Strategic voting in multi-winner elections with approval balloting: a theory for large electorates

Abstract

We propose a theory of strategic voting in multi-winner elections with approval balloting. With a tiny probability that any vote might be misrecorded, best responses involve voting by pairwise comparisons. Two candidates play a critical role: the weakest expected winner and the strongest expected loser. Expected winners are approved if and only if they are preferred to the strongest expected loser and expected losers are approved if and only if they are preferred to the weakest expected winner. At equilibrium, if any, a candidate is elected if and only if he is approved by at least half of the voters. With single-peaked preferences, an equilibrium always exists, in which the first candidates according to the majority tournament relation are elected. The theory is applied to individual data from the 2011 Regional Government election in Zurich.

Appendix

1.1 Proof of Lemma 1

Take as given the profile of strategies (ballots) of the voters, $s=\left( s_{i}\right) _{i=1,\ldots ,N}$. For any two candidates c and $c^{\prime },$

$$\begin{aligned} \Pr \left[ S(c)=S(c^{\prime })\right] =\sum _{t=0}^{t=N}\Pr \left[ S(c)=S(c^{\prime })=t\right] , \end{aligned}$$

and, by independence:

$$\begin{aligned} \Pr \left[ S(c)=S(c^{\prime })\right] =\sum _{t=0}^{N}\left( \Pr \left[ S(c)=t\right] \cdot \Pr \left[ S(c^{\prime })=t\right] \right) . \end{aligned}$$

Without loss of generality, assume that $\widehat{s}(c)\ge \widehat{s}(c^{\prime })$.

Consider first the case where $t>\widehat{s}(c)$. The first order probability of the event $S(c)=t$ is

$$\begin{aligned} \left( \begin{array}{c} N-\widehat{s}(c) \\ t-\widehat{s}(c)\end{array}\right) \varepsilon ^{t-\widehat{s}(c)}. \end{aligned}$$

(4)

Indeed, as one can easily check, the event $S(c)=t$ requires at least $t-\widehat{s}(c)$ mistakes, and can indeed result from that precise number of mistakes. One can and must pick $t-\widehat{s}(c)$ individuals who voted against c, among the $N-\widehat{s}(c)$ who voted against c, and change their votes to a YES vote in favor of candidate c. Thus the probability (4). A similar argument holds for the probability that $c^{\prime }$ get t votes, therefore, the first order probability of the event $S(c)=S(c^{\prime })=t$ is:

$$\begin{aligned} \left( \begin{array}{c} N-\widehat{s}(c) \\ t-\widehat{s}(c)\end{array}\right) \left( \begin{array}{c} N-\widehat{s}(c^{\prime }) \\ t-\widehat{s}(c^{\prime })\end{array}\right) \varepsilon ^{2t-\widehat{s}(c)-\widehat{s}(c^{\prime })}. \end{aligned}$$

Similarly, when $t<\widehat{s}(c^{\prime })$, the first order probability of the event $S(c)=t$ is

$$\begin{aligned} \left( \begin{array}{c} \widehat{s}(c) \\ \widehat{s}(c)-t\end{array}\right) \varepsilon ^{\widehat{s}(c)-t} \end{aligned}$$

(pick $\widehat{s}(c)-t$ individuals among the $\widehat{s}(c)$ who voted for c, and change their votes to a NO vote for candidate c). Therefore the first order probability of the event $S(c)=S(c^{\prime })=t$ is:

$$\begin{aligned} \left( \begin{array}{c} \widehat{s}(c) \\ \widehat{s}(c)-t\end{array}\right) \left( \begin{array}{c} \widehat{s}(c^{\prime }) \\ \widehat{s}(c^{\prime })-t\end{array}\right) \varepsilon ^{\widehat{s}(c)+\widehat{s}(c^{\prime })-2t}. \end{aligned}$$

Last, when $\widehat{s}(c^{\prime })\le t\le \widehat{s}(c)$, the first order probability of the event $S(c)=S(c^{\prime })=t$ is :

$$\begin{aligned} \left( \begin{array}{c} \widehat{s}(c) \\ \widehat{s}(c)-t\end{array}\right) \left( \begin{array}{c} N-\widehat{s}(c^{\prime }) \\ t-\widehat{s}(c^{\prime })\end{array}\right) \varepsilon ^{\widehat{s}(c)-\widehat{s}(c^{\prime })}. \end{aligned}$$

