An Influence-Based Theory of Strategic Voting in Large Elections∗ Guilherme Carmona† University of Surrey May 9, 2014

Abstract We consider a voting game where voters’ preferences depend both on the identity of the winning candidate and on the vote casted, the latter reflecting a social or ethical component of voters’ preferences. We show that, in all sufficiently large voting games, all symmetric equilibria are sincere. We then consider a version of the same voting game with pre-election actions (e.g. voters communicate, observe the outcome of polls, or make costly campaign contributions) that endogenously determine voters’ social component of preferences. We then show that in all sufficiently large voting games with pre-election actions, all symmetric equilibria are strategic. Thus, in sufficiently large elections, symmetric equilibrium voting is strategic if and only if there is pre-election interaction. ∗

I wish to thank Nathana¨el Berestycki, Marco Mariotti, Jaideep Roy, Hamid Sabourian, Francesco

Squintani and seminar participants at the Universities of Cambridge, Essex, Manchester, St. Andrews and Surrey, the 20th European Workshop on General Equilibrium and the 2013 RES conference for very helpful suggestions. I also thank Girija Bahety for excellent research assistance. Financial support from Funda¸c˜ ao para a Ciˆencia e a Tecnologia is gratefully acknowledged. Any remaining errors are, of course, mine. † Address: University of Surrey, School of Economics, Guildford, GU2 7XH, UK; email: [email protected].

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1

Introduction

During election times, it is not uncommon to hear voters arguing that, although they prefer candidate a, they will vote for candidate b to prevent candidate c from winning. This voting behavior may, in fact, be expected. As Duverger (1954) has argued: In cases where there are three parties operating under the simple-majority single-ballot system the electors soon realize that their votes are wasted if they continue to give them to a third party: whence their natural tendency to transfer their vote to the less evil of its two adversaries in order to prevent the success of the greater evil. This argument provides a possible justification for the occurrence of strategic voting. By definition, voting is sincere if each voter votes for his favorite candidate and is strategic when it is not sincere. While there is ample empirical evidence supporting strategic voting (see Fujiwara (2011) and the references therein), Duverger’s (1954) argument above is less convincing. A difficulty with this argument when applied to large elections is that it is unlikely that individuals vote in the way suggested in it since the effect of a single vote on who wins is small. Several authors have expressed this view recently. For instance, Meirowitz and Shotts (2009) showed, in a model where voters can signal their policy preferences, that this signalling effect, as opposed to a pivot effect (i.e. the impact each vote has on the winning probabilities of the different candidates), drives equilibrium voting behavior in large elections. Furthermore, Feddersen, Gailmard, and Sandroni (2009) showed, theoretically and experimentally, “that voters in large elections tend to vote against their material self-interest and to vote for a morally or ethically appealing alternative”. If voting is strategic but voters do not determine their vote based on its impact on the candidates’ winning probabilities, why do voters vote strategically and what determines their vote? Besides being an interesting question in its own right, this question is relevant for the evaluation of the simple-majority single-ballot voting system as a mechanism for collective decision making. 2

We answer the above question as follows. We consider a qualified-majority voting game similar to that of Fey (1997) and Myatt (2007) where voters vote for one of two candidates in an attempt to defeat a disliked status quo candidate.1 This means that each voter prefers each of these two candidates over the status quo (and this is commonly known), but how each voter ranks the two candidates is her own private information. We depart from the setting considered in those papers as follows. First, we assume, as in Feddersen, Gailmard, and Sandroni (2009), that the payoff each voter receives depends, besides on the identity of the winner, on the ballot that she submits. This latter aspect of voters’ payoffs reflects a social or ethical component of their preferences. The social/ethical component is weighted (arbitrarily) less than the individual/material component of preferences (i.e. the idiosyncratic component determined by the identity of the winner of the election). For this reason, the sincerity of a vote is defined with respect to the individual/material preference: voting is sincere if each voter votes for his favorite candidate according to his individual/material preferences. Second, we introduce to the model described above the possibility that voters influence others before the election takes place and let voters’ social/ethical component of their preferences be determined endogenously. We formalize influence abstractly as a costly activity that each voter can take and whose potential consequence is that it determines how each voter feels about the vote he casts. More concretely, we introduce an influence payoff for each voter that depends on his material preferences, the influence actions of all voters and on the ethical view he chooses to adopt regarding his vote, i.e. on his social/ethical component of preferences. The ethical part of voters’ preferences is therefore a social component of preferences precisely because it is determined by the (pre-election) actions of all voters. In contrast, the material part of voters’ preferences, which can be interpreted as measuring how desirable the policies of each candidate are for each voter, is an individual component of preferences as it 1

Fey’s (1997) model has three candidates, A, B and C, but the only non-trivial decision in the

equilibria considered in that paper is taken by the group of voters who dislike candidate C and consists on deciding whether to vote for A or B.

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does not depend on the action of other voters. In this light, a benchmark case of our setting arises when there is no pre-election interaction and where the two components of voters’ preferences coincide (i.e. a voter prefers that candidate a wins the election if and only if he received a higher payoff from voting a): This case is interpreted as describing an autarkic situation where no social notion of ethics arise and, thus, voters have no reason to have an ethical component of their preferences different from their material component. The benchmark case differs from a standard voting model such as that in Fey (1997) only due to the introduction of the social/ethical component of preferences, which works as an equilibrium refinement device. Indeed, no matter how small the weight attached to the vote casted is, it will eventually be bigger than the probability of each voter affecting the outcome of the election and, consequently, all symmetric equilibria of all sufficiently large such voting games will be sincere. The above result is useful to serve as a contrasting benchmark for our strategic result, which is obtained when pre-election interaction (i.e. influence) is allowed and the ethical/social component of preferences is determined endogenously. In this case, we show that in all sufficiently large voting games with pre-election actions, all symmetric equilibria are strategic.2 This result is in sharp contrast with the one obtained when influence is not present and, together, they yield an influence-based theory of strategic voting in large elections: In our framework, strategic voting is prevalent if and only if influence is allowed. Intuitively, this result holds because, when the number of voters is large and influence is not allowed, each voter rarely affects the outcome of the election. In contrast, when influence is allowed and because it is assumed to be costly, a voter will chose to influence only if this choice has a sufficiently large effect on the outcome of the election, in particular, by having some voters voting strategically. Thus, the possibility of influence brings the strategic nature present in elections with a small number of 2

Roughly, a strategy is symmetric in our setting if voters with the same material preferences

(their type in our model) and the same information about the pre-election actions vote in the same candidate.

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voters to elections featuring a large number of voters. Several authors have emphasized the importance of pre-election interaction. Preelection interaction can take the form of focal manipulation and candidate presentation (see Myerson and Weber (1993)), future promises by candidates (Myerson (1993)), effort made by leaders (Shachar and Nalebuff (1999)), be a determinant of voters’ perceptions of the candidates’ relative chances of winning the election (Myerson and Weber (1993) and Cox (1994)), can occur through polls (McKelvey and Ordeshook (1985), Cukierman (1991) and Fey (1997)) and through repeated elections (Piketty (2000), Castanheira (2003) and Meirowitz and Shotts (2009)), be a coordination device when the popularity of the candidates is not common knowledge (Myatt (2007)), be direct advertising by candidates or word-of-mouth communication (Galeotti and Mattozzi (2011)) and so on. Our contribution to this literature can be seen as providing, in a simple model, an abstract formalization of influence. Besides permitting several interpretations of influence, our approach allows us, more importantly, to identify general properties that are sufficient for pre-election actions taken by voters in an attempt to influence others to lead to the prevalence of strategic voting in large elections. The flexibility of our formalization of influence allows us to readily build examples of equilibria with strategic voting in settings with important forms of pre-election actions. This is illustrated by an example in Section 6.1 where voters can make costly campaign contributions (either one monetary unit or zero). In the context of such example, we build symmetric equilibria where each voter chooses to contribute to her favorite candidate with a strictly positive probability and to vote for the candidate that receives the highest total contribution. We then show that, in any such equilibrium, the probability of each voter contributing to the campaign of some candidate converges to zero, implying that the fraction of those who contribute is close to zero with a probability close to 1 in large elections. Furthermore, we show that the ability of campaign contributions to coordinate voters’ votes critically depends on the paraments: The status quo wins with a probability close to 1 in large elections for some parameter values, and loses with a probability close to 1 in large elections for other 5

parameter values. Furthermore, the abstract nature of our formalization of influence has the advantage that we do not need to adopt a particular influence payoff function, i.e. particular determinants of voters’ ethical views. Nevertheless, in our conditions on influence, we do assume that relative influence matters for the choice of voters’ ethical ranking of candidates in the following sense: If one candidate receives considerable more support than the other, then voters will be inclined to regard the former as the ethical candidate. In particular, the assumption that relative influence matters is satisfied when voters like to be part of a majority, an assumption we use in our examples and which has been used in a voting context (without influence) by Callander (2008). Justifications for the assumption that voters (and individuals more generally) like to be part of a majority includes information cascades and social pressure to conform; furthermore, we present an example in Section 6.2 where such desire to be part of a majority arises due to homophily in the sense of individuals providing favors to those having the same ethical views, and where the dependence of voters’ payoff on the ballot submitted can be interpreted as a cost of lying about one’s voting intentions.3 Our result on strategic voting, as well as the one on sincere voting, are generally in line with the experimental evidence summarized in Rietz (2003) on elections with three candidates where one of them is a Condorcet loser, the latter being interpreted as the undesirable status quo to fit our framework. Specifically, in these experiments, there are six voters whose favorite candidate is the Condorcet loser and eight voters for whom the Condorcet loser is the least preferred candidate; however, for each of the candidates other than the Condorcet loser there are four out of these eight voters for whom that candidate is the favorite one. Note that the Condorcet loser wins if voting is sincere and one of the remaining candidate wins if all voters that dislike the Condorcet loser coordinate their votes, in particular, by having four of them voting strategically.4 Experimental evidence from these elections show that the 3

See Callander (2008) for more detail on the former justifications. Our example in Section 6.2 is

based on Bramoull´e and Goyal (2013); see also Currarini, Jackson, and Pin (2009) for a model of homophily. 4 The eight voters who dislike the Condorcet loser correspond to the voters in our framework.

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Condorcet loser tends to win in the absence of pre-election interaction. However, when the election is preceded by some form of interaction (for instance, non-binding pre-election polls or costly campaign contributions), then the non-Condorcet-loser candidate that receives the strongest support tends to win. Thus, as in our model, the outcome implied by sincere voting occurs if and only if there is no pre-election interaction. The paper is organized as follows. Section 2 discusses further related work. The model is presented in Section 3 and the sincere voting result when influence is not allowed appears in Section 4. Section 5 starts with some motivating examples of forms of pre-election interaction between the voters and then presents two strategic voting results. These results differ in that one allows for the presence of public information regarding the pre-election actions whereas the other does not. The extension of these results to the case of mixed strategies is also presented in Section 5. Our model and results are illustrated by two examples in Section 6. Section 7 contains some concluding remarks. The proofs of our results are in the Appendix.5

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Related literature

Other authors have obtained strategic voting results. Feddersen (1992) showed that, in a model where voting is costly and voters’ preferred positions on the policy space (which is a ball in an Euclidean space) are dispersed, all equilibria have only two winning positions. This result is driven by pivotal considerations: Voting cannot be sincere roughly because those voters whose favorite position is not a winning position would prefer either to vote for the closest winning position or to abstain. Another model yielding strategic voting where voting behavior is driven by a pivot effect is that of Fey (1997). There it is shown that there exist both Duvergerian and non-Duvergerian voting symmetric equilibria, the former being a strategic voting 5

The intuition behind our results is relatively simple and is, therefore, presented next to them. In

contrast, the proofs are somewhat complicated due to the need to deal with (a) measure theoretical arguments and (b) off-the-equilibrium-path behavior.

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equilibrium and the latter being, in the limit as the number of voters converges to infinity, a sincere voting equilibrium. Fey (1997) does provide a theory of strategic voting by showing that the non-Duvergerian asymptotic-sincere equilibrium is not robust to a refinement of the equilibrium concept based on stability. Our sincere voting result can be contrasted with Fey’s (1997) since we can regard the introduction of preferences for the vote casted as a refinement device. In fact, no matter how small is the weight given to this element of preference, symmetric equilibria will be sincere in large elections. Another strategic voting result based on a pivot effect is that of Myatt (2007). In his framework, as in ours, voters need to coordinate their votes on one of two candidates to beat a disliked status quo. Voters receive private signals of the relative popularity of each of the candidates. Myatt (2007) showed the existence of strategic symmetric equilibria where each voter will vote strategically whenever he receives a strong enough signal that his least favorite candidate is the most popular. While formally quite different from our strategic voting result, the idea that voters use signals to coordinate their voting behavior in Myatt’s (2007) model is generally consistent with the formalization of influence in our model. Finally, we note that further strategic voting results have been obtain in the context of the different but related problem of voting in binary agendas. See, for instance, Ordeshook and Palfrey (1988) and, more recently, Ahn and Oliveros (2012).

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The model

We present our framework in this section and use its elements to define a sequence of dynamic games with incomplete information. In short, our goal is to determine whether or not all symmetric sequential equilibria in all sufficiently large games in such a sequence are sincere or strategy; the notions of a symmetric equilibrium as well as those of a sincere symmetric equilibrium and of a strategic symmetric equilibrium are also introduced in this section.

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3.1

Timing and the election

There are n voters who can vote for one of two candidates, a or b; the set of candidates is denoted by C = {a, b} and the set of voters by In = {1, . . . , n}. There are two periods. In the first period, each voter chooses whether to influence others or not and, in case he decides to influence, how to influence. The effect of influence is that it may change the social ranking of candidates of those voters who are being influenced. Accordingly, each voter decides on whether to keep or change his social ranking in period 2. In period 2, there is also an election and voters vote for a candidate (abstention is not allowed). The outcome of the election is determined by a qualified majority election rule as in Myatt (2007): Candidate c ∈ C wins if he receives strictly more than n¯ x votes, where 1/2 ≤ x¯ < 1 (note that there can be no ties between candidates a and b since x¯ ≥ 1/2). Furthermore, if neither candidates receives more than n¯ x votes, then the outcome is the status quo (e.g. a third, unmodeled, candidate wins). Apart from the tie-breaking rule, the case x¯ = 1/2 corresponds to the standard simple majority election rule. The case x¯ > 1/2 can be interpreted as follows. In addition to the n voters, there is a strictly positive number r of (unmodelled) voters, each of whom votes for an (unmodeled) third candidate, and the outcome of the election is determined by the simple majority rule with ties allocated to this third candidate. Letting x¯ = r/n, the assumption that x¯ > 1/2 then means that 2r > n and the assumption that x¯ < 1 means r < n. More importantly, candidate c ∈ C wins if he receives strictly more than n¯ x = r votes.

3.2

Actions and payoffs

Each voter’s payoff depends, in particular, on two rankings of candidates, namely the voter’s individual ranking θ and his social ranking θ0 . Both rankings are an element of a finite subset Θ of R2 . A generic element θ = (θa , θb ) of Θ lists, for each c ∈ C, the payoff each voter with ranking θ obtains in the case candidate c is elected. If the outcome is the status quo, then each voter receives a zero payoff. We assume that Θ

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is symmetric (in the sense that if (α, β) ∈ Θ, then (β, α) ∈ Θ), θa 6= θb and θc > 0 for all θ ∈ Θ and c ∈ C. The following notation will be useful. We let θ∗ = maxθ∈Θ θa = maxθ∈Θ θb . Furthermore, for each θ ∈ Θ, let t(θ) = (β, α) if θ = (α, β) and let f (θ) denote the favorite candidate given θ, i.e. f (θ) = c if θc = max{θa , θb }. For each n ∈ N and voter i ∈ In , i’s action set in period 1 is Xin . The interpretation of Xin is that it includes all the actions voter i can take to influence the decision of other voters on whom to vote. Each voter can choose to abstain from influencing others and such action is denoted by e; thus, e ∈ Xin for all i ∈ In . For each c ∈ C, let Xin (c) be the set of actions that are interpreted as support for candidate c. Thus, Xin (a) ∪ Xin (b) ⊆ Xin \ {e} but we do not require that Xin (a) ∪ Xin (b) = Xin \ {e}. In period 2, voter i simultaneously chooses her social ranking of candidates θ0 ∈ Θ and vote c ∈ C.6 Each voter’s influence payoff depends on the actions chosen in the pre-election period and on the choice of the social ranking of candidates. For each n ∈ N and Q i ∈ In , letting X n = j∈In Xjn , voter i’s influence payoff function is u˜ni : X n ×Θ2 → R with the interpretation that u˜ni (x, θ0 , θ) is voter i’s influence payoff when x ∈ X is chosen in the pre-election period, voter i chooses social ranking θ0 and voter i’s individual ranking is θ. Each voter’s election payoff depends on his vote, on the distribution of votes and on his social and individual ranking of candidates. Let ∆(C) be the set of probability distributions over C and, for each c ∈ C, let Wc = {π ∈ ∆(C) : πc > x¯} be the set of distributions in which candidate c wins (indeed, if π ∈ Wc and c0 6= c, then πc0 < πc since πc > x¯ ≥ 1/2). The election payoff of voter i depends on his vote, weighted by the corresponding weight in the social ranking θ0 , and on the identity of the winner, weighted by the corresponding weight in the individual ranking θ. The weight on the identity of the winner is 1, while that on the vote casted is λ ∈ (0, 1). Specifically, the election payoff v(c, π, θ0 , θ) of a voter with individual ranking θ and social ranking 6

The assumption that this choice is made simultaneously is not important, it means only that

voter j 6= i does not observe the choice of i’s social ranking of candidates before the election.

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θ0 when he votes for candidate c and the outcome of the election is described by π is defined by setting, for all c ∈ C, π ∈ ∆(C) and θ, θ0 ∈ Θ, v(c, π, θ0 , θ) = λθc0 + θa 1Wa (π) + θb 1Wb (π).7 Each voter’s total payoff is the sum of her influence and election payoffs. We let uni : X n × C × ∆(C) × Θ2 → R denote voter i’s payoff function, which satisfies uni (x, c, π, θ0 , θ) = u˜ni (x, θ0 , θ) + v(c, π, θ0 , θ) for all x ∈ X n , c ∈ C, π ∈ ∆(C) and θ, θ0 ∈ Θ.

3.3

Information

For all n ∈ N and i ∈ In , voter i’s individual ranking of candidates θ is determined by a distribution µ on Θ with µ(θ) > 0 for all θ ∈ Θ. Voters’ individual rankings of candidates are independent, i.e., letting µni = µ for all i ∈ In , we assume that {µni }ni=1 are independent. For convenience, we use µn to denote the corresponding product measure on Θn . We also assume that the election is such that voters need some coordination to defeat the status quo, in the following sense. If each voter votes in his favorite candidate f (θ) according to his individual ranking of candidates θ, then the fraction of votes each candidate c ∈ C receives (approximately and with a probability close P to 1) in large elections is µ({θ ∈ Θ : f (θ) = c}) = θ∈Θ:f (θ)=c µ(θ). We assume that µ({θ ∈ Θ : f (θ) = c}) < x¯ for each c ∈ C. Each voter knows his own individual ranking of candidates but not the individual ranking of candidates of any other voter. His decision to influence is made in the first period after observing is own individual ranking of candidates. The choice of his social ranking of candidates and his vote is made simultaneously in the second period after observing the influence choice of a subset of voters and additional public information on the entire vector of such choices. For each n ∈ N and i ∈ In , the set of voters whose 7

In this formula, we use the following standard notation: For all subsets A of ∆(C), the function

1A : ∆(C) → {0, 1} is defined by 1A (π) = 1 if π ∈ A and 1A (π) = 0 otherwise, for all π ∈ ∆(C).

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influence action is observed by voter i is denoted by In (i) and satisfies i ∈ In (i) ⊆ In . The additional public information about the vector of influence actions is described by a function gn , mapping X n onto some finite set Γn . A special and trivial case of such public information occurs when, effectively, there is no public information, in which case we can let Γn = {0} and gn (x) = 0 for all x ∈ X n . Another special case is when influence consists on monetary contributions to each candidate’s campaigns and each voter can observe the total contribution each candidate receives. This case can P be modeled by letting Xin be a finite subset of R2+ for all i ∈ In , Γn = i∈In Xin ⊂ R2+ P and gn (x) = i∈In xi for all x ∈ X n .

3.4

Sequences of games

The above elements are used to define a sequence of dynamic games with incomplete information as follows. For each n ∈ N, the set of players is the set of voters In . Players’ individual rankings of candidates θn = (θin )i∈In are drawn according to µn . n Each player i ∈ In observes θin but not θ−i . We often call θin player i’s type.

