent s normative preference is identified with his preference over singleton menus. Formally, let Δ denote the set of probability distributions over a finite set of prizes, call β Δ a lottery. Let X denote the set of closed nonempty subsets of Δ call x X a menu. We endow X with the Hausdorff topology define the mixture operation λx + (1 λ)y {λβ + (1 λ)β : β x β y} for λ [0 1]. Our primitive is a binary relation over X which represents the agent s preference. Consider the following axioms. AXIOM 1 Weak Order: is complete transitive. AXIOM 2 Continuity: The sets {x : x y} {x : y x} are closed. 2 See Noor (2007) for a model that relaxes this assumption allows an agent to be tempted at the time of choosing menus.

4 MULTIPLE TEMPTATIONS 351 AXIOM 3 Independence: If x y, then for every z X λ (0 1], λx + (1 λ)z λy + (1 λ)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 definition. First, for any menu x, let conv(x) denote its convex hull. DEFINITION 1: x conv(x) is critical for x if for all y such that x conv(y) conv(x), wehavey x. Observe that if x is critical for x, then x x. We think of a critical set as stripping away the irrelevant alternatives of a menu. That is, suppose x is critical for x let β x \ x.thensincex is critical, we have x {β} x. That is, adding β to x does not affect the agents ranking of x.sinceβ x,we conclude then that β is not important to the decision maker when evaluating x. AXIOM 4 Finiteness: Every menu has a finite 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 effects of temptation on preference. 3 AXIOM 5 Exclusion: If {β} {α} for every β x, then x x {α}. If every alternative in x normatively dominates α, then Axiom 5 (exclusion) states that the agent would prefer to exclude α from his menu. We can think of α as a bad alternative being added to a good menu x.ifα tempts the agent, then this temptation will conflict with his normative preference. If the agent thinks he might choose α, then he would be choosing α over an alternative that is normatively superior. Thus adding α can only make the menu less desirable. AXIOM 6 Inclusion: If {α} {β} for every β x, then x {α} x. If α normatively dominates every alternative in x, then Axiom 6 (inclusion) states that the agent would prefer to include α in his menu. We can think of α as a good alternative being added to a bad menu x. Ifα tempts the agent, 3 Inclusion was proposed independently by Nehring (2006) under the name singleton monotonicity. Chrasekher (2009) considered an axiom labeled A1 that is similar to, though distinct from, exclusion.

5 352 JOHN E. STOVALL there is no conflict with his normative preference. If the agent thinks he might choose α, then he would be choosing α over an alternative that is normatively inferior. Thus adding α 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 satisfies weak order, continuity, independence, finiteness, exclusion, inclusion if only if has a temptation representation. The proof is given in Appendix B. REMARK 1: Finiteness is independent of our other axioms. As an example, preferences represented by { } V(x)= [u(β) + v(β)] v(β) μ(dv) β x β x where μ is a measure (with possibly infinite support) over the set of von Neumann Morgenstern expected-utility functions, would satisfy all our axioms but finiteness. 4 This is in contrast to GP s model. Though GP did not assume finiteness, it is implied by their main axiom, set betweenness. REMARK 2: Though exclusion inclusion are necessary, the proof of sufficiency does not actually use their full force. Specifically, 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 first 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. 4 In recent work, Dekel Lipman (2007) adapted our proof of Theorem 1 to characterize such preferences. Besides dropping finiteness, their set of axioms differs 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.

6 MULTIPLE TEMPTATIONS 353 GP s main theorem states that satisfies weak order, continuity, independence, set betweenness if only if has the following representation. DEFINITION 2: A self-control representation is a function V GP such that (2) V GP (x) = [u(β) + v(β)] v(β) β x β x 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 {b} {c} {p} The agent thinks he will experience either a salt craving or a sugar craving. If he has a salt craving, then he is better off not having potato chips as an option. Hence {b} {b p} {b c} {b c p} If he has a sugar craving, then he is better off not having chocolate cake as an option. Hence {b} {b c} {b p} {b c p} This violates set betweenness since {b c} {b p} {b c} {b p}. However, it is consistent with exclusion inclusion. GP argued that, for preferences with a self-control representation, certain behavior (described below) reveals that an agent anticipates exerting selfcontrol. However this interpretation is not justified for preferences with a temptation representation. Uncertainty about temptation is key to this difference. 5 5 Dekel Lipman (2007) argued that GP s definition can have interpretations other than self-control, even for preferences with a self-control representation. Roughly, they showed that a self-control representation can be interpreted alternatively as allowing uncertainty in temptation.

