The Leader Rule: A model of strategic approval voting in a large electorate Jean-Fran¸cois Laslier∗ ´ ´ Laboratoire d’Econom´ etrie, Ecole Polytechnique 1 rue Descartes, 75005 Paris [email protected] April 30, 2008

Abstract The paper considers Approval Voting for a large population of voters. It is supposed that voters evaluate the relative likelihood of pairwise ties among candidates based on statistical information about candidate scores. This leads them to vote sincerly and according to a simple behavioral rule we call the “Leader Rule”. At equilibrium, if a Condorcet-winner exists, this candidate is elected. ∗

Thanks for their remarks to Jean Baratgin, Steve Brams, Micael Castanheira, Nicolas Gravel, Fran¸cois Maniquet, Remzi Sanver and Karine Van der Straeten and to anonymous reviewers of this journal. Errors are mine.

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1

Introduction

Approval Voting (AV) is the method of election according to which a voter can vote for as many candidates as she wishes, the elected candidate being the one who receives the most votes. In this paper two results about AV are reached in the case of a large electorate : the sincerity of individual behavior (voters choose sincere ballots) and the Condorcet-consistency of the choice function defined by approval voting (whenever a Condorcet winner exists, it is the outcome of the vote). Under AV, a ballot is a subset of the set candidates. A ballot is said to be sincere, for a voter, if it shows no “hole” with respect to the voter’s preference ranking; if the voter sincerely approves of a candidate x she also approves of any candidate she prefers to x. Therefore, under AV, a voter has several sincere ballots at her disposal: she can vote for her most preferred candidate, or for her two, or three, or more most preferred candidates.1 It has been found by Brams and Fishburn (1983) that a voter should always vote for her most-preferred candidate and never vote for her leastpreferred one. Notice that this observation implies that strategic voting is sincere in the case of three candidates. The debate about strategic voting under AV was made vivid by a paper by Niemi (1984). Niemi argued that, because there is more than one sincere approval ballot, the rule “almost begs the voter to think and behave strategically, driving the voter away from honest behavior” (Niemi’s emphasis, p. 953). Niemi then gave some examples showing that an approval game cannot be solved in dominant strategies. Brams and Fishburn (1985), responded to this view, but the debate was limited by the very few results available about equilibria of voting games in general and strategic approval voting in particular. For instance in one chapter of their book, Brams and Fishburn discuss the importance of pre-election polls. They give an example to prove that, under AV, adjustment caused by continual polling can have various effect and lead to cycling even when a Condorcet winner exists (example 7, p.120). But with no defined notion of rational behavior they have to postulate specific (and changing) adjustment behavior from the voters. With the postulate that voters use sincere and undominated (“admissible”) strategies but can use any of these, Brams and Sanver (2003, 2006) describe the set of possible winners of an AV election. They conclude that 1 While some scholars see this feature as a drawback (Saari and Van Newenhizen, 1988), observation shows that people appreciate to have this degree of freedom: see Laslier and Van der Straeten (2008) for an experiment, and the survey Brams and Fishburn (2005) on the practice of AV.

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a plethora of candidates pass this test. This literature assumes that voters are only interested in the elected candidate and not on how many votes are obtained by the other candidates (no expressive voting) and the present paper will not depart from this standard assumption. But none of the above approaches uses an equilibrium theory of approval voting. The rationality hypothesis in a voting situation without expressive voting may have odd implications. For instance it implies that a voter is indifferent between all her strategies as soon as her votes cannot change the result of the election. As a consequence, in an election held under plurality or AV rule, any situation in which one candidate is slightly (three votes) ahead of the others is trivially a Nash equilibrium; thus it is fair to say that with many voters, in most voting situations, the Nash equilibrium concept does not refine the set of outcomes very much.2 Taking a standard game-theoretical point of view, De Sinopoli, Dutta and Laslier (2006) have studied in details examples of AV games. They showed that even powerful refinement considerations such as strategic stability or regularity could not guarantee the election of a Condorcet winner when it exists nor exclude insincere behavior at equilibrium. All these results are essentially negative ones, in the sense that this literature makes almost no prediction for AV games. A breakthrough in the rational theory of voting occurred when it was realized that considering large numbers of voters was technically possible and offered a more realistic account of political elections. This approach was pioneered by Myerson and Weber (1993). In the same paper, AV and other rules are studied on an example with three types of voters and three candidates (a Condorcet cycle). Subsequent papers by Myerson improved the techniques and tackled several problems in the theory of voting (Myerson 1998, 2000, 2002). Using the so-called Poisson-Myerson model of voter participation, Myerson (2002) obtained that approval voting guarantees the choice of the majoritarian outcome in the case where there are only two types of voters, that is when more than half of the population share the same most-prefered outcome. This result may seem very basic, but it is fundamental for the theory of majority voting since, as Myerson shows, other natural voting rules do not satisfy this requirement. The present paper applies similar, but not identical, techniques to approval voting with no restriction on the number of candidates or voter types. 2

Sequential voting shemes are often solvable by iterative elimination of dominated strategies (Moulin 1979, Peleg 1984, Peress 2004, Bag et al. 2007). Such is not the case here.