Players choose actions simultaneously in each period. For each i ∈ In , player i chooses xi ∈ Xin in period 1, and (θi0 , ci ) ∈ Θ × C in period 2. For simplicity, player i’s action space in period 1 is type-independent. However, we allow for player i’s action set in period 2 to be type-dependent by letting Yi (θ) ⊆ Θ denote the set of player i’s possible social rankings of candidates when he is of type θ. This is done to capture the case where each voter’s social ranking of candidates must be equal to her individual ranking (this will be part of our definition of the no influence case), a case in which Yi (θ) = {θ} for all i ∈ In and θ ∈ Θ. In period 1, player i ∈ In chooses xi ∈ Xin after observing θin . After the first period choice x = (x1 , . . . , xn ) is made, player i observes xj for all j ∈ In (i) and also gn (x). He then chooses θi0 ∈ Θ and ci ∈ C. Player i’s payoff is then uni (x, ci , π(¯ c), θi0 , θin ), where c¯ = (cj )j∈In ∈ C n and π(¯ c) ∈ ∆(C) consists on the fraction of votes each candidate receives, i.e. πc (¯ c) = |{j ∈ In : cj = c}|/n for each c ∈ C. We denote the game just defined by Gn = (In , C, Θ, µn , gn , (Xin , uni , In (i), Yi )i∈In ). We call a sequence of games {Gn }∞ n=1 as defined above a sequence of election games with influence. 12

3.5

Strategies

We next define the notion of a pure strategy. We first consider the case of pure strategies, the case of mixed strategies being considered in Section 5.4. For all n ∈ N, a strategy profile (a strategy, for short) is s = (sti )i∈In ,1≤t≤3 such that, for all i ∈ In , (a) Q  n 2 s1i maps Θ into Xin , (b) s2i maps X j ×Γn ×Θ into Θ with si (x, γ, θ) ∈ Yi (θ) j∈In (i)   Q Q n 3 n × Γn × Θ for all x ∈ j∈In (i) Xj , γ ∈ Γn and θ ∈ Θ, and (c) si maps j∈In (i) Xj into C. A strategy s and a profile of types θn determine a sequence of actions (an outcome path) a(s, θn ) in the following way. For all i ∈ In , a1i (s, θn ) = s1i (θin ), a2i (s, θn ) = s2i ((a1j (s, θn ))j∈In (i) , gn (a1 (s, θn )), θin ) and a3i (s, θn ) = s3i ((a1j (s, θn ))j∈In (i) , gn (a1 (s, θn )), θin ). We say that a strategy s is sincere if a3i (s, θn ) = f (θin ) for all i ∈ In and θn ∈ Θn . Thus, a strategy is sincere if each voter votes for his favorite candidate according to his individual ranking of candidates. A strategy s is strategic if s is not sincere. Players’ payoffs are also determined by a strategy s and a profile of types θn . We let uni (s, θn ) denote the corresponding payoff for player i and we have that uni (s, θn ) = uni (a1 (s, θn ), a3i (s, θn ), π(a3 (s, θn )), a2i (s, θn ), θin ), where, as above, πc (a3 (s, θn )) = |{j ∈ In : a3j (s, θn ) = c}|/n for each c ∈ C. We use sequential equilibrium as our equilibrium concept (a generic belief system will be denoted by ν) and focus on the following class of “symmetric” strategies. A strategy is symmetric if for all i, j ∈ In , γ ∈ Γn and θ ∈ Θ, s3i ((xl )l∈In (i) , γ, θ) = s3j ((xl )l∈In (j) , γ, θ) whenever xl = e for all l ∈ (In (i) ∪ In (j)) \ (In (i) ∩ In (j)). The set (In (i)∪In (j))\(In (i)∩In (j)) is the set of players whose pre-election action is observed by only one of the players i and j; hence a strategy is symmetric if each voter’s voting behavior depends only on his type θ (i.e. his individual ranking of candidates), on the public information γ and on the influence actions of those who do not abstain to influence. 13

4

Sincere voting without influence

Our first result concerns the symmetric equilibria of the voting problem that occurs when voters (effectively) cannot influence others and their social ranking of candidates cannot differ from their individual one. Such no influence case consists of a modification of a standard voting problem, the modification being that each voter’s preferences depend on the vote she casts. The motivation for this variation on the standard voting problem is that it works as an equilibrium selection device that allows us to obtain a sharp benchmark result to which the result for voting problems with influence will be compared to. Indeed, as Proposition 1 shows, sincere voting is the unique symmetric equilibrium of all sufficiently large election games without influence. An election game without influence is formalized by restricting the pre-election interaction, forcing each player to choose the no-influence action e and a social ranking of candidates θ0 equal to her individual ranking θ. Given a sequence of election games ∞ with influence {Gn }∞ n=1 as defined in Section 3, we say that {Gn }n=1 is a sequence of

influence-free election games if Xin = {e} and Yi (θ) = {θ} for all n ∈ N, i ∈ In and θ ∈ Θ. Before we state our result for sequences of influence-free election games, note that several simplifications of our setting arise for such sequences. First, for all n ∈ N and i ∈ In , there is only one strategy s1i and only one strategy s2i . Hence, voter i’s strategy is fully described by s3i . Second, s3i is fully described by a function mapping Θ into Q C since j∈In (i) Xjn and Γn are singletons. Hence, for each game Gn is a sequence of influence-free election games, s3i is regarded as a function from Θ into C. Third, and consequently, a strategy s in such a game Gn is symmetric if and only if each voter’s voting behavior depends only on his type, since all voters (are forced to) abstain to influence and, consequently, there is no public information on influence (i.e. Γn is a singleton). More formally, s is symmetric if and only if there exists a function σ : Θ → C such that s3i (θ) = σ(θ) for all i ∈ In and θ ∈ Θ. Fourth, and finally, Gn is effectively a static game with incomplete information. Hence, the set of sequential

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equilibrium strategies of Gn coincides with the set of Bayes-Nash equilibria of Gn . Our main result for the case of sequences of influence-free election games is that sincere voting is the unique symmetric equilibrium of all sufficiently large such games. This result can be understood as follows (see Section A.3 for its proof). When a symmetric strategy is being played, each voter’s type determines his vote and, as types are iid, each player’s faces a sequence of iid choices by the other players. This implies that for each symmetric strategy, using either a law of large number or a local central limit theorem, that there is a population size such that it is unlikely that a voter has an impact on the outcome of the election whenever the number of voters exceed that size. Thus, since each voter has a cost of voting for the least preferred candidate according to his individual ranking of candidates, then it is optimal for each voter to vote for his favorite candidate. As there are only finitely many symmetric strategies, we obtain the following result. Proposition 1 Let {Gn }∞ n=1 be a sequence of influence-free election games. Then there exists N ∈ N such that, for all n ≥ N , (s, ν) is a symmetric sequential equilibrium of Gn if and only if s is sincere.

5

Strategic voting with influence

Proposition 1 implies that some modification of the influence-free election game is needed in order to obtain strategic voting in equilibrium. Thus, in this section, we seek conditions on influence that imply strategic voting in equilibrium. We start by considering some examples of influence and use then to highlight the conditions we use in our main results (the proof of the claims made in these examples are in Section A.8).

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5.1 5.1.1

Examples Opinion leaders

In the example of this section, there is at least one widely heard opinion leader, which can also be thought of as one of the candidates, a newspaper, a tv station, and so on. For concreteness, we assume that voter 1 is such an opinion leader and assume that 1 ∈ In (i) for all n ∈ N and i ∈ In , i.e. every voter observes the opinion leader’s pre-election actions. The above assumption requires the existence of at least one opinion leader; in particular, it allows for more than one opinion leader. The assumption that all voters observe the opinion leader’s pre-election actions will be weakened below. Furthermore, the presence of an opinion leader dispenses with public information. Thus, we assume that Γn = {0} and gn (x) = 0 for all n ∈ N and x ∈ X n and, hence, we simply ignore the function gn . In this example, each voter’s pre-election action is simply which candidate to support, if any. Thus, Xin = {a, b, e} for all n ∈ N and i ∈ In , where e stands for no influence. Each voter’s influence payoffs is given by u˜ni (x, θ0 , θ) = θfβ(θ0 ) |{j ∈ In (i) \ {i} : xj = f (θ0 )}| − ε1C (xi ),

(1)

for all n ∈ N, i ∈ In , x ∈ X n and θ, θ0 ∈ Θ, where β ≥ 0 and 0 < ε < θ∗ (recall that θ∗ = maxθ∈Θ θa = maxθ∈Θ θb ). Thus, each voter is willing to have a favorite candidate of her social ranking of candidates different from her favorite candidate of her social ranking of candidates if the former has sufficiently more support in the group of voters she observes prior to the election. However, she has a bias to keep her favorite candidate if the support of each candidate is sufficiently close. Let {Gn }∞ n=1 be a sequence of election games with influence (as specified in Section 3) satisfying the assumptions made above. It will follow from our main result for sequences of election games with influence, Proposition 2 below, that, for all n sufficiently large, all symmetric sequential equilibria of Gn are strategic. Four properties of the assumptions made in this example are important for this result. First, influence is costly as we have that u˜ni ((e, x−i ), θ0 , θ) − u˜ni (x, θ0 , θ) = ε for 16

all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e. Thus, there is a positive cost of influencing. Second, influence is not too costly since ε < θ∗ = maxθ∈Θ θa = maxθ∈Θ θb . In particular, for a voter with type θ such that θa = θ∗ , it would pay to incur such cost in order to change the outcome of the election from the status quo to candidate a. Third, there is a visible player, which is the opinion leader, voter 1. In fact, the fraction of those voter who observe voter 1’s pre-election action equals 1, i.e. |{j ∈ In : 1 ∈ In (j)}|/n = 1. We note that, more generally, it suffices to assume that there is a visible player relative to the tightness of the election. This point is illustrated by assuming that the distribution of individual rankings of candidates is such that µ({θ ∈ Θ : f (θ) = a}) = µ({θ ∈ Θ : f (θ) = b}) = 1/2 and that x¯ = 1/2. Given these extra assumptions, we only need that voter 1 be heard by a fraction of voters which is bounded away from zero, i.e. |{j ∈ In : 1 ∈ In (j)}|/n ≥ δ for some δ > 0. Forth, relative influence matter in the choice of the social component θ0 of preferences. In particular, if voter 1 supports candidate b before the election when no one else influences, then the influence payoff of each voter other than voter 1 will rank those θ0 in which b is the favorite candidate higher than those θ0 in which a is the favorite candidate. 5.1.2

Influence through a network

This example is similar to the previous one in that voters care about how many people are influencing for a given candidate. However, we now assume that voter are connected via a network and that each voter gives more weight to the pre-election actions of those who are closer to him. For each n ∈ N, let hn be an undirected network, i.e. hn = (hnij )i,j∈In with hnij = hnji and hnij ∈ {0, 1} for all i, j ∈ In . Whenever hnij = 1, this means that voters i and j are connected. This may be interpreted as meaning that they are friends, that they communicate, etc. We assume that hn is a connected network, which means that there is a path between every pair of voters. Let d(i, j; hn ) denote the length 17

of the shortest path between voters i and j in hn . We assume that the networks are “uniformly connected” in the sense that there exists m ∈ N such that d(i, j; hn ) ≤ m for all n ∈ N and i, j ∈ In . Furthermore, for simplicity, we assume that In (i) = In for all i ∈ In so that each voter i observes the pre-election action of any other voter j, with the interpretation that such observation is direct if hnij = 1 and indirect if hnij = 0. As in the previous example, each voter’s pre-election action is simply which candidate to support, if any (i.e. Xin = {a, b, e} for all n ∈ N and i ∈ In ), and we omit the function gn . Furthermore, each voter’s influence payoffs is given by X n u˜ni (x, θ0 , θ) = β d(i,j;h ) − ε1C (xi ),

(2)

j∈In \{i}:xj =f (θ0 )

for all n ∈ N, i ∈ In , x ∈ X n and θ, θ0 ∈ Θ, where 0 < ε < θ∗ and 0 < β < 1. Note that here we do not assume that voters have a bias to keep their favorite candidate but this element can be added as in the previous example. As in the previous example, influence is costly but not too costly. Each player is visible and relative influence matters. Thus, by Proposition 2 below, if {Gn }∞ n=1 is a sequence of election games with influence satisfying the assumptions made in this section, then, for all n sufficiently large, all symmetric sequential equilibria of Gn are strategic. 5.1.3

Polls

In the example of this section we assume that all players observe the outcome of a poll before the election and that the pre-election action each voter can take is to express her intention to vote for a given candidate, if any. Thus, Xin = {a, b, e} for all n ∈ N and i ∈ In . The outcome of the poll is the number of those who did support some candidate (e.g., those who accepted to participate in the poll) and, for each candidate, the fraction of those that support it. Thus, for each n ∈ N, we let Γn = gn (X n ) ⊂ {0, . . . , n} × (∆(C) ∪ {(0, 0)}) and gn (x) = (gn1 (x), gn2 (x)), where gn1 (x) = |{i ∈ In : xi 6= e}| and     |{i∈In1:xi =a}| , |{i∈In1 :xi =b}| if gn1 (x) > 0, gn (x) gn (x) gn2 (x) =  (0, 0) if gn1 (x) = 0. 18

(3)

2 2 for all n ∈ N, i ∈ In and x ∈ X n . We write gn2 (x) = (gn,a (x), gn,b (x)).

Regarding each voter’s influence payoffs, we assume that 2 u˜ni (x, θ0 , θ) = gn,f (θ0 ) − ε1C (xi ),

(4)

for all n ∈ N, i ∈ In , x ∈ X n and θ, θ0 ∈ Θ, where 0 < ε < θ∗ . Thus, voters prefer to adopt a social ranking of candidates whose favorite candidate is the one who scores better in the poll. As in the example in Section 5.1.1, this assumption can be made less extreme by assuming that each voter has a bias towards keeping his favorite candidate of its individual ranking of candidates as the favorite candidate of his social ranking of candidates. As in the previous two examples, influence is costly but not too costly. However, this example differs from the first two in that there is public influence via the poll, i.e. the function gn . In terms of our main results, this will mean that the presence of a visible player is not longer needed. The assumption that relative influence matter is still needed and is, in fact, strengthened by requiring, roughly, that the public information is informative enough for voters to recognize when relative influence is sufficiently high. Given a sequence {Gn }∞ n=1 of election games with influence satisfying the assumption made in this section, Proposition 3 below establishes that, for all n sufficiently large, all symmetric sequential equilibria of Gn are strategic. 5.1.4

Campaign contributions

In our final example, we consider that each voter’s pre-election action consists of a monetary contribution to the campaign of each candidate. We assume that such contributions are bounded by m ∈ N so that, for each n ∈ N and i ∈ In , Xin = {(xi,a , xi,b ) ∈ {0, . . . , m}2 : xi,a + xi,b ≤ m}. The total amount contributed to each campaign is public information. To this P P end, let Γn = i∈In Xin and gn (x) = i∈In xi for all x ∈ X n . We write gn (x) = (gn,a (x), gn,b (x)) for all x ∈ X n . Each voter’s influence payoffs depends on his own contribution and on the total contribution for each campaign as follows: u˜ni (x, θ0 , θ) = ζθf (θ0 ) + βgn,f (θ0 ) (x) − xi,a − xi,b , 19

(5)

for all n ∈ N, i ∈ In , x ∈ X n and θ, θ0 ∈ Θ, where 1 − θ∗ < β < 1, 0 < ζ < β/θ∗ and θ∗ = minθ∈Θ |θa − θb |. As in the previous example, influence is costly but not too costly. Furthermore, relative influence matters. Thus, by Proposition 3 below, if {Gn }∞ n=1 is a sequence of election games with influence satisfying the assumptions made in this section, then, for all n sufficiently large, all symmetric sequential equilibria of Gn are strategic.

5.2

Sufficient conditions for strategic voting: Private influence

We start by presenting sufficient conditions for strategic voting when there is a visible player, as it is the case in the examples of Sections 5.1.1 and 5.1.2. A first condition is that exerting influence is costly, in the sense that, ceteris paribus, the influence payoff of each voter when he abstains to influence is higher (and bounded away) from his influence payoff when he actively influences. Condition 1 There exists ε > 0 such that, for all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e, u˜ni ((e, x−i ), θ0 , θ) − u˜ni (x, θ0 , θ) ≥ ε. Intuitively, this condition is required for a result stating that all symmetric equilibria are strategic because it bounds the fraction of those voters that influence and, thus, allows these voters to have an impact on the winner of the election. Indeed, it only makes sense to pay the cost of influence if this brings about some benefit in terms of the outcome of the election. While influence should be costly for all symmetric equilibria to be strategic, it cannot be too costly. Indeed, if exerting influence is too costly for each voter, then all voters abstain from influencing. If, in addition, each voter choose her social ranking of candidates equal to her individual one when no one influences (a plausible scenario), then we would be effectively in the setting of Proposition 1, where all symmetric equilibria of all sufficiently large election games are sincere. The condition requiring

20

that influence is not too costly formally requires each voter to have a pre-election action supporting some candidate (her favorite according to his individual ranking of candidates) such that, ceteris paribus, the loss of influence payoff of doing so relative to choosing to abstain from influencing is less than (and bounded away) from θ∗ , which is the maximum a voter can gain from the election. Condition 2 There exists η ∈ [ε, θ∗ ) such that, for all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e, u˜ni ((e, x−i ), θ0 , θ) − u˜ni ((x∗i , x−i ), θ0 , θ) ≤ η for some x∗i ∈ Xin (f (θ)). A third condition requires the existence of a visible player relative to the tightness of the election. For our purpose, the tightness of the election is measured as follows. Suppose that a has a priori more support in the sense that µ({θ ∈ Θ : f (θ) = a}) ≥ µ({θ ∈ Θ : f (θ) = b}); by assumption, we have x¯ > µ({θ ∈ Θ : f (θ) = a}) and, as discussed previously, sincere voting in large elections implies that the status quo wins. Suppose instead that a fraction δ > 0 of the voters vote for candidate a independently of their own type, whereas each one in the remaining fraction votes sincerely and let δ be such that δ+(1−δ)µ({θ ∈ Θ : f (θ) = a}) > x¯. In this case, a wins the election in all sufficiently large elections. Then δ 0 = inf{δ > 0 : δ+(1−δ)µ({θ ∈ Θ : f (θ) = a}) > x¯} is a measure of how easy it is to defeat the status quo, i.e. it is a measure of the tightness of the election. Our condition 3 requires that there is a player whose preelection action is seem by a fraction of voters strictly above δ 0 . Condition 3 There exists δ > 0 such that δ + (1 − δ) max{µ({θ ∈ Θ : f (θ) = a}), µ({θ ∈ Θ : f (θ) = b})} > x¯ and, for all n ∈ N, i ∈ In such that |{j ∈ In : i ∈ In (j)}|/n ≥ δ. Our forth condition requires that relative influence matters. For each voter i ∈ In , the relative influence for a is

|{j∈In \{i}:xj ∈Xjn (a)}| , |{j∈In \{i}:xj ∈Xjn (b)}|

21

i.e. it is the number of those voters

other than i who support a relative to the number of those who support b. Formally, the condition requires that if the relative influence for a is sufficiently high, then, ceteris paribus, choosing a social ranking of candidates θ0 that ranks a above b yields an higher (and bounded away) influence payoff than the one obtained by choosing a comparable social ranking θˆ that ranks b above a (comparable in the sense of

θa0 θb0

=

θˆb ). θˆa

Condition 4 There exists ψ > 0 such that, for all n ∈ N, i ∈ In , x ∈ X n , θ, θ0 , θˆ ∈ Θ, and c, c0 ∈ C with c 6= c0 , there exists M ∈ N such that: |{j ∈ In \ {i} : xj ∈ Xjn (c)}| > M |{j ∈ In \ {i} : xj ∈ Xjn (c0 )}|, θc0 > θc0 0 and

θˆc0 θc0 = θc0 0 θˆc

ˆ θ) + ψ. implies u˜ni (x, θ0 , θ) > u˜ni (x, θ, We note that we can always choose M = n in Condition 4, in which case |{j ∈ In \ {i} : xj ∈ Xjn (c)}| > M |{j ∈ In \ {i} : xj ∈ Xjn (c0 )}| is equivalent to |{j ∈ In \ {i} : xj ∈ Xjn (c)}| ≥ 1 and |{j ∈ In \ {i} : xj ∈ Xjn (c0 )}| = 0. Hence, to verify Condition 4, it suffices to consider x ∈ X n satisfying these latter conditions. We turn to the statement of our main result for election games with private influence. Given a sequence {Gn }∞ n=1 of election games with influence, we say that Gn is a private influence election game if Γn = {0}, gn (x) = 0 for all x ∈ X n and Conditions 1–4 hold. We say that {Gn }∞ n=1 is a sequence of election games with private influence if Gn is a private influence election game for all n ∈ N. Proposition 2 below states that, for all sufficiently large private influence games, all symmetric equilibria are strategic. Proposition 2 Let {Gn }∞ n=1 be a sequence of election games with private influence. Then there exists N ∈ N such that, if n ≥ N and (s, ν) is a symmetric sequential equilibrium of Gn , then s is strategic. Furthermore, there exists B > 0 such that, for all n ≥ N , µn ({θn ∈ Θn : a3i (s, θn ) 6= f (θin ) for some i ∈ In }) ≥ B. Proposition 2 provides an explanation for why voters vote strategically. In fact, in sufficiently large elections, all equilibria are strategic and, in fact, are such that 22

the probability of having some voter voting strategically is bounded away from zero. Together with Proposition 1, it forms an influence-based theory of strategic voting since it is the possibility of voters to influence each other before the election that changes the nature of voting from sincere to strategic. An intuition for Proposition 2 is as follows (see Section A.4 for its proof). Suppose that there exists an equilibrium (s, ν) where all voters vote sincerely. Given that no candidate has a large enough support to defeat the status quo (since µ({θ ∈ Θ : f (θ) = c}) < x¯ for each c ∈ C), then the status quo will be the outcome of the election with a probability close to one whenever n is large. This then implies that no voter influences under s is played. Indeed, if some voter were to choose to influence when s is played, then he could profitable deviate by not influencing as this would save the cost of doing so (as specified in Condition 1). Furthermore, although stoping to influence may change the outcome of the election, given that the status quo wins with a probability close to one and is the worst possible outcome, only with a small probability can this change result in a decrease in the payoff received from the outcome of the election. As a result of the above, any visible voter (in the sense of Condition 3) can change the outcome of the election and make one of the candidates win. In fact, by influencing for that candidate (say, candidate a) when no one else is influencing, such voter creates enough relative influence in favor of a and a gain for all other voters of choosing a social ranking of candidates where a is the top candidate (as specified in Condition 4). That this is indeed the optimal response of other voters will be established by showing that such gain is greater than any possible loss from the election payoff, which in turn holds because each voter affects the outcome of the election with a probability close to zero. Since influence is not too costly in the sense of Condition 2, then such visible voter has a profitable deviation from (s, ν). This contradiction to the assumption that (s, ν) is an equilibrium shows that any equilibrium must be strategic.