7 354 JOHN E. STOVALL DEFINITION 3: has self-control at z if there exists x y such that z = x y x x y y. As GP (2001, pp ) explained, [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. 6 The intuition motivating this definition 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 supported their interpretation of this behavior with the following theorem. 7 First, for any continuous f : Δ R,define c(x f) arg β x f(β) THEOREM 2 Gul Pesendorfer (2001, Theorem 2): Suppose has a self-control representation given by (2). The following statements 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 ) = This statement captures the intuition behind GP s definition 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 definition 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, α ω: u v 1 v 2 α ω Notation has been changed to be consistent with ours. 7 This is not the full statement of GP s theorem. We have omitted a portion that is not related to the present discussion.

8 MULTIPLE TEMPTATIONS 355 Suppose also that he ranks menus by the temptation representation V(x)= 2 i=1 1 { } 2 [u(β) + v i(β)] v i(β) β x β x Then {α} {α ω} {ω}. However, in no state does the agent anticipate exerting self-control. In state 1, the agent is most tempted by α (since v 1 (α) > v 1 (ω)), but he also expects to choose α (since u(α) + v 1 (α) > u(ω) + v 1 (ω)). Similarly in state 2, the agent is most tempted by ω, he expects to choose ω. So even though the agent has self-control (according to GP s definition) at {α ω}, condition ( ) does not hold. Though {α} {α ω} captures the fact that ω is tempting in some state {α ω} {ω} captures the fact that the agent expects to choose α in some state, those two states are not the same. Hence GP s definition is not enough to characterize ( ) for a temptation representation. In fact, Example 2 shows a little bit more. If has a temptation representation {α} {ω}, then there are only three possible orderings of {α}, {ω},{α ω}: (i) {α} {α ω} {ω}. (ii) {α} {α ω} {ω}. (iii) {α} {α ω} {ω}. One can show that {α} {α ω} in ranking (i) implies α c({α ω} u+ v i ) c({α ω} v i ) for every i. 8 Similarly, {α ω} {ω} in ranking (ii) implies ω c({α ω} u+ v i ) c({α ω} 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 Dekel, Lipman, Rustichini (2009) Using Example 1 as part of their motivation to weaken set betweenness, DLR proposed their own axiom to capture temptation. AXIOM 8 Desire for Commitment (DFC): For every x, there exists α x such that {α} x. 8 If not, then there exists i such that v i (ω) > v i (α), which implies u(α) > u(α) + v i (α) v i (ω), which implies V T ({α})>v T ({α ω}), a contradiction.

9 356 JOHN E. STOVALL DLR showed that satisfies weak order, continuity, independence, finiteness, DFC, a technical axiom called approximate improvements are chosen if only if has the following representation. DEFINITION 4: A DLR temptation representation is a function V DLR such that V DLR (x) = I i=1 [ q i { u(β) + ] v j (β) } v j(β) β x β x j Ji j Ji where q i > 0 for every i, i q i = 1, where u each v j are von Neumann Morgenstern expected-utility functions. 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 affected 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 ranking of menus {c} {p} {c p} This ranking is consistent with DFC but not with inclusion. 9 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 conflict with his normative preference. 10 This example suggests a need for a stronger set of axioms. DLR proposed their own strengthening of DFC. AXIOM 9 Weak Set Betweenness: If {α} {β} for all α x β 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 9 This ranking is also consistent with DLR s other axiom, approximate improvements are chosen. 10 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)).

10 MULTIPLE TEMPTATIONS 357 inclusion? It should be obvious that weak set betweenness implies exclusion inclusion. 11 The following lemma shows that in the presence of our other axioms, the other direction holds as well. LEMMA 1: If satisfies weak order, continuity, exclusion, inclusion, then satisfies weak set betweenness. PROOF: We show that the conclusion holds for finite menus. The result then follows from continuity the fact that, in the Hausdorff topology, any menu is the limit of a sequence of finite menus (see GP, Lemma 0). Let x ={α 1 α M } y ={β 1 β N } satisfy {α 1 } {α 2 } {α M } {β 1 } {β 2 } {β N } By repeatedly applying exclusion, we obtain {α 1 α M } {α 1 α M } {β 1 β N } or x x y. By repeatedly applying inclusion, we obtain {α 1 α M } {β 1 β N } {β 1 β N } or x y y. Q.E.D. 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. 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, to denote von Neumann Morgenstern expected-utility functions as well as K-vectors of payoffs of pure outcomes, so that u(β) = β u,...foranyf R K,define c(x f) arg β x β f H f {g R K : g f = 0} 11 In the statement of weak set betweenness, take y as a singleton to get exclusion take x as a singleton to get inclusion.

