Satisficing with a variable threshold

Journal Pre-proof Satisficing with a variable threshold Matthew Kovach, Levent Ülkü

PII: DOI: Reference:

S0304-4068(20)30003-3 https://doi.org/10.1016/j.jmateco.2019.12.005 MATECO 2368

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Journal of Mathematical Economics

Received date : 13 January 2019 Revised date : 2 October 2019 Accepted date : 28 December 2019

Please cite this article as: M. Kovach and L. Ülkü, Satisficing with a variable threshold. Journal of Mathematical Economics (2020), doi: https://doi.org/10.1016/j.jmateco.2019.12.005. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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Satis…cing with a Variable Threshold Matthew Kovachy

Levent Ülküz

September 30, 2019

Abstract

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We axiomatize a model of satis…cing which features random thresholds and the possibility of choice abstention. Given a menu, the decision maker …rst randomly draws a threshold. Next, using a list order, she searches the menu for alternatives which are at least as good as the threshold. She chooses the …rst such alternative she …nds, and if no such alternative exists, she abstains. Since the threshold is random, so is the resulting behavior. We characterize this model using two simple axioms. In general the revelation of the model’s primitives is incomplete. We characterize a specialization of the model for which the underlying preference and list ordering are uniquely identi…ed by choice frequencies. We also show that our model is a special Random Utility Model. J.E.L. codes: D0. Keywords: Stochastic Choice, Satis…cing, Random Thresholds, Stochastic Consideration

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We would like to thank Yuhta Ishii, Lina Lukyantseva, Pietro Ortoleva, Romans Pancs and audiences at Cal Poly SLO, COLMEX, Rutgers, UCSB, BRIC 2017 and D-TEA 2017, as well as the editor and two referees for helpful comments. Levent Ülkü acknowledges …nancial support from the Asociación Mexicana de Cultura. y Department of Economics, Virginia Tech, 880 West Campus Drive, Blacksburg, VA 24061; [email protected]. z Centro de Investigación Económica, Department of Economics, ITAM, Río Hondo 1, Ciudad de México 01080, Mexico; [email protected].

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Introduction

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We study a stochastic variation of the celebrated satis…cing model due to Simon (1955). A satis…cing decision maker is commonly modeled using three psychological primitives: (1) a utility function, (2) a threshold utility level, and (3) a list ordering of alternatives. In any given menu, the decision maker’s choice is the …rst alternative in the list order whose utility beats the threshold. If the search process yields no such alternative, the decision maker chooses a maximal utility alternative (which is often explicitly assumed to be unique) or the last alternative in the list.1 Our variation features two important departures from this basic model. First, we operate in an environment with choice abstention. We interpret choice abstention simply as walking away from the menu, a commonly observed behavior when choice is not forced, as is often the case outside of laboratory settings.2 The second departure pertains to the nature of the threshold. We operate in an ordinal environment and our decision maker is endowed with a preference rather than a utility function. Concordantly, we assume that the threshold is given by an alternative rather than a utility level. We further assume that the threshold alternative is randomly drawn and its realization is only observable to the decision maker. Hence our model allows for the decision maker to have di¤erent thresholds for choices made from di¤erent menus, or for choices made from the same menu at di¤erent times. Hence our primitives are (1) a preference, (2) a threshold probability for every alternative, and (3) a list order. Faced with any menu, the decision maker randomly draws a threshold, and chooses the …rst listed alternative which is at least as good as the threshold. If there exists no such alternative, she abstains from choice. Our main results are as follows. In Theorem 1, we show that our model is a special case of the famous Random Utility Models. In Theorem 2, together with Lemmas 4 and 6, we characterize our model using two simple axioms. The revelation of the primitives of our model is generally incomplete. In Theorem 3, we characterize a special case which rules

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This idea of satis…cing has proved in‡uential in and beyond economics, and the literature is vast. For examples of deterministic satis…cing models in economics, see Tyson (2008), Papi (2012) and Barberà and Neme (2016). See also the (u; v) procedure in Kalai et al. (2002) and Example 2 in Rubinstein and Salant (2006). Examples outside economics include Schwartz et al. (2002) and Ward (1992). 2 See for instance Gerasimou (2018), Aguiar (2017), Kovach (2016) and Manzini and Mariotti (2014) for recent examples of abstract choice models which allow for abstention.

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out zero-probability thresholds. The underlying preference and list of this specialization are uniquely identi…able from choice data. Section 5 contains results which study the consequences of menu-dependence in thresholds and correlation between the underlying primitives. At the core of our model is the idea that thresholds used by decision makers in determining good enough alternatives may be random. There is empirical evidence for this in Caplin et al. (2011). We would like to suggest two potential reasons for variation in thresholds. First, thresholds may capture a decision maker’s changing psychological state or mood. Whether or not an alternative is good enough may be subject to changes in a variety of variables such as the decision maker’s ambition level or past choices, which may not be observable. Second, many decisions take place in the presence of an unobservable outside option, which renders any alternative inferior to it unacceptable.3 A combination of these two reasons may be in e¤ect in certain situations, whereby a decision maker’s mood depends on her outside option. Furthermore these reasons apply even when stochastic choice is meant to capture a population’s behavior, as moods and outside options may change across individuals. Our axiomatic analysis fully accounts for the behavioral consequences of randomness in thresholds within a satis…cing framework, without taking a stance on its sources.

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Relation to the literature Aguiar et al. (2016) study a di¤erent stochastic variation of satis…cing. In their model the decision maker is endowed with a …xed utility, a …xed threshold, and a random search order. They leave the analysis of satis…cing models where stochastic behavior arises out of randomness due to other primitives as an open question.4 We take up precisely this suggestion in our study of random thresholds. One important di¤erence between our work and Aguiar et al. (2016) is that we allow for choice abstention. This leads to a very di¤erent revelation exercise for primitives and, consequently, the axioms used in the two papers are not similar. Aguiar and Kimya (2019) study a generalization of the model in Aguiar et al. (2016) that allows for menu-dependence in thresholds. They show that this generalization can This type of status quo bias has been modeled in Masatlioglu and Ok (2005). See also Kahneman et al. (1991) for a review of related experiments and empirical studies. 4 In particular, they show that for a class of satis…cing models with random search orders, an additional randomness in the threshold is inconsequential in terms of the scope of the resulting behavior.

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explain the attraction and compromise e¤ects. In contrast, our model is a Random Utility Model, as is the model in Aguiar et al. (2016). Therefore we cannot explain these behavioral phenomena. Our model can produce choice frequency reversals and stochastically intransitive behavior, however, as we show via examples in Section 3.1. Stochastic satis…cing of the type we consider in this paper is essentially a model of stochastic consideration. Indeed in our model the decision maker only considers satisfactory alternatives for choice and chooses among them using the list order. Whether or not an alternative is satisfactory is determined by a random threshold. In this sense our paper is closely related to the stochastic consideration model in Manzini and Mariotti (2014). The relationship is quite clear from an axiomatic viewpoint, as our Impact Asymmetry axiom is directly borrowed from them. Despite this, neither model is a special case of the other. However, as we discuss in Section 5.3, a variation of our model in which the thresholds are allowed to change during search yields the model in Manzini and Mariotti (2014). Hence their stochastic consideration model could be interpreted as one of satis…cing as well.5 Our paper is also related to models of choice from lists. Rubinstein and Salant (2006) study a model of choice from observable lists. Their main focus is on rational choice from lists and how the ordering of alternatives a¤ects choice. They do however consider stochastic choice functions from lists, though in their model randomness come from variation in the list, as in Aguiar et al. (2016). Finally, Yildiz (2016) also considers random choice from lists. In his paper choice from lists is determined by sequential comparisons and randomness in choice is due to randomness in the preference used to compare alternatives. Consequently the behavioral implications of both of these papers are quite distinct from our model of satis…cing.