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The abstract theory of the refinement of Nash equilibrium considers small perturbations in the choice of strategies or in the payoffs (“trembling hand” and other perfection criteria, strategic stability). The PoissonMyerson model introduces uncertainty at the level of the number of players. In the present paper, uncertainty is introduced in an other way. We imagine that there is some small but strictly positive probability that each vote is wrongly recorded. This probability is supposed to be independent of the identity of the voter, of the candidates, and of the other voters’ vote. It should be interpreted as the probability of a technical mistake which is unavoidable when recording each individual vote. Referring to the controversial election of the US president in November 2000, the term “Florida tremble” was suggested to describe this hypothesis. This term rightly conveys the idea that material mistakes exist in elections. Notice however that in our model, with independence and a large number of voters, the mistakes introduce no bias in the election. Rationality implies that a voter can decide of her vote by limiting her conjecture to those events in which her vote is pivotal. In a large electorate, this is a very rare event, and it may seem unrealistic that actual voters deduce their choices from implausible premises. One can wish that a positive theory of the voter be more behavioral and less rational. In the present paper, voters are supposed to be partly but not fully rational. More exactly it will be supposed that they base they reasoning (maximization of expected utility) on the possibility of pairwise ties but neglect the possibility of ties between three, or more, candidates.3 Voters are rational in the sense that they maximize expected utility for a well-defined utility function and subjective belief.4 We shall prove that, in the case of approval voting, the rational response is very simple. It can be described as follows. Let x1 be the candidate the voter thinks is the most likely to win (the “Leader”). The voter will approve of any candidate she prefers to x1 . She will never approve of a candidate she prefers x1 to. To decide whether she will approve of x1 or not, she compares x1 to the second most likely winner (the “Challenger”, the most serious contender of the Leader). This behavior recommends a sincere ballot; it satisfies the requirement of relative sincerity defined by Dellis and Oak (2006). and its implementation does not require sophisticated computations, it only requires that the voters holds a conjecture about which two candidates are 3

Appendix 2 discusses the question of three-way ties. Coming back to Florida, the final certified margin of Bush against Gore was 537 votes. Some of the 97,421 voters who voted for Ralph Nader may have failed to reason on the basis of the pivotal event “Gore almost tie with Bush”. 4

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to receive the most and second-most votes. We can call this behavioral rule the “Leader Rule”. For reasons that will be transparent in the sequel, this rule can also be seen as a “lexicographic” response of the decision maker to subjective uncertainty Unknown contingencies being considered in a particular order rather than weighted by probabilities. We will also show that the lexicographic response is, in this case, identical to a well known “boundedly rational” but psychologically sound behavior: the choice by sequetial elimination theory of Tversky (1997a). The rational and the behavioral models are here equivalent. Note that under Approval Voting, the scores of different candidates may be corelated since the same individuals may vote for several candidates5 . But in the model presented here, the individual behavior can be computed knowing only the agregate scores of the various candidates and without knowing these corelations. Therefore our model can be seen as the description of voters reacting to polls which would simply report candidate scores. The model here differs from Myerson-Poisson models, in which the population of voters is uncertain. If uncertainty about candidate scores were to arise from uncertainty about the population of voters, then errors could be correlated. Nunez (2007) studies, on an example, possible consequences of this point. In most models of probabilistic voting, payoff disturbances are introduced in such a way that the probabilistic assumptions allow for expressive voting, so that agregate voter behavior can, in some cases, maximize the sum of individuals utilities. (See Linbeck and Weibull 1987, Coughlin 1992, McKelvey and Patty 2006.) Here the stochastic element we introduce is purely mechanical: by mistake some votes are not counted. This perturbation introduces no kind of inter-individual comparison of utility. Indeed, all our results can be phrased in terms of individual ordinal preferences. It should also be mentionned that, although we also solve a strategic game, our approach in this paper is simpler, from the game-theoretical point of view than the one of McKelvey and Patty. We have a game with payoff uncertainty (information is not perfect but all voters have the same information) whereas the model of McKelvey and Patty rests on the existence of private information of the voters, leading to the study of a Bayesian game of incomplete information. 5

If we were to study standard plurality rule instead of approval voting then our approach would not provide new insights compared to what is known from Cox or Myerson work. There is a multiplicity of equilibria and in each equilibrium two candidates receive all the votes.

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In the present paper, the rational players are the voters, not the candidates. The same kind of model (large electorate with uncertainty) is often used to study the strategic interaction between parties chosing their platforms. The recent literature has focussed on the question of party objectives. Do they maximize their probability of being elected or their expected vote share ? (Laffond et al., 1994, Laslier 2000, Duggan 2000, Patty 2007a, Mandler, 2007.) This question is not tackled here. The paper is organized as follows. After this introduction, Section 2 describes all the essential features of the model. Section 3 informally describes the main point in the argument. Section 4 contains the formal statements and proofs of the results: the description of rational voting (Theorem ??) from which we deduce sincerity (Corollary ??), and the description of equilibrium approval scores (Theorem ??) from which we deduce Condorcet consistency (Corollary ??). Some computations are provided in an Appendix.

2

The model

Candidates and voters Let X = {1, ..., K} denote the finite set of candidates. We follow Myerson and Weber (1993) in considering that there exists a finite number of voter types τ ∈ T . A voter of type τ evaluates the utility of the election of candidate x ∈ X according to a von Neumann and Morgenstern utility index uτ (x). A preference on X is a transitive and complete binary relation. In this paper all preferences are supposed to be strict: no voter is indifferent between two candidates.6 Preferences are denoted in the usual way: x Pτ y means that τ -voters prefer candidate x to candidate y, thus: x Pτ y ⇐⇒ uτ (x) > uτ (y). Notice that one has to considerVNM utility functions besides preference relations, because voters take decisions under uncertainty. But it turns out that the obtained results can all be phrased in terms of preferences. It will be proven that rational behavior in the considered situation only depends on preferences. 6

Preferences are exogeneous. In view of the convergence results obtained in the literature on electoral competition (see Downs 1957, Cox 1987, Banks and Duggan 2004, McKelvey and Patty 2006) and in particular about Approval Voting (Cox 1985, Dellis and Oak 2006) the assumption of strict preferences would not be reasonable if candidate positions, and hence voter preferences, were endogeneous.