11 358 JOHN E. STOVALL In particular, H 1 ={g R K : g 1 = 0} is the set of vectors whose coordinates sum to zero. Let Δ denote the relative interior of Δ, that is, Δ {β = (β k ) Δ :0<β k < 1 k= 1 K} APPENDIX B: PROOF OF THEOREM 1 We prove the sufficiency part of the theorem in Section B.1 followed by the necessity part in Section B.2. B.1. Sufficiency of Axioms The proof that the axioms are sufficient will proceed as follows. In Section B.1.1, we present some useful results for linear functionals. In Section B.1.2, we define a key intermediate representation, the finite additive expected-utility (EU) representation. We then prove some results for this representation our axioms. In Section B.1.3, we use these results to finish the proof. B.1.1. Some Results for Linear Functionals We present the first two lemmas of this section without proof. LEMMA 2: Suppose u H 1. Suppose x is a sphere in H u Δ f g H 1 \{0}. Write f = au + f g = bu + g, where f g 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 0, then c(x f) is a singleton. We will use Lemma 2 often, especially the cases (i) u = 0, (ii) f = g, (iii) f g H u. DEFINITION 5: A set of vectors F is not redundant if for every f g F, f is not a positive scalar multiple of g. be two finite non- LEMMA 3: Suppose u H 1, let {f i } I {g i=1 j} J j=1 redundant sets of vectors in H u H 1 \{0}. Then I i=1 β x β f i = J j=1 β x β g j for all closed x H u Δ 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.

12 i=1 MULTIPLE TEMPTATIONS 359 LEMMA 4: Suppose u H 1, let {f i } I {g i=1 j} J j=1 be two finite sets of vectors in H u H 1. Then I J β f i = β g j for all closed x H u Δ β x β x j=1 if only if there exist (i) I {1 I} J {1 J}, (ii) I 1 I N, a partition of I, J 1 J N, a partition of J, (iii) positive scalars {p i } i I {q j } j J, where i I n p i = j J n q j = 1 for every n {1 N}, (iv) {h 1 h N } H u H 1 \{0} not redundant, such that f i = 0 i / I g j = 0 j / J such that for every n, f i = p i h n i I n g j = q j h n j J n PROOF: The if part is straightforward, so we just prove the only if part. Define I {i : f i 0} J {j : g j 0}. Partition I into I 1 I N,where i i I n if only if f i is a positive scalar multiple of f i. Similarly, partition J into J 1 J M,wherej j J m if only if g j is a positive scalar multiple of g j.thenwehave N n=1 ( ) β f i = β x i I n M m=1 ( ) β g j β x j J m for all closed x H u Δ where { i I n f i } N { n=1 j J m g j } M are finite sets of vectors in m=1 Hu H 1 \{0}, each of which is not redundant. Lemma 3 then implies that N = M, that i I n f i = j J n g j for every n. Defineh n i I n f i for every n. Observe that h n H 1 H u \{0} for every n that {h 1 h N } is not redundant. For every i I,letn(i) denote the n such that i I n. Observe that for every i I, f i is a positive scalar multiple of h n(i).soforeveryi I,definep i > 0by the equation f i = p i h n(i). Similarly, for every j J,letn(j) denote the n such that j J n. For every j J,defineq j > 0 by the equation g j = q j h n(j). Hence i I n p i = j J n q j = 1 for every n. Q.E.D.

13 360 JOHN E. STOVALL B.1.2. Finite Additive EU Representation Some Preliminary Results In their appendix, DLR proved the following theorem. THEOREM 3 Dekel, Lipman, Rustichini (2009, Theorem 6): The preference satisfies weak order, continuity, independence, finiteness if only if has the representation (3) V(x)= I i=1 β x w i(β) J j=1 β x v j(β) 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 finite additive EU representation. It is a modified version of the set of preferences studied by Dekel, Lipman, Rustichini (2001). For such a representation, define (4) u i w i j v j which represents preference over singleton menus. DEFINITION 6: A finite additive EU representation given by (3)is in reduced form if {w 1 w I v 1 v J } H 1 \{0} is not redundant. LEMMA 5: If has a finite additive EU representation, then it has a reduced form finite additive EU representation. The proofoflemma 5 is straightforward, so we omit it. The following lemma shows some implications of exclusion inclusion. 12 LEMMA 6: Suppose has a reduced form finite additive EU representation, satisfies 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 first 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 Δ.Sincew i H 1 \{0}, Lemma 2 implies c(x w i ) ={β i } 12 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.