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This two-way relationship between limited consideration and satis…cing extends to deterministic environments. In models of two-stage choice such as Manzini and Mariotti (2007) and Bajraj and Ülkü (2015), the decision maker uses one binary relation to shortlist alternatives in the …rst stage, and chooses from the shortlist using a di¤erent binary relation in the second stage. If we assume that the …rst binary relation is a preference, and the second a search order, the behavior becomes one of satis…cing: the …rst stage identi…es a set of satisfactory alternatives, and the second stage chooses the …rst such alternative in the search order.

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Notation and basic de…nitions

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Binary relations and probability functions Let X be a …nite set. A binary relation % on X is complete if for all x; y 2 X, either x % y or y % x; transitive if for all x; y; z 2 X, x % z whenever x % y and y % z; and a weak order if it is complete and transitive. If % is a weak order on X and S X, max(S; %) := fx 2 S : x % y for every y 2 Sg and min(S; %) := fx 2 S : y % x for every y 2 Sg. Given a weak order %, we de…ne its strict part and its indi¤erence part in the usual way: (1) x y if and only if x % y and not y % x, (2) x y if and only if x % y and y % x. A weak order % is a linear order if it is also antisymmetric, i.e., if for all x; y 2 X, x = y whenever x % y and y % x. If % is a linear order and S X is nonempty, then max(S; %) and min(S; %) are singletons. In this case we abuse notation and refer to the unique maximal (minimal) element of S according to % as max(S; %) (min(S; %)) as well. For any x and S, by x % ( )S, we mean x % ( )y for every y 2 S. Similarly S % ( )T means x % ( )y for every (x; y) 2 S T . P A probability on X is a map : X ! [0; 1] such that x2X (x) = 1. A probability is strict if (x) > 0 for every x 2 X.

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Stochastic choice functions with abstention Let X be a …nite set of alternatives and D be the set of menus, i.e., the set of all nonempty subsets of X. There exists some a 62 X which denotes choice abstention. Abstention is implicitly feasible in every menu and its choice is observable. We denote the set of possible choices available to the decision maker when faced with menu A 2 D by A := A [ fa g. A stochastic choice function is a P map p : X D ! [0; 1] satisfying, for every A, (1) x2A p(x; A) = 1, and (2) p(x; A) = 0 for every x 62 A . In other words, p de…nes a probability on A for every menu A, and p(x; A) is the frequency with which the decision maker chooses x when faced with A. A stochastic choice function p satis…es Regularity if whenever A B and x 2 A , it follows that p(x; B) p(x; A). Regularity says that the choice frequency of an alternative (or abstention) cannot increase when more options become available. Let L denote the set of linear orders on X . A stochastic choice function p is a Random Utility Model (RUM) if there exists a probability on L such that X p(x; A) = (%) %2L: x=max(A ;%)

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for every menu A and every alternative x 2 A . Hence the choice frequency of x in A is given by the cumulative probability of drawing a preference (a linear order % on X ) according to which x is the best alternative in A . Note that a RUM treats abstention as an alternative, in the sense that a is in the domain of preferences and is chosen if and only if it is preferable to every member of the menu. It is well known that Regularity is necessary but not su¢ cient for a stochastic choice function to be a RUM.6

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Random Satis…cing Rules

We are now ready to de…ne the class of stochastic choice functions of our interest.

p(x; A) =

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De…nition 1 A stochastic choice function p is a Random Satis…cing Rule (RSR) if there exists a triple (%; L; ) where % is a weak order on X, L is a linear order on X and is a probability on X such that for every A and every x 2 A; X

(a):

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a2X : x=max(fy2A:y%ag;L)

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The underlying primitives of a RSR have the following interpretations. The weak order % on X gives the decision maker’s preferences over alternatives. The linear order L on X is a list which determines the sequence with which the decision maker peruses through any given menu. For every a 2 X , (a) is the probability that the threshold used by the decision maker is a. Faced with a given menu A the decision maker operates as follows. 1. She randomly draws a threshold alternative a 2 X . The threshold a and her preference % together determine the set fy 2 A : y % ag of satisfactory feasible alternatives. This set may be empty.

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2. She considers alternatives in A sequentially using the restriction of list L to menu A. If A contains a satisfactory alternative, she chooses the …rst such alternative on the list. Hence if fy 2 A : y % ag = 6 ;, her choice is max(fy 2 A : y % ag; L). See Chambers and Echenique (2016).

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3. If A contains no satisfactory alternative, i.e., if fy 2 A : y % ag = ;, she abstains. This happens if a 2 X but a A, or if a = a . Hence p(a ; A) = (a ) +

X

(a).

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a2X:a A

Behavioral patterns

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P P For instance, p(x; fxg) = a-x (a), for every x 2 X. In general, p(x; A) a-x (a), as some thresholds which render x acceptable may lead to the choice of a di¤erent acceptable member of A which precedes x in the list order. We emphasize that % and L are de…ned on X and not on X . This means that a decision maker using a RSR does not compare a to any x 2 X in terms of preference or list precedence. The decision maker abstains not because abstention is preferable to other available options, but because the alternatives she faces are inferior to the realization of her random threshold. Consequently the choice of abstention is behavioral rather than rational. This is an important di¤erence between RSRs and RUMs. We also note that the psychological primitives %, L and do not depend on the menu the decision maker faces. In particular even though the decision maker typically uses di¤erent thresholds in di¤erent menus (or in the same menu at di¤erent times) the distribution of the thresholds is menu-independent. In Section 5.2 we study the consequences of relaxing this assumption. As we noted in the Introduction, RSRs may be interpreted as models of stochastic consideration similar to those studied in Manzini and Mariotti (2014), Brady and Rehbeck (2016) and Aguiar (2017). In this interpretation, the linear order L assumes the role of a deterministic preference. In every menu A, the decision maker uses the binary relation % and the alternative a to form a consideration set fy 2 A : y % ag A, and then maximizes her preference L on this set. Since the threshold alternative a is random, so are the resulting consideration set and the decision maker’s choice.

We begin with a preliminary result. Lemma 1 Let p be a Random Satis…cing Rule de…ned by primitives (%; L; ) and x 2 A 2 D. 7

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1. If x = max(A; L), then p(x; A) = p(x; fxg) 2. If yLx and y % x for some y 2 Anfxg, then p(x; A) = 0.

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3. If x 6= max(A; L), x fy 2 Anfxg : yLxg and z 2 max(fy 2 Anfxg : yLxg; %), then p(x; A) = p(x; fxg) p(z; fzg).

as desired.