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We wish to consider a large electorate. To do so, we consider a fixed finite number v of voters and replicate this set n times in the following way. Let pτ denotes the fraction of type-τ voters, with: X pτ = 1. τ ∈T

In the n-fold replicate economy the number of type-τ voters is nvpτ and the total number of voters is nv. Voting An approval voting ballot is just a subset of the set of candidates. The relative score s(x) of candidate x is the fraction of the electorate that approves of x. If all type-τ voters choose ballot Bτ ⊆ X, the relative score is the sum of pτ over types τ such that x ∈ Bτ : X s(x) = pτ . τ : x∈Bτ

Here the absolute score of x, Pthat is number of votes in favor of x is nvs(x). Of course 0 ≤ s(x) ≤ 1, but x∈X s(x) is usually larger than 1. The elected candidate is one with highest score. Ties are resolved by a fair lottery. The following simple lemma is just the description of the voter’s best responses when there is no uncertainty and in the case where no more than two candidates are in the position of being elected. Lemma 1 Suppose a given voter, with type τ , knows the other voters’ vote. Let S max be the highest absolute score computed from the other voters’ vote. The voter’s best response is such that: 1. If exactly one candidate has the score S max and no candidate has the score S max − 1 then any ballot is a best response. 2. If exactly two candidates have the score S max and no candidate has the score S max − 1, call x and x0 these two candidates, with xPτ x0 , then ballot B is a best response if and only if x ∈ B and x0 ∈ / B. 3. If exactly one candidate, say x, has the high score S max and exactly one candidate, say x0 , has the score S max − 1, with xPτ x0 , then ballot B is a best response if and only if x ∈ B or x0 ∈ / B. 4. If exactly one candidate, say x, has the high score S max and exactly one candidate, say x0 , has the score S max − 1, with x0 Pτ x, then ballot B is a best response if and only if x0 ∈ B and x ∈ / B. 7

Proof. By definition of approval voting, the voter can increase by one unit the score of any candidate. In the first case, the voter thus cannot influence the outcome of the election. In the second case: If the voter votes for x and not for x0 , as in the lemma, x will be elected, providing utility uτ (x). If the voter votes for both x and x0 or for neither of them, then the outcome of the election will be a random choice between x and x0 , providing utility (uτ (x) + uτ (x0 )) /2. If the voter votes for x0 and not for x, x0 will be elected, providing utility uτ (x0 ). Since uτ (x) > (uτ (x) + uτ (x0 )) /2 > uτ (x0 ) the first alternative is preferred. In the third and fourth cases the voter can only induce a tie between x and x0 , by voting for x0 and not voting for x. In the third case this is what she must avoid to do because she prefers x to a tie between x and x0 . In the fourth case this is what she must do because she prefers the tie. Trembling Ballots We consider the following perturbations. For any voter and for each candidate approved by this voter there is a probability ε > 0 that this vote is not recorded. This probability is supposed to be small. Precisely, it will be sufficient to suppose that ε < 1/v (independently of n). We also suppose that these mistakes occur independently of the voter, of the candidate, and of the voter approving or not other candidates. For candidate x and voter i, let ηi,x be equal to 1 with probability ε and to 0 with the complement probability. When the (intended) number of votes for candidate x is nvs(x), the realized number of votes is a random variable S n (x). Denote by AV (x) the set of voters who approve of x, then: X S n (x) = (1 − ηi,x ). i∈AV (x)

There are nvs(x) voters in AV (x) so the random variable S n (x) is binomial, with expected value and variance: E[S n (x)] = (1 − ε)nvs(x) V[S n (x)] = ε(1 − ε)nvs(x). Given the intended scores ns(x), the realized scores are independent random variables. This is an important difference with the Myerson-Poisson model, in which candidate scores are not statistically independent. The interpretation here is that the vote of a voter for a candidate is not recorded either because of a mistake by the voter or because of a mistake in the vote count. If mistakes where done for instance by losing simultaneously all the 8

votes of a voter, the model would be almost identical to the Myerson-Poisson one, with Binomial instead of Poisson variables, and independence would be lost. Then our results would not hold anymore, as prove by Nunez (2007). To get intuition about this model, denote by bn (x) the realized relative score: S n (x) bn (x) = . nv For n large, the central limit theorem implies that the random variable bn (x) is approximately normal. One can write: bn (x) Ã a(x)

=

var

=

n→∞

N (a(x), var) ,

(1 − ε)s(x), ε(1 − ε)s(x) . nv

The expectation of bn (x), denoted a(x), is just a linear transformation of the relative score s(x), and a and s rank candidates the same way.7 The variance of bn (x) is decreasing as 1/n.

3

The argument

In this section the main argument for the proof is informally stated. More or less serious races Given a strategy profile and its score vector s, the most probable event is that the candidate with highest score wins, but it may be the case that mistakes are such that another candidate does, and it may be the case that two (or more) candidates are so close that one vote can be decisive. In what follows, it will be needed to evaluate the probabilities of some of these events when the number of voters is large. One ballot may have consequences on the result of the election only if the first ranked candidates have scores that are within one vote of each other. The probability of such a pivotal event is small if n is large, but some of these events are even much less probable than others. Neglecting three (or more)way ties in front of two-way ties, it will be proved that different two-way ties 7

It is therefore tempting to directly use this continuous model and to define as “approval scores” independent Gaussian variables N a(x), ε(1−ε)s(x) . This was done in early nv version of this paper (Laslier 2004) but it is not true that the normal approximation is valid for event of small probabilities as those to be taken into account here (“large deviation” events).