23

5.3

Sufficient conditions for strategic voting: Public influence

We next consider the case of a non-trivial public information. Whenever the public information is sufficiently informative, in the sense of gn distinguishing from preelection action profiles that are considerably different, there is no longer the need to assume the existence of a visible player. Thus, as Proposition 3 below states, the prevalence of strategic voting in large elections holds when Condition 3 is replaced by a condition on the informativeness of public information. The condition on the informativeness of public information we require is as follows: If a profile of pre-election actions x that conveys the same information of a preelection action profile x0 where one voter supports b, say, and all the other voters abstain from influencing (i.e. gn (x) = gn (x0 )), then the number of those who, in the profile x, choose to influence cannot exceed some fixed number m, independently of the population size n. Furthermore, if a social ranking of candidates θ0 is preferred to ˆ θ)), then the same is true when another θˆ when x0 is played (i.e. u˜i (x0 , θ0 , θ) > u˜i (x0 , θ, x is played. Condition 5 combines these requirements with those of Condition 4 on relative influence to yield a slightly weaker condition. In it statement, e¯ denotes the pre-election action profile (e, . . . , e) ∈ X n , i.e. the pre-election action profile where every voter abstains from influencing. Condition 5 There exist ψ > 0 and m ∈ N such that, for all n ∈ N, c ∈ C, i ∈ In , x ∈ X n and x0i ∈ Xin (c), gn (x) = gn (x0i , e¯−i ) implies that |{j ∈ In : xj 6= e}| ≤ m and ˆ θ) + ψ for all j ∈ In \ {i}, θ, θ0 , θˆ ∈ Θ and c0 ∈ C \ {c} such that u˜nj (x, θ0 , θ) > u˜nj (x, θ, θc0 > θc0 0 and

θc0 θc0 0

=

θˆc0 . θˆc

Given a sequence {Gn }∞ n=1 of election games with influence, we say that Gn is a public influence election game if Conditions 1, 2 and 5 hold. We say that {Gn }∞ n=1 is a sequence of election games with public influence if Gn is a public influence election game for all n ∈ N. Proposition 3 below states that, for all sufficiently large public influence games, all symmetric equilibria are strategic.

24

Proposition 3 Let {Gn }∞ n=1 be a sequence of election games with public influence. Then there exists N ∈ N such that, if n ≥ N and (s, ν) is a symmetric sequential equilibrium of Gn , then s is strategic. Furthermore, there exists B > 0 such that, for all n ≥ N , µn ({θn ∈ Θn : a3i (s, θn ) 6= f (θin ) for some i ∈ In }) ≥ B.

5.4

Mixed strategies

The contrasting nature of equilibria with or without influence relies, in part, on the ability of a single voter to impact on the outcome of the election. In fact, when influence is not allowed, this impact is very small in large elections. In contrast, when influence is allowed, this impact can be sufficiently large to change the winner of the election; this happens, in particular, when all voters but one choose not to influence. In this sense, allowing for influence transforms a large election into a small one. As a result of the above, existence of an equilibrium in pure strategies becomes an issue. More precisely, while existence of an equilibrium in pure strategies is not problematic in the second period, due to the presence of a large number of voters (Proposition 1 exemplifies this point), it is an issue in the first period whenever the number of voters choosing to influence is small. For this reason, we allow voters to use mixed strategies in this section and extend Propositions 2 and 3 to equilibria in which the second period strategy is pure but where the first period strategy is allowed to be mixed. For all finite sets F , ∆(F ) denotes the set of probability distributions over F . For all σ ∈ ∆(F ), let supp(σ) denote the support of σ, i.e. the set of x ∈ F such that σ(x) > 0. For all n ∈ N, a mixed strategy profile (a mixed strategy, for short) in Gn is σ = (σi1 , σi2 )i∈In such that (a) σi1 maps Θ into ∆(Xin ), (b) σi2 Q  n maps X × Γn × Θ into ∆(Θ × C) with σi2 (x, γ, θ) ∈ ∆(Yi (θ) × C) for all j j∈In (i) Q x ∈ j∈In (i) Xjn , γ ∈ Γn and θ ∈ Θ (strictly speaking, σ is a behavior strategy). We focus on mixed strategy equilibrium of Gn with the property that σi2 (x, γ, θ) is pure Q for all i ∈ In , x ∈ j∈In (i) Xjn , γ ∈ Γn and θ ∈ Θ. The set of such strategies is denoted 25

by Σn . Given σ ∈ Σn , we write σi2 = (s2i , s3i ) for all i ∈ In where, as in Section 3, s2i  Q Q n × Γn × Θ into Θ with s2i (x, γ, θ) ∈ Yi (θ) for all x ∈ j∈In (i) Xjn , maps X j j∈In (i)  Q n × Γn × Θ into C. X γ ∈ Γn and θ ∈ Θ, and s3i maps j j∈In (i) We turn to the mixed strategy version of Propositions 2 and 3. It requires the following notation and terminology. Given n ∈ N, σ ∈ Σn , θn ∈ Θn and x ∈ X n , let Q σ ˜ 1 (x|θn ) = i∈In σi1 (xi |θin ) be the probability of x occurring when σ is being played and θn describes players’ types. Furthermore, for all i ∈ In , let xi = (xj )j∈In (i) and a2i (σ, x, θn ) = s2i (xi , gn (x), θin ) and a3i (σ, x, θn ) = s3i (xi , gn (x), θin ). We say that a strategy σ ∈ Σn is sincere if σ ˜ 1 ({x ∈ X n : a3i (σ, x, θn ) = f (θin ) for all i ∈ In }|θn ) = 1 for all θn ∈ Θn . A strategy σ ∈ Σn is strategic if σ is not sincere. Having mixed strategies does not change the conclusion reached in Propositions 2 and 3 according to which, when influence is allowed, all equilibria of sufficiently large games have some players voting strategically. This follows from Lemma 1 below which shows that all non-pure equilibria must be strategic. The intuition for the result is as follows. Let θn ∈ Θn . If the probability of strategic voting is zero, then voting is sincere for all x ∈ supp(˜ σ 1 (·|θn )), i.e. a3i (σ, x, θn ) = f (θin ) for all i ∈ In . If σ 1 is not pure, there exists i ∈ In such that supp(σi1 (θin )) is not a singleton. A zero probability of strategic voting implies that the continuation is the same independently of the choice of the element in supp(σi1 (θin )). But then player i strictly prefers the element in supp(σi1 (θin )) with the lowest influence cost, i.e. not to influence. Lemma 1 states a stronger result, namely, that the probability of strategic voting is bounded away from zero. Its statement requires the following notation. Given a strategy σ ∈ Σn , let µn ⊗ σ 1 ∈ ∆(Θn × X n ) be defined by setting, for each θn ∈ Θn and x ∈ X n , 1

n

(µn ⊗ σ )(θ , x) = µn (θ)

n Y

σi1 (xi |θin ).

i=1

Lemma 1 Let {Gn }∞ n=1 be a sequence of election games with influence such that Gn satisfies Condition 1 for all n ∈ N. Then there exist N ∈ N and B 0 > 0 such that if 26

n ≥ N , σ ∈ Σn is such that (µn ⊗ σ 1 )({(θn , x) ∈ Θn × X n : a3i (σ, x, θn ) 6= f (θin ) for some i ∈ In }) < B 0 and (σ, ν) is a symmetric sequential equilibrium of Gn , then σ is a pure strategy. As a consequence of Lemma 1 and Propositions 2 and 3, we obtain that all symmetric sequential equilibria are strategic in sufficiently large elections. Proposition 4 Let {Gn }∞ n=1 be a sequence of election games with influence. Then there exists N ∈ N such that: 1. If Gk is a private influence election game for all k ∈ N, n ≥ N , σ ∈ Σn and (σ, ν) is a symmetric sequential equilibrium of Gn , then σ is strategic and there exists B ∗ > 0 such that, for all n ≥ N , (µn ⊗ σ 1 )({(θn , x) ∈ Θn × X n : a3i (σ, x, θn ) 6= f (θin ) for some i ∈ In }) ≥ B ∗ . 2. If Gk is a public influence election game for all k ∈ N, n ≥ N , σ ∈ Σn and (σ, ν) is a symmetric sequential equilibrium of Gn , then σ is strategic and there exists B ∗ > 0 such that, for all n ≥ N , (µn ⊗ σ 1 )({(θn , x) ∈ Θn × X n : a3i (σ, x, θn ) 6= f (θin ) for some i ∈ In }) ≥ B ∗ .

6

Further examples

In this section, we present an example of a symmetric equilibrium in a sequence of election games with influence. Furthermore, we give a simple example leading to influence preferences satisfying the sufficient conditions for strategic voting.

6.1

Symmetric equilibria with campaign contributions

We consider a simplified version of the setting with campaign contributions described in Section 5.1.4 and compute one symmetric equilibrium to illustrate our framework and results. Besides the assumptions made in that section, we assume m = 1, x¯ > 1/2, 27

¯ θ), (θ, θ)}, ¯ with θ¯ > θ, µ(θ) = 1/2 for each θ ∈ Θ and In (i) = {i} for all i ∈ In Θ = {(θ, and n ∈ N. Respectively, these additional assumptions mean that: (1) each voter’s total campaign contribution does not exceed 1, (2) each candidate needs more that ¯ θ) prefers 50% of the votes to win, (3) there are only two types, where type θa := (θ, ¯ prefers candidate b, (4) both types are equally likely candidate a and type θb := (θ, θ) and (5) each voter has no information about others’ campaign contribution other than that conveyed by the total contribution to each campaign. Let {Gn }∞ n=1 be the sequence of games just defined. The symmetric equilibrium we constructed is such that voters use the total campaign contributions to coordinate their votes. Specifically, each voter votes for the candidate with the highest support; furthermore, in case of a tie in total contributions, she votes sincerely. Thus,    a if γa > γb ,   s3i (x, γ, θ) = b if γa < γb ,     f (θ) otherwise. Each voter aligns her social ranking of candidates with her vote, i.e.    θa if γa > γb ,   s2i (x, γ, θ) = θb if γa < γb ,     θ otherwise. In the first period, each voter contributes to candidate a’s campaign when of type θa with probability p ∈ (0, 1) and contributes to candidate b’s campaign when of type θb with the same probability p; she chooses not to contribute with the remaining probability. Thus, for each i ∈ In , σi1 ((1, 0)|θa ) = σi1 ((0, 1)|θb ) = p and σi1 ((0, 0)|θa ) = σi1 ((0, 0)|θb ) = 1 − p. Let σ(p) be the strategy just defined and ν(p) the corresponding belief system (since p ∈ (0, 1), all information sets are reached with a strictly positive probability and, therefore, ν(p) is unique and computed using Bayes rule).

28

Note that σ(p) is symmetric in the sense of Section 3 since s3i ((0, 0), γ, θ) = s3j ((0, 0), γ, θ) for all i, j ∈ In , γ ∈ Γn and θ ∈ Θ. Furthermore, each voter’s second period strategy is pure and, hence, σ(p) ∈ Σn . Therefore, by Proposition 4, σ(p) must be strategic in order to be an equilibrium in some sufficiently large election. In particular, this implies that σ(p) can only be an equilibrium if p > 0 since, otherwise, voting would be sincere. Before turning to the properties of p, we argue that σ(p) specifies optimal choices of social rankings of candidates and votes. Indeed, if γ such that γa > γb is observed by voters, then each voter chooses θa as his social ranking of candidates and votes for a. Thus, a wins the election and continues to win even if one single voter deviates from σ(p) by voting b, provided that 1 − 1/n > x¯ (hence, for all sufficiently large n). Thus, voting for a is optimal and so is choosing θa to be his social ranking of candidates since γa > γb . Similarly, it is optimal for each voter to choose θb to be his social ranking of candidates and vote for b when γb > γa . When γa = γb , all voters vote sincerely and it is the case that one voter can affect the outcome of the election only with a small probability in large elections. Thus, ignoring this nearly negligible impact on the outcome of the election, failing to vote sincerely costs ζ(θ¯ − θ) > 0 if the voter aligns his social ranking of candidates with his vote and λ(θ¯ − θ) > 0 if he does not. Consequently, we obtain the following lemma (see Section A.9 for its proof). Lemma 2 There exists N ∈ N such that, for all n ≥ N , p ∈ (0, 1), x ∈ X n , θ ∈ Θ, i ∈ In and strategies σ ˜i for player i, uni (σ(p), ν(p)|θ, xi , gn (x)) ≥ uni (˜ σi , σ−i (p), ν(p)|θ, xi , gn (x)). It follows by Lemma 2 that, to show that σ is an equilibrium, it suffices to establish that there is no profitable unilateral deviation from σ 1 when s2 and s3 are as specified as above. Fixing s2 and s3 defines the following static game with incomplete information which has the property that there is no profitable unilateral deviation from σ 1 if σ 1 is one of its (mixed strategy) Bayes-Nash equilibrium: Each player’s

29

action set is {(0, 0), (1, 0), (0, 1)}, each player’s type space is Θ, the probability distribution on Θn is µn and each player’s payoff function is, for each i ∈ In , x ∈ Xn and θn ∈ Θn , vin (x, θn ) = ζθf (s2i (xi ,gn (x),θin )) + βgn,f (s2i (xi ,gn (x),θin )) (x) − xi,a − xi,b + λθ¯ + θa 1A (gn (x)) + θb 1B (gn (x)) + 1(A∪B)c (gn (x)) (θa 1A˜ (θn ) + θb 1B˜ (θn )) , where A = {γ ∈ Γn : γa > γb }, B = {γ ∈ Γn : γa < γb }, A˜ = {θn ∈ Θn : |{j ∈ In : ˜ = {θn ∈ Θn : |{j ∈ In : θjn = θb }| > n¯ x}. x} and B θjn = θa }| > n¯ In Proposition 5 below (its proof being presented in Section A.10) we first show that, for each n sufficiently large, there exists pn ∈ (0, 1) such that σ(pn ) is an equilibrium of Gn . In light of the above discussion, pn must be such that each voter of, say, type θa is indifferent between contributing to a’s campaign (i.e. choose (1, 0)) or abstain to contribute (i.e. choose (0, 0)) and that each of these two options is (weakly) better than to contribute to b’s campaign (i.e. choose (0,1)). For each n ∈ N and p ∈ [0, 1], let Fn (p) be the payoff difference between choosing (1, 0) and (0, 0). When no one contributes (p = 0 and the voter in question chooses (0, 0)), then voting is sincere and, since x¯ > 1/2 = µ(θa ) = µ(θb ), then, with a probability close to 1 in large elections, neither a nor b win. In this case, it pays to contribute to a’s campaign as this would make a be the most supported candidate and make a win the election. Thus, Fn (0) > 0. In contrast, when everyone else contributes (p = 1), either a or b win with a probability close to 1 which does not change much depending on whether the voter in question chooses (1, 0) or (0, 0). As a result, choosing (0, 0) is better as it saves the cost of contributing, i.e. Fn (1) < 0. Furthermore, a voter with type θa prefer to contribute to a’s campaign than to b’s campaign; in fact, the payoff difference of choosing (1, 0) and (0, 1) is proportional to θ¯ − θ. Hence, the existence of pn ∈ (0, 1) such that σ(pn ) is an equilibrium of Gn follows from the intermediate value theorem since this theorem implies that there exists pn ∈ (0, 1) such that Fn (pn ) = 0. Proposition 5 There exists N ∈ N and, for each n ≥ N , pn ∈ (0, 1) such that (σ(pn ), ν(pn )) is a symmetric equilibrium of Gn . Furthermore, if {pn }∞ n=1 is such that (σ(pn ), ν(pn )) is a symmetric equilibrium of Gn for all n sufficiently large, then: 30

1. limn→∞ pn = 0. 2. For all ε > 0, there exist N ∈ N and δ > 0 such that, for all n ≥ N , 1 − β < θ¯ < 1 − β + δ and 0 < ζ < δ, the probability that either a or b win the election under (σ(pn ), ν(pn )) is less than ε. 3. For all ε > 0, there exist N ∈ N and B > 0 such that, for all n ≥ N and θ¯ > B, the probability that either a or b win the election under (σ(pn ), ν(pn )) is greater than 1 − ε. Proposition 5 also establishes some asymptotic properties of σ(pn ). It shows that pn → 0, which means that each voter contributes with a probability converging to zero and, by the law of large numbers, that the fraction of those contributing is close to zero with a probability close to 1. Furthermore, in terms of the outcome of the election, it is possible that the probability that either a or b win be close to 1 but it is also possible, with different parameter values, that the probability that neither a nor b win be close to 1. In particular, influence in the form of campaign contributions can be very effective in coordinating voters in defeating the status quo. However, despite having all voters contributing to some of the two campaigns, such coordination effort may fail and, in fact, the status quo can win with a probability close to 1. Roughly, the intuition for the above asymptotic properties is as follows. The benefit for a voter of type θa of choosing to contribute to a’s campaign instead of not contributing (i.e. Fn (pn )) is that a wins when such voter chooses (1, 0) and both candidates receive equal support from the other voters; in contrast, were such voter to choose (0, 0), then voting would be sincere when both candidates receive equal support from the other voters and, in large elections, the probability that neither a nor b win is close to 1. Hence, when both candidates receive equal support from the ¯ Choosing (1, 0) instead of other voters, the voter in question gains approximately θ. (0, 0) costs (at least) 1−β regardless of relationship of each candidates’ total campaign contributions.8 Letting ρn denote the probability that both candidates receive equal 8

While is costs 1 monetary unit to contribute, contributing also increases by β a voter’s influence

payoff. Thus, effectively, the cost of contributing is only 1 − β.