When $t>\widehat{s}(c)$, $2t-\widehat{s}(c)-\widehat{s}(c^{\prime })>\widehat{s}(c)-\widehat{s}(c^{\prime })$ and when $t<\widehat{s}(c^{\prime }),$ $\widehat{s}(c)+\widehat{s}(c^{\prime })-2t>\widehat{s}(c)-\widehat{s}(c^{\prime })$. Therefore, one can see that the event $S(c)=S(c^{\prime })$ has first order probability:

$$\begin{aligned} \left( \sum _{t=\widehat{s}(c^{\prime })}^{\widehat{s}(c)}\left( \begin{array}{c} \widehat{s}(c) \\ \widehat{s}(c)-t\end{array}\right) \left( \begin{array}{c} N-\widehat{s}(c^{\prime }) \\ t-\widehat{s}(c^{\prime })\end{array}\right) \right) \cdot \varepsilon ^{\widehat{s}(c)-\widehat{s}(c^{\prime })}, \end{aligned}$$

so that the requirement of the event $S(c)=S(c^{\prime })$ is $\widehat{s}(c)-\widehat{s}(c^{\prime })$. $\square$

1.2 Proof of Lemma 2

Given a profile of strategies $\left( s_{i}\right) _{i=1,\ldots ,N}$, denote by $S_{M}$ the random variable describing the M-th largest score obtained from the realized votes of all voters. Formally: for any vector of realized scores $(S(c))_{c\in \mathfrak {C}}$, let $S_{M}$ be the unique number which satisfies the following two conditions:

• 1. $\left| \left\{ c\in \mathfrak {C}:S(c)>S_{M}\right\} \right| \le M-1$,

• 2. $\left| \left\{ c\in \mathfrak {C}:S(c)\ge S_{M}\right\} \right| \ge M$.

Candidates with scores strictly larger than $S_{M}$ are elected, candidates with scores strictly smaller are not elected, and a candidate with score $S_{M}$ is elected either for sure (if he is the only candidate with realized score $S_{M}$) or with some probability (in case of a tie with other candidates).

The event “Candidate c is caught in an exact tie for election” is the event “$S(c)=S_{M}$ and there exists at least one other $c^{\prime }\ne c$ such that $S(c)=S(c^{\prime })=S_{M}$”.

Consider first the case where $k\le M$. Let us show that the requirement of the event “$S(c_{k})=S_{M}$ and there exists at least another $k^{\prime }\ne k$ such that $S(c_{k})=S(c_{k^{\prime }})=S_{M}$” is $\widehat{s}(c_{k})-\widehat{s}(c_{M+1})$.

Note that $\widehat{s}(c_{k})-\widehat{s}(c_{M+1})$ mistakes (from reference scores $\widehat{s}$) are sufficient to reach this outcome. Indeed, if out of the $\widehat{s}(c_{k})$ voters who did vote for $c_{k}$, one picks $\widehat{s}(c_{k})-\widehat{s}(c_{M+1})$ of them and change their votes (no other mistake being made), the resulting scores are $S(c)=\widehat{s}(c)$ for all $c\ne c_{k}$ and $S(c_{k})=\widehat{s}(c_{M+1})=S(c_{M+1})$. Note that this situation involves a two-way tie between candidate $c_{k}$ and candidate $c_{M+1}$.

One can also check that any other vector of mistakes inducing that candidate $c_{k}$ is caught in an exact tie for election implies at least as many mistakes.

Therefore the requirement of the event “$S(c_{k})=S_{M}$ and there exists at least one other $k^{\prime }\ne k$ such that $S(c_{k})=S(c_{k^{\prime }})=S_{M}$” is exactly $\widehat{s}(c_{k})-\widehat{s}(c_{M+1})$.

Besides, one may check that the event “$S(c_{k})=S_{M}$ and there exists at least one other $k^{\prime }\notin \left\{ k,M+1\right\} \$such that $S(c_{k})=S(c_{k^{\prime }})=S_{M}$” (that is, not having candidate $c_{M+1}$ part of the tie for the Mth position) involves strictly more mistakes.¹¹

Consider now the case $k\ge M+1$. Let us show that the requirement of the event “$S(c_{k})=S_{M}$ and there exists at least another $k^{\prime }\ne k$ such that $S(c_{k})=S(c_{k^{\prime }})=S_{M}$” is $\widehat{s}(c_{M})-s(\widehat{s}_{k})$.