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are negligible one in front of the other according to a simple lexicographic pattern: The most probable one is a tie {x1 , x2 } between candidates x1 and x2 : that is the most “serious race”. The second most serious race is {x1 , x3 }. The race {x1 , x4 }, as well as the race {x2 , x3 } between x2 and x3 (ahead of x1 ) are also negliglible in front of the race {x1 , x3 }, etc. Lemma ?? states this point precisely. This observation turns out to be sufficient to infer rational behavior. Our rational voter will obey a simple heuristic by considering in a sequential way the different occurrences of her being pivotal, between these pairs of candidates. Rational behavior and the Law of Lexicographic Maximization: We will show that this rational behavior can be deduced from a simple heuristic, which can be described as follows in very general terms. Let D be a finite set of possible decisions and Ω = {ω1 , ..., ωN } a finite partition of events, with π the probability measure on Ω and u a von NeumannMorgenstern utility. The maximization problem of a rational decision-maker can be written N X π(ωk ) Eu(d, ωk ), max d∈D

k=1

where Eu(d, ωk ) denotes the expected utility of decision d conditional on event ωk . Suppose that π is such that, for 1 ≤ k < k 0 ≤ N , the probability π(ωk ) is large compared to the probability π(ωk0 ): π(ω1 ) À π(ω2 ) À ... À π(ωN ). Then the above maximization problem is solved recursively by the following algorithm: • D0 = D. • For k = 1 to N , Dk = arg maxd∈Dk−1 Eu(d, ωk ). For instance, if the utility Eu(d, ω1 ) in the most probable event ω1 is maximized at a unique decision, that is to say if D1 is a singleton set, then this decision is the optimal one. If D1 contains several elements, then searching for the best decision can proceed by leaving aside all decisions which are not in D1 and going to the next most probable event in order to distinguish between the elements of D1 and finding D2 . The algorithm proceeds until a single decision is reached, or until all events have been 10

considered and thus the remaining decisions give the same utility in any event. Best responses With the above “trembling ballot” model and a large enough number of voters, and if voters neglect thre-way or more ties, it will be proved that the law of lexicographic maximization applies, so that the rational behavior of such a voter is the “Leader Rule” described in the introduction: all the other candidates are compared to the announced winner (the Leader), and the Leader himself is compared to his most serious contender, the Challenger. This is the content of Theorem ??. The most probable event (ω1 ) is that there is no tie and that a vote makes no difference; this event leaves available all possible decisions (ballots) and D1 = D0 is the set of all possible ballots. The second most probable event (ω2 ) is a tie between the two first-ranked candidates, the Leader x1 and the Challenger x2 ; in this case, the voter knows which are the good decisions to take: depending on her preference, she approves of x1 or x2 but not both. For this voter, this defines D2 , which is a strict subset of D1 but not yet a singleton because she has not decided yet whether she approves of the other candidates or not. It will be seen that for all k ≥ 2, candidate xk appears in the algorithm to be compared with the Leader x1 , thereby defining a unique best response, a ballot in which all candidates xk are compared with the Leader. Note that this is not an equilibrium effect: this strategy actually describes the best response to any strategy profile which generates no ties in the scores. To give an example, suppose that there are four candidates with relative scores: s(x1 ) > s(x2 ) > s(x3 ) > s(x4 ). A voter with preference x3 Pτ x1 Pτ x4 Pτ x2 will consider in turn the ordered comparisons: (1) I approve x1 and not x2 because I prefer x1 to x2 . (2) I approve x3 because I prefer x3 to x1 . (3) I do not approve x4 because I prefer x1 to x4 . Such a voter will therefore cast the ballot: B = {x1 , x3 }. Likewise, a voter with preference x3 Pτ x2 Pτ x4 Pτ x1

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will cast the ballot: B = {x2 , x3 , x4 }. Comments The above algorithm is sound from the behavioral point of view. Behavioral theories of individual choice often incorporates the idea that actual choices are obtained after a qualitative and discrete process of simplification and elimination, rather than after the weighting of probabilities and utilities which characterizes rational behavior. We refer the interested reader to the second volume of the classic book Suppes et al. (1989), pages 436-457, and to Roy (1943), Tversky (1972a, 1972b) and Tversky and Sattah (1979) for early references; more recent references include Colman and Stirk (1999), Newell, Weston and Shanks (2003), Katsikopoulos and Martignon (2006). As to voting, the same ideas have been recently used by Patty (2007b), who takes up Tversky’s model in order to decribe how a voter ranks, rather than weights, the various political issues. In the present model model, voters do so for the possible events (close races between candidates) and not for different issues. But indeed this is similar as to the logic of individual decision: the point is that individuals consider their different reasons to act in a lexicographic way by defining priorities, when rational behavior would command to weight these reasons . The main theoretical contibution of the present paper is that this “boundedly rational” rule of behavior is, in a sense, rational if the number of voters is large.