31

support from the other voters, a rough upper bound on Fn (pn ) is ρn θ¯ − (1 − β) (the upper bound on Fn (pn ) we use in the proof differs from this but has the same limit ¯ properties). In equilibrium, we have that Fn (pn ) = 0 and, hence, ρn ≥ (1 − β)/θ. Thus, it must be that pn → 0 for the probability that both candidates receive equal support from the other voters not to converge to zero, i.e. if pn does not converge to zero, then some subsequence of ρn converges to zero. When θ¯ is close to 1 − β and ζ to zero, it can be shown that the probability that both candidates receive equal support from the other voters must indeed be close to 1, which makes the probability of either a or b winning the election be close to zero in large elections. In contrast, when θ¯ is large, the probability that both candidates receive equal support from the other voters will be close to zero (but will not converge to zero), and this makes the probability of either a or b winning the election be close to 1 in large elections.

6.2

Influence payoff: An example

In our formalization of the influence payoff function, the assumption that relative influence matter reflects a desire for social conformity with a majority of individuals. This preference for conformity can arise due to a desire for status as in Bernheim (1994) or fashion as in Karni and Schmeidler (1990). Here is an alternative economic story, based on Bramoull´e and Goyal (2013). Suppose that one (and only one) individual in In has an idea that he cannot use. He can pass it to any other individual, yielding the latter a (high) payoff of M > 0. Each individual has the idea with probability 1/n. If j has the idea, then he can pass it to only one other individual, who is interviewed by j. Each individual i 6= j is called to the interview with probability 1/(n − 1) and is asked to reveal his social ranking of candidates. Individual j passes the idea to i if and only if, for some c ∈ C, f (θi0 ) = c if xj ∈ Xjn (c), and f (θi0 ) = c if xj 6∈ Xjn (a) ∪ Xjn (b) and f (θj ) = c. For this choice of j to be well-defined, when the dependence of the election payoff on θ0 is interpreted as a preference for voting ethically, we may assume that either j observes i’s choice of the social ranking of candidates or that it is very costly for i to lie about his social ranking of candidates so that, in equilibrium, each voter truthfully 32

reveals his social ranking of candidates. A more natural interpretation of this example is to regard θ0 as the reported social ranking of candidates and interpreted the parameter λ in the election payoff as (being proportional to) the cost of lying about one’s voting intentions as follows: If a voter votes a and chooses to say that his social ¯ where θ¯ > θ, then his election payoff will be λθ plus ranking of candidates is (θ, θ), the term concerning the outcome of the election; in contrast, if he says that his social ¯ θ), then the election payoff is λθ¯ plus the term concerning ranking of candidates is (θ, the outcome of the election. Hence, the difference in election payoffs, λ(θ¯ − θ) > 0, can be interpreted as the cost of lying about his voting intentions. For simplicity, assume as in Section 6.1 that Θ = {θa , θb } and µ(θa ) = µ(θb ) = 1/2. Then, for each i ∈ In , x ∈ X n and θ ∈ Θ, u˜ni (x, θa , θ) =

M |{j 6= i : xj ∈ Xjn (a)}| + α(x−i ), n(n − 1)

u˜ni (x, θb , θ) =

M |{j 6= i : xj ∈ Xjn (b)}| + α(x−i ). n(n − 1)

and

where α(x−i ) =

M |{j 6= i : xj 6∈ Xjn (a) ∪ Xjn (b)}| . 2n(n − 1)

In particular, by adding an ε > 0 cost of choosing xi ∈ Xin (a) ∪ Xin (b), we obtain the influence payoff function in the example of Section 5.1.1 (up to α and a constant) for the case β = 0.

7

Concluding remarks

We have built a voting model where voters’ preferences have an ethical component as they depend on the vote casted and not just, as standard, on a material component arising from the outcome of an election. Voters’ ethical component of preferences is, our model, determined endogenously by voters’ pre-election actions, interpreted as actions taken to influence others. We then showed that allowing for influence considerably changes the nature of voting behavior in large elections. In particular,

33

when influence is allowed, voting behavior is largely determined by voters’ pre-election interaction. The above point has been illustrated by analysing whether voting is sincere or strategic in large elections. Since the weight given to the ethical component of voters’ preferences is smaller (and arbitrarily so if desired) than the weight given to material component, we say that a voter votes sincerely if she votes for her favorite candidate according to the latter component (otherwise the voter is said to vote strategically). We have then showed that, in large elections, all symmetric equilibria are strategic when influence is allowed and are sincere when influence is not allowed. Thus, in our model, strategic voting occurs if and only if influence is allowed. A more precise way of describing our strategic voting result is that we have provided conditions on pre-election actions (i.e. influence) for the above statement to hold. These conditions require influence to be costly (intuitively, to be exercised only when there is a small number of voters doing so) but not too costly (so that it is not the case that is never exercised). Furthermore, we also require that the relative influence received by the candidates matter for how voters determine their ethical component of preferences and that influence be visible enough so that such relative influence can be determined by voters. Furthermore, we have particularized our general framework to the case costly campaign contributions and have constructed symmetric strategic equilibria with the following properties: There is a small fraction of voters choosing to influence and, depending on some parameter values, such influence may or may not be successful in coordinating votes to defeat the status quo. This paper leaves several open questions. In particular, it would be interesting to know whether our strategic voting result (a) holds with alternative sufficient conditions, namely in terms of the influence payoff function (e.g. by weakening or dispensing with the assumption that relative influence matters) and (b) extends to richer setting, for instance, with more than two candidates or in a framework that encompasses several voting models in the literature. Furthermore, it would be interesting to analyse additional concrete examples of the model in this paper (possibly 34

with some variations) to address voting issues other than strategic voting.

35

A A.1

Online Appendix Notation

We introduce additional notation which is used throughout this appendix. 1. Recall that, for each finite set F , ∆(F ) denotes the set of all probability distributions over F . Given ν ∈ ∆(Θ) and σ : Θ → C, let ν ◦ σ −1 ∈ ∆(C) be defined by ν ◦ σc−1 = ν({ω ∈ Ω : σ(θ) = c}) for each c ∈ C. 2. Let S denote the set of functions σ : Θ → C. Since both Θ and C are finite, then S is also a finite set. 3. Let θ∗ = minθ∈Θ |θa − θb |. 4. Let µ = minθ∈Θ µ(θ). 5. For each n ∈ N and θn ∈ Θn , let τ (θn ) ∈ ∆(Θ) be defined by τθ (θn ) = |{i ∈ In : θin = θ}|/n for all θ ∈ Θ. 6. Let e¯ = (e, . . . , e) ∈ X n . Furthermore, for any x ∈ X n and i ∈ In , let xi = (xj )j∈In (i) . 7. For all n ∈ N and Y ⊆ Θn , Y c denotes the complement of Y in Θn .

A.2

Auxiliary results

The first lemma in this section deals with the following setting. Fix n ∈ N, m ∈ N and m + 1 distinct players, i0 , . . . , im ∈ In . For each 0 ≤ l ≤ m, player il has type θl ∈ Θ and votes for candidate cl ∈ C. The set of the remaining players, In \ {i0 , . . . , im }, is partitioned into two groups, In1 and In2 , such that all players in In1 vote according to σ ∈ S whereas all players in In2 vote according to σ 0 ∈ S. Let δn ∈ [0, 1] be such that |In1 | = nδn (and, hence, |In2 | = n(1 − δn ) − m − 1). Assume that, for the players in In1 ∪ In2 , types are identically and independently distributed with common distribution µ.

36

If the realized types of players other than i0 , . . . , im are θn−m−1 ∈ Θn−m−1 , then the empirical type-distribution of the players in In1 is given by τ 1 (θn−m−1 ) ∈ ∆(Θ) defined by τθ1 (θn−m−1 ) = |{l ∈ In1 : θln−m−1 = θ}|/nδn for all θ ∈ Θ. The empirical type-distribution of the players in In2 is τ 2 (θn−m−1 ) ∈ ∆(Θ) defined by τθ2 (θn−m−1 ) = |{l ∈ In2 : θln−m−1 = θ}|/[n(1 − δn ) − m − 1] for all θ ∈ Θ. Thus, the fraction of votes candidate c ∈ C receives is πc (θ

n−m−1

1

) := δn τ (θ

n−m−1

)◦σc−1 +(1−δn −(m+1)/n)τ 2 (θn−m−1 )◦σc0−1 +

Pm

1c (cl ) . n

l=0

For convenience, let z = (i0 , . . . , im , θ0 , . . . , θm , c0 , . . . , cm , σ, σ 0 ). We seek to define a high-probability subset Bzn−m−1 ⊆ Θn−m−1 of realizations of the types of players other than players i0 , . . . , im such that player i0 cannot affect the outcome of the election given (θ0 , . . . , θm , c0 , . . . , cm , σ, σ 0 ). This is achieved as follows. For any c ∈ C, let An−m−1 = {θn−m−1 ∈ Θn−m−1 : n|πc (θn−m−1 ) − x¯| ≤ 1}. z,c n−m−1 Furthermore, let An−m−1 = An−m−1 ∪ Az,b and Bzn−m−1 be the complement of z z,a

An−m−1 . z Let oz (θn−m−1 ) be the outcome of the election and, for any c 6= c0 , let oz,c (θn−m−1 ) be the outcome of the election when player i0 changes his vote from c0 to c. Lemma 3 For each γ > 0, m ∈ N and {δn }∞ n=1 ⊆ [0, 1], there exists N ∈ N such that ) < γ for all n ≥ N and z ∈ Inm+1 × Θm+1 × C m+1 × S 2 . µn−m−1 (An−m−1 z Furthermore, oz (θn−m−1 ) = oz,c (θn−m−1 ) for any z ∈ Inm+1 × Θm+1 × C m+1 × S 2 , θn−m−1 ∈ Bzn−m−1 and c 6= c0 . Proof. We first relabel players so that i0 = 1, . . . , im = m+1. Moreover, note that it suffices to show that for each γ > 0, m ∈ N, θ0 , . . . , θm ∈ Θ, , c0 , . . . , cm ∈ C and σ, σ 0 ∈ S, there exists Nz ∈ N, where z = (1, . . . , m + 1, θ0 , . . . , θm , c0 , . . . , cm , σ, σ 0 ), such that µn−m−1 (An−m−1 ) < γ for all n ≥ Nz . This is so because Θ, C and S are z finite. Hence, fix γ > 0, m ∈ N, θ0 , . . . , θm ∈ Θ, c0 , . . . , cm ∈ C and σ, σ 0 ∈ S. Note that it n−m−1 suffices to show that, for each c ∈ C, there is Nc ∈ N such that µn−m−1 (Az,c ) < γ/2

for all n ≥ Nc since then Nz = max{Na , Nb }. Fix c ∈ C. 37

Fix n ∈ N. Roughly, the idea is to show that if either In1 or In2 is large, then the conclusion follows. To this end, let rn = max{|In1 |, |In2 |} and note that rn ≥ (n − m − 1)/2. For concreteness, assume that rn = |In1 |; the case where rn = |In2 | is analogous. Define, for all k ∈ {1, . . . , rn }, Yk : Θrn → {0, 1} by Yk (θrn ) = 1 if σ(θkrn ) = c and Yk (θrn ) = 0 if σ(θkrn ) 6= c. Then, Y1 , . . . , Yrn are iid with mean µ◦σc−1 and variance α2 := µ◦σc−1 (1−µ◦σc−1 ). Furthermore, Y1 takes values on Z and there does not exist a1 and h1 > 1 such that Y1 takes values on {a1 + h1 k : k ∈ Z} with probability 1; indeed, a1 and h1 must be such that, for some k1 , k2 ∈ Z, a1 + k1 h1 = 1 and a1 + k2 h1 = 0, which implies h1 = 1/(k1 − k2 ) ≤ 1. In the terminology of Petrov (1975, p. 2), 1 is a maximal span of the distribution of Y1 . Furthermore, for each θrn ∈ Θrn ,

Prn

k=1

Yk (θrn ) = rn τ (θn−m−1 ) ◦ σc−1 . Thus,

if and only if θn−m−1 ∈ An−m−1 z,c m rn X X 1c (cl )] ≤ 1. Yk (θrn ) − [n¯ x − (n − rn − m − 1)τ (θn−m−1 ) ◦ σc0−1 − l=0

k=1

Note that

Prn

k=1

Yk (θrn ) ∈ {0, . . . , rn } and (n−rn −m−1)τ (θn−m−1 )◦σc0−1 ∈ {0, . . . , n−

rn − m − 1}. Thus, for each l ∈ {0, . . . , n − rn − m − 1}, there exists β(l) ∈ N such that

m rn X X rn Yk (θ ) − [n¯ x−l− 1c (cl )] ≤ 1 k=1

implies that

Prn

l=0

rn

k=1

Yk (θ ) ∈ {β(l) − 1, β(l), β(l) + 1}.

Let ϕ denote the density of the standard normal distribution. Let η > 0 and let √ N1 ∈ N be such that, for all n ≥ N1 , (ϕ(0) + η)/α rn < γ/6. It follows from the local central limit theorem (see Petrov (1975, Theorem 1, p. 187)) that there is N2 ∈ N such that, for all n ≥ N2 and all β ∈ Z, µrn ({θ

rn

rn

∈Θ

σ

ϕ( β−rαn√πrcn(µ) ) + η ϕ(0) + η ≤ . : Yk (θ ) = β}) < √ √ α rn α rn k=1 rn X

rn

Thus, define Nc = max{N1 , N2 }. Note that, for each θn−m−1 ∈ Θn−m−1 , letting θn−rn −m−1 be the vector of types of the players in In2 , then τ 2 (θn−m−1 ) = τ 2 (θn−rn −m−1 ). Hence, we will regard τ 2 as a 38

function on Θn−rn −m−1 . Then, for all n ≥ Nc , letting rn0 = n − rn − m − 1, we have that 0

µn−m−1 (An−m−1 ) z,c

≤

rn X

0

0

0

µrn0 ({θrn ∈ Θrn : rn0 τ (θrn ) ◦ σc0−1 = l}) ×

l=0

×µrn ({θrn ∈ Θrn :

rn X

Yk (θrn ) ∈ {β(l) − 1, β(l), β(l) + 1}}) < γ/2

k=1

since µrn ({θ

rn

rn

∈Θ

:

rn X

Yk (θrn ) ∈ {β(l) − 1, β(l), β(l) + 1}}) < 3γ/6 = γ/2.

k=1

This completes the proof of the first part. We turn now to the second part of the lemma. Let θn−m−1 ∈ Bzn−m−1 and c 6= c0 . We consider three cases: (1) πc (θn−m−1 ) > x¯, (2) πc0 (θn−m−1 ) > x¯ and (3) πc (θn−m−1 ) ≤ x¯ and πc0 (θn−m−1 ) ≤ x¯. In case (1), oz (θn−m−1 ) = c and we also have oz,c (θn−m−1 ) = c since πc (θn−m−1 ) + 1/n > x¯. In case (2), we have that oz (θn−m−1 ) = c0 and also that πc (θn−m−1 ) < x¯ (since x¯ ≥ 1/2). Since θn−m−1 ∈ Bzn−m−1 , then πc0 (θn−m−1 ) − 1/n > x¯ and πc (θn−m−1 ) + 1/n < x¯. Hence oz,c (θn−m−1 ) = c0 . Finally, in case (3) we have that oz (θn−m−1 ) is the status quo. We have that πc0 (θn−m−1 ) − 1/n < x¯ and hence oz,c (θn−m−1 ) 6= c0 . Since θn−m−1 ∈ Bzn−m−1 , then πc (θn−m−1 ) + 1/n < x¯. Hence oz,c (θn−m−1 ) is also the status quo. It then follows from the above three cases that oz (θn−m−1 ) = oz,c (θn−m−1 ), as desired. The following corollary of Lemma 3 will be used later. It assumes m = 1 and considers a set of types of all players other than player i0 such that player i0 has no impact on the election; i.e. the type of player i1 is no longer fixed. Players’ types are still independent with player i1 ’s type being distributed according to some ρ ∈ ∆(Θ) and, for any l ∈ In1 ∪In2 , player l’s type being distributed according to µ. Furthermore, player i1 votes according to σ ˆ ∈ S. Let w = (i0 , i1 , θ0 , ρ, c0 , σ ˆ , σ, σ 0 ) ∈ In2 × Θ × ∆(Θ) × C × S 3 and πc (θn−1 ) := δn τ 1 (θn−2 ) ◦ σc−1 + (1 − δn − 2/n)τ 2 (θn−2 ) ◦ σc0−1 + 39

1c (c0 ) + 1c (ˆ σ (θin−1 )) 1 . n

n−1 Define, analogously as before, Aw,c = {θn−1 ∈ Θn−1 : n|πc (θn−1 ) − x¯| ≤ 1} for any c

n−1 n−1 n−1 = (Aw ) . Define µρn−1 by µρn−1 (θn−1 ) = = An−1 c ∈ C, An−1 w,a ∪ Aw,b and Bw w Q ρ(θin−1 ) j6∈{i0 ,i1 } µ(θjn−1 ) for all θn−1 ∈ Θn−1 ; however, in the special case where 1

ρ = µ, we simply write µn−1 instead of µρn−1 . Also analogously as before, let ow (θn−1 ) be the outcome of the election and, for any c 6= c0 , let ow,c (θn−1 ) be the outcome of the election when player i0 changes his vote from c0 to c. Corollary 1 For each γ > 0 and {δn }∞ n=1 ⊆ [0, 1], there exists N ∈ N such that 2 3 µρn−1 (An−1 w ) < γ for all n ≥ N and w ∈ In × Θ × ∆(Θ) × C × S .

Furthermore, ow (θn−1 ) = ow,c (θn−1 ) for any w ∈ In2 × Θ × ∆(Θ) × C × S 3 , θn−1 ∈ Bwn−1 and c 6= c0 . Proof. Let γ > 0 and let N ∈ N be as in Lemma 3. Let also n ≥ N and w = (i0 , i1 , θ0 , ρ, c0 , σ ˆ , σ, σ 0 ) ∈ In2 × Θ × ∆(Θ) × C × S 3 . For each θ1 ∈ Θ, define z(θ1 ) = (i0 , i1 , θ0 , θ1 , c0 , σ ˆ (θ1 ), σ, σ 0 ). Then µρn−1 (An−1 w ) =

X

  ρ(θ1 )µn−2 An−2 z(θ1 ) < γ.

θ1 ∈Θ

Regarding the second part of the Corollary, let w = (i0 , i1 , θ0 , ρ, c0 , σ ˆ , σ, σ 0 ) ∈ In2 × n−1 Θ×∆(Θ)×C ×S 3 , θn−1 ∈ Bwn−1 and c 6= c0 be given. Let z = z(θin−1 ) and θn−2 = θ−i . 1 1

Then, we have that ow (θn−1 ) = oz (θn−2 ), ow,c (θn−1 ) = oz,c (θn−2 ) and θn−2 ∈ Bzn−2 . Hence, by Lemma 3, oz (θn−2 ) = oz,c (θn−2 ) and, therefore, ow (θn−1 ) = ow,c (θn−1 ). We present in this section a variation of Lemma 3 which will be used in the proof of Lemma 2. It consider the case where δn = 1 − (m + 1)/n and In2 = ∅. We require that m to be small relative to n so that n − m − 1 converges to infinity to obtain a statement that does not depend on m. Lemma 4 For each γ > 0, ξ > 0 and ξ 0 ∈ R, there exists N ∈ N such that µn−m−1 (An−m−1 ) < γ for all n ≥ N , m ∈ N such that n − m − 1 ≥ ξn + ξ 0 and z z ∈ Inm+1 × Θm+1 × C m+1 × S 2 . Furthermore, oz (θn−m−1 ) = oz,c (θn−m−1 ) for any z ∈ Inm+1 × Θm+1 × C m+1 × S 2 , θn−m−1 ∈ Bzn−m−1 and c 6= c0 . 40

Proof. The proof is analogous to the one of Lemma 3. Let γ > 0, ξ > 0 and ξ 0 ∈ R be given and, since S and C are finite, fix also σ, σ 0 ∈ S and c ∈ C. Define, for all k ∈ N, Yj : Θk → {0, 1} by Yj (θk ) = 1 if σ(θjk ) = c and Yk (θk ) = 0 if σ(θjk ) 6= c. Then, Y1 , . . . , Yk are iid with mean µ ◦ σc−1 and variance α2 := µ ◦ σc−1 (1 − µ ◦ σc−1 ). Let ϕ denote the density of the standard normal distribution. Let η > 0 and let √ N1 ∈ N be such that, for all n ≥ N1 , (ϕ(0) + η)/α ξn + ξ 0 < γ/6. As in the proof of Lemma 3, it follows from the local central limit theorem (see Petrov (1975, Theorem 1, p. 187)) that there is N2 ∈ N such that, for all n ≥ N2 and all β ∈ Z, σ

(µ) √c ϕ( β−kπ )+η ϕ(0) + η α k √ √ . µk ({θ ∈ Θ : Yj (θ ) = β}) < ≤ α k α k j=1 k

k

k X

k

Let N3 = (N2 − ξ 0 )/ξ and define N = max{N1 , N3 }. Let n, m ∈ N be such that n ≥ N and n − m − 1 ≥ ξn + ξ 0 . Then p √ (ϕ(0) + η)/α n − m − 1 ≤ (ϕ(0) + η)/α ξn + ξ 0 < γ/6 and n − m − 1 ≥ ξn + ξ 0 ≥ ξN3 + ξ 0 = N2 . Thus, letting rn = n − m − 1, we have that µrn ({θ

rn

rn

∈Θ

:

rn X

Yj (θrn ) = β}) <

j=1

ϕ(0) + η √ α rn

for each β ∈ Z. The remainder of the proof is now precisely analogous to that of Lemma 3.