14 MULTIPLE TEMPTATIONS 361 is a singleton that {β i }=c(x u) or {β i }=arg min β x β u. Sincev j is not a positive scalar multiple of w i nonzero for every j, Lemma 2 also implies β x β v j >β i v j j Since β i Δ u H 1, there exists ε>0 such that β β i εu Δ satisfies (5) β x β v j >β v j j Observe that β w i >β i w i since ( εu) w i = aεu u>0. Also observe that β u<β i u, which implies that {β} {β } for every β x. Now consider the menu x {β }.Inequality(5) implies β x {β } β v j = β x β v j j We also have β x {β } β w i β x β w i i with a strict inequality for i. Therefore, V(x {β })>V(x). But this contradicts exclusion since {β} {β } for every β x. Q.E.D. Next we introduce a key axiom that will be useful for proving an intermediate result later (Lemma 9). DEFINITION 7: A menu x is constant if for every β β x, {β} {β }. AXIOM 10 Constant Menus Are Not Tempted (CMNT): For every constant menu x, for every α x, {α} x. Intuitively, CMNT states that a constant menu cannot tempt since there can be no conflict between an agent s normative preference a temptation. The following lemma shows that CMNT is implied by our axioms. LEMMA 7: Suppose satisfies weakorder, continuity, exclusion, inclusion. Then satisfies 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 finite additive EU representation satisfies CMNT. If u = 0, then I = J = 0.

15 362 JOHN E. STOVALL PROOF: Ifu = 0, then {α} {β} for every α β Δ. Hence, for every x X, x is a constant menu. CMNT then implies V(x)= 0 for every x X or i β x β w i = j β x β v j x X But since {w 1 w I } H 1 \{0} is not redundant {v 1 v J } H 1 \{0} is not redundant, Lemma 3 implies I = J w i = v i for every i. But since {w 1 w I v 1 v J } is not redundant, it must be that I = J = 0. Q.E.D. The following lemma shows the implications of CMNT. LEMMA 9: Suppose has a reduced form finite additive EU representation satisfies CMNT. Then there are (i) scalars a 1 a I b 1 b J, where I a i=1 i J b j=1 j = 1, (ii) I {1 I} J {1 J}, (iii) I 1 I N, a partition of I, J 1 J N, a partition of J,(iv)positive scalars {p i } i I {q j } j J, where i I n p i = j J n q j = 1 for every n {1 N}, (v) {f 1 f N } H u H 1 \{0} not redundant, such that w i = a i u i / I v j = b j u j / J such that for every n {1 N}, w i = a i u + p i f n i I n v j = b j u + q j f n j J n PROOF: Ifu = 0, then Lemma 8 implies that the result is trivial. So assume u 0. Observe that u H 1. Observe also that for every i, there is w i H u H 1 scalar a i such that (6) w i = a i u + w i for every j, there is ṽ j H u H 1 scalar b j such that (7) v j = b j u + ṽ j Then u = ( i a i j b j)u + i w i j ṽj. Sinceu 0 w i ṽ j H u for every i j, this means i a i j b j = 1.

16 MULTIPLE TEMPTATIONS 363 Let x be any constant menu. Set ū β u for any β x. Then CMNT implies that ū = I i=1 β x β w i J j=1 β x β v j Hence, using (6)(7), ū = I i=1 = ū + β x [a iū + β w i ] I i=1 β x β w i J j=1 J j=1 β x [b jū + β ṽ j ] β x β ṽ j which implies I β x β w i = J β x β ṽ j for any constant menu x. By Lemma 4, there are (i) I {1 I} J {1 J}, (ii) I 1 I N, a partition of I,J 1 J N, a partition of J, (iii) positive scalars {p i } i I {q j } j J, where i I n p i = j J n q j = 1 for every n {1 N}, (iv) f 1 f N H 1 H u \{0}, such that w i = 0 i / I ṽ j = 0 j / J such that for every n, w i = p i f n i I n ṽ j = q j f n j J n Inserting these into (6)(7) gives us our result. Q.E.D. B.1.3. Finishing the Proof of Sufficiency We now show that our axioms are sufficient for a temptation representation. Let satisfy weak order, continuity, independence, finiteness, exclusion inclusion. By Theorem 3 Lemma 5, has a reduced form finite additive EU representation V of the form given in (3). Define u by equation (4). By Lemma 7, satisfies CMNT. We apply Lemma 9 to get (i) scalars a 1 a I b 1 b J,where I a i=1 i J b j=1 j = 1, (ii) I {1 I}