4

Results

The next lemma describes how small the probability of a pairwise pivotal event is. To do so, some notation is helpful. Notation. For each non-empty subset Y of candidates, denote by pivot(n, Y ) the event: ∀ y ∈ Y , S n (y) ≥ max S n (x) − 1 x∈X

0

n

0

∀y ∈ / Y , S (y ) < max S n (x) − 1. x∈X

Denote by T3 the event of a three-way (or more) tie: [ T3 = pivot(n, Y ) Y ⊂X,#Y ≥3

T3c

and by the complementary event of no three-way or more ties. Given the voters’ strategies, denote by Pr [·] the probability distribution defined 12

by the above “trembling ballot” model. For instance, according to Pr [·] the random variables S n (x) are binominal and independent. When Pr[T3c ] > 0 we denote by P [·] the probability Pr conditioned on the non-occurence of three-way or more ties: P = Pr | T3c (1) Definition 1 Given two pairs of candidates {x, x0 } and {y, y 0 }, {x, x0 } is a more serious race than {y, y 0 } if P [pivot(n, {y, y 0 })] = 0. n→+∞ P [pivot(n, {x, x0 })] lim

This is denoted {y, y 0 } ¿ {x, x0 }. Lemma 2 Suppose that there are no ties in the candidate scores and label the candidates such that s(xK ) < s(xK−1 ) < ... < s(x2 } < s(x1 ), then the two-candidate races involving a given candidate xi are ordered: {xi , xK } ¿ {xi , xK−1 } ¿ ... ¿ {xi , xi+1 } ¿ {xi , xi−1 } ¿ ... ¿ {xi , x1 }. (This lemma is proved in the appendix.) Remark that the lemma does not say, for instance, which one of the two races {x1 , x4 } and {x2 , x3 } is most serious. The lemma says that {x1 , x4 } is the most serious among all the races that involve candidate x4 , and this is all what will be needed about candidate x4 , and similarly for the other candidates. We can now state and prove the key result in this paper. Theorem 1 (Best responses) Let s be a score vector with two candidates at the two first places and no tie: x1 and x2 such that s(x1 ) > s(x2 ) > s(y) for y ∈ X, y 6= x1 , x2 . There exists n0 such that, for all n > n0 and for all type τ , any type-τ voter has a unique best-response ballot Bτ∗ . This ballot is described by the following rule (the “lexicographic response” to s): • for τ such that uτ (x1 ) > uτ (x2 ), Bτ∗ = {x ∈ X : uτ (x) ≥ uτ (x1 )} , • for τ such that uτ (x1 ) < uτ (x2 ), Bτ∗ = {x ∈ X : uτ (x) > uτ (x1 )} .

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Proof. Suppose first that uτ (x1 ) > uτ (x2 ). To prove that Bτ∗ is the unique best response, we prove that any other ballot B is not. More precisely, given B 6= Bτ∗ , we will find a strictly better response B 0 which differs from B only on one candidate: B 0 will be of the form either B 0 = B ∪ {x} or B 0 = B \ {x}. Hence the following simple observation will be useful. First point. Let E [uτ (B)] and E [uτ (B 0 )] denote the expected utility of strategies B and B 0 for a voter i of type τ , given the other voters’ strategies. Suppose that B 0 = B ∪{x} for some x ∈ / B. Then if the vote of i in favor of x is not recorded, which was denoted as ηi,x = 0 and happens with probability ε, voting B 0 is the same as voting B, thus: £ ¤ £ ¤ E uτ (B 0 ) = εE [uτ (B)] + (1 − ε)E uτ (B 0 ) | ηi,x = 1 where E [uτ (B 0 ) | ηi,x = 1] denotes the expected utility of casting ballot B 0 knowing that there is no mistake in the recording of the approval of candidate x by voter i. It follows that £ ¤ £¡ ¢ ¤ E uτ (B 0 ) − uτ (B) = (1 − ε)E uτ (B 0 ) − uτ (B) | ηi,x = 1 , which means that the comparison between B and B 0 = B ∪ {x} for a voter can be made under the assumption that a vote in favor of x would be recorded for sure. Of course, the same is true when comparing B and B 0 = B \ {x} for x ∈ B. To lighten notation, in the sequel we will forget the condition and write E [·] for E [· | ηi,x = 1]. Second point. This first point being made, we shall distinguish three cases, depending where x1 , the leading candidate, stands in the voter’s preference and ballot. (i) Suppose that there exists x ∈ X such that uτ (x) > uτ (x1 ) and x ∈ / B. 0 Let then B = B ∪ {x}. To compare ballots B and B 0 , the voter computes the difference X £ ¤ ∆= P[pivot(n, Y )] E uτ (B 0 , Y ) − uτ (B, Y ) , Y

where E[uτ (B, Y )] denotes the expected utility for her of choosing ballot B conditional on the event pivot(n, Y ). By definition of P, P[pivot(n, Y )] = 0 for #Y ≥ 3. The events pivot(n, Y ) with #Y = 1 or x ∈ / Y have no consequences therefore the sum in ∆ can run over the subsets Y of the form {x, y} with y 6= x. Among these is {x, x1 }. From Lemma ??, it follows that in the event of this race the voter’s utility is strictly larger approving of x than not. Thus E[uτ (B 0 , {x, x1 }) − uτ (B, {x, x1 })] > 0. 14

As proved in Lemma ??, the probability of pivot(n, {x, y}) is decreasing in n in such a way that for all y 6= x1 , lim

n→∞

P[pivot(n, {x, y})] = 0. P[pivot(n, {x, x1 })]