A.3

Proof of Proposition 1

Recall that, as discussed in Section 4, if s is symmetric in Gn , with n ∈ N, then the voting behavior is described by a function σ : Θ → C. Furthermore, recall that Gn is effectively a static game with incomplete information and, thus, its set of sequential equilibrium strategies of Gn coincides with its set of Bayes-Nash equilibria. Let θ˜ = minθ∈Θ min{θa , θb } and recall that θ∗ = minθ∈Θ |θa − θb |. Define γ = ˜ > 0 and let N ∈ N be given by Corollary 1 corresponding to δn = 1 for λθ∗ /(λθ∗ + θ) all n ∈ N and γ. 41

We first establish the necessity part. Fix a non-sincere σ ∈ S. Since σ is nonsincere, let θ ∈ Θ be such that σ(θ) 6= f (θ). We will apply Corollary 1 to the case w = (1, 2, θ, µ, σ(θ), σ, σ, σ), i.e. the player who may change his action is player 1, and every player other than player 1 uses σ and has his type determined by µ (in particular, player 2 plays no special role). Hence, by Corollary 1, µn−1 (An−1 w ) < γ and ow (θn−1 ) = ow,c (θn−1 ) for all n ≥ N , θn−1 ∈ Bwn−1 and c ∈ C. Fix n ≥ N . The distribution of votes when all players vote according to σ is τ (θ, θn−1 ) ◦ σ −1 . We consider player 1 voting for f (θ) instead of σ(θ) and let π(θn−1 ) denote the resulting distribution on C when all the players other than player 1 vote according to σ and their types are given by θn−1 . To show that σ is not a Bayes-Nash equilibrium, it suffices to show that X

  µn−1 (θn−1 ) v(f (θ), π(θn−1 ), θ) − v(σ(θ), τ (θ, θn−1 ) ◦ σ −1 , θ) > 0.

θn−1 ∈Θn−1

For all θn−1 ∈ Bwn−1 , we have that ow (θn−1 ) = ow,f (θ) (θn−1 ) and, hence,
v(f (θ), π(θn−1 ), θ) − v(σ(θ), τ (θ, θn−1 ) ◦ σ −1 , θ) = λ θf (θ) − θσ(θ) ≥ λθ∗ . n−1 Furthermore, for all θn−1 ∈ An−1 ) = σ(θ) w , the worst case scenario is when ow (θ

and ow,f (θ) (θn−1 ) is the status quo. Thus, ˜ v(f (θ), π(θn−1 ), θ) − v(σ(θ), τ (θ, θn−1 ) ◦ σ −1 , θ) ≥ −θσ(θ) ≥ −θ. Therefore, X

  µn−1 (θn−1 ) v(f (θ), π(θn−1 ), θ) − v(σ(θ), τ (θ, θn−1 ) ◦ σ −1 , θ) >

θn−1 ∈Θn−1

> (1 − γ)λθ∗ − γ θ˜ = 0. Hence, σ is not a Bayes-Nash equilibrium as desired. We now establish the sufficiency part. To this end, consider player 1 (without loss of generality) and θ ∈ Θ. Let, similarly to above, w = (1, 2, θ, µ, f (θ), f, f, f ); n−1 Corollary 1 implies that µn−1 (Aw ) < γ and ow (θn−1 ) = ow,c (θn−1 ) for all n ≥ N ,

θn−1 ∈ Bwn−1 and c ∈ C. 42

Fix n ≥ N . The distribution of votes when all players vote according to f is τ (θ, θn−1 ) ◦ f −1 . We consider player 1 voting for c 6= f (θ) instead of f (θ) and let π(θn−1 ) denote the resulting distribution on C when all the players other than player 1 vote according to f and their types are given by θn−1 . As above, for all θn−1 ∈ Bwn−1 , we have that ow (θn−1 ) = ow,c (θn−1 ) and, hence,
v(f (θ), τ (θ, θn−1 ) ◦ f −1 , θ) − v(c, π(θn−1 ), θ) = λ θf (θ) − θc ≥ λθ∗ . n−1 , Furthermore, and also as above, for all θn−1 ∈ Aw

˜ v(f (θ), τ (θ, θn−1 ) ◦ f −1 , θ) − v(c, π(θn−1 ), θ) ≥ −θc ≥ −θ. Therefore, X

  µn−1 (θn−1 ) v(f (θ), τ (θ, θn−1 ) ◦ f −1 , θ) − v(c, π(θn−1 ), θ) >

θn−1 ∈Θn−1

> (1 − γ)λθ∗ − γ θ˜ = 0. Hence, f is a Bayes-Nash equilibrium as desired.

A.4

Proof of Proposition 2

We assume that µ ◦ fa−1 = max{µ ◦ fa−1 , µ ◦ fb−1 } for concreteness. Recall that θ∗ = maxθ∈Θ θa = maxθ∈Θ θb and θ∗ = minθ∈Θ |θa − θb |. Let ψ > 0 be given be Condition 4, 0 < γ < min{ψ/θ∗ , λθ∗ /θ∗ } and, for each n ∈ N, using Condition 3, let in ∈ In be such that |{j ∈ In : in ∈ In (j)}|/n ≥ δ, where δ > 0 is such that δ + (1 − δ)µ ◦ fa−1 > x¯. Since x¯ < 1, we may assume that δ < 1. Let dn be the smallest integer greater or equal to nδ and δn = dn /n. Then limn n(1 − δn ) = ∞ since dn − 1 < nδ for all n ∈ N implies that n(1 − δn ) = n − dn > n(1 − δ) − 1 → ∞. Let N1 ∈ N be given by Corollary 1 corresponding to {δn }∞ n=1 and γ. Let β > 0 be such that µ ◦ fa−1 + β < x¯ and δ + (1 − δ)µ ◦ fa−1 − β > x¯. Let ε and η be as in Conditions 1 and 2 respectively. Furthermore, let 0 < γ 0 < ε/θ∗ be such that η < (1 − γ 0 )θ∗ − γ 0 θ∗ . Recall that, for each n ∈ N and θn ∈ Θn , τ (θn ) ∈ ∆(Θ) is defined by τθ (θn ) = |{i ∈ In : θin = θ}|/n for all θ ∈ Θ. For all θ ∈ Θ, let ˜ n−1 = {θn−1 ∈ Θn−1 : |τθ (θ0 , θn−1 ) − µ(θ)| ≤ β/|Θ|}. B θ 43

˜ n−1 = ∩θ∈Θ B ˜ n−1 . It follows by Kalai (2004, Lemma 5) that Furthermore, let B θ ˜ n−1 )c ) ≤ 2|Θ|e−2 µn−1 ((B

(n(β/|Θ|)−1)2 n−1

.

Similarly, let  ˜ n(1−δn ) = θn(1−δn ) ∈ Θn(1−δn ) : |τθ (θn(1−δn ) ) − µ(θ)| ≤ β/|Θ| B θ ˜ n(1−δn ) = ∩θ∈Θ B ˜ n(1−δn ) . It then follows by Kalai (2004, Lemma 5) that and B θ 2

˜ n(1−δn ) )c ) ≤ 2|Θ|e−2( Θβ ) µn(1−δn ) ((B

n(1−δn )

.

˜ n−1 )c ) < γ 0 /2 and Hence, let N2 ∈ N be such that, for all n ≥ N2 , µn−1 ((B ˜ n(1−δn ) )c ) < γ 0 . µn(1−δn ) ((B Let N = max{N1 , N2 } and n ≥ N . Let (s, ν) be a symmetric sequential equilibrium of Gn and, in order to reach a contradiction, assume that s is such that µn ({θn ∈ Θn : ai (s, θn ) = f (θin ) for all i ∈ In }) > 1 −

µγ 0 , 2

where, recall, µ = minθ∈Θ µ(θ). Let E n = {θn ∈ Θn : ai (s, θn ) = f (θin ) for all i ∈ In } and, for each i ∈ In and n−1 θ ∈ Θ, let Ei,θ = {θ−i ∈ Θn−1 : (θ, θ−i ) ∈ E n }. We have that

µγ 0 X n−1 n−1 1− µ(θ)µn−1 (Ei,θ < ) ≤ µ min µn−1 (Ei,θ 0 ) + 1 − µ, 0 θ 2 θ∈Θ hence, n−1 min µn−1 (Ei,θ 0 ) > 1 − 0 θ

γ0 . 2

First note that, for all i ∈ In and θ ∈ Θ, player i’s beliefs after observing θ, denoted νi,θ , is a probability distribution over the other players’ types. Since the information set corresponding to the observation of θ is reached with strictly positive probability, then νi,θ is determined by Bayes’ rule and equals µn−1 . Indeed, for each θ−i ∈ Θn−1 , νi,θ (θ−i ) = P

µn (θ, θ−i )

θˆ−i ∈Θn−1

µn (θ, θˆ−i )

=

Y

µ(θj ) = µn−1 (θ−i ).

j6=i

Hence, it follows that ˜ n−1 ∩ E n−1 ) > 1 − γ 0 . νi,θ (B i,θ 44

˜ n−1 ∩ E n−1 , then the outcome of the election Furthermore, note that if θn−1 ∈ B i,θ when the other players’ types are θn−1 is the status quo, i.e. ow (θn−1 ) is the status quo when, in the notation of Corollary 1, w = (i, j, θ, µ, f (θ), f, f, f ) and i 6= j, i.e. player i has type θ and votes f (θ), and all other voters (including an arbitrary voter j) have their type determined by µ and vote according to f . Indeed, the distribution of votes is given by τ (θ, θn−1 ) ◦ f −1 and we have that, for each c ∈ C, X

τ (θ, θn−1 ) ◦ fc−1 =

τθˆ(θ, θn−1 ) <

ˆ ˆ θ∈Θ:f (θ)=c

X

≤

X

ˆ + µ(θ)

ˆ ˆ θ∈Θ:f (θ)=c

β ˆ ˆ = c}| |{θ ∈ Θ : f (θ) |Θ|

ˆ + β = µ ◦ f −1 + β < x¯. µ(θ) c

ˆ ˆ θ∈Θ:f (θ)=c

Claim 1 s1i (θ) = e for all i ∈ In and θ ∈ Θ. Proof of Claim 1.

Suppose not; then s1i (θ) 6= e for some i ∈ In and θ ∈ Θ.

Consider sˆi defined by sˆ1i (θ) = e, sˆ2i ((e, y), θ) = s2i ((s1i (θ), y), θ) and sˆ3i ((e, y), θ) = Q s3i ((s1i (θ), y), θ) for all y ∈ l∈In (i)\{i} Xl , and sˆi = si elsewhere. We next show that, given the belief system ν, the expected payoff for player i of the strategy (ˆ si , s−i ) is higher than that of s. Fix θn−1 ∈ Θn−1 and, for convenience, let θ0 = s2i (xi , θ) where x = a1 (s, (θ, θn−1 )). By definition, sˆ2i ((e, xi−i ), θ) = s2i (xi , θ) = θ0 . Condition 1 implies that u˜ni ((e, x−i ), θ0 , θ) − u˜ni (x, θ0 , θ) ≥ ε. Let π(θn−1 ) ∈ ∆(C) denote the distribution of votes when the type profile of players other than i is θn−1 and players play s, and π ˆ (θn−1 ) when players play (ˆ si , s−i ). ˜ n−1 ∩ E n−1 , then, by the argument above, the outcome of the election is If θn−1 ∈ B i,θ the status quo. Since the payoff received when the status quo is the outcome of the election is the lowest a player can receive (due to min{θa , θb } > 0), it then follows that, regardless of outcome of the election when players’ types are (θ, θn−1 ) and players play (ˆ si , s−i ), the resulting payoff is at least as high as the payoff resulting from the ˜ n−1 ∩ E n−1 , outcome of the election when players play s. Thus, for each θn−1 ∈ B i,θ v(f (θ), π ˆ (θn−1 ), θ0 , θ) − v(f (θ), π(θn−1 ), θ0 , θ) ≥ 0. 45

˜ n−1 ∩ E n−1 , a lower bound on the above difference of payoffs If, however, θn−1 6∈ B i,θ is obtained as follows. The worst case scenario is when ow (θn−1 ) = f (θ), i.e. player i’s favorite candidate wins if s is played, and the status quo is the outcome under (ˆ si , s−i ). In this case, there is a payoff change of −θf (θ) ≥ −θ∗ . Hence, for each ˜ n−1 ∩ E n−1 , θn−1 6∈ B i,θ v(c, π ˆ (θn−1 ), θ0 , θ) − v(c, π(θn−1 ), θ0 , θ) ≥ −θ∗ , where c = a3i (s, (θ, θn−1 )). ˜ n−1 ∩ E n−1 ) > 1 − γ 0 , then Since µn−1 (B i,θ uni (ˆ si , s−i , ν|θ) − uni (s, ν|θ) ≥ ε − γ 0 θ∗ > 0, a contradiction to the assumption that (s, ν) is a sequential equilibrium. This contradiction establishes that s1i (θ) = e for all i ∈ In and θ ∈ Θ, as claimed. Let x ∈ X be such that xin ∈ Xin (a) and xl = e for all l 6= in . Since s is symmetric, then there exists σ : Θ → C such that, for each j ∈ In such that in ∈ In (j), s3j (xj , θ) = σ(θ). Furthermore, for any j ∈ In such that in 6∈ In (j) and θˆ ∈ Θ, we have that ˆ = f (θ) ˆ (recall that xj = (xl )l∈I (j) for all j ∈ In and x ∈ X n and that s3j (¯ ej , θ) n e¯ = (e, . . . , e)). This can be seen as follows. Fix such j and θˆ and let θ−j ∈ Θn−1 ˆ = be such that θ−j ∈ Ej,n−1 ; since µn−1 (Ej,n−1 ) > 0, such θ−j exists. Hence, f (θ) θˆ θˆ n−1 j ˆ ˆ θ−j ))l∈I (j) , θ) ˆ = s3 (¯ s3j ((a1l (s, θ, n j e , θ), where the first equality follows from θ−j ∈ Ej,θˆ

and the second by Claim 1. Claim 2 σ(θ) = a. Proof of Claim 2. Fix j ∈ In \ {in } such that in ∈ In (j); since |{l ∈ In : in ∈ In (l)}| ≥ nδ such j exists. We first characterize player j’s beliefs after observing his Q type θ and xj ∈ l∈In (j) Xl , denoted by νj,θ,xj . Note that νj,θ,xj is a probability meaQ sure over Θn−1 × ( l∈In (j)c Xl ). Let {sk }∞ k=1 be a sequence of totally mixed strategies converging to s such that ν is the limit of the beliefs obtain from sk via Bayes’ rule. We have that s1,k l (θ) is the mixed action that player l plays in period 1 when of type 46

θ in the kth element of the sequence; we let the probability of action xl ∈ Xl being 9 played be s1,k ej , θ−j ) is l (xl |θ) for all k ∈ N, l ∈ In , θ ∈ Θ and xl ∈ Xl . Then νj,θ,xj (¯

obtained as follows. Let dk =

X

Y

µ(θˆl )s1,k l (e|θl )

ˆ |In (j)−2| l∈In (j)\{j,in } θ∈Θ 1 and note that limk dk = 1 since s1,k l (e|θl ) → sl (e|θl ) = 1 for all l ∈ In and θl ∈ Θ.

Then Q 1,k µ(θ)sj1,k (e|θ)µ(θin )s1,k in (xin |θin ) l6=j,in µ(θl )sl (e|θl ) P  νj,θ,xj (¯ e , θ−j ) = lim = 1,k k ˆ ˆ µ(θ)s1,k (e|θ) µ( θ)s (x | θ) d ˆ in k j in θ∈Θ ! Y µ(θin )s1,k in (xin |θin ) = lim P µ(θˆl ). 1,k ˆ ˆ k (x | θ) µ( θ)s ˆ in in θ∈Θ l6=j,in j

Let, for each θ˜ ∈ Θ, ˜ 1,k (xin |θ) ˜ µ(θ)s in ˜ ρ(θ) = lim P . ˆ ˆ 1,k (xin |θ) k µ(θ)s ˆ in θ∈Θ Then, in the notation of Corollary 1, νj,θ,xj (¯ ej , θ−i ) = µρn−1 (θ−i ). Furthermore, since P ρ ¯j . θ−i ∈Θn−1 µn−1 (θ−i ) = 1, then νj,θ,xj (y, θ−i ) = 0 for all y 6= e Suppose, in order to reach a contradiction, that σ(θ) 6= a; then σ(θ) = b. Let θ0 = s2j (xj , θ) and consider two cases. The first case is when θa0 > θb0 . In this case, let sˆ3j (xj , θ) = a and sˆj = sj elsewhere. Note that, when s is played, players in {j 0 ∈ In : in ∈ In (j 0 )} vote according to σ, while the remaining players vote according to f . Let, in the notation of Corollary 1, w = (j, in , θ, ρ, σ(θ), σ, σ, f ) with In1 = {j 0 ∈ In : in ∈ In (j 0 )} \ {j, in }, i.e. voter j has type θ and votes σ(θ), voter in ’s type is determined by ρ and he votes according to σ, voters in In1 vote according to σ and those in In2 = In \ (In1 ∪ {j, in }) vote according to f . Hence, Corollary 1 implies that µρn−1 (Bwn−1 ) > 1 − γ and recall that player j does not affect the outcome of the election whenever θn−1 ∈ Bwn−1 . If, instead, θn−1 6∈ Bwn−1 , then the worst it can happen is that f (θ) is elected when s is played whereas the status quo is elected when (ˆ sj , s−j ) is played. Hence, unj (ˆ sj , s−j , ν|θ, xj ) − unj (s, ν|θ, xj ) ≥ λ(θa0 − θb0 ) − γθ∗ ≥ λθ∗ − γθ∗ > 0, 9

The assumption that sk is totally mixed implies that s1,k l (xl |θ) > 0 and the assumption that sk

1 converges to s implies that limk s1,k l (xl |θ) = sl (xl |θ).

47

a contradiction. The second case is when θa0 < θb0 . Recall that t(θ0 ) = (θb0 , θa0 ). Since |{l ∈ In (j) \ {j} : xl ∈ Xln (a)}| = 1 and |{l ∈ In (j) \ {j} : xl ∈ Xln (b)}| = 0, Condition 4 implies that u˜nj (x, t(θ0 ), θ) > u˜nj (x, θ0 , θ) + ψ. Let sˆj defined by sˆ2j (xj , θ) = t(θ0 ), sˆ3j (xj , θ) = a and sˆj = sj elsewhere. As in the first case above, player j loses at most θ∗ due to a change in the outcome of the election when (ˆ sj , s−j ) is played instead of s but only when θn−1 6∈ Bwn−1 . Because sˆ2j (xj , θ) = t(θ0 ), we obtain that unj (ˆ sj , s−j , ν|θ, xj ) − unj (s, ν|θ, xj ) ≥ ψ − γθ∗ > 0, a contradiction. This contradiction and the one above establish that σ(θ) = a as claimed. We next show that player in has a profitable deviation when his type is θ such e, θ). First, recall that the status quo wins for all θn−1 ∈ that θa = θ∗ . Let θ0 = s2in (¯ ˜ n−1 ∩ E n−1 . Hence, B in ,θ unin (s, ν|θ) = u˜nin (¯ e, θ0 , θ) + v(a, τ (θ, θn−1 ) ◦ f −1 , θ0 , θ) < u˜nin (¯ e, θ0 , θ) + λθa0 + γ 0 θ∗ . Consider x∗in ∈ Xi∗n (a) as in Condition 2 and let x = (x∗in , e¯−in ). Let sˆin defined by sˆ1in (θ) = x∗in , sˆ2in (xin , θ) = θ0 and sˆin = sin elsewhere. e, θ0 , θ) − η. By Claim 2, we have that, Condition 2 implies that u˜nin (x, θ0 , θ) ≥ u˜nin (¯ ˆ = σ(θ) ˆ = a. Furthermore, for all j ∈ In \{in } such that in ∈ In (j) and θˆ ∈ Θ, s3j (xj , θ) since s is symmetric, sˆ3in (xin , θ) = s3in (xin , θ) = σ(θ) = a. As we have shown above, ˆ = f (θ) ˆ for any j ∈ In such that in 6∈ In (j) and θˆ ∈ Θ. we also have that s3j (¯ ej , θ) Let π ˆ (θn−1 ) be the distribution of votes for each θn−1 ∈ Θn−1 when (ˆ sin , s−in ) is played. For convenience, let Iˆn (in ) be the set of j ∈ In such that in ∈ In (j) and, renaming players if needed, assume that Iˆn (in )c ⊆ {1, . . . , n(1 − δn )}. Then, it follows

48

˜ n(1−δn ) . In fact, that π ˆa (θn−1 ) > x¯ (i.e. a wins the election) when θn(1−δn ) ∈ B π ˆa (θ

n−1

ˆ |{j ∈ Iˆn (in )c : θjn−1 = θ}| n

X |Iˆn (in )| ) = + n ˆ ˆ

θ∈Θ:f (θ)=a n(1−δn )

≥ δn + (1 − δn )

|{j ∈ {1, . . . , n(1 − δn )} : θj n(1 − δn )

X ˆ ˆ θ∈Θ:f (θ)=a

≥ δ + (1 − δ)

X

ˆ = θ}|

ˆ −β µ(θ)

ˆ ˆ θ∈Θ:f (θ)=a

= δ + (1 − δ)µ ◦ fa−1 − β > x¯. Hence, X

sin , s−in , ν|θ) = u˜nin (x, θ0 , θ) + unin (ˆ >

e, θ0 , θ) u˜nin (¯

−

µn−1 (θn−1 )v(a, π ˆ (θn−1 ), θ0 , θ)

θn−1 ∈Θn−1 η + λθa0 +

(1 − γ 0 )θ∗ .