17 364 JOHN E. STOVALL J {1 J}, (iii) I 1 I N, a partition of I,J 1 J N, a partition of J, (iv) positive scalars {p i } i I {q j } j J,where i I n p i = j J n q j = 1 for every n {1 N},(v){f 1 f N } H u H 1 \{0} not redundant, such that w i = a i u i / I v j = b j u j / J such that for every n {1 N}, (8) (9) w i = a i u + p i f n v j = b j u + q j f n i I n j J n To interpret this geometrically, for every n, ±u f n define 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 /p i as describing the angle that w i makes with u f n. Similarly, we can think of the ratio b j /q j as describing the angle that v j makes with u f n.figure1 illustrates. FIGURE 1. Half-planes for f 1 f 2.Soi I 1 w i = a i u + p i f 1.Also,j J 2 v j = b j u + q j f 2.

18 MULTIPLE TEMPTATIONS 365 The following notation will be useful. For i I,letn(i) denote the n such that i I n. Similarly, for j J,letn(j) denote the n such that j J n.forany n {1 N} for any θ R, define the sets { I n (θ) i I n : a } i θ p i { J n (θ) j J n : b } j <θ q j Thus I n (θ) is the set of the w i s that make an angle with f n weakly less than arctan(θ u / f n ). Similarly, J n (θ) is the set of v j s that make an angle with f n strictly less than arctan(θ u / f n ). Before proceeding, we outline the rest of the proof to help the reader follow the argument. Using the fact that {w 1 w I v 1 v J } is not redundant, Lemma 10 shows that we can slightly perturb θ without changing the sets I n (θ) J n (θ). Lemma 11 is the key step in the proof. It shows the relationship between the p i s q j 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. 13 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 finite additive EU representation gives the result. LEMMA 10: For every n {1 N} for every θ R, there exists an interval [θ θ] θ, where θ < θ, such that I n (θ) = I n (θ ) θ [θ θ] J n (θ) = J n (θ ) θ [θ θ] Figure 2 illustrates Lemma 10. PROOF: Fixn θ.since{w 1 w I v 1 v J } is not redundant, a i /p i b j /q j for every i I n j J n. Hence, it cannot be that a i /p i = b j /q j = θ for some i I n j J n. So we consider two overlapping cases. 13 Alternatively, one could prove Lemma 11 by constructing a lottery menu that would otherwise violate inclusion.

19 366 JOHN E. STOVALL FIGURE 2. θ = min j Jn\J n(θ ) b j /q j θ = i In(θ ) a i /p i. It is easy to see that I n (θ) = I n (θ ) J n (θ) = J n (θ ) for any θ [θ θ]. Case 1 a i /p i θ i I n. By definition, I n (θ ) ={i I n : a i /p i θ }.But since a i /p i θ for every i I n, then I n (θ ) ={i I n : a i /p i <θ }. So there exists ε>0such that I n (θ ε) = I n (θ ) J n (θ ε) = J n (θ ).Setθ θ ε θ θ. Case 2 b j /q j θ j J n. Observe that J n \ J n (θ ) ={j J n : b j /q j θ }. But since b j /q j θ for every j J n, then J n \ J n (θ ) ={j J n : b j /q j >θ }.So there exists ε>0 such that J n \ J n (θ + ε) = J n \ J n (θ ) I n \ I n (θ + ε) = I n \I n (θ ), which implies J n (θ +ε) = J n (θ ) I n (θ +ε) = I n (θ ).Setθ θ θ θ + ε. Q.E.D. 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 {1 N} for every θ R, we have p i q j 0 i I n(θ) j J n(θ) PROOF: Ifu = 0, then I = J = by Lemma 8 the result is vacuous. So assume u 0