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Proof. 1. Suppose that x = max(A; L). Then the decision maker chooses x if and only if the threshold renders x acceptable. In other words, any threshold a 2 X such that x % a P leads to the choice of x in A. This gives p(x; A) = a-x (x) = p(x; fxg). 2. Suppose that there exists some y 2 Anfxg such that yLx and y % x. Then any threshold a which renders x acceptable, also renders y acceptable and leads to the choice of y, or another alternative which precedes y. Hence there exists no threshold which leads to the choice of x, giving p(x; A) = 0. 3. To …nish, suppose that x is not the …rst listed alternative in A, but that x is strictly preferred to all of its predecessors. Let z be a %-maximal alternative in fy 2 Anfxg : yLxg. Then any threshold a such that z % a leads to the choice of z or one of the predecessors of z. It follows that x is chosen if and only if the threshold a is such that x % a z. This gives X X X p(x; A) = (a) = (a) (a) = p(x; fxg) p(z; fzg);

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Lemma 1 calculates the choice frequency of an alternative x in menu A in three mutually exclusive and exhaustive scenarios. 1. If x is the …rst alternative the decision maker sees in A, then the remaining alternatives in A do not reduce the choice frequency of x and therefore p(x; A) = p(x; fxg). This follows because every threshold a - x leads to the choice of x in A as well as in fxg. 2. If there exists an alternative y in A which precedes x in the list order and which is at least as good as x in the preference order, then p(x; A) = 0. This follows because every threshold a - x leads to the choice of y or a di¤erent alternative which precedes y. 3. Finally, suppose that x is not the …rst alternative seen by the decision maker, but is strictly better than every alternative which the decision maker sees before x. Let z be a preference-maximal alternative among those alternatives. Then x z and a 8

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threshold a leads to the choice of x if and only if x % a p(x; fxg) p(z; fzg).

z. Consequently p(x; A) =

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Lemma 1 helps in proving the following important result. Theorem 1 Every Random Satis…cing Rule is a Random Utility Model.

Proof. Let p be a RSR de…ned by the primitives (%; L; ). Enumerate the equivalence classes of alternatives fIi gni=1 produced by % on X such that In I1 . Let (Ii ) = P Pn a2Ii (a) be the probability that the threshold belongs to Ii , so that i=1 (Ii )+ (a ) = 1. Consider the following inductively de…ned n + 1 linear orders %1 ; : : : ; %n+1 on X .

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%1 preserves the ordering of L and places a last. Hence: (1) x %1 a for all x 2 X, and (2) for x; y 2 X, x %1 y i¤ xLy.

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For all i = 2; : : : ; n, %i preserves the ordering of %i 1 , except moves all alternatives S which are strictly inferior to Ii below a . Hence: (1) x %i a %i y for all x 2 nj=i Ij , S and y 2 ij=11 Ij , and (2) for all x; y 2 X, x %i y i¤ xLy. Finally, let %n+1 be such that (1) a %n+1 x for every x 2 X, and (2) for all x; y 2 X, x %n+1 y i¤ xLy.

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Hence if x 2 Ii then x %j a for all j = 1; : : : ; i. Now let (%i ) = (Ii ) for all i = 1; : : : n and (%n+1 ) = (a ) so that is a probability on the set of all linear orders on X . Let q be the RUM de…ned by . We will show that q = p. It su¢ ces to show that q(x; A) = p(x; A) for all A and for all x 2 A. Take any menu A and any x 2 A. Suppose x 2 Ii . Using the calculations in Lemma 1, the following three exhaustive cases complete the proof. Case 1: x = max(A; L). In this case x = max(A ; %j ) if and only if j i. It follows that X X X q(x; A) = (%j ) = (Ij ) = (a) = p(x; fxg) = p(x; A)

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where the last equality follows from the …rst part of Lemma 1. Case 2: yLx and y % x for some y 2 Anfxg. In this case x = max(A ; %j ) for no j and q(x; A) = 0 = p(x; A), where the second equality is due to the second part of Lemma 2. 9

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Case 3: yLx for some y 2 Anfxg, but x y for every such y. Let z 2 arg maxy2Anfxg:yLx p(y; fyg). Let Ii be the indi¤erence class of x and Ij be the indi¤erence class of z, with j < i, as x z. Then x tops all linear orders %k with j < k i and consequently X q(x; A) = (Ij ) = p(x; fxg) p(z; fzg) = p(x; A); j
where the last equality follows from the third part of Lemma 1.

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Hence, even though RSRs and RUMs are de…ned by di¤erent sets of primitives and, as we made note above, di¤er markedly in the nature of abstention decisions, RSRs form a subclass of RUMs. An important consequence of this observation is that RSRs satisfy Regularity. Hence RSRs cannot explain behavioral phenomena such as the attraction e¤ect and the compromise e¤ect, which are violations of Regularity. As the next example shows, RSRs can explain the phenomenon of choice frequency reversal, whereby the introduction of a new alternative to a menu changes the choice frequency ordering of two extant alternatives.7

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Example 1 Let X = fx; y; zg and % and L be as follows: x z y, yLzLx. Suppose (y) = (z) = 2=5 and (x) = 1=5. The resulting RSR p is such that p(x; fx; yg) = 3=5 > 2=5 = p(y; fx; yg) but p(x; fx; y; zg) = 1=5 < 2=5 = p(y; fx; y; zg).

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In general if the decision maker is using a RSR, a new alternative (z; in Example 1) may decrease the choice frequency of a strong alternative (x) which it precedes, while leaving unchanged that of a weaker alternative (y) which precedes it. Consequently, choice frequencies in any given menu cannot form the basis for preference revelation for RSRs, as they may lead to contradicting revelations in di¤erent sets. This is in line with our calculations in Lemma 1, which shows that choice frequencies are determined by the decision maker’s preferences, list order and threshold probabilities jointly. A second behavioral pattern which can be explained by stochastic thresholds in satis…cing is intransitive choice frequencies, as discussed in Tversky (1969). The following condition conceptualizes transitivity in choice frequencies. See, for instance, Simonson (1989) and Tsetsos et al. (2010) for more on the behavioral phenomena mentioned in this paragraph.

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Stochastic Transitivity: If p(x; fx; yg) 1=2.8

1=2 and p(y; fy; zg)

1=2, then p(x; fx; zg)

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This axiom is reasonable in models where choice frequencies in any given menu represent the decision maker’s transitive preferences. Since this is not the case for RSRs, it should not be surprising that RSRs may fail the axiom. Example 2 Let X = fx; y; zg and % and L be as follows: y x z, zLxLy. Suppose (z) = (y) = 2=5 and (x) = 1=5. The resulting RSR p is such that p(x; fx; yg) = 3=5, p(y; fy; zg) = 3=5 but p(x; fx; zg) = 1=5.

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Revelation

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We now visit the question of revelation. The following result delineates the extent to which choice frequencies reveal the underlying primitives % and L of a RSR.

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Lemma 2 Let p be a Random Satis…cing Rule de…ned by primitives (%; L; ); and x and y be distinct alternatives. 1. If p(x; fxg) > p(y; fyg), then x

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2. If p(y; fyg) > p(y; fx; yg), then xLy.

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P Proof. The …rst implication follows from the observation that p(x; fxg) = a-x (a) P and p(y; fyg) = a-y (a). Hence, if p(x; fxg) > p(y; fyg), then fa 2 X : x % ag fa 2 X : y % ag, implying x y. To see the second implication, suppose that p(y; fyg) > p(y; fx; yg) but yLx. It follows that y = max(fx; yg; L), and by the …rst part of Lemma 1, that p(y; fyg) = p(y; fx; yg), a contradiction. Since L is a linear order, we conclude that xLy.

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The idea behind these revelations is straightforward. Suppose that a decision maker using a RSR abstains more often when faced with menu fyg than when faced with fxg. This indicates that it is strictly more likely that the decision maker …nds x satisfactory A stronger condition is to require p(x; fx; zg) maxfp(x; fx; yg); p(y; fy; zg)g whenever p(x; fx; yg) 1=2 and p(y; fy; zg) 1=2: See, for instance, Rieskamp et al. (2006).