It then follows that unin (ˆ sin , s−in |ν) − unin (s|ν) ≥ −η + (1 − γ 0 )θ∗ − γ 0 θ∗ > 0. Hence, player in has a profitable deviation, contradicting the assumption that (s, ν) is a sequential equilibrium. This contradiction establishes that µn ({θn ∈ Θn : ai (s, θn ) = f (θin ) for all i ∈ In }) ≤ 1 −

µγ 0 , 2

and, hence, n

n

n

µn ({θ ∈ Θ : ai (s, θ ) 6= Thus, to conclude the proof, let B =

A.5

f (θin )

µγ 0 for some i ∈ In }) ≥ . 2

µγ 0 . 2

Proof of Proposition 3

We assume that µ ◦ fa−1 = max{µ ◦ fa−1 , µ ◦ fb−1 } for concreteness. Recall that θ∗ = maxθ∈Θ θa = maxθ∈Θ θb and θ∗ = minθ∈Θ |θa − θb |. Let 0 < γ < min{ψ/θ∗ , λθ∗ /θ∗ } and let m ∈ N be as in Condition 5. Let δn = (n − m)/n for all n ∈ N and N1 ∈ N be given by Corollary 1 corresponding to {δn }∞ n=1 and γ. 49

Let β > 0 be such that µ ◦ fa−1 + β < x¯. Let ε and η be as in Conditions 1 and 2 respectively. Furthermore, let 0 < γ 0 < ε/θ∗ be such that η < θ∗ − γ 0 θ∗ . For all θ ∈ Θ, let ˜ n−1 = {θn−1 ∈ Θn−1 : |τθ (θ0 , θn−1 ) − µ(θ)| ≤ β/|Θ|}. B θ ˜ n−1 = ∩θ∈Θ B ˜ n−1 . It follows by Kalai (2004, Lemma 5) that Furthermore, let B θ ˜ n−1 )c ) ≤ 2|Θ|e−2 µn−1 ((B

(n(β/|Θ|)−1)2 n−1

. Hence, let N2 ∈ N be such that, for all n ≥ N2 ,

˜ n−1 )c ) < γ 0 /2. µn−1 ((B Let N = max{N1 , N2 } and n ≥ N . Let (s, ν) be a symmetric sequential equilibrium of Gn and, in order to reach a contradiction, assume that s is is such that µn ({θn ∈ Θn : ai (s, θn ) = f (θin ) for all i ∈ In }) > 1 −

µγ 0 , 2

where, recall, µ = minθ∈Θ µ(θ). Let E n = {θn ∈ Θn : ai (s, θn ) = f (θin ) for all i ∈ In } and, for each i ∈ In and n−1 θ ∈ Θ, let Ei,θ = {θ−i ∈ Θn−1 : (θ, θ−i ) ∈ E n }. As in the proof of Proposition 2, we

˜ n−1 ∩ E n−1 ) > 1 − γ 0 and that, θn−1 ∈ B ˜ n−1 ∩ E n−1 , then the outcome have that νi,θ (B i,θ i,θ of the election when the other players’ types are θn−1 is the status quo, i.e. ow (θn−1 ) is the status quo when w = (i, j, θ, µ, f (θ), f, f, f ) and i 6= j. Claim 3 s1i (θ) = e for all i ∈ In and θ ∈ Θ. The proof of Claim 3 is analogous to the proof of Claim 1. Let x ∈ X n be such that x1 ∈ X1 (a) and xl = e for all l 6= 1. Since s is symmetric, then there exists σ : Θ → C such that, for each j ∈ In such that 1 ∈ In (j), s3j (xj , gn (x), θ) = σ(θ). Furthermore, there exists σ 0 : Θ → C such that, for each j ∈ In such that 1 6∈ In (j), s3j (¯ ej , gn (x), θ) = σ 0 (θ). Claim 4 If there exists j ∈ In \ {1} such that 1 ∈ In (j), then σ(θ) = a. Proof of Claim 4.

This proof is analogous to the proof of Claim 2 with the

following modifications.

50

First, note that νj,θ,xj ,gn (x) is a probability measure over Θn−1 × A, where A = Q {y ∈ l∈In (j)c Xln : gn (xj , y) = gn (x)}. Second, the formula for νj,θ,xj ,gn (x) (¯ ej , θ−j ) is the same as in the proof of Claim 2 except that now dk =

X X Y

Y

ˆ µ(θˆl )s1,k l (yl |θl )

ˆ n−1 y∈A l6∈In (j) θ∈Θ

ˆ µ(θˆl )s1,k l (e|θl );

l∈In (j)\{j,1}

note, however, that we still have dk → 1. Third, and finally, we use Condition 5 instead of Condition 4 in the case where θa0 < θb0 . Claim 5 If there exists j ∈ In \ {1} such that 1 ∈ In (j), then σ 0 (θ) = a. Proof of Claim 5.

Fix j ∈ In such that 1 6∈ In (j). We start by compute

player j’s beliefs. Let {sk }∞ k=1 be a sequence of totally mixed strategies converging to s such that ν is the limit of the beliefs obtain from sk via Bayes’ rule. Then, for Q each y ∈ l∈In (j)c Xln such that gn (xj , y) = gn (x) and θ−j ∈ Θn−1 , νj,θ,xj ,gn (x) (y, θ−j ) is obtained as follows. Let     Y Xln : gn (xj , y 0 ) = gn (x) A = y0 ∈   c l∈In (j)

and dk =

X X Y

0 ˆ µ(θˆl )s1,k l (yl |θl )

ˆ n−1 y 0 ∈A l6∈In (j) θ∈Θ

Y

ˆ µ(θˆl )s1,k l (e|θl );

l∈In (j)\{j}

then Q νj,θ,xj ,gn (x) (y, θ−j ) = lim

l6∈In (j)

µ(θl )s1,k l (yl |θl )

Q

l∈In (j)\{j}

µ(θl )s1,k l (e|θl )

dk

k

.

Fix y ∈ A and θ−j ∈ Θn−1 such that νj,θ,xj ,gn (x) (y, θ−j ) > 0. By Condition 5, there exist i1 , . . . , im ∈ In (j)c such that yl = e for all l 6∈ {i1 , . . . , im }. Then, as we next show, νj,θ,xj ,gn (x) (θn−m−1 |y, θi1 , . . . , θim ) = µn−m−1 (θn−m−1 ).

51

In fact, νj,θ,xj ,gn (x) (θn−m−1 |y, θi1 , . . . , θim ) = P

νj,θ,xj ,gn (x) (y, θi1 , . . . , θim , θn−m−1 )

θˆn−m−1 ∈Θn−m−1

Qm

νj,θ,xj ,gn (x) (y, θi1 , . . . , θim , θˆn−m−1 )

1,k µ(θil )s1,k il (yil |θil ) l∈In \{j,i1 ,...,im } µ(θl )sl (e|θl ) = lim P Q Qm 1,k ˆ 1,k ˆ k l∈In \{j,i1 ,...,im } µ(θl )sl (e|θl ) l=1 µ(θil )sil (yil |θil ) θˆn−m−1 ∈Θn−m−1 Q 1,k l∈In \{j,i1 ,...,im } µ(θl )sl (e|θl ) = lim P Q ˆ 1,k ˆ k n−m−1 n−m−1 ˆ l∈In \{j,i1 ,...,im } µ(θl )sl (e|θl ) θ ∈Θ Y = µ(θl ), l=1

Q

l∈In \{j,i1 ,...,im }

˜ ˜ the last equality holding because limk s1,k l (e|θ) = 1 for all l ∈ In and θ ∈ Θ by Claim P Q 3 and because θˆn−m−1 ∈Θn−m−1 l∈In \{j,i1 ,...,im } µ(θˆl ) = 1. Suppose, in order to reach a contradiction, that σ 0 (θ) 6= a; then σ 0 (θ) = b. Let θ0 = s2j (xj , gn (x), θ) and consider two cases. The first case is when θa0 > θb0 . In this case, let sˆ3j (xj , gn (x), θ) = a and sˆj = sj elsewhere. For each y ∈ A, let i1 (y), . . . , im (y) ∈ In (j)c be such that yl = e for all l 6∈ {i1 , . . . , im }. Note first that, for each y ∈ A, the influence payoff is the same under s and (ˆ sj , s−j ) since in both cases it equals u˜nj ((xj , y), θ0 , θ). Furthermore, since νj,θ,xj ,gn (x) (θn−m−1 |y, θi1 (y) , . . . , θim (y) ) = µn−m−1 (θn−m−1 ) for all y ∈ A, θi1 (y) , . . . , θim (y) ∈ Θm and θn−m−1 ∈ Θn−m−1 , we have that unj (ˆ sj , s−j , ν|θ, xj , gn (x)) − unj (s, ν|θ, xj , gn (x)) = X X νj,θ,xj ,gn (x) (y, θi1 (y) , . . . , θim (y) )× (6)

y∈A θi1 (y) ,...,θim (y)

X

 µn−m−1 (θn−m−1 ) v(a, π ˆ (θn−m−1 ), θ0 , θ) − v(b, π(θn−m−1 ), θ0 , θ)

! 

,

θn−m−1 ∈Θn−m−1

where, given y ∈ A and θi1 (y) , . . . , θim (y) ∈ Θm , π(θn−m−1 ) ∈ ∆(C) is the distribution of votes when s is played and π ˆ (θn−m−1 ) ∈ ∆(C) is the distribution of votes when (ˆ sj , s−j ) is played, both when the type profile of those players in the complement of {j, i1 (y), . . . , im (y)} is θn−m−1 . Note that, when s is played, players in {j 0 ∈ In : 1 ∈ In (j 0 )} vote according to σ, while the remaining players vote according to σ 0 . Fix y ∈ A and θi1 (y) , . . . , θim (y) ∈ Θm 52

and let z = (j, i1 (y), . . . , im (y), θ, θi1 (y) , . . . , θim (y) , σ 0 (θ), σ(θi1 (y) ), . . . , σ(θim (y) ), σ 0 , σ) with In1 = {j 0 ∈ In : 1 ∈ In (j 0 )}\{j}. Hence, Lemma 3 implies that µn−m−1 (Bzn−m−1 ) > 1 − γ and recall that player j does not affect the outcome of the election whenever θn−m−1 ∈ Bzn−m−1 . If, instead, θn−m−1 6∈ Bzn−m−1 , then the worst it can happen is that f (θ) is elected when s is played whereas the status quo is elected when (ˆ sj , s−j ) is played. Hence, X

  µn−m−1 (θn−m−1 ) v(a, π ˆ (θn−m−1 ), θ0 , θ) − v(b, π(θn−m−1 ), θ0 , θ)

θn−m−1 ∈Θn−m−1 ≥ λ(θa0 − θb0 ) −

γθ∗

and, therefore, unj (ˆ sj , s−j , ν|θ, xj , gn (x)) − unj (s, ν|θ, xj , gn (x)) ≥ λ(θa0 − θb0 ) − γθ∗ ≥ λθ∗ − γθ∗ > 0, a contradiction. The second case is when θa0 < θb0 . Then Condition 5 implies that, for all y ∈ A, u˜nj ((xj , y), t(θ0 ), θ) > u˜nj ((xj , y), θ0 , θ) + ψ. Let sˆj defined by sˆ2j (xj , gn (x), θ) = t(θ0 ), sˆ3j (xj , gn (x), θ) = a and sˆi = si elsewhere. Fix y ∈ A and θi1 (y) , . . . , θim (y) ∈ Θm and let z = (j, i1 (y), . . . , im (y), θ, θi1 (y) , . . . , θim (y) , σ 0 (θ), σ(θi1 (y) ), . . . , σ(θim (y) ), σ 0 , σ) with In1 = {j 0 ∈ In : 1 ∈ In (j 0 )} \ {j}. As in the first case above, player j loses at most θ∗ due to a change in the outcome of the election when (ˆ sj , s−j ) is played instead of s but this can happen only when θn−m−1 6∈ Bzn−m−1 . Because sˆ2j (xj , gn (x), θ) = t(θ0 ), we obtain, using (6) above, that unj (ˆ sj , s−j , ν|θ, xj , gn (x)) − unj (s, ν|θ, xj , gn (x)) ≥ ψ − γθ∗ > 0, a contradiction. This contradiction and the one above establish that σ 0 (θ) = a as claimed. We next show that player 1 has a profitable deviation when his type is θ such that θa = θ∗ . Let θ0 = s21 (¯ e, gn (¯ e), θ). First, recall that the status quo wins for all ˜ n−1 ∩ E n−1 . Hence, θn−1 ∈ B 1,θ un1 (s, ν|θ) < u˜n1 (¯ e, θ0 , θ) + λθa0 + γ 0 θ∗ . 53

Consider x∗1 ∈ X1∗ (a) as in Assumption 2 and let x = (x∗1 , e¯−1 ). Let sˆ1 defined by sˆ11 (θ) = x∗1 , sˆ21 (x1 , gn (x1 ), θ) = θ0 , sˆ31 (x1 , gn (x1 ), θ) = a and sˆ1 = s1 elsewhere. e, θ0 , θ) − η. By Claims 4 and 5, we have Condition 2 implies that u˜n1 (x, θ0 , θ) ≥ u˜n1 (¯ ˆ = a. that, for all j ∈ In \ {1} and θˆ ∈ Θ, s3j (xj , θ) Letting π ˆ (θn−1 ) be the distribution of votes for each θn−1 ∈ Θn−1 when (ˆ s1 , s−1 ) is played, it follows that π ˆa (θn−1 ) = 1 (i.e. a wins the election) for all θn−1 ∈ Θn−1 . Hence, X

un1 (ˆ s1 , s−1 , ν|θ) = u˜n1 (x, θ0 , θ) + >

u˜n1 (¯ e, θ0 , θ)

−

µn−1 (θn−1 )v(a, π ˆ (θn−1 ), θ0 , θ)

θn−1 ∈Θn−1 η + λθa0 +

θ∗ .

It then follows that u1 (ˆ s1 , s−1 |ν) − u1 (s|ν) ≥ −η + θ∗ − γ 0 θ∗ > 0. Hence, player 1 has a profitable deviation, contradicting the assumption that (s, ν) is a sequential equilibrium. This contradiction establishes that n

n

n

µn ({θ ∈ Θ : ai (s, θ ) =

f (θin )

µγ 0 for all i ∈ In }) ≤ 1 − , 2

and, hence, µn ({θn ∈ Θn : ai (s, θn ) 6= f (θin ) for some i ∈ In }) ≥ Thus, to conclude the proof, let B =

A.6

µγ 0 . 2

µγ 0 . 2

Proof of Lemma 1

We assume that µ ◦ fa−1 = max{µ ◦ fa−1 , µ ◦ fb−1 } for concreteness. Recall that θ∗ = maxθ∈Θ θa = maxθ∈Θ θb . Let β > 0 be such that µ ◦ fa−1 + β < x¯. Let ε > 0 be as in Assumption 1 and let 0 < γ 0 < ε/θ∗ . For all θ ∈ Θ, let ˜ n−1 = {θn−1 ∈ Θn−1 : |τθ (θ0 , θn−1 ) − µ(θ)| ≤ β/|Θ|}. B θ ˜ n−1 = ∩θ∈Θ B ˜ n−1 . Then there exists N ∈ N be such that, for all Furthermore, let B θ ˜ n−1 )c ) < γ 0 /2. n ≥ N , µn−1 ((B 54

Let n ≥ N . Define B 0 =

µγ 0 . 2

Let (σ, ν) be a symmetric sequential equilibrium of

Gn such that σ belongs to Σn and be such that (µn ⊗ σ 1 )({(θn , x) ∈ Θn × X n : ai (σ, x, θn ) 6= f (θin ) for some i ∈ In }) < B 0 . Let Eˆ n = {(θn , x) ∈ Θn × X n : ai (σ, x, θn ) = f (θin ) for all i ∈ In } and, for each n−1 n i ∈ In , θ ∈ Θ and xi ∈ Xin , let Eˆi,θ,x = {(θ−i , x−i ) ∈ Θn−1 ×X−i : ((θ, θ−i ), (xi , x−i )) ∈ i

Eˆ n }. We have that 1−

X µγ 0 n−1 n−1 1 )(Eˆi,θ,x < ) ≤ µ min µ(θ)σi1 (xi |θ)(µn−1 ⊗ σ−i µn−1 (Eˆi,θ 0 ,x0 ) + 1 − µ, i i θ0 ,x0i 2 (θ,xi )

hence, n−1 min µn−1 (Ei,θ 0 ,x0 ) 0 0 i θ ,xi

γ0 >1− . 2

n ˆ n−1 = B ˜ n−1 × X−i Recall that, for all i ∈ In and θ ∈ Θ, νi,θ = µn−1 . Letting B , we 1 ˆ n−1 ∩ Eˆ n−1 ) > 1 − γ 0 for each xi ∈ Xin . Furthermore, as in have that (µn−1 ⊗ σ−i )(B i,θ,xi

ˆ n−1 ∩ Eˆ n−1 Claim 1, the outcome of the election is the status quo for all (θn−1 , x−i ) ∈ B i,θ,xi for each xi ∈ Xin . We now show that σi1 (θ) = e for all i ∈ In and θ ∈ Θ, i.e. σ 1 is pure. The argument is analogous to that of Claim 1. Suppose, in order to reach a contradiction, that σi1 (θ) 6= e for some i ∈ In and θ ∈ Θ. Consider σ ˆi defined by σ ˆi1 (θ) = e, and σ ˆi2 (θ0 , c|(e, x−i )i , gn (e, x−i ), θ) =

X

σi1 (xi |θ)σi2 (θ0 , c|(xi , x−i )i , gn (xi , x−i ), θ)

xi ∈Xin n for all θ0 ∈ Θ, c ∈ C, and x−i ∈ X−i , and σ ˆi = σi elsewhere.

We next show that, given the belief system ν, the expected payoff for player i of the strategy (ˆ σi , σ−i ) is higher than that of σ. Assumption 1 implies that n . u˜ni ((e, x−i ), θ0 , θ) − u˜ni ((xi , x−i ), θ0 , θ) ≥ ε for all xi ∈ Xin and x−i ∈ X−i

Let π(θn−1 , x−i , xi ) ∈ ∆(C) denote the distribution of votes when the type profile of players other than i is θn−1 , the first period action profile is (xi , x−i ) and players play σ, and π ˆ (θn−1 , x−i , xi ) when players play (ˆ σi , σ−i ).