Note that $\widehat{s}(c_{M})-\widehat{s}(c_{k})$ mistakes (from reference scores $\widehat{s}$) are sufficient to reach the outcome $S(c_{k})=S(c_{M})=S_{M}$. Indeed, if out of the $N-\widehat{s}(c_{k})$ voters who did not vote for $c_{k}$, one picks $\widehat{s}(c_{M})-\widehat{s}(c_{k})$ and change their votes (no other mistakes being made), the resulting scores are $S(c)=\widehat{s}(c)$ for all $c\ne c_{k}$ and $S(c_{k})=\widehat{s}(c_{M})=S(c_{M})$.

One can check that any other vector of mistakes inducing this outcome implies at least as many mistakes, therefore the requirement of the event “$S(c_{k})=S_{M}$ and there exists at least one other $k^{\prime }\ne k$ such that $S(c_{k})=S(c_{k^{\prime }})=S_{M}$” is exactly $\widehat{s}(c_{M})-\widehat{s}(c_{k})$.

Besides, one may check that the event “$S(c_{k})=S_{M}$ and there exists at least one other $k^{\prime }\notin \left\{ k,M\right\}$ such that $S(c_{k})=S(c_{k^{\prime }})=S_{M}$” involves strictly more mistakes.¹²

$\square$

1.3 Proof of Proposition 3

Consider a profile of strategies (ballots) from voters other than voter i: $s_{-i}=\left( s_{j}\right) _{j\ne i}$. Let $\widehat{s}_{-i}$ denote the vector of expected scores obtained by the candidates from the votes of all the voters except voter i. Let the candidates be labelled in such a way that:

$$\begin{aligned} \widehat{s}_{-i}(c_{1})>\widehat{s}_{-i}(c_{2})>\cdots> \widehat{s}_{-i}(c_{M})>\widehat{s}_{-i}(c_{M+1})>\cdots >\widehat{s}_{-i}(c_{K}). \end{aligned}$$

Assume that the expected vote difference between any two candidates is at least 3, that is, $\widehat{s}_{i}(c_{k})-\widehat{s}_{i}(c_{k+1})\ge 3$ for all $k=1,\ldots ,K-1$.

To start the proof, consider a voter who contemplates any ballot $s_{i}$ she could cast. Given the strategies $s_{-i}$ of the other voters, the ex post utility that voter i derives from ballot $s_{i}$ depends on the realization of the random variable $\omega$ describing the mistakes made when recording the ballots (remember $\omega _{j,c}=1$ means that a mistake is made when recording voter j’s vote about candidate c, see Sect. 3). Denote this ex post utility by $U_{i}(s_{i},s_{-i},\omega )$. The expected utility derived from strategy $s_{i}$ is $\sum _{\omega }U_{i}(s_{i},s_{-i},\omega )\Pr [\omega ].$

Consider two ballots, $s_{i}$ and $s_{i}^{\prime }$,the voter prefers $s_{i}$ to $s_{i}^{\prime }$ if and only if

$$\begin{aligned} \Delta =\sum _{\omega }U_{i}(s_{i},s_{-i},\omega )\Pr [\omega ]-\sum _{\omega }U_{i}(s_{i}^{\prime },s_{-i},\omega )\Pr [\omega ]\ge 0. \end{aligned}$$

Obviously all the elementary events $\omega$ such that $U_{i}(s_{i},s_{-i},\omega )=U_{i}(s_{i}^{\prime },s_{-i},\omega )$ cancel in this inequality so that the sum can run over elementary events such that $U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega )$. This remark, with the fact that the probabilities $\Pr [\omega ]$ are polynomials in $\varepsilon$ (the requirement of event $\omega$ being $\left| \omega \right|$), provides the technique for finding best responses to an expected score vector $\widehat{s}_{-i}$ when $\varepsilon$ is small. Let m be the requirement of the event $U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega )$. Then:

$$\begin{aligned} \Delta= & {} \sum _{\begin{array}{c} \omega :U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega ) \\ \left| \omega \right| =m \end{array}}\left[ U_{i}(s_{i},s_{-i},\omega )-U_{i}(s_{i}^{\prime }, s_{-i},\omega )\right] \Pr [\omega ] \\&+\sum _{\begin{array}{c} \omega :U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega ) \\ \left| \omega \right| >m \end{array}}\left[ U_{i}(s_{i},s_{-i},\omega )-U_{i}(s_{i}^{\prime },s_{-i},\omega )\right] \Pr [\omega ]. \end{aligned}$$

The first part, where the sum runs over elementary events $\omega$ with requirement m, is a polynomial in $\varepsilon$ of leading term $G\varepsilon ^{m}$, where

$$\begin{aligned} G=\sum _{\begin{array}{c} \omega :U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega ) \\ \left| \omega \right| =m \end{array}}\left[ U_{i}(s_{i},s_{-i},\omega )-U_{i}(s_{i}^{\prime },s_{-i},\omega )\right] \end{aligned}$$

does not depend on $\varepsilon$.