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than she does y, which in returns means that x is strictly better than y. If, on the other hand, the removal of x from menu fx; yg increases the choice frequency of y, then some thresholds leading to the choice of fxg in fx; yg must be leading to the choice of y in fyg. This is only feasible if x precedes y in the list order. We also note that the revelations in Lemma 2 only rely on small menus, where the decision maker chooses between abstention and at most two alternatives. The following example demonstrates that neither implication in Lemma 2 can, in general, be reversed. Example 3 Suppose that X = fx; yg, x y and yLx. If (x) = 0, then p(x; fxg) = (y) = p(y; fyg). If (y) = 0 on the other hand, then p(x; fxg) = (x) = p(x; fx; yg).

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In Lemma 7, we will see that the implications in Lemma 2 can be reversed for RSRs de…ned by strictly positive threshold probabilities. The revelation of the threshold probabilities underlying a RSR is also delicate. The following result indicates that we may only hope to identify the probability that the threshold belongs to any given indi¤erence class of the decision maker’s preferences.

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Lemma 3 Let % be a weak order on X with indi¤erence classes I1 ; :::; In , and and 0 P P be two probabilities on X such that a2Ii (a) = a2Ii 0 (a) for every i = 1; :::; n. For every linear order L on X, the primitives (%; L; ) and (%; L; 0 ) de…ne the same Random Satis…cing Rule.

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Proof. Let p and p0 be the RSRs de…ned by (%; L; ) and (%; L; 0 ) respectively. For any P P x 2 X, let I x be the indi¤erence class of x in %, so that a-I x (a) = a-I x 0 (a) for every x. Take any A and any x 2 A. It su¢ ces to show that p(x; A) = p0 (x; A). Since both p and p0 are de…ned by the same preference and list order, using Lemma 1, there are three cases to consider. If x = max(A; L), then X X 0 p(x; A) = p(x; fxg) = (a) = (a) = p0 (x; fxg) = p0 (x; A). a0 -I x

a0 -I x

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If y % x and yLx for some y 2 Anfxg, then p(x; A) = 0 = p0 (x; A). Finally, suppose that there exists some y 2 Anfxg such that yLx, but x y for every such y, and let z be a preference maximal alternative among the predecessors of x in A. Then X X 0 p(x; A) = (a) = (a) = p0 (x; A). I z a0 -I x

I z a0 -I x

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This completes the proof.

Characterization

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To characterize the class of RSRs we will need two axioms. The …rst has appeared in Manzini and Mariotti (2014). Impact Asymmetry (IA): For every distinct x; y 2 X, if p(y; fyg) > p(y; fx; yg); then p(x; fxg) = p(x; fx; yg).9

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In view of Lemma 2, IA says that the revealed list order should be asymmetric: if the removal of x from fx; yg boosts the probability that y is chosen, then x is revealed to precede y, and the removal of y should not have an impact on the choice frequency of x. Indeed, RSRs satisfy this property. Lemma 4 If p is a Random Satis…cing Rule, then p satis…es Impact Asymmetry.

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Proof. Suppose that p(y; fyg) > p(y; fx; yg) for distinct x; y 2 X. By Lemma 2, xLy. By part 1 of Lemma 1, p(x; fxg) = p(x; fx; yg). We next introduce the following strengthening of Regularity.

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Strong Regularity (SR): For every A and B and every x 2 A \B , if p(x; A) then p(x; A) = p(x; A [ B).10

p(x; B);

SR tells us how choice frequencies in a menu are determined by choice frequencies in submenus. For instance, sequential application of SR gives p(x; A) =

min p(x; fx; yg) for every x 2 A, and

y2Anfxg

p(a ; A) = min p(a ; fyg). y2A

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Their statement is slightly di¤erent: if p(y; fyg) 6= p(y; fx; yg), then p(x; fxg) = p(x; fx; yg). Our statement is equivalent to theirs under Regularity. 10 If we write A x B whenever p(x; A) p(x; B) and let x denote the indi¤erence part of x , this axiom can be restated as follows: If B x A, then A x A [ B. This is a form of Set Betweenness which appears in the literature on menu preferences under temptation. See, for instance, Gul and Pesendorfer (2001).

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We make note that SR implies Regularity.

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Lemma 5 If p satis…es Strong Regularity, then it also satis…es Regularity. Proof. Take any menu A, x 2 A and y 62 A . If p(x; A) p(x; fx; yg), then p(x; A [ fyg) = p(x; A). If p(x; fx; yg) < p(x; A), then p(x; A [ fyg) = p(x; fx; yg) < p(x; A). Hence p(x; A [ fyg) p(x; A). Regularity follows from this observation. As we show next, RSRs satisfy SR.

Lemma 6 If p is a Random Satis…cing Rule, then p satis…es Strong Regularity.

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Proof. Suppose that p is a RSR de…ned by (%, L; ). For any menu A and any x 2 A, let T (x; A) = fa 2 X : x = max(fy 2 A : y % ag; L). Similarly, let T (a ; A) = fa 2 X : a y for every y 2 Ag [ fa g. Using this notation, for every A and every x 2 A , P p(x; A) = a2T (x;A) (a). Suppose that p(x; A) p(x; B). Case 1: x = a . Choose y A 2 max(A; %), y B 2 max(B; %) and y A[B 2 max(A[B; %). We have X X p(a ; A) p(a ; B) , (a) (a) a2T (a ;A)

(a ) +

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,

(a)

a2T (a ;B)

(a ) +

a yA

, yA % yB , y A[B

X

X

(a)

a yB

yA

, p(a ; A [ B) = p(a ; A)

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where the last implication follows from a similar reasoning used in the preceding steps. Case 2: x 2 A \ B. By Theorem 1, p is a RUM and therefore it satis…es Regularity. Hence p(x; A) p(x; A [ B). We will establish the reverse inequality. If p(x; A) = 0, there is nothing to show. Suppose that p(x; A) > 0. This implies that p(x; B) > 0 as well, as we have supposed that p(x; B) p(x; A). Hence T (x; A) and T (x; B) are nonempty. Choose ax;A 2 min(T (x; A); %). Take any a such that x % a % ax;A . No predecessor of x in A is acceptable when the threshold is a, since any such predecessor 14

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would also be acceptable when the threshold is ax;A , contradicting ax;A 2 T (x; A). Hence a 2 T (x; A) as well. We conclude that T (x; A) = fa : x % a % ax;A g. Similarly, choosing ax;B 2 min(T (x; B); %) and ax;A[B 2 min(T (x; A [ B); %), T (x; B) = fa : x % a % ax;B g and T (x; A [ B) = fa : x % a % ax;A[B g. Since p(x; A) p(x; B), it follows that ax;A % ax;B , i.e., T (x; A) T (x; B), which is the key observation. Now take any a 2 T (x; A). Then a 2 T (x; B) as well. This implies that there is no predecessor of x in A or in B which is acceptable when the threshold is a. Hence a 2 T (x; A [ B). It follows that T (x; A) T (x; A [ B) and therefore p(x; A) p(x; A [ B), as desired. Lemmas 4 and 6, together with our next result, jointly characterize the class of RSRs.

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Theorem 2 A stochastic choice function is a Random Satis…cing Rule if it satis…es Impact Asymmetry and Strong Regularity. Proof. See the Appendix.

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Unfortunately, as the following example shows, SR and IA do not identify the underlying primitives of a RSR uniquely. Example 4 (Unidenti…ed % and L for a RSR) Let X = fx; yg and consider the following stochastic choice function:

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p(x; fxg) = p(x; fx; yg) = 0,

pfy; fyg) = p(y; fx; yg) = 0,

p(a ; fxg) = p(a ; fyg) = p(a ; fx; yg) = 1:

It can easily be checked that p satis…es IA and SR. Hence, by Theorem 2, p is a RSR. Since the decision maker abstains surely in every menu, it must be the case that (a ) = 1 and therefore is identi…ed. However neither %, not L can be identi…ed.