55

ˆ n−1 ∩ Eˆ n−1 , then the outcome of the election is Fix xi ∈ Xin . If (θn−1 , x−i ) ∈ B i,θ,xi the status quo in both cases and, hence, v(c, π ˆ (θn−1 , x−i , xi ), θ0 , θ) − v(c, π(θn−1 , x−i , xi ), θ0 , θ) ≥ 0, ˆ n−1 ∩ Eˆ n−1 , then for each (c, θ0 ) ∈ C × Θ. Moreover, if (θn−1 , x−i ) 6∈ B i,θ,xi v(c, π ˆ (θn−1 , x−i , xi ), θ0 , θ) − v(c, π(θn−1 , x−i , xi ), θ0 , θ) ≥ −θ∗ , for each (c, θ0 ) ∈ C × Θ. We have that X

uni (σ, ν|θ) =

1 (µn−1 ⊗ σ−i )(θn−1 , x−i )

n (θn−1 ,x−i )∈Θn−1 ×X−i

XX X

σi1 (xi |θ)σi2 (θ0 , c|(xi , x−i )i , gn (xi , x−i ), θ)

c∈C θ0 ∈Θ xi ∈X n

i  n  u˜i ((xi , x−i ), θ0 , θ) + v(c, θ0 , θ, θn−1 )

and X

uni (ˆ σi , σ−i , ν|θ) =

1 (µn−1 ⊗ σ−i )(θn−1 , x−i )

n (θn−1 ,x−i )∈Θn−1 ×X−i

XX

σ ˆi2 (θ0 , c|(e, x−i )i , gn (e, x−i ), θ)

c∈C θ0 ∈Θ

 n  u˜i ((e, x−i ), θ0 , θ) + v(c, θ0 , θ, θn−1 ) Since σ ˆi2 (θ0 , c|(e, x−i )i , gn (xi , x−i ), θ) =

X

σi1 (xi |θ)σi2 (θ0 , c|(xi , x−i )i , gn (xi , x−i ), θ)

xi ∈Xin n 1 ˆ n−1 ∩ Eˆ n−1 ) > 1 − γ 0 , then for all θ0 ∈ Θ, c ∈ C, and x−i ∈ X−i , and (µn−1 ⊗ σ−i )(B i,θ

σi , σ−i , ν|θ) − uni (σ, ν|θ) ≥ uni (ˆ X 1 ˆ n−1 ∩ Eˆ n−1 ))] > ε − γ 0 θ∗ > 0, σi1 (xi |θ)[ε − γ 0 (1 − (µn−1 ⊗ σ−i )(B i,θ xi ∈Xin

a contradiction to the assumption that (σ, ν) is a sequential equilibrium. This contradiction establishes that σi1 (θ) = e for all i ∈ In and θ ∈ Θ, as claimed. 56

A.7

Proof of Proposition 4

We establish part 1 of Proposition 4, part 2 being analogous. Let N1 and B be as in Proposition 2 and N2 and B 0 be as in Lemma 1. Define N = max{N1 , N2 } and B ∗ = min{B, B 0 }. Let n ≥ N , σ ∈ Σn and (σ, ν) be a symmetric sequential equilibrium of Gn , and suppose that (µn ⊗ σ 1 )({(θn , x) ∈ Θn × X n : a3i (σ, x, θn ) 6= f (θin ) for some i ∈ In }) < B ∗ . Since B ∗ ≤ B 0 , Lemma 1 implies that σ is pure. Hence, letting a(σ, θn ) be as in Section 3 for each θn ∈ Θn , it follows by Proposition 2 that (µn ⊗ σ 1 )({(θn , x) ∈ Θn × X n : a3i (σ, x, θn ) 6= f (θin ) for some i ∈ In }) = µn ({θn ∈ Θn : a3i (σ, θn ) 6= f (θin ) for some i ∈ In }) ≥ B ≥ B ∗ , a contradiction.

A.8 A.8.1

Claims made in the examples Opinion leaders

We show that Conditions 1-4 hold in the example of Section 5.1.1. We have that u˜ni ((e, x−i ), θ0 , θ) − u˜ni (x, θ0 , θ) = ε for all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e. Hence, Condition 1 holds since ε > 0 and so does Condition 2 because ε < θ∗ . Condition 3 holds with δ = 1 since 1 ∈ In (j) for all j ∈ In . Regarding Condition 4, letting n ∈ N, i ∈ In , x ∈ X n and c, c0 ∈ C be such that c 6= c0 , |{j ∈ In \ {i} : xj ∈ Xjn (c)}| = 1, |{j ∈ In \ {i} : xj ∈ Xjn (c0 )}| = 0, then, for all θ, θ0 , θˆ ∈ Θ such that θc0 > θc0 0 and

θc0 θc0 0

=

θˆc0 , θˆc

ˆ θ) = θβ . u˜ni (x, θ0 , θ) − u˜ni (x, θ, c Recalling that Θ is symmetric, we have that minθ∈Θ θa = minθ∈Θ θb and, hence, let ψ = minθ∈Θ θaβ > 0.

57

A.8.2

Influence through a network

We consider the example of Section 5.1.2 and show that Conditions 1-4 hold. We have that u˜ni ((e, x−i ), θ0 , θ) − u˜ni (x, θ0 , θ) = ε for all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e. Hence, Condition 1 holds since ε > 0 and so does Condition 2 because ε < θ∗ . Condition 3 holds with δ = 1 since 1 ∈ In (j) for all j ∈ In . Regarding Condition 4, letting n ∈ N, i ∈ In , x ∈ X n and c, c0 ∈ C be such that c 6= c0 , |{j ∈ In \ {i} : xj ∈ Xjn (c)}| = 1, |{j ∈ In \ {i} : xj ∈ Xjn (c0 )}| = 0, then, for all θ, θ0 , θˆ ∈ Θ such that θc0 > θc0 0 and

θc0 θc0 0

=

θˆc0 , θˆc

ˆ θ) ≥ β m > 0 u˜ni (x, θ0 , θ) − u˜ni (x, θ, since d(i, j; hn ) ≤ m for all j ∈ In and β > 0. Hence, let ψ = β m . A.8.3

Polls

Conditions 1 and 2 hold in the example of Section 5.1.3 exactly as in the two previous examples. We now show that Condition 5 holds. Let ψ = m = 1 and let n ∈ N, c ∈ C, i ∈ In , x ∈ X n and x0i ∈ Xin (c) be such that gn (x) = gn (x0i , e¯−i ). For concreteness, assume that c = a. Then gn (x) = gn (x0i , e¯−i ) = (1, 1, 0) and, thus, there exists l ∈ In such that x = (a, e¯−l ). It then follows that |{j ∈ In : xj 6= e}| = 1 = m. Furthermore, if θ, θ0 , θˆ ∈ Θ are such that θa0 > θb0 and

θa0 θb0

=

θˆb , θˆa

ˆ θ) = 1 = ψ then u˜nj (x, θ0 , θ) − u˜nj (x, θ,

for all j ∈ In . A.8.4

Campaign contributions

We show that Conditions 1, 2 and 5 hold in the example of Section 5.1.4. Here it is key to make the following definitions. For all n ∈ N and i ∈ In , let Xin (a) = {xi ∈ Xin : xi,a > 0 and xi,b = 0}, Xin (b) = {xi ∈ Xin : xi,b > 0 and xi,a = 0} and e = (0, 0). For all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e, we have that u˜ni ((e, x−i ), θ0 , θ) − u˜ni (x, θ0 , θ) = (1 − β)(xi,a + xi,b ) ≥ 1 − β. Hence, Condition 1 holds with ε = 1 − β > 0. 58

Regarding Condition 2, set η = 1 − β < θ∗ . Then, for all n ∈ N, i ∈ In , θ, θ0 ∈ Θ and x ∈ X n with xi 6= e and x∗i ∈ Xin (f (θ)) defined by x∗i,f (θ) = 1 and x∗i,c = 0 for c 6= f (θ), we have that u˜ni ((e, x−i ), θ0 , θ) − u˜ni ((x∗i , x−i ), θ0 , θ) = 1 − β = η. We next show that Condition 5 holds. Let n ∈ N, c ∈ C, i ∈ In , x ∈ X n and x0i ∈ Xin (c) be such that gn (x) = gn (x0i , e¯−i ). For concreteness, assume that c = a. Then gn (x) = gn (x0i , e¯−i ) = (x0i,a , 0) and, thus, there exist m0 voters, i1 , . . . , im0 ∈ In Pm0 0 such that, for all l = 1, . . . , m0 , xil = (xil ,a , 0), xil ,a > 0 and l=1 xil ,a = xi,a . Since xil ,a ≥ 1 for all l, then m0 = |{j ∈ In : xj 6= e}| ≤ m. Furthermore, if j 6= i, θ, θ0 , θˆ ∈ Θ are such that θa0 > θb0 and

θa0 θb0

=

θˆb , θˆa

ˆ θ) = then u˜nj (x, θ0 , θ) − u˜nj (x, θ,

βx0i,a − ζ(θf (θ) ˆ − θf (θ0 ) ) ≥ β − ζθ∗ > 0. Hence, set ψ = β − ζθ∗ .

A.9

Proof of Lemma 2

Let N1 ∈ N be such that 1 − 1/n > x¯ for all n ≥ N1 . Let 0 < γ < θ∗ min{λ, ζ}/θ∗ (in ¯ ξ = x¯ − 1/2 > 0 and ξ 0 = −1. Furthermore, let this example, θ∗ = θ¯ − θ and θ∗ = θ), N2 ∈ N be given by Lemma 4 corresponding to γ, ξ and ξ 0 , and N = max{N1 , N2 }. Let n ≥ N , p ∈ (0, 1), x ∈ X n , θ ∈ Θ, i ∈ In and a strategy σ ˜i for player i be given. We first compute player i’s beliefs after observing (θ, xi , gn (x)), denoted n : by νi,θ,xi ,gn (x) . First note that νi,θ,xi ,gn (x) is a probability distribution on {y ∈ X−i

gn (xi , y) = gn (x)} × Θn−1 . Moreover, since p ∈ (0, 1), all information sets are reached with a strictly positive probability when σ(p) is played. Hence, beliefs νi,θ,xi ,gn (x) are computed using Bayes rule. n Fix y ∈ X−i such that gn (xi , y) = gn (x). Let J(y) = {j ∈ In \ {i} : yj = (0, 0)},

L(y) = In \ (J(y) ∪ {i}) and, for each l ∈ L(y), θ˜l (y) = θa if yl = (1, 0) and θ˜l (y) = θb if yl = (0, 1). We claim that, for each θn−1 ∈ Θn−1 ,   2−|J(y)| if θn−1 = θ˜ (y) for all l ∈ L(y), l l νi,θ,xi ,gn (x) (θn−1 |y) =  0 otherwise. Indeed, using Bayes rule, µ(θjn−1 )s1j (yj |θjn−1 ) |y) = P . Q ˆn−1 )s1 (yj |θˆn−1 ) µ( θ n−1 n−1 ˆ j j j j6=i θ ∈Θ Q

νi,θ,xi ,gn (x) (θ

n−1

j6=i

59

(7)

We have that, for each j 6= i, and θ0 ∈ Θ,    1 − p if j ∈ J(y),   s1j (yj |θ0 ) = 0 if j ∈ L(y) and θ0 6= θ˜j (y),     p if j ∈ L(y) and θ0 = θ˜j (y). Thus, we have that νi,θ,xi ,gn (x) (θn−1 |y) = 0 if θln−1 6= θ˜l (y) for some l ∈ L(y). Fur˜ n−1 = {θˆn−1 ∈ Θn−1 : θˆn−1 = θ˜l (y) for all l ∈ L(y)}, we have that thermore, letting Θ l ˜ n−1 | = 2|J(y)| and |Θ Q νi,θ,xi ,gn (x) (θ

n−1

|y) = P

p j∈L(y) 2 Q p 1−p Q j∈L(y) 2 j∈J(y) 2

j∈J(y)

˜ n−1 θˆn−1 ∈Θ

1−p 2

Q

=

1 = 2−|J(y)| . n−1 ˜ |Θ |

We next establish the lemma by showing that uni (σ(p), ν(p)|θ, xi , gn (x)) ≥ uni (˜ σi , σ−i (p), ν(p)|θ, xi , gn (x)). We consider three cases. The first case is when gn,a (x) > gn,b (x). In this case, for all j ∈ In , xˆj ∈ Xjn and θˆ ∈ ˆ ˆ = θa and s3 (ˆ xj , gn (x), θ) Θ, s2j (ˆ j xj , gn (x), θ) = a. Hence, a wins the election regardless of whether player i deviates from σi (p) or not (since 1 − 1/n > x¯). Therefore, X

uni (σ(p), ν(p)|θ, xi , gn (x)) =

νi,θ,xi ,gn (x) (y, θn−1 )[˜ uni ((xi , y), θa , θ) + λθ¯ + θa ],

(y,θn−1 )

and for some θ0 ∈ Θ and c ∈ C, uni (˜ σi , σ−i (p), ν(p)|θ, xi , gn (x)) =

X

νi,θ,xi ,gn (x) (y, θn−1 )[˜ uni ((xi , y), θ0 , θ) + λθc0 + θa ].

(y,θn−1 )

Since, for all θ0 ∈ Θ and c ∈ C, θc0 ≤ θ¯ and u˜ni ((xi , y), θa , θ) − u˜ni ((xi , y), θb , θ) = n ζ(θa −θb )+β(gn,a (x)−gn,b (x)) ≥ β−ζθ∗ > 0 for all y ∈ X−i such that gn (xi , y) = gn (x),

it follows that uni (σ(p), ν(p)|θ, xi , gn (x)) − uni (˜ σi , σ−i (p), ν(p)|θ, xi , gn (x)) ≥ 0. The second case is when gn,a (x) < gn,b (x) and is analogous to the one above. The third and final case is when gn,a (x) = gn,b (x). In this case, for all j ∈ In , ˆ = θˆ and s3 (ˆ ˆ ˆ xˆj ∈ Xjn and θˆ ∈ Θ, s2j (ˆ xj , gn (x), θ) j xj , gn (x), θ) = f (θ). Suppose that 60

θ = θa , the case θ = θb being analogous. We may assume that either s˜2i (xi , gn (x), θ) 6= s2i (xi , gn (x), θ) or s˜3i (xi , gn (x), θ) 6= s3i (xi , gn (x), θ) or both. If s˜3i (xi , gn (x), θ) = s3i (xi , gn (x), θ) then uni (σ(p), ν(p)|θ, xi , gn (x)) − uni (˜ σi , σ−i (p), ν(p)|θ, xi , gn (x)) = ζθ∗ > 0. Hence, we may assume that s˜3i (xi , gn (x), θ) 6= s3i (xi , gn (x), θ), i.e. s˜3i (xi , gn (x), θ) = b. n Fix y ∈ X−i such that gn (xi , y) = gn (x) and, also, θn−1 ∈ Θn−1 such that

νi,θ,xi ,gn (x) (θn−1 |y) > 0. Then, by (7), θln−1 = θ˜l (y) for all l ∈ L(y). Since θ = θa , P player i votes for a under σi (p) and, given y, j6=i yj,a players in In \ {i} vote for a as well (since θln−1 = θ˜l (y) for all l ∈ L(y)). For convenience, for each xi ∈ Xin , let   0 if x = (1, 0), i a(xi ) =  1 if x = (0, 0) or x = (0, 1), i i and

  0 if x = (1, 0), i b(xi ) =  1 if x = (0, 0) or x = (0, 1), i i

P Hence, 1 + j6=i yj,a = gn,a (x) + a(xi ) voters are certain to vote for a. Likewise, P j6=i yj,b = gn,b (x) − b(xi ) = gn,a (x) − b(xi ) voters are certain to vote for b. Each of the remaining n − a(xi ) + b(xi ) − 2gn,a (x) voters, i.e. those in J(y), vote for a if and only if his type is θa . Conditional on y, for the change of vote by voter i to change the outcome of the election, it must be that either a wins when voter i votes a (i.e. when he plays s3i (xi , gn (x), θ)) or b wins when voter i votes b (i.e. when he plays s˜3i (xi , gn (x), θ)). The former can happen only if |J(y)| > n¯ x − a(xi ) − gn,a (xi ) and the latter only if |J(y)| > n¯ x + b(xi ) − gn,a (xi ) − 1. Since gn,a (x) + gn,b (x) ≤ n and gn,a (x) = gn,b (x), then gn,a (x) ≤ n/2. Hence, n¯ x − a(xi ) − gn,a (xi ) ≥ n(¯ x − 1/2) − 1 = nξ + ξ 0 and n¯ x + b(xi ) − gn,a (xi ) − 1 ≥ n(¯ x − 1/2) − 1 = nξ + ξ 0 . Thus, letting z = (i0 , i1 , . . . , i|L(y)| , θ0 , . . . , θ|L(y)| , c0 , . . . , c|L(y)| , f, f ), i0 = i, θ0 = θ, c0 = a, L(y) = {i1 , . . . , i|L(y)| }, θl = θ˜l (y) and cl = f (θ˜l (y)) for all l ∈ L(y), and m = |L(y)| so that n − m − 1 = |J(y)| ≥ nξ + ξ 0 , it follows by Lemma 61

4 that uni (σ(p), ν(p)|θ, xi , gn (x)) − uni (˜ σi , σ−i (p), ν(p)|θ, xi , gn (x)) ≥ min{ζ, λ}θ∗ − θ∗ γ > 0. This concludes the proof of the lemma.

A.10

Proof of Proposition 5

Fix n ∈ N and i ∈ In . For each p ∈ [0, 1], let Pnp (x−i , θ−i ) be the probability P 1 and Pnp (x−i ) = θ−i Pnp (x−i , θ−i ) (see Claim 6 of (x−i , θ−i ) induced by µn−1 and σ−i below for more detail on these probabilities). We will show that there exists pn ∈ (0, 1) such that X (x−i ,θ−i )

X

X

Pnpn (x−i , θ−i )vin ((1, 0), x−i , θa , θ−i ) =

Pnpn (x−i , θ−i )vin ((0, 0), x−i , θa , θ−i ), (8)

(x−i ,θ−i )

X

Pnpn (x−i , θ−i )vin ((1, 0), x−i , θa , θ−i ) ≥

(x−i ,θ−i )

X

(x−i ,θ−i )

Pnpn (x−i , θ−i )vin ((0, 1), x−i , θb , θ−i )

X

=

(x−i ,θ−i )

X

Pnpn (x−i , θ−i )vin ((0, 1), x−i , θa , θ−i ), (9) Pnpn (x−i , θ−i )vin ((0, 0), x−i , θb , θ−i ),(10)

(x−i ,θ−i )

X

Pnpn (x−i , θ−i )vin ((0, 1), x−i , θb , θ−i ) ≥

(x−i ,θ−i )

Pnpn (x−i , θ−i )vin ((1, 0), x−i , θb , θ−i ).(11)

(x−i ,θ−i )

We will show that (8) and (9) hold, (10) and (11) being analogous. Let Fn (p) =

X

Pnp (x−i , θ−i ) (vin ((1, 0), x−i , θa , θ−i ) − vin ((0, 0), x−i , θa , θ−i ))

(x−i ,θ−i )

for each p ∈ [0, 1]. We compute Fn (p) in what follows, first by computing some relevant probabilities. Let Γ0n = {γ ∈ {0, . . . , n − 1}2 : γa + γb ≤ n − 1} and, for each P n γ ∈ Γ0n , let Pnp (γ) = Pnp ({x−i ∈ X−i : j6=i xj = γ}). Claim 6 The following holds for each p ∈ [0, 1]: n 1. For all x−i ∈ X−i such that

Pnp ({θ−i

n−1

∈Θ

P

j6=i

xj,a =

P

˜ −i ) = : (θ , θ−i ) ∈ A}|x a

j6=i

xj,b , 1

2n−2γa −1

X k>n¯ x−γa −1



 n − 2γa − 1 , k (12)

62

and Pnp ({θ−i

n−1

∈Θ

˜ −i ) = : (θ , θ−i ) ∈ B}|x

1

a

2n−2γa −1

X n − 2γa − 1 , (13) k k>n¯ x−γ a

where γa =

P

j6=i

xj,a .

n such that 2. For all x−i ∈ X−i

Pnp ({θ−i

n−1

∈Θ

P

j6=i

xj,a + 1 =

˜ −i ) = : (θ , θ−i ) ∈ A}|x

P

j6=i

xj,b ,

1

a

X

2n−2γa −2

k>n¯ x−γa

  n − 2γa − 2 k −1 (14)

and Pnp ({θ−i

n−1

∈Θ

˜ −i ) = : (θ , θ−i ) ∈ B}|x a

1 2n−2γa −2

X n − 2γa − 2 , (15) k k>n¯ x−γ a

where γa =

P

j6=i

xj,a .