One can thus factor out P[pivot(n, {x, x1 })] in ∆ and write X ∆ =a+ o{x,y} (n) P[pivot(n, {x, x1 })] y6=x,x1

for

£ ¤ a = E uτ (B 0 , {x, x1 }) − uτ (B, {x, x1 })

and o{x,y} (n) =

¤ P[pivot(n, {x, y})] £ E uτ (B 0 , {x, y}) − uτ (B, {x, y}) , P[pivot(n, {x, x1 })]

with a being strictly positive and with o{x,y} (n) tending to 0 when n tends to infinity. It follows that, for n large enough, ∆ > 0. This establishes that B is not a best response, more exactly there exists a number nB,τ depending on B and τ such that, for all n > nB,τ , B is not a best response. Because there is a finite number of ballots and types, nB,τ can be chosen independently of B and τ . (ii) Suppose that there exists x ∈ X such that uτ (x) < uτ (x1 ) and x ∈ B. Let then B 0 = B \ {x}. The reasoning is the same as in the previous case: The relevant events are again pivot(n, Y ) for x ∈ Y, the voter’s utility is strictly larger not voting for x, and the relevant race is again {x, x1 }. (iii) Suppose that x1 ∈ / B. The conclusion follows considering B 0 = B∪{x1 }; the race {x1 , x2 } is here relevant and is the most serious one. From items (i), (ii) and (iii) it follows that the voter’s best response must satisfy (i) uτ (x) ≥ uτ (x1 ) if x ∈ B, (ii) uτ (x) ≤ uτ (x1 ) if x ∈ / B, and (iii) x1 ∈ B. Therefore Bτ∗ = {x ∈ X : uτ (x) ≥ uτ (x1 )} as stated. The argument is identical in the case uτ (x1 ) < uτ (x2 ). Notice that the previous result implies that, for n large enough, all voters of a given type use the same strategy when responding to a score vector that satisfy the mentioned properties. The next definition is standard in the study of approval voting.

15

Definition 2 A ballot B is sincere for a type-τ voter if uτ (x) > uτ (y) for all x ∈ B and y ∈ / B. As a direct consequence of the previous theorem, one gets the following corollary. Corollary 1 (Sincerity) For a large electorate, in the absence of tie the best response is sincere. It should be emphasized that the previous theorem and corollary are true even out of equilibrium; they just describe the voter’s response to a conjecture she holds about the candidate scores. The voter just takes into account that she estimates the candidate scores with a statistical disturbance of order 1/n and neglects three-way or more ties. We now turn to equilibrium considerations. An equilibrium here is simply a fixed point of the correspondence which associates to a score vector the score vector obtained after all voters have chosen a best response (in the sense defined above) to this score. The piece of notation p[x, y] will be used to denote the fraction of voters who prefer x to y: X p[x, y] = pτ , τ : x Pτ y

so that for all x 6= y, p[x, y] + p[y, x] = 1. The number of voters who prefer x to y is nvp[x, y]. Theorem 2 (Equilibrium) Let s be a vector with two candidates at the two first places and no tie: x1 and x2 such that s(x1 ) > s(x2 ) > s(y) for y ∈ X, y 6= x1 , x2 . There exists n0 such that, for all n > n0 , if s is the score vector of an equilibrium of the game with n voters, then • the score of the first-ranked candidate is his majoritarian score against the second-ranked candidate: s(x1 ) = p [x1 , x2 ] , • the score of any other candidate is his majoritarian score against the first-ranked candidate: x 6= x1 ⇒ s(x) = p [x, x1 ] .

16

Proof. This is a direct consequence of Theorem ??. Each voter approves of x1 if and only of she prefers x1 to x2 . For x 6= x1 , she approves of x if and only if she prefers x to x1 . Definition 3 A Condorcet winner at profile p is a candidate x1 such that p [x1 , x] > 1/2 for all x 6= x1 . A best contender to a candidate x1 is a candidate x2 6= x1 such that p [x2 , x1 ] ≥ p [x, x1 ] for all x 6= x1 . As it is well known, the existence of a Condorcet winner is a substantial assumption. On the contrary the assumption that there is a unique best contender to a candidate x is an innocuous one, inasmuch as one can assume away exact ties in the preference profile.8 Corollary 2 (Condorcet consitency) For a large electorate, if there is an equilibrium with no tie, the winner of the election is a Condorcet winner. If the preference profile admits a Condorcet winner and the Condorcet winner has a unique best contender then the game has a unique equilibrium. In this equilibrium the Condorcet winner is elected. Proof. Notice that, as a consequence of Theorem ??, no voter votes simultaneously for x1 and x2 , and each of them votes for either x1 or x2 . Here, s(x2 ) = 1 − s(x1 ) < s(x1 ) implies s(x2 ) < 1/2, thus, for x 6= x1 1 p [x, x1 ] = s(x) ≤ s(x2 ) < . 2

These results have a non-trivial implication for the case of a preference profile with no Condorcet winner, although they do not allow for a complete characterization of equilibrium in that case. In that case, at equilibrium, the score vector must exhibit a tie between the two top candidates or between several second-ranked candidates. Approval voting is not a solution to the so-called Condorcet paradox. Indeed, approval voting retains the basic disequilibrium property of majority rule: if there is no Condorcet winner then any announced winner will be defeated, according to approval voting, by another candidate, preferred to the former by more than half of the population. 8

Things would be different in a model of party competition with endogeneous candidate positions.

17

A

Appendix

A.1

Proof of Lemma ??

When a voter, in the n-replicate population, wonders whom to approve of, the pivotal events she has to take into account are the almost ties arising from the other voters strategies. These events are described either by random variables of the form Sxn à B(nvs(x), 1 − ε), obtained from nvs(x) votes, or by random variables Sexn à B(nvs(x) − 1, 1 − ε), obtained from nvs(x) − 1 votes; the second case arises if the voters of her type approve of x at the considered profile. In the first part of the proof (“Computation in the binomial model”) we will consider that all variables are of the form Sxn . In the second part of the proof (“Environmental invariance”) we will check that the proof is valid in the general case because the ties arising from variables Sxn or Sexn have the same order of magnitude. A.1.1