20 MULTIPLE TEMPTATIONS 367 Now suppose the result is not true, that is, there exist n θ such that (10) p i q j > 0 i I n (θ ) j J n (θ ) By Lemma 10, there exists a nonsingleton interval [θ θ] θ such that (11) (12) I n (θ) = I n (θ ) J n (θ) = J n (θ ) θ [θ θ] θ [θ θ] In the remainder of the proof, we construct a menu x {β } lottery β that violate exclusion given (10). We show this by determining where each expected-utility function (i.e., w 1 w I v 1 v J ) attains its imum on x {β } then again on x {β } {β }.Figure3 collects these results. As Figure 3 shows, the only functions whose imum changes when β is added to the menu are those associated with I n (θ ) J n (θ ). This allows us to use (10) to show that V(x {β })<V(x {β } {β }), violating exclusion. Let x be a sphere in H u Δ. Observe that x is a constant menu. Since f n H u H 1 \{0}, Lemma 2 implies that c(x f n ) ={β n } is a singleton that (13) β x β w i = β n w i i I n Also, since {f 1 f N } H u H 1 \{0} is not redundant, Lemma 2 implies (14) β x β w i >β n w i i I \ I n FIGURE 3. Where the imum for the respective expected-utility function is attained. For example, the third row first column show β x {β } β w i = β x β w i for i I \ I n.

21 368 JOHN E. STOVALL Similarly, we can show (15) (16) β x β v j = β n v j β x β v j >β n v j j J n j J \ J n Observe that (θ/(f n f n ))f n (1/(u u))u H 1 since u f n H 1. Hence, β n Δ, inequalities (14)(16) imply that there exists ε>0 such that ( θ β β n + ε f n 1 ) f n f n u u u Δ (17) β w i >β w i i I \ I n β x (18) β x β v j >β v j j J \ J n Observe that β u = β n u ε, which implies β u = min β x {β } β u since x is constant. Geometrically, β is a lottery that is an ε move from β n in the direction of (θ/(f n f n ))f n (1/(u u))u. Consider the menu x {β }. To fill in the column x {β } infigure3, we must determine the imizers over x {β } for the expected-utility functions {w 1 w I v 1 v J }. 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 β w i = β n w i + εp i (θ a i /p i ) for i I n.since ε>0p i > 0, this implies β w i β n w i if only if i I n (θ). Hence equations (11)(13)imply β w β x {β i = β w i i I n (θ ) } β w β x {β i = β w i } β x i I n \ I n (θ ) Similarly, using equations (12)(15), we obtain β v β x {β j = β v j j J n (θ ) } β v β x {β j = β v j } β x j J n \ J n (θ )

22 MULTIPLE TEMPTATIONS 369 Now we construct the lottery β that will lead to a contradiction of exclusion. The idea is that we will take the lottery β move a small distance in the direction of f n.sincef 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 β x {β } β w i >β w i i I \ I n β x {β } β v j >β v j j J \ J n Since f n H 1 β Δ, there exists ˆε > 0 such that β +ˆε f n Δ (19) (20) β x {β } β w i >(β +ˆε f n ) w i β x {β } β v j >(β +ˆε f n ) v j i I \ I n j J \ J n Define ˆε ε(( θ θ)/(f n f n )) > 0. Observe that by the definition of β, ( β +ˆε f n = β n + ε θ f n f n f n 1 ) u u u Hence, using (8), (β +ˆε f n ) w i = β n w i + εp i ( θ a i /p i ) for i I n.this implies (β +ˆε f n ) w i β n w i if i I n \ I n ( θ). Hence by (11), (21) β n w i (β +ˆε f n ) w i i I n \ I n (θ ) Similarly, using (12), we obtain (22) β n v j >(β +ˆε f n ) v j j J n \ J n (θ ) Set ˆε min{ˆε ˆε } β β +ˆεf n. Hence β Δ, so consider the menu x {β } {β }. First, observe that β u = β u = min β x {β } β u. Hence, {β} {β } for every β x {β }. To fill in the column x {β } {β } in Figure 3, we must determine the imizers over x {β } {β } for the expected-utility functions {w 1 w I v 1 v J }. 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 \ I n J \ J n.)