3. For all γ ∈ Γ0n , Pnp (γ)

=

1 2γa +γb



 n−1 pγa +γb (1 − p)n−1−γa −γb . γa , γb , n − 1 − γa − γb

(16)

Proof. We start by noting that player i’s beliefs, after observing his type θa , are given by νi,θa (θ−i ) = 2−(n−1) for all θ−i ∈ Θ−i . Furthermore, the probability of Q n x−i ∈ X−i given θ−i ∈ Θn−1 is Pnp (x−i |θ−i ) = j6=i σj1 (xj |θj ). Thus, Pnp (θ−i , x−i ) = Q 2−(n−1) j6=i σj1 (xj |θj ). n Recall the notation used in Lemma 2. For each x−i ∈ X−i , J(x−i ) = {j ∈ In \{i} :

xj = (0, 0)}, L(x−i ) = In \ (J(x−i ) ∪ {i}) and, for each l ∈ L(x−i ), θ˜l (x−i ) = θa if P P xl = (1, 0) and θ˜l (x−i ) = θb if xl = (0, 1); let also γa = j6=i xj,a and γb = j6=i xj,b . Since |J(x−i )| = n − 1 − γa − γb , (7) implies that, for each θ−i ∈ Θn−1 ,   2−(n−1−γa −γb ) if θn−1 = θ˜ (x ) for all l ∈ L(x ), l −i −i l Pnp (θ−i |x−i ) =  0 otherwise.

(17)

n Consider now x−i ∈ X−i with γa = γb . Then Pnp (θ−i |x−i ) > 0 and (θa , θ−i ) ∈ A˜

if and only if γa + 1 + |{j ∈ J(x−i ) : θj = θa }| > n¯ x, i.e. |{j ∈ J(x−i ) : θj = θa }| > n¯ x − γa − 1. Since |J(x−i )| = n − 2γa − 1, (12) follows. Similarly, Pnp (θ−i |x−i ) > 0 63

˜ if and only if γb + |{j ∈ J(x−i ) : θj = θb }| > n¯ and (θa , θ−i ) ∈ B x, i.e. |{j ∈ J(x−i ) : θj = θb }| > n¯ x − γa (since γa = γb ). Thus, (13) follows. n with γb = γa +1. Then Pnp (θ−i |x−i ) > 0 and (θa , θ−i ) ∈ A˜ Consider next x−i ∈ X−i

if and only if γa + 1 + |{j ∈ J(x−i ) : θj = θa }| > n¯ x, i.e. |{j ∈ J(x−i ) : θj = θa }| > n¯ x − γa − 1. Since |J(x−i )| = n − γa − γb − 1 = n − 2(γa + 1), (14) follows. Equation (15) follows by an analogous argument. n such that Finally, given γ ∈ Γ0n , the number of x−i ∈ X−i
n−1 . Hence, (16) follows. γa ,γb ,n−1−γa −γb

P

j6=i

xj = γ is

For convenience, let ζn (α) =

θ¯

  X n − 2α − 1 n − 2α − 1 θ + n−2α−1 k 2 k k>n¯ x−α−1 k>n¯ x−α X

2n−2α−1

for each α ∈ N such that 2α + 1 ≤ n, and X n − 2α − 2 X n − 2α − 2 θ¯ θ 0 ζn (α) = n−2α−2 + n−2α−2 2 2 k k k>n¯ x−α−1 k>n¯ x−α for each α ∈ N such that 2α + 2 ≤ n. Claim 7 For all n ∈ N and p ∈ [0, 1], Fn (p) =

n−1 X

Pnp (α, α)(θ¯ − ζn (α)) −

Pnp (α, α + 1)(θ − ζn0 (α) − ζ(θ¯ − θ))

α=0

α=0

−1 + β − β

n−1 X

X

Pnp (γ).

γ:γa +1≤γb

P P n and let γ = ( j6=i xj,a , j6=i xj,b ). For convenience, let Proof. Fix x−i ∈ X−i vin (xi , x−i , θa ) =

X

Pnp (θ−i |x−i )vin (xi , x−i , θa , θ−i )

θ−i

for each xi ∈ Xin . We consider four cases. The first case is when γb = γa . In this case, vin ((1, 0), x−i , θa ) = ζ θ¯ + β(γa + ¯ v n ((0, 0), x−i , θa ) = ζ θ¯ + βγa + ζn (γa ) + λθ¯ (using 12 and 13) and 1) − 1 + θ¯ + λθ, i vin ((1, 0), x−i , θa ) − vin ((0, 0), x−i , θa ) = β − 1 + θ¯ − ζn (γa ). The second case is when γb = γa + 1. In this case, vin ((1, 0), x−i , θa ) = ζ θ¯ + β(γa + 1) − 1 + ζn0 (γa ) + λθ¯ (using 14 and 15), vin ((0, 0), x−i , θa ) = ζθ + β(γa + 1) + θ + λθ¯ and vin ((1, 0), x−i , θa ) − vin ((0, 0), x−i , θa ) = ζ(θ¯ − θ) − θ − 1 + ζn0 (γa ). 64

The third case is when γb > γa +1. In this case, vin ((1, 0), x−i , θa ) = ζθ+βγb −1+θ+ ¯ v n ((0, 0), x−i , θa ) = ζθ+βγb +θ+λθ¯ and v n ((1, 0), x−i , θa )−v n ((0, 0), x−i , θa ) = −1. λθ, i i i Finally, the forth case is when γb < γa . In this case, vin ((1, 0), x−i , θa ) = ζ θ¯ + ¯ v n ((0, 0), x−i , θa ) = ζ θ¯ + βγa + θ¯ + λθ¯ and v n ((1, 0), x−i , θa ) − β(γa + 1) − 1 + θ¯ + λθ, i i vin ((0, 0), x−i , θa ) = β − 1. P Since Fn (p) = x−i Pnp (x−i ) (vin ((1, 0), x−i , θa ) − vin ((0, 0), x−i , θa )), the claim follows. Claim 8 There exists N ∈ N such that, for all n ≥ N , Fn (0) > 0. Proof. Fix n ∈ N. When p = 0, then Pn (0, 0) = 1. Hence, Fn (0) = θ¯ − (1 − β) − ζn (0). Since θ¯ > 1 − β, it suffices to show that limn ζn (0) = 0. Note that ¯ n−1 ({θn−1 : |{j : θn−1 = θa }| + 1 > n¯ ζn (0) = θµ x}) j +θµn−1 ({θn−1 : |{j : θjn−1 = θb }| > n¯ x}). Let 0 < ε < x¯ − 1/2 (recall that x¯ > 1/2) and note that [ {θn−1 : |{j : θjn−1 = θa }| + 1 > n¯ x} {θn−1 : |{j : θjn−1 = θb }| > n¯ x} ( ) [ |{j : θjn−1 = θc }| 1 ⊆ θn−1 : > +ε n−1 2 c∈C for all n sufficiently large. Since ( )! n−1 c |{j : θ = θ }| 1 2 j µn−1 ∪c∈C θn−1 : > +ε ≤ 4e−2ε (n−1) n−1 2 for all n (see e.g. Kalai (2004, Lemma 2)) and 4e−2ε

2 (n−1)

→ 0, it follows that

limn ζn (0) = 0. The following construction will be used later. Given a sequence {qn }∞ n=1 ⊆ [0, 1], for each n ∈ N and j ∈ {1, . . . , n}, let Yj,n be a random variable taking value 1 with probability qn /2, −1 with probability qn /2 and 0 with probability 1 − qn . The interpretation is that Yj,n−1 = 1 when player j 6= i chooses xj = (1, 0), Yj,n−1 = 0 when he chooses xj = (0, 0) and Yj,n−1 = −1 when he chooses xj = (0, 1). Let P Tn = nj=1 Yj,n . 65

Lemma 5 If {qn }∞ n=1 ⊆ [0, 1] is such that limn qn > 0, then limn P (Tn = 0) = 0 and limn P (Tn = −1) = 0. Proof. We will apply Theorem 1.2 in Davis and McDonald (1995). To this end, √ let Qn = n min{2(1 − qn ), qn }, an = 0 and bn = nqn . We have that limn bn = ∞ 2

since limn qn > 0. Since Qn ≤ nqn , then lim supn Qbnn ≤ 1. Furthermore, note that P 2 E(Yj,n ) = 0 and b2n = nj=1 E(Yj,n ). Since, for each ε > 0, the set {|Yj,n | ≥ εbn } is empty for all sufficiently large n, then Z n X 1 2 lim Yj,n dP = 0. 2 n→∞ b |Y |≥εb n n j,n j=1 Thus, by the central limit theorem (e.g. Theorem 27.2 in Billingsley (2012)), (Tn − an )/bn converges in distribution to the standard normal. Thus, letting ϕ denote the density of the standard normal distribution, Theorem 1.2 in Davis and McDonald (1995) imply that |bn P (Tn = 0) − ϕ(0)| → 0 and |bn P (Tn = −1) − ϕ(−1/bn )| → 0. Therefore, both bn P (Tn = 0) and bn P (Tn = −1) converge to ϕ(0). Since bn → ∞, then P (Tn = 0) → 0 and P (Tn = −1) → 0, as claimed. Claim 9 There exists N ∈ N such that, for all n ≥ N , Fn (1) < 0. Proof. We have that ζn0 (α) ≤ θ¯ for each α and (θ¯ − θ)ζ < β < 1. Thus, ! n−1 X ¯ Fn (1) ≤ (1 + θ) (P 1 (α, α) + P 1 (α, α + 1)) − (1 − β). n

n

α=0

Since 1 − β > 0, it suffices to show that limn

Pn−1

1 1 α=0 (Pn (α, α) + Pn (α, α + 1))

follows from Lemma 5 by letting qn = 1 for all n ∈ N and noting that P 1 P (Tn−1 = 0) and n−1 α=0 Pn (α, α + 1) = P (Tn−1 = −1).

Pn−1

α=0

= 0. This

Pn1 (α, α) =

Let N1 ∈ N be as in Claim 8, N2 ∈ N be as in Claim 9, N3 ∈ N be as in Lemma 2 and N = max{N1 , N2 , N3 }. Fix n ≥ N and note that p 7→ Fn (p) is continuous by Claim 7 and (16). It then follows by the intermediate value theorem (e.g. Theorem 4.23 in Rudin (1976)), together with Claims 8 and 9, that there exists pn ∈ (0, 1) such that Fn (pn ) = 0. Hence, (8) holds.

66

We next show that (9) holds for all n ∈ N. Let X

Fˆn (p) =

Pnp (x−i , θ−i ) [vin ((0, 1), x−i , θa , θ−i ) − vin ((0, 0), x−i , θa , θ−i )]

(x−i ,θ−i )

for each p ∈ [0, 1]; it is then enough to show that Fn (p) − Fˆn (p) ≥ 0 for all n ∈ N and p ∈ [0, 1]. This is shown in the following claim. Claim 10 For all n ∈ N and p ∈ [0, 1], Fn (p) − Fˆn (p) ≥ 0. P P n and let γ = ( j6=i xj,a , j6=i xj,b ). Similarly to the proof of Proof. Fix x−i ∈ X−i Claim 7, we have that    β − 1 + θ − ζn (γb ) − ζ(θ¯ − θ)      −θ¯ − 1 + ζ 0 (γ ) n b n a n a ui ((0, 1), x−i , θ )−ui ((0, 0), x−i , θ ) =  −1       β−1

if γa = γb , if γa = γb + 1, if γa > γb + 1, if γa < γb .

It is clear from (16) that Pnp (γa , γb ) = Pnp (γb , γa ) for all γ ∈ {0, . . . , n − 1}2 with γa + γb ≤ n − 1. It then follows that Fˆn (p) =

n−1 X

Pnp (α, α)(θ

− ζn (α) − ζ(θ¯ − θ)) −

α=0

n−1 X

Pnp (α + 1, α)(θ¯ − ζn0 (α))

α=0

−1 + β − β

X

Pnp (γ)

γ:γb +1≤γa

=

n−1 X

Pnp (α, α)(θ

− ζn (α) − ζ(θ¯ − θ)) −

Pnp (α, α + 1)(θ¯ − ζn0 (α))

α=0

α=0

−1 + β − β

n−1 X

X

Pnp (γ)

γ:γa +1≤γb

Thus, Fn (p) − Fˆn (p) = (θ¯ − θ)(1 + ζ)

n−1 X

(Pnp (α, α) + Pnp (α, α + 1)) ,

α=0

and hence, Fn (p) − Fˆn (p) ≥ 0. We now turn to the last part of Proposition 5. Let {pn }∞ n=1 be such that (σ(pn ), ν(pn )) is a symmetric equilibrium of Gn for all n sufficiently large. We first show that limn→∞ pn = 0. Suppose not; then, there exists a subsequence {pnk }∞ k=1 such that limk pnk > 0. 67

Since (σ(pn ), ν(pn )) is a symmetric equilibrium of Gn for all n sufficiently large, then

n−1 X ¯ 0 = Fn (pn ) ≤ (1 + θ) (Pnpn (α, α) + Pnpn (α, α + 1))

! − (1 − β)

α=0

and, hence, n−1 X

(Pnpn (α, α) + Pnpn (α, α + 1)) ≥

α=0

1−β 1 + θ¯

for all such n. However, we will show using Lemma 5 that

Pnk −1 α=0

pn

(Pnk k (α, α) +

pn

Pnk k (α, α + 1)) → 0, which is a contradiction. To apply Lemma 5, let qnk −1 = pnk for all k ∈ N and, for each n ∈ [nk −1, nk+1 −1], P k −1 pnk P k −1 pnk Pnk (α, α + Pnk (α, α) = P (Tnk −1 = 0) → 0 and nα=0 let qn = qnk −1 . Then, nα=0 1) = P (Tnk −1 = −1) → 0 by Lemma 5. We now turn to the last two claims of Proposition 5 and, to simplify the notation, write Pn for Pnpn . Let wn denote the probability that neither a nor b win the election under (σ(pn ), ν(pn )). We have that wn =

n−1 1 − pn X Pn (α, α)[1 − µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α + 1 > n¯ x}) 2 α=0

−µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α > n¯ x})] n−1 1 − pn X + Pn (α, α)[1 − µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α > n¯ x}) 2 α=0 −µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α + 1 > n¯ x})] n−1 pn X + Pn (α, α + 1)[1 − µn−2α−2 ({θn−2α−2 : |{j : θj = θa }| + α + 1 > n¯ x}) 2 α=0 −µn−2α−2 ({θn−2α−2 : |{j : θj = θb }| + α + 1 > n¯ x})] n−1 pn X + Pn (α + 1, α)[1 − µn−2α−2 ({θn−2α−2 : |{j : θj = θa }| + α + 1 > n¯ x}) 2 α=0 −µn−2α−2 ({θn−2α−2 : |{j : θj = θb }| + α + 1 > n¯ x})], where the first term refers to the case where voter 1, say, is of type θa and chooses (0, 0), the second to the case where voter 1 is of type θb and chooses (0, 0), the third to the case where voter 1 is of type θa and chooses (1, 0), and the forth to the case where voter 1 is of type θb and chooses (0, 1). Since pn → 0, µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α + 1 > n¯ x}) = µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α + 1 > n¯ x}) 68

and µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α > n¯ x}) = µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α > n¯ x}), then limn wn = limn wn0 , where wn0 =

Pn−1

α=0

Pn (α, α) [1 − µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α + 1 > n¯ x}) −µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α > n¯ x})]. Pn−1

Pn (α, α)[µn−2α−1 ({θn−2α−1 : |{j : θj = θa }|+α +1 > n¯ x})+ P : |{j : θj = θb }| + α > n¯ x})] = limn→∞ n−1 α=0 Pn (α, α)ζn (α) =

Claim 11 limn→∞

α=0

µn−2α−1 ({θn−2α−1 P 0 limn→∞ n−1 α=0 Pn (α, α + 1)ζn (α) = 0.

Proof. For each α ∈ N, |{j ∈ {1, . . . , n − 2α − 1} : θj = θa }| + α + 1 ≤ n − 2α − 1 + α + 1 = n − α. Hence, |{j ∈ {1, . . . , n − 2α − 1} : θj = θa }| + α + 1 > n¯ x implies α < n(1 − x¯). Hence, µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α + 1 > n¯ x}) = 0 for all α ≥ n(1 − x¯). Similarly, µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α > n¯ x}) = 0 for all α ≥ n(1 − x¯) − 1. Fix α < n(1 − x¯) and let 0 < ε < x¯ − 1/2 (recall that x¯ > 1/2). We have that {θn−2α−1 : |{j : θjn−2α−1 = θa }| + α + 1 > n¯ x} ( ) n−2α−1 c [ |{j : θ = θ }| 1 j ⊆ θn−2α−1 : > +ε n − 2α − 1 2 c∈C for all n sufficiently large since, whenever |{j : θjn−2α−1 = θa }| + α + 1 > n¯ x, then |{j : θjn−2α−1 = θa }| n¯ x−α−1 n¯ x−1 > ≥ → x¯. n − 2α − 1 n − 2α − 1 n−1 Similarly, we have that {θn−2α−1 : |{j : θjn−2α−1 = θb }| + α > n¯ x} ( ) [ |{j : θjn−2α−1 = θc }| 1 n−2α−1 ⊆ θ : > +ε . n − 2α − 1 2 c∈C Furthermore, we have that ( µn−2α−1

∪c∈C

θn−2α−1

|{j : θjn−2α−1 = θc }| 1 : > +ε n − 2α − 1 2

69

)! ≤ 4e−2ε

2 (n−2α−1)

for all n (see e.g. Kalai (2004, Lemma 2)). Since 2 (n−2α−1)

4e−2ε it follows that

Pn−1

α=0

2 (n−2(n(1−¯ x))−1)

≤ 4e−2ε

2 (2n(¯ x−1/2)−1)

= 4e−2ε

,

Pn (α, α)[µn−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α + 1 > n¯ x}) + 2 (2n(¯ x−1/2)−1)

µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α > n¯ x})] ≤ 4e−2ε

→ 0.

¯ n−2α−1 ({θn−2α−1 : |{j : θj = θa }| + α + 1 > n¯ Note that ζn (α) ≤ θ[µ x}) + µn−2α−1 ({θn−2α−1 : |{j : θj = θb }| + α > n¯ x})] for all α ∈ N and, hence, we have that Pn−1 ¯ n−2α−2 ({θn−2α−2 : |{j : θj = limn α=0 Pn (α, α)ζn (α) = 0. Furthermore, ζn0 (α) ≤ θ[µ θa }| + α + 1 > n¯ x}) + µn−2α−2 ({θn−2α−2 : |{j : θj = θb }| + α > n¯ x})] for all α ∈ N and, Pn−1 similarly to what was done above, we obtain that limn→∞ α=0 Pn (α, α+1)ζn0 (α) = 0. We next show that for all ε > 0, there exist N ∈ N and δ > 0 such that, for all n ≥ N 1 − β < θ¯ < 1 − β + δ and 0 < ζ < δ, the probability that either a or b win the election under (σ(pn ), ν(pn )) is less than ε. In light of Claim 11 and the fact that Pn−1 Pn (α, α) as n → ∞, limn wn = limn wn0 , it suffices to show that the limit of α=0 θ¯ → 1 − β and ζ → 0 equals 1. We have that 0 = Fn (pn ) ≤

n−1 X

Pn (α, α)θ¯ +

n−1 X

Pn (α, α + 1)ζn0 (α) + ζ(θ¯ − θ) − 1 + β

α=0

α=0

and, by Claim 11, it follows that, as n → ∞, θ¯ → 1 − β and ζ → 0, P n−1 0 X ¯ 1 − β − n−1 α=0 Pn (α, α + 1)ζn (α) + ζ(θ − θ) Pn (α, α) ≥ 1≥ → 1. ¯ θ α=0 Finally, we show that for all ε > 0, there exist N ∈ N and B > 0 such that, for all n ≥ N and θ¯ > B, the probability that either a or b win the election under (σ(pn ), ν(pn )) is greater than 1 − ε. In light of Claim 11 and the fact that limn wn = P limn wn0 , it suffices to show that limn→∞ n−1 α=0 Pn (α, α) = 0. From Fn (pn ) = 0 it follows that n−1 n−1 X X ¯ ¯ Pn (α, α)(θ−ζn (α)) = Pn (α, α+1)(θ−ζn0 (α)−ζ(θ−θ))+1−β+β α=0

α=0

X γ:γa +1≤γb

Thus, by Claim 11, we have that, as n → ∞ and θ¯ → ∞, Pn−1 n−1 X 1 + θ + α=0 Pn (α, α)ζn (α) 0≤ Pn (α, α) ≤ → 0. ¯ θ α=0 70

Pn (γ) ≤ θ+1.

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