Computation in the binominal model

For an integer n, let Sxn , for x = 1, ..., K, be independent binomial random variables of parameters nvs(x) and 1 − ε: Sxn à B(nvs(x), 1 − ε) for i = 0, ..., nvs(x). To lighten notation we write nx ≡ nvs(x) so that for i = 0, ..., nx : Pr [Sxn = i] = Cni x (1 − ε)i εnx −i First step. We must consider the occurence of ties up to one vote. Consider the event of a tie at the top between two candidates y and z. With the notation used in section 4, Y = {y, z} with Syn = Szn or Syn = Szn + 1, and for all x 6= y, z, Sxn + 1 < Syn . In a first step, look at the event of an exact tie between y, z and only those, that is ∀x 6= y, z , Syn = Szn > Sxn + 1 . The probability of this event is: P(n, y, z) =

nz X

Cni y Cni z (1 − ε)2i

i=0 ny +nz −2i

×ε

Pr [∀x 6= y, z, Sxn < i − 1]

18

(2)

under the probability Pr. Under the probability P, that is conditionally on the non-occurence of a three-way tie, it is simply P(n, y, z) Pr[T3c ] since we consider the event of a tie between y and z and only those. Now consider three candidates y, z, t with scores ordered as: s(y) > s(z) > s(t). I will prove that the probability of the race between y and t is very small compared to the probability of the race between y and z. Write P(n, y, t) in the following form: P(n, y, t) =

nt X i−2 X

Cni y Cni t Cnj z (1 − ε)2i+j

i=0 j=0 ny +nz +nt −2i−j

×ε

Pr [∀x 6= y, z, t, Sxn < i − 1] .

To decipher this formula: the index i is the value of Syn = Stn and j the value of Szn . I claim that, for j < i, Cni t Cnj z < Cnj t Cni z . To see this, write Cni t Cnj z Cnj t Cni z

=

nt (nt − 1) ... (nt − i + 1) nt (nt − 1) ... (nt − j + 1) nz (nz − 1) ... (nz − j + 1) nz (nz − 1) ... (nz − i + 1) (nt − j) ... (nt − i + 1) (nz − j) ... (nz − i + 1)

× =

and notice that for each of the i − j remaining term, nt − u < nz − u, hence the ratio is smaller than 1. It follows that if we take again the above formula for P(n, y, t) and exchange i and j in Cni z and Cnj t and if we denote Bn =

nt X i−2 X

Cni y Cni z Cnj t (1 − ε)2i+j

i=0 j=0 ny +nz +nt −2i−j

×ε

Pr [∀x 6= y, z, t, Sxn < i − 1] ,

we obtain: P(n, y, t) < Bn . 19

Notice that Bn is a part of the summation that defines P(n, y, z): P(n, y, z) =

nz X i−2 X

Cni y Cni z Cnj t (1 − ε)2i+j

i=0 j=0 ny +nz +nt −2i−j

×ε

Pr [∀x 6= y, z, t, Sxn < i − 1]

= Bn + Cn , where Cn is the same sum for i going from nt + 1 to nz . We will now prove that this partial sum up to nt is negligible compared to the whole sum. One can write: Bn = Cn =

nt X

£ ¤ Pr Syn = Szn = i Pr [∀x 6= y, z, Sxn < i − 1] ,

i=0 nz X

(3)

£ ¤ Pr Syn = Szn = i Pr [∀x 6= y, z, Sxn < i − 1] .

i=nt +1

Using that Pr [∀x 6= y, z, Sxn < i − 1] is obviously increasing with i, Bn < Pr [∀x 6= y, z, Sxn < nt − 1] Cn > Pr [∀x 6= y, z, Sxn < nt − 1]

nt X

£ ¤ Pr Syn = Szn = i ,

i=0 nz X

£ ¤ Pr Syn = Szn = i .

i=nt +1

£ ¤ £ ¤ By independence, Pr Syn£= Szn =¤ i = Pr Syn = i · Pr [Szn = i]. For ε such that nz < (1 − ε)ny , Pr Syn = i is increasing with i for i ∈ {0, ..., nz }: To check this point, note that for i ≤ (1 − ε)ny − 1, £ ¤ Pr Syn = i + 1 = Cni+1 (1 − ε)i+1 εny −i−1 y and £ ¤ Pr Syn = i + 1 (n − i) (1 − ε) ny − i εny + 1 £ ¤ = y ≥ ≥ > 1. n (i + 1) ε εny εny Pr Sy = i The inequality nz < (1 − ε)ny can be written: ε < (vs(y) − vs(z))/vs(y), and vs(y) and vs(z) are two different integers smaller than the initial number of voters v. Therefore this inequality is true for all n for all ε < 1/v. Under

20

this condition, it follows: nt £ ¤X Bn < Pr [∀x 6= y, z, Sxn < nt − 1] Pr Syn = nt Pr [Szn = i] , i=0

Cn

nz £ ¤ X Pr [Szn = i] > Pr [∀x 6= y, z, Sxn < nt − 1] Pr Syn = nt i=nt +1

and thus

Bn Pr [Szn ≤ nt ] < . Cn Pr [Szn > nt ] When n tends to infinity, the variable Szn /n is approximately normal, with mean µ = (1 − ε)vs(z) and variance ε(1 − ε)vs(z)/n. Like previously, ε < 1/v implies that vs(t) < µ. The weak law of large number implies that the probability that Szn /n is less than vs(t) tends to 0 when n tends to infinity, which means here that Pr [Szn ≤ nt ] tends to 0 and thus Pr [Szn > nt ] n tends to 1. Therefore B Cn tends to 0. Recall that P(n, y, t) < Bn and P(n, y, z) = Bn + Cn , it follows that P(n, y, t) = 0. n→∞ P(n, y, z) lim