23 370 JOHN E. STOVALL Inequality (21) implies β n w i β w i i I n \ I n (θ ) which implies β w β x {β } {β i = β w } β x {β i } i I n \ I n (θ ) Inequality (22) implies β n v j >β v j j J n \ J n (θ ) which implies β v β x {β } {β j = β v } β x {β j } j J n \ J n (θ ) Using (8), observe that β w i = β w i +ˆεp i f n f n for i I n.butˆεp i f n f n > 0, so this implies β w i >β w i for every i I n. Hence, this holds for every i I n (θ ) or β w i >β w i i I n (θ ) which implies β x {β } {β } β w i = β w i i I n (θ ) since β x {β } β w i = β w i for i I n (θ ). Similarly, β v j >β v j j J n (θ ) which implies β x {β } {β } β v j = β v j j J n (θ ) since β x {β } β v j = β v j for j J n (θ ). As is evident from viewing Figure 3, the only expected-utility functions that increase by going from x {β } to x {β } {β } are those associated with the sets I n (θ ) J n (θ ). Hence, V(x {β } {β }) V(x {β }) = (β β ) w i (β β ) v j i I n (θ ) = i I n (θ ) ˆεf n w i j J n (θ ) j J n (θ ) ˆεf n v j

24 = i I n (θ ) MULTIPLE TEMPTATIONS 371 ˆεp i f n f n ( =ˆε(f n f n ) > 0 i I n (θ ) j J n (θ ) p i j J n (θ ) ˆεq j f n f n since ˆε(f n f n )>0 p i I n (θ ) i q j J n (θ ) j > 0by(10). This implies x {β } {β } x {β }, which contradicts exclusion since {β} {β } for every β x {β }. Q.E.D. LEMMA 12: For every n, there exist nonnegative scalars {c ij } i In j J n, {d ij } i In j J n, {e ij } i In j J n, where j J n d ij = 1 for all i I n i I n e ij = 1 for all j J n, such that q j ) d ij w i = c ij u + e ij v j i I n j J n PROOF: The proof is by construction. Fix n. For every i I n j J n,define { } ˆL ij min p i q j i I n : a i /p i a i /p i j J n : b j /q j b j /q j { } p i q j i I n : a i /p i >a i /p i j J n : b j /q j >b j /q j L ij {0 ˆL ij } Hence, L ij 0 for every i I n j J n. One can show that for every i I n, j J n L ij = p i, that for every j J n, i I n L ij = q j Now for every i I n j J n,defined ij L ij /p i, e ij L ij /q j,c ij a i d ij b j e ij. (These are well defined since p i q j are strictly positive.) Fix i I n j J n. Observe that d ij 0, e ij 0, j J n d ij = 1, i I n e i 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 c ij = b ) j L ij p i q j

25 372 JOHN E. STOVALL Hence, c ij 0ifonlyifa i /p i b j /q j.sincel ij > 0, it must be that q j > j J n : b j /q j b j /q j But Lemma 11 implies p i i I n : a i /p i >b j /q j Together, these imply p i > i I n : a i /p i >b j /q j i I n : a i /p i >a i /p i p i j J n : b j /q j b j /q j q j i I n : a i /p i >a i /p i p i which can only be true if a i /p i >b j /q j. Q.E.D. Now we finish the proof of sufficiency. Observe that we can write V as V(x)= N { n=1 i I n β x β w i j J n β x β v j + i/ I β x β w i j/ J β x β v j } Using Lemma 12,weget V(x)= = N { ( )} β d ijw i β e ijv j β x β x n=1 i I n j J n + i/ I β x β w i j/ J β x β v j N { ( )} [β c iju + β e ij v j ] β e ijv j β x β x n=1 i I n j J n + i/ I β x β w i j/ J β x β v j If c ij = 0, then d ij = e ij = 0sincew i v j are not redundant. So let M denote the number of (i j) pairs in I J such that n(i) = n(j) c ij 0. Let m(i j) denote a distinct element of {1 M}. Define ˆv m(i j) (e ij /c ij )v j

26 ĉ m(i j) c ij. Hence we can write V(x)= M m=1 MULTIPLE TEMPTATIONS 373 } ĉ m { [β u + β ˆv m] β ˆv m β x β x + i/ I β x β w i j/ J β x β v j For i/ I, recall that Lemma 6 the fact that w i 0 imply that w i = a i u, where a i > 0. So for i/ I,defineṽ i 0. Similarly, for j/ J, Lemma 6 the fact that v j 0 imply that v j = b j u,whereb j > 0. So for j/ J,define v j u. Hence (23) V(x)= M m=1 } ĉ m { [β u + β ˆv m] β ˆv m β x β x + } a i { [β u + β ṽ i] β ṽ i i/ I β x β x + } b j { [β u + β v j] β v j j/ J β x β x Finally, observe that u = w i v j + w i v j i I j J i/ I j/ J N { } = (d ij w i e ij v j ) + w i v j n=1 i I n j J n i/ I j/ J N { } = c ij u + a i u + b j u n=1 i I n j J n i/ I j/ J ( M = ĉ m + a i + b j )u i/ I j/ J m=1 which implies M m=1 ĉm + i/ I a i + j/ J b j = 1. So {ĉ 1 ĉ m } {a i } i/ I {b j } j/ J is a set of positive scalars that sum to 1. Hence (23) is a temptation representation.