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As Example 4 hints, the issue that arises in the identi…cation of and L is the possibility of zero-probability thresholds. In fact, as we saw in Example 3, zero-probability thresholds are the reason why the revelation results in Lemma 2 are potentially incomplete. This motivates the following specialization of our model.

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De…nition 2 A stochastic choice function p is a Strict Random Satis…cing Rule (SRSR) if it is a Random Satis…cing Rule and the underlying probability function is strict.

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If p is a SRSR, then every alternative x 2 X has a strictly positive probability of being drawn as the threshold. An important consequence of this restriction is that the implications in Lemma 2 can be reversed. Lemma 7 Let p be a Strict Random Satis…cing Rule de…ned by primitives (%; L; ). For every x; y 2 X : 1. p(x; fxg) > p(y; fyg) if and only if x

y,

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2. p(y; fyg) > p(y; fx; yg) if and only if xLy.

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Proof. In light of Lemma 2, we need only show the if directions. Let p be a SRSR. P (x) > 0. If x y, then First note that for every x 2 X, p(x; fxg) = a-x (a) p(x; fxg) p(y; fyg) + (x) > p(y; fyg), as desired. Now suppose xLy. If x % y as well, then p(y; fx; yg) = 0 < p(y; fyg), where the equality follows from part 2 of Lemma 1. If y x, on the other hand, p(y; fx; yg) = p( y; fyg) p(x; fxg) < p(y; fyg), where the equality follows from part 3 of Lemma 1. In order to characterize the class of SRSRs we will need the following axiom:

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No Sure Abstention (NSA): For every x 2 X, p(a ; fxg) < 1.11 NSA says that the decision maker should not abstain for sure in any singleton menu. SRSRs satisfy this axiom. To see this, note that x % x since % is a weak order, and (x) > 0 since is a strict probability. Hence p(x; fxg) (x) > 0 for all x.

11

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Theorem 3 A stochastic choice function is a Strict Random Satis…cing Rule if and only if it satis…es Impact Asymmetry, Strong Regularity and No Sure Abstention. Furthermore, if p satis…es these three axioms, then the underlying preference and list order are uniquely identi…ed. NSA may actually be extended to all menus A due to Regularity. To see this, consider any menu A. If p(a ; fxg) < 1 for every x 2 A then by Regularity p(a ; A) p(a ; fxg) < 1.

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Proof. See the Appendix.

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We …nish this section by establishing the independence of our three axioms IA, SR and NSA. Example 5 Let X = fx; yg and consider the following stochastic choice function: p(x; fx; yg) = p(x; fxg) = 3=4; p(y; fx; yg) = p(y; fyg) = 0;

p(a ; fx; yg) = p(a ; fxg) = 1=4 and p(a ; fyg) = 1: Note that p fails NSA since p(a ; fyg) = 1, but satis…es IA and SR.

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Example 6 Let X = fx; yg and consider the following stochastic choice function: p(x; fx; yg) = 1=2; p(x; fxg) = 3=4; p(y; fx; yg) = 1=2; p(y; fyg) = 1;

p(a ; fx; yg) = p(a ; fyg) = 0 and p(a ; fxg) = 1=4.

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Note that p fails IA since p(x; fxg) > p(x; fx; yg) and p(y; fyg) > p(y; fx; yg). However p satis…es SR and NSA. Example 7 Let X = fx; yg and consider the following stochastic choice function: p(x; fx; yg) = p(x; fxg) = 1=2;

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p(y; fx; yg) = 1=2; p(y; fyg) = 7=8; p(a ; fx; yg) = 0; p(a ; fxg) = 1=2 and p(a ; fyg) = 1=8: Note that p fails SR since p(a ; fx; yg) < p(a ; fyg) < p(a ; fxg). However p satis…es IA and NSA.

5.1

Additional observations

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5

Preference and List Correlation

Until this point we have made no assumptions about how the preference and list may relate to one another. An interesting special case arises when they are perfectly correlated; 17

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x % y if and only if xLy for distinct x and y. The characterization of this special case is straightforward, since we may directly impose coincidence of % and L. Consider the following axiom.

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Preference-List Consistency (PLC): For all x; y 2 X,

p(y; fyg) > p(y; fx; yg) ) p(x; fxg) > p(y; fyg).

PLC states that if combining x with y reduces the probability of choosing y, then x must have a strictly greater probability of being chosen than y in singleton menus. Theorem 4 A Strict Random Satis…cing Rule satis…es Preference-List Consistency if and only if %= L.

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Proof. Suppose p is a SRSR de…ned by (%; L; ) and that it satis…es PLC. Consider any x; y and, since L is complete, suppose without loss that xLy. It follows that p(y; fyg) > p(y; fx; yg), and since PLC holds p(x; fxg) > p(y; fyg). It follows, by Lemma 6, that x % y and not y % x, hence %= L. Now suppose %= L and pick some x; y suppose p(y; fyg) > p(y; fx; yg). By Lemma 6, it follows that xLy. Since %= L it follows, again by Lemma 6, that p(x; fxg) > p(y; fyg).

5.2

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We note that when the preference and list coincide (1) the decision maker’s preference is a linear order, and (2) the decision maker will either choose the preference maximal alternative in A or abstain. Indeed, suppose that %= L. and let x = max(A; %). Then P p(x; A) = a-x (a). For every y 2 Anfxg, xLy and x % y, and by part 2 of Lemma 1, p(y; A) = 0: It follows that p(a ; A) = 1 p(x; A).

Menu-dependence

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We introduce a natural generalization of our model which allows for menu-dependence of the threshold distribution. We believe that in some instances menu-dependence is quite sensible. For example, if a decision maker receives a noisy signal about the utility values of a menu’s alternatives and then forms a threshold, then the distribution of her thresholds will naturally depend on the feasible menu. In its most general formulation, the Menu-dependent Random Satis…cing Rule involves the following straightforward modi…cation of our model. 18

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De…nition 3 A stochastic choice function p is a Menu-dependent Random Satis…cing Rule (MDRSR) if there exists a triple (%; L; f A gA2D ) where % is a weak order on X, L is a linear order on X and A is a probability on X for every A 2 D such that for every A and every x 2 A; X p(x; A) = A (a): a2X : x=max(fy2A:y%ag;L)

A Menu-dependent Random Satis…cing Rule is strict (SMDRSR) if A.

A

is strict for every

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Our …rst result is that when A is allowed to assign zero probabilities to some thresholds, menu-dependence can generate any stochastic choice function and thus it lacks any testable implications. Theorem 5 Every stochastic choice function is a Menu-dependent Random Satis…cing Rule.

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Proof. Consider an arbitrary stochastic choice function p. Let L be any linear order and de…ne a binary relation as follows: for every x; y 2 X, x % y if and only if yLx. Note that % is also a linear order. Then for any A 2 D, de…ne A : X ! [0; 1] as follows, for each x 2 A let A (x) = p(x; A). Given % and L, for every A and x 2 A , a threshold of x always leads to the choice of x. That is, x = max(fy 2 A : y % xg; L) for every x 2 A. Hence X p(x; A) = A (x) = A (a). a2X : x=max(fy2A:y%ag;L)

Therefore p is represented by the MDRSR generated by (%; L; f

A gA2D ).