This means that the exact tie {y, t} is negligible in front of the exact tie {y, t} acoording to probability distribution Pr. Dividing by Pr[T3c ] one sees that this is true as well conditionally on the non-occurence of two-way ties. Second step. Consider next the event pivot(n, {y, z}) of an almost exact tie at the top between the two candidates y and z with y < z. This is the union of the three disjoint events: ∀x 6= y, z , Syn = Szn > Sxn + 1 or ∀x 6= y, z , Syn − 1 = Szn > Sxn or ∀x 6= y, z , Szn − 1 = Syn > Sxn Denote by P(n, y, z), P0 (n, y, z), and P00 (n, y, z) the probabilities of these three events. For t with s(y) > s(z) > s(t), the same reasoning which led in P(n,y,t) = 0 can be made and leads to the two same the first step to limn→∞ P(n,y,z) other conclusions limn→∞

P0 (n,y,t) P0 (n,y,z)

= 0 and limn→∞

P00 (n,y,t) P00 (n,y,z)

= 0. Therefore

P(n, y, t) + P0 (n, y, t) + P00 (n, y, t) =0 n→∞ P(n, y, z) + P0 (n, y, z) + P00 (n, y, z) lim

and we obtain that

P[pivot(n, {y, t})] = 0. n→∞ P[pivot(n, {y, z})] lim

21

(4)

A.1.2

Environmental invariance

Let

Sexn à B(nvs(x) − 1, 1 − ε).

Consider a strategy profile in which type-τ voters approve of candidate x, for a type τ voter, the number of votes in favor of x is Sexn if this particular voter decides not to approve of x,.or Sxn if she decides to approve of x. We now check that this feature has no consequence on the voter’s decision because the ties arising from variables Sxn or Sexn have the same order of magnitude. To see this, consider for instance the event of an exact tie at the top between y and z when the score of y is Seyn à B(nvs(y) − 1, 1 − ε) and the score of z is Szn à B(nvs(z), 1 − ε). The event

∀x 6= y, z , Seyn = Szn > Sxn

has the following probability: nz X

Q(n, y, z) =

Cni y −1 Cni z (1 − ε)2i

i=0 ny −1+nz −2i

×ε

Pr [∀x 6= y, z, Sxn < i]

(we still denote ny = ns(y) and nz = ns(z)), to be compared with P(n, y, z) in formula (??). Because Cni y −1 < Cni y , it is easy to see that Q(n, y, z) < (1/ε)P(n, y, z). Using the fact that Cni y =

ny ny Cni y −1 ≤ Ci , ny − i ny − nz ny −1

one can also see that P(n, y, z) < ε

ny Q(n, y, z). ny − nz

We thus obtain that, for all n, ε<

P(n, y, z) s(y) <ε . Q(n, y, z) s(y) − s(z) 22

(5)

We leave to the reader the easy verification of similar formulas for all events of the form ∀x 6= y, z , Sbyn = Sbzn > Sbxn , where the variables Sbxn are equal either to Sxn or to Sexn , as well as for the almost tie events of the form ∀x 6= y, z , Sbyn + 1 = Sbzn > Sbxn and

∀x 6= y, z , Sbzn + 1 = Sbyn > Sbxn .

From the observation (??) one can deduce that the conclusion (??) that the race {y, t} is negligible compared to the race {y, z} holds for all the voters, independently of who this voter is voting for. This point plays the role of the “environmental invariance” property that Myerson obtains with Poisson random variables9 . With binomial variables, different voters do not have exactly the same information, but the difference is so small that it does not matter for finding best responses.

A.2

About three-way ties

It may seem intuitively plausible that, in the binomial model studied here and in other similar models, three-way ties are negligible compared to twoway ties. Myerson and Weber (1993) make this assumption in their seminal paper, followed by Bouton and Castanheira (2008) in their study of Approval Voting under imperfect information. McKelvey and Patty (2006) present this point as a result obtained (see their lemma 13 point b) by the same arguments which prove that the probability of a tie goes to zero. We explicitely made the hypothesis that a voter neglects three-way (or more) ties (see (??)) but in early versions of this paper, this point was also announced as a result10 . Certainly this assumption is reasonable as a behavioral (or rather: a psychological) one but “large deviations” in Probability Theory provide many counter-intuitive results, and the following simple example will explain why neglecting three way ties among binomial variables cannot be obtained as a result. There are two binomial variables, S1n à B(n1 , 1 − ε) S2n à B(n2 , 1 − ε) 9 10

Milchtaich (2004) offers for a more systematic treatment of this question. Laslier (2004). Thanks to a referee of this journal who spot the mistake.

23

with n1 = nvs(x1 ), n2 = nvs(x2 ) and a third variable S3n which we suppose is not random: S3n = nt ∈ N Then compare the probability of the two events S1n = S3n > S2n (typically, a two-way tie between 1 and 3) and S1n = S2n = S3n (typically, a three-way tie between 1, 2, and 3). For t < vs(x2 ) < vs(x1 ), by independence: Pr[S1n = nt] · Pr[S2n = nt] Pr[S2n = nt] Pr[S1n = S2n = S3n ] = = . Pr[S1n = S3n > S2n ] Pr[S1n = nt] · Pr[S2n < nt] Pr[S2n < nt] The numerator of this fraction is the probability that the random variable exactly equals the (large) integer nt. Intuition may lead to think that this probability is infinitesimal compared to the denominator, which is the probability that S2n is any integer smaller than nt. But such is not the case. When n tends to infinity this ratio tends to a strictly positive value, as proved in Timashev (1997, corollary 4). Taking into account this phenomenon would lead to an untractable model in general since then the orders of magnitude of the likelihood of various events would not be sufficient anymore to derive the individual best responses. For this reason we explicitely made the assumption that voters reason conditionally on the non-occurence of three-way ties. S2n

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