27 374 JOHN E. STOVALL 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 X α Δ be such that {β} {α} for every β x. Thenmin β x u(β) u(α). We will show that for every i, β x [u(β) + v i(β)] β x v i(β) β x {α} [u(β) + v i(β)] β x {α} v i(β) thus proving that V(x) V(x {α}).fixi. Case 1 u(α) + v i (α) β x [u(β) + v i (β)]. Since min β x u(β) u(α),it must be that v i (α) β x v i (β). Hence β x {α} [u(β) + v i(β)] β x {α} v i(β) = u(α) min u(β) [u(β) + v i(β)] v i(β) β x β x β x where the second inequality can easily be verified by making the substitution w i u v i. Case 2 β x [u(β) + v i (β)] > u(α) + v i (α).then β x {α} [u(β) + v i(β)] β x {α} v i(β) = β x [u(β) + v i(β)] β x {α} v i(β) β x [u(β) + v i(β)] β x v i(β) Q.E.D. APPENDIX 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 α x such that x {α}. Fix any vectors w 1 f H 1 \{0} such that f w 1.Setv 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)= β x w 1(β) + β x w 2(β) β x v 1(β) β x v 2(β)

28 MULTIPLE TEMPTATIONS 375 which is a finite additive EU representation, hence satisfies 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 1 2a { u(β) + β x 2 2a [u(β) + ˆv 1(β) + ˆv 2 (β)] β x } ˆv 2 (β) β x ˆv 1 (β) β x where ˆv 1 = ((2 2a)/(1 2a))v 1 ˆv 2 = ((2 2a)/(1 2a))v 2,whichisa DLR temptation representation. Hence, this preference satisfies 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 satisfies DFC. But this implies that the preference represented by V satisfies desire for better alternatives. (This preference also satisfies DLR s other axiom approximate improvements are chosen as well as an axiom symmetric to this, so even adding these axioms would be insufficient 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 sufficiency for Theorem 1.) Fix any β Δ.Letε>0 be such that α β + εu Δ. Observe then that α w 1 >β w 1, β w 2 >α w 2, β v 1 >α v 1,α v 2 >β v 2.Letε > 0be such that β β + ε f Δ α v 2 >β v 2.ThenV({α β β })>V({α β}) since w 2 is the only vector whose imum changes when β is added to {α β}. But this violates exclusion since β u = β u<α u. A violation of inclusion could be constructed in a similar manner. REFERENCES CHANDRASEKHER, M. (2009): A Theory of Local Menu Preferences, Working Paper, Arizona State University. [351] DEKEL, E., AND B. LIPMAN (2007): Self-Control Rom Strotz Representations, Working Paper, Boston University. [352,353] DEKEL, E., B. LIPMAN, AND A. RUSTICHINI (2001): Representing Preferences With a Unique Subjective State Space, Econometrica, 69, [351,360] (2009): Temptation-Driven Preferences, Review of Economic Studies, 76, [350,355,360] DEKEL, E.,B.LIPMAN, A.RUSTICHINI, AND T. SARVER (2007): Representing Preferences With a Unique Subjective State Space: Corrigendum, Econometrica, 75, [352] GUL, F., AND W. PESENDORFER (2001): Temptation Self-Control, Econometrica, 69, [350,352,354] KOPYLOV, I. (2009): Perfectionism Choice, Working Paper, UC Irvine. [356] NEHRING, K. (2006): Self-Control Through Second-Order Preferences, Working Paper, UC Davis. [351]

29 376 JOHN E. STOVALL NOOR, J. (2007): Commitment Self-Control, Journal of Economic Theory, 135, [350] SARVER, T. (2008): Anticipating Regret: Why Fewer Options May Be Better, Econometrica, 76, [356] Dept. of Economics, Harkness Hall, University of Rochester, Rochester, NY 14627, U.S.A.; jstovall@mail.rochester.edu. Manuscript received August, 2008; final revision received September, 2009.