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When % and L are reverse orderings, the unique threshold that leads to the choice of x is x, for every x 2 A. Then since A may be zero for some alternatives we may match the probability of choosing x with the probability of drawing x as a threshold. The result thus hinges on the ‡exibility generated from allowing some thresholds to have zero probability and allowing threshold distributions to vary arbitrarily across menus. If instead we require A (a) > 0 for every threshold a and menu A, we will not be able to match all stochastic choice functions. The characterization of SMDRSR comes from 19

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combining NSA with two new axioms. The …rst is a weaker version of SR and the second is a weak form of transitivity. We …rst introduce the weakening of SR.

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Dominance: If p(y; fx; yg) = 0, then for every A that contains x, p(y; A) = 0. Dominance simply requires that if y is never chosen against x in the binary menu then it should never be chosen from any menu containing x. Necessity of this axiom is straightforward, since, with strictly positive threshold probabilities, if p(y; fx; yg) = 0, then x is preferred to y and precedes y in the list order. Since the list and preference are menu-independent, x will always block the choice of y in any larger menu. It is also straightforward to see that Dominance is implied by SR.

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Finally, consider the following transitivity condition.

No Zero Cycles (NZC): If p(y; fx; yg) = 0 and p(z; fy; zg) = 0, then p(z; fx; zg) = 0.

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NZC is a relatively weak form of transitivity that only applies to binary menus when one alternative has a zero probability. The argument for necessity of this axiom is similar to the argument for necessity of Dominance. That is, recall that the revealed preference implications of p(y; fx; yg) = 0 are that x is preferred to y and precedes y. Since % and L are transitive, it must be the case that the sub-relation formed by agreement of % and L must also be transitive. This is precisely what NZC requires; that there are no cycles when the preference and the list order agree. We also note that NZC is implied jointly by IA and SR. Theorem 6 A stochastic choice function is a Strict Menu-dependent Random Satis…cing Rule if and only if it satis…es Dominance, No Zero Cycles and No Sure Abstention.

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Proof. Necessity has already been argued for Dominance and NZC. NSA follows since A is strict for every A. In the reverse direction, suppose that p satis…es Dominance, NZC and NSA. Now, de…ne B as follows, x B y if and only if p(x; fx; yg) > 0 and p(y; fx; yg) = 0. Clearly this is antisymmetric and transitivity follows immediately from NZC. We then de…ne % and L by completing B two di¤erent ways. Let % be any linear order that is a completion of B, and let L be the completion such that for any x; y 2 = B, xLy if and only if y % x. That is, whenever there is a pair of alternatives that cannot be compared 20

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A (a)

=

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according to B, we assume % and L order them di¤erently. Then for any A and x 2 A, let T (x; A) denote the set of thresholds that lead to x in A, as in the proof of Lemma 6. We then de…ne a menu-dependent threshold distribution as follows. Fix an x 2 A for which p(x; A) > 0. Then for each a 2 T (x; A), let p(x; A) . jT (x; A)j

The fact that A (a) is a valid probability measure taking strictly positive values follows from that fact that the collection T (x; A) for p(x; A) > 0 partitions X . Finally, let q be the satis…cing rule de…ned by (%; L; f A gA2D ). Then X

A (a)

a2T (x;A)

and the proof is complete.

X

=

a2T (x;A)

p(x; A) = p(x; A), jT (x; A)j

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q(x; A) =

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It is worth noting that any stochastic choice function p for which p(x; A) > 0 for all x 2 A will trivially satisfy Dominance and NZC, and hence any such p is a SMDRSR. As a corollary, the class of SMDRSRs contains all full support stochastic choice rules. We also note that menu-dependence of the threshold distribution can create violations of regularity and thus SMDRSRs allow for non-RUM behavior.

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Example 8 Let X = fx; y; zg with x y z and zLyLx. Consider two menus, 1 A1 = fyg and A2 = fy; zg. Let A1 (a) = 3 for all a 2 X, while A2 (a) = 21 if a = x and A2 (a) = 41 if a 2 fy; zg. That is, the agent becomes “more ambitious" in the larger menu. It is then simple to verify that p(a ; fy; zg) = 12 > 13 = p(a ; fyg), and hence the decision maker is increasingly likely to walk away in the larger menu.

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It would be very interesting, though outside the scope of this paper, to consider imposing additional restrictions on the menu-dependent distribution of thresholds. For example, one could consider restricting A to return a threshold within A, thereby eliminating the need for the outside option, or require it to belong to some class of functions. Alternatively, one could require that the distribution changes in a systemic way to capture some behavioral feature. One such example, illustrated in Example 9, is assuming that the decision maker is more ambitious in larger menus. That is, A is more likely to draw a 21

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5.3

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high threshold than B whenever B A. This feature of thresholds is consistent with the experimental …ndings in Caplin et al. (2011) and may generate the choice overload phenomena. We leave these extensions for future work.

Sequential draws of thresholds

We introduce a variation of our model which allows the threshold to change during search.

x%a

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De…nition 4 A stochastic choice function p is a Random Satis…cing Rule with Sequential Draws (RSRSD) if there exist (1) a weak order % on X, (2) a linear order L on X and (3) a probability on X , such that for every A and every x 2 A, 0 1 0 1 X X Y @1 p(x; A) = @ (a)A . (a)A y2Anfxg:yLx

y%a

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The interpretation is as follows. Given a menu A, the decision maker …rst considers the alternative max(A; L). She draws a threshold. If the alternative considered is at least as good as the threshold, then it is the choice. Otherwise the decision maker passes on max(A; L), moves down on the list to max(Anfmax(A; L)g; L) and draws another threshold. The process is repeated until either a considered alternative is at least as good as the corresponding threshold, or all alternatives are exhausted, in which case the choice is a . Importantly, there is no recall, in the sense that once the decision maker passes on an alternative, she cannot go back to it afterwards. The threshold draws are statistically independent and they follow . Hence p(x; A) is the joint probability that (1) x is at least as good as the threshold drawn when the decision maker has considered it, and (2) every time the decision maker considered an alternative before x, the corresponding threshold has been strictly better than the considered alternative. This is precisely the expression in the ultimate display, under the assumption of statistically independent threshold draws. P Let (y) = y%a (a) for every y 2 X, to get p(x; A) = (x)

Y

(1

(y)) .

y2Anfxg:yLx

If (a) > 0 for every a 2 X, i.e., if p is a SRSR, then takes values in (0; 1). Interpreting (x) as an alternative-speci…c consideration probability and L as the decision maker’s 22

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Conclusion

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6

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preference, this expression yields the model characterized in Manzini and Mariotti (2014). In their model the decision maker is equipped with two psychological primitives: a linear order giving her preference, and a map : X ! (0; 1) which gives the probability (x) that she considers any alternative x. She chooses the best alternative according to her preference among the ones she considers. If no feasible alternative falls in her consideration set, she abstains. The resulting behavior is captured precisely by the equation in the ultimate display. Hence the stochastic consideration model of Manzini and Mariotti (2014) can also be interpreted as a model of stochastic satis…cing with repeated independent draws of thresholds during search. This possibility of multiple interpretations has important consequences with respect to the revelation of primitives. To wit, suppose that choice data satis…es the axioms in Manzini and Mariotti (2014). If we believe that the data is generated by a stochastic consideration model, then p(y; fyg) > p(x; fx; yg) implies x is better than y. If we believe that the data is generated by a satis…cing type choice rule, however, this same observation merely means x precedes y in the search order.

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In this paper we characterize a model of stochastic satis…cing with random thresholds. The decision maker is equipped with a preference and a list ordering of alternatives. She draws a random threshold alternative and chooses the …rst alternative on the list which is at least as good as the threshold. If no such alternative exists, she abstains. We …rst show that this model is a Random Utility Model. Next, we identify two simple axioms which together characterize the resulting behavior. In general the underlying preference and list ordering of a Random Satis…cing Rule cannot be uniquely identi…ed by choice frequencies. However, we show that if every alternative assumes a strictly positive probability of being the threshold, then both primitives are uniquely identi…ed. Our axioms have bites, and the revelations of our model can be made, using menus where the decision maker chooses between abstention and at most two alternatives. A key feature of our model is that choice is not forced. It may be desirable to understand the e¤ects on our satis…cing model of removing the no-choice option and forcing choice from every menu. Our model may be modi…ed in two distinct ways to capture this. First, as in Aguiar et al. (2016) we may assume that if no alternative is acceptable 23

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at a given threshold, the decision maker chooses the best available alternative. Since the preferences of our decision maker is given by a weak order, this would entail some randomization over possibly multiple best alternatives. A second way of removing the no-choice option is to take conditional probabilities of choice as data, and to assume that the decision maker has already drawn a threshold for which some alternative has been deemed acceptable. This is analogous to the way in which Brady and Rehbeck (2016) modify the stochastic consideration model of Manzini and Mariotti (2014). In future work, it would be interesting to analyze the consequences of removing the outside option from our model in one of these two distinct ways.

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References

[1] Aguiar, Victor H. (2017) “Random Categorization and Bounded Rationality”, Economics Letters, 159:46-52.

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[2] Aguiar, Victor H., Boccardi, M. J. and Dean, M. (2016) “Satis…cing and Stochastic Choice”, Journal of Economic Theory, 166:445-482. [3] Aguiar, V. and Kimya, M. (2019) “Adaptive Stochastic Search”, Journal of Mathematical Economics, 81:74-83.

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[4] Bajraj, G. and Ülkü, L. (2015) “Choosing Two Finalists and the Winner”, Social Choice and Welfare, 45:729-744. [5] Barberà, S. and Neme, A. (2016) “Ordinal Relative Satis…cing Behavior”Barcelona GSE working paper. [6] Brady, R. and Rehbeck, J. (2016) “Menu Dependent Stochastic Feasibility”, Econometrica, 84(3):1203-1223.

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[7] Caplin, A., Dean, M., Martin, D. (2011) “Search and Satis…cing”, American Economic Review, 2899-2922. [8] Chambers, C. and Echenique, F. (2016) Revealed Preference Theory, Econometric Society Monographs, Cambridge University Press. 24

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[9] Gerasimou, G. (2018) “Indecisiveness, Undesirability and Overload Revealed Through Rational Choice Deferral”, Economic Journal, 128:2450-2479.

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[10] Gul, F. and Pesendorfer, W. (2001) “Temptation and Self-Control”, Econometrica, 69(6): 1403-1435. [11] Kahneman, D., Knetsch, J., Thatler, R. (1991) “Rationalizing Choice Functions by Multiple Rationales”, Journal of Economic Perspectives, 5(1):193-206. [12] Kalai, G., Rubinstein, A. and Spiegler, R. (2002) “Anomalies: The Endowment E¤ect, Loss Aversion, and Status Quo Bias”, Econometrica, 70(6):2481-2488.

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[13] Kovach, M. (2016) “Thinking Inside the Box: Status Quo Bias and Stochastic Consideration”, unpublished manuscript. [14] Manzini, P., and Mariotti, M. (2007) “Sequentially Rationalizable Choice”, American Economic Review, 97(5):1824-1839.

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[15] Manzini, P., and Mariotti, M. (2014) “Stochastic Choice and Consideration Sets”, Econometrica, 82 (3):1153-76. [16] Masatlioglu, Y., Nakajima, D. and Ozbay, E. (2012) “Revealed Attention”American Economic Review, 102(5):2183-2205.

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[17] Masatlioglu, Y. and Ok, E. (2005) “Rational Choice with Status Quo Bias”Journal of Economic Theory, 121:1-29. [18] Papi, M. (2012) “Satis…cing Choice Procedures” Journal of Economic Behavior and Organization, 84(1):451-462.

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[19] Rieskamp, J., Busemeyer, J. R. and Mellers, A. (2006) “Extending the Bounds of Rationality: Evidence and Theories for Preferential Choice” Journal of Economic Literature, 44:631-661. [20] Rubinstein, A. and Salant, Y. (2006) “A Model of Choice from Lists” Theoretical Economics, 1:3-17.

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[21] Schwartz, B., Ward, A., Monterosso, J., Lyubomirsky, S., White, K. and Lehman, D. R. (2002) “Maximizing versus satis…cing: happiness is a matter of choice”Journal of Personality and Social Psychology, 83(5):1178.

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[22] Simon, H. (1955) “A Behavioral Model of Rational Choice”, Quarterly Journal of Economics, 99-118. [23] Simonson, I. (1989) “Choice based on reasons: The case of attraction and compromise e¤ects”, Journal of Consumer Research, 16(2):158-174. [24] Tsetsos, K., Usher, M. and Chater, N. (2010) “Preference Reversal in Multiattribute Choice”Psychological Review, 117:1275-1293.

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[25] Tversky, A. (1969) “Intransitivity of Preferences”Psychological Review, 76:31-48. [26] Tyson, C. (2008) “Cognitive Constraints, Contraction Consistency and the Satis…cing Criterion”, Journal of Economic Theory, 138(1):51-70.

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[27] Ward, D. (1992) “The Role of Satis…cing in Foraging Theory”Oikos, 312-317. [28] Yildiz, K. (2016) “List Rationalizable Choice”Theoretical Economics, 11:587-599.

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Appendix: Proofs of Theorems 2 and 3

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Proof of Theorem 2. Suppose that p satis…es SR and IA. We will show that p is a RSR. Note that, by Lemma 4, p satis…es Regularity as well. Divide X into two disjoint subsets X+ = fx 2 X : p(x; fxg) > 0g and X0 = fx 2 X : p(x; fxg) = 0g. If X+ = ?, then p(a ; A) = 1 for all A by SR. Since any RSR for which (a ) = 1 replicates this behavior, p is a RSR. In the remainder of the proof we will suppose that X+ is nonempty. De…ne a binary relation % on X as follows: for any x; y 2 X , x % y i¤ p(x; fxg)

p(y; fyg).

Note that % is complete and transitive, hence a weak order on X.

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Next de…ne a binary relation L+ as follows: for any x; y 2 X+ , xL+ y i¤ x = y or p(y; fyg) > p(y; fx; yg):

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We will show that L+ is a linear order on X+ . It helps to start with proving the following claim. Claim: For every distinct x; y 2 X+ , if p(x; fxg) = p(x; fx; yg), then p(y; fyg) > p(y; fx; yg). The claim provides a partial converse to IA which holds only for alternatives in X+ . To prove it, suppose that p(x; fxg) = p(x; fx; yg) for distinct x; y 2 X+ . By Regularity p(y; fyg) p(y; fx; yg). Suppose towards a contradiction that p(y; fyg) = p(y; fx; yg) and without loss of generality that p(x; fxg) p(y; fyg), i.e., p(a ; fxg) p(a ; fyg). By SR, p(a ; fx; yg) = 1 p(x; fxg) and p(y; fyg) = p(y; fx; yg) = 1

p(x; fx; yg)

p(a ; fx; yg) = 0:

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This contradicts y 2 X+ and proves the claim. Going back to the properties of L+ , completeness and antisymmetry are easily established using the axioms, the de…nition of L+ and the claim. To see that L+ is transitive, pick any x; y; z 2 X+ and suppose xL+ y and yL+ z but not xL+ z. Since L+ is complete, it follows that zL+ x. It is now without loss of generality to suppose p(x; fxg) p(y; fyg) p(z; fzg). Consider the menu fx; yg. Since xL+ y, p(y; fx; yg) < p(y; fyg) and by IA, p(x; fx; yg) = p(x; fxg). Note also that by SR, p(a ; fx; yg) = p(a ; fxg) = 1 p(x; fxg). Summing probabilities, we get p(y; fx; yg) = 0. Similarly p(z; fy; zg) = 0 and p(x; fx; zg) = p(x; fxg) p(z; fzg)). Using SR again, we get p(a ; fx; y; zg) = 1 p(x; fxg), p(x; fx; zg) = p(x; fxg) p(z; fzg), while p(y; fx; y; zg) = p(z; fx; y; zg) = 0. Summing probabilities, we get p(z; fzg) = 0 which contradicts z 2 X+ . This establishes that L+ is also transitive, hence a linear order on X+ . Now let L be any linear order on X which extends L+ , in the sense that if x; y 2 X+ and xL+ y, then xLy. To construct the threshold probabilities, let I1 ; : : : ; In be the indi¤erence classes in X produced by % and ordered such that In I1 . Hence p(x; fxg) = p(y; fyg) if x; y 2 Ii . De…ne i = p(x; fxg) for any x 2 Ii . Let 0 = 0, and 4 i = i i 1 for every 4 i i = 1; : : : ; n. Now de…ne a function : X ! [0; 1] as follows: (x) = jIi j if x 2 Ii and (a ) = 1 maxx2X p(x; fxg). It follows that is a probability on X . 27

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To …nish, let q be the RSR de…ned by the primitives (%; L; ). As we have established in the …rst part of the proof, q satis…es SR (hence regularity) and IA. We will show that p(x; A) = q(x; A) for every A and every x 2 A. First note that if A = fxg, then X p(x; fxg) = (a) = q(x; fxg) . a-x

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The …rst inequality is by the construction of . The second equality is by the de…nition of a RSR. Hence p and q coincide on singleton menus. Now suppose x 2 X0 , i.e., p(x; fxg) = 0. By Regularity of p , then, p(x; A) = 0 as well, for any menu A. We also have q(x; A) q(x; fxg) = 0, where the inequality is by Regularity of q and the equality is by the fact that q(x; fxg) = p(x; fxg). Hence p(x; A) = q(x; A) for any A and for any x 2 X0 \ A. For the remainder of the proof suppose that x 2 X+ . We will …rst show that p(x; fx; yg) = q(x; fx; yg) for any y 6= x. There are three cases to consider regarding the value of p(x; fx; yg). Case 1: p(x; fx; yg) = 0. Since x 2 X+ , p(x; fxg) > p(x; fx; yg). We will argue that yLx and y % x, so that, by part 2 of Lemma 1, q(x; fx; yg) = 0 as well. To show that yLx, it su¢ ces to show that y 2 X+ and use the de…nition of L+ . To that end, suppose towards a contradiction that p(y; fyg) = 0. Then p(a ; fxg) < p(a ; fyg) and by SR, p(a ; fx; yg) = p(a ; fxg). Furthermore p(y; fx; yg) = 0 by Regularity. Summing up choice frequencies in fx; yg gives p(x; fx; yg) + 1 p(x; fxg) < 1, an impossibility. We conclude that y 2 X+ and therefore that yL+ x and consequently yLx. Now we will show that p(x; fxg) p(y; fyg) so that y % x. Towards a contradiction suppose p(y; fyg) < p(x; fxg). By SR, then, p(a ; fx; yg) = p(a ; fxg) = 1 p(x; fxg). Note also that p(y; fyg) = p(y; fx; yg) by IA and Regularity of p. Summing up choice frequencies in fx; yg yields 1 p(x; fxg) + p(y; fyg) < 1, an impossibility. Hence p(y; fyg) p(x; fxg) and y % x as desired. Case 2: p(x; fx; yg) = p(x; fxg). Suppose that y 2 X+ . It follows that xLy by the claim above and the de…nition of L. Hence q(x; fx; yg) = q(x; fxg) by Lemma 1. Since q and p agree on singletons, q(x; fx; yg) = p(x; fx; yg) as desired. Now suppose that y 2 X0 . Then p(y; fx; yg) = 0 since p is regular, q(y; fyg) = 0 since p and q coincide on singletons and q(y; fx; yg) = 0 since q is regular. By SR of q and p, q(a ; fx; yg) = q(a ; fxg) and p(a ; fx; yg) = p(a ; fxg). Since q(a ; fxg) = p(a ; fxg) as well, it follows that p(a ; fx; yg) = q(a ; fx; yg) and consequently p(x; fx; yg) = q(x; fx; yg). 28

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Case 3: 0 < p(x; fx; yg) < p(x; fxg). We will show that yLx, x y and p(x; fx; yg) = p(x; fxg) p(y; fyg). That p(x; fx; yg) = q(x; fx; yg) will then follow from part 3 of Lemma 1. First note that p(y; fx; yg) = p(y; fyg) by IA and Regularity. Next note that p(y; fyg) > 0. For if p(y; fyg) = 0, then (1) p(y; fx; yg) = 0 by Regularity, and (2) p(a ; fx; yg) = 1 p(x; fxg) by SR, implying that the sum of choice frequencies in fx; yg is strictly less than 1, an impossibility. Hence y 2 X+ as well and yLx. Now suppose that p(x; fxg) p(y; fyg). Then p(a ; fx; yg) = 1 p(y; fyg) by SR and the sum of choice frequencies in fx; yg exceeds 1, an impossibility. Hence p(x; fxg) > p(y; fyg) and x y. Furthermore, by SR, p(a ; fx; yg) = 1 p(x; fxg). Now p(x; fx; yg) = 1 p(a ; fx; yg) p(y; fx; yg) = p(x; fxg) p(y; fyg), as desired. Now that we have established p(x; A) = q(x; A) for any singleton or doubleton A, take any A containing x and at least two other alternatives. Note p(x; A) = min p(x; fx; yg) = min q(x; fx; yg) = q(x; A), y2Anfxg

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where the …rst equality follows from SR of p, the second from the fact that p and q coincide on doubletons, and the third from the fact that q is a RSR and therefore satis…es SR. This …nishes the proof.

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Proof of Theorem 3. Suppose p is a SRSR. We argued that p must satisfy NSA in the main text. Since every SRSR is a RSR, p satis…es SR and IA as well by Lemmas 4 and 6. In the reverse direction, suppose that p satis…es the three axioms. The argument mimics the proof of Theorem 2 almost verbatim, with two slight di¤erences. The …rst di¤erence is that X+ = X by NSA, and it follows that L = L+ . In particular, the arguments regarding alternatives belonging to X0 can be dropped. The second di¤erence pertains the construction of probability . As de…ned in the proof of Theorem 2, 4 1 > 0 by NSA, and 4 i > 0 for i > 1 by construction. Now let (x) = 4jIi ji if x 2 Ii and (a ) = 1 maxx2X p(x; fxg). It follows that is a probability on X and (x) > 0 for every x 2 X. To see that the preference and list order are uniquely identi…ed, suppose that p satis…es the axioms and is a SRSR de…ned by the primitives (%; L; ) and also by (%0 ; L0 ; 0 ). If x % y and y 0 x, then, by Lemma 6, p(x; fxg) p(y; fyg) > p(x; fxg), a contradiction. Hence no such x and y can exist and %=%0 . If xLy and yLx for distinct x and y, then, again by Lemma 6, p(y; fyg) > p(y; fx; yg) and p(x; fxg) > p(x; fx; yg), which violates 29

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IA. Hence no such x and y can exist and L = L0 .

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