The leading term of the second part has a strictly higher exponent, hence $G=\lim _{\varepsilon \rightarrow 0}\Delta \varepsilon ^{-m}$.

It follows that, for $\varepsilon$ small enough, the sign of $\Delta$ is the sign of G if $G\ne 0$. This implies that, in order to know whether $s_{i}$ yields larger expected utility than $s_{i}^{\prime }$, one can restrict attention to those events which realize $U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega )$ with the smallest number of mistakes. Those events will involve ties (or near ties, with a one vote margin) for election of some candidates.

Given $s_{-i}$, what are the ballots $s_{i}$ and $s_{i}^{\prime }$ and the events $\omega$ which realize $U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega )$?

A necessary condition is that the ballots $s_{i}$ and $s_{i}^{\prime }$ differ on a candidate which is caught in a tie (or a near tie) for election. Under our assumption that the voters in their computation of best responses neglect the possibility of three-way ties, we will focus on ties and near ties which involve exactly two candidates. Two candidates are said to be caught in an exact tie for election if the realized scores, given the votes of all the voters other than i, are such that both candidates receive the Mth highest score; they are said to be caught in a near tie for election if realized scores given the votes of all the voters other than i, are such that one of the candidate get the M-th highest score and the other candidate exactly one less vote. In both types of events, by voting for one of these candidates but not for the other, voter i can change the outcome of the election. Note that the difference between requirement of a tie and requirement of a near tie, for any given two candidate, is at most two.

Now, what are the events and ballots which realize $U_{i}(s_{i},s_{-i},\omega )\ne U_{i}(s_{i}^{\prime },s_{-i},\omega )$ with the smallest number of mistakes?

Lemma 2 provides the answer. A straightforward adaptation of Lemma 2 states that, given the strategies $s_{-i}$ of all voters but i, the requirement of the event “Candidate $c_{k}$ is caught in an exact tie for election (not taking into account the vote of voter i)” is $\widehat{s}_{-i}(c_{k})-\widehat{s}_{-i}(c_{M+1})$ if $k\le M$ and $\widehat{s}_{-i}(c_{M})-\widehat{s}_{-i}(c_{k})$ if $k\ge M+1$. Therefore, the most likely exact tie for election occurs between candidate $c_{M}$ (the weakest expected winner) and candidate $c_{M+1}$ (the strongest expected loser), since the requirement of this event is $\widehat{s}_{-i}(c_{M})-\widehat{s}_{-i}(c_{M+1})$. Given our assumption that the expected vote difference between any two candidates are at least 3, the most likely near tie (that is, with a one vote margin) for election also occurs between candidates $c_{M}$ and $c_{M+1}$. Therefore, if voter i is pivotal, it will most likely be in deciding who between candidate $c_{M}$ and candidate $c_{M+1}$ will be elected. Therefore, if she prefers candidate $c_{M}$ to candidate $c_{M+1}$ ($u_{i}(c_{M})>u_{i}(c_{M+1})$), she should vote for candidate $c_{M}$ and not vote for candidate $c_{M+1}$. Similarly, if $u_{i}(c_{M})<u_{i}(c_{M+1})$, she should vote for candidate $c_{M+1}$ and not vote for candidate $c_{M}$. Her choice about candidates $c_{M}$ and $c_{M+1}$ is thus decided by this pairwise comparison between the two candidates.

What is the next most likely pivot-type event, involving at least one candidate other than candidate $c_{M}$ and candidate $c_{M+1}$?

Again, Lemma 2 provides the answer. It will be either a tie (or near tie) for election between $c_{M-1}$ and $c_{M+1}$, or a tie (or near tie) for election between $c_{M}$ and $c_{M+2}$, depending on whether $\widehat{s}_{-i}(c_{M-1})-\widehat{s}_{-i}(c_{M+1})$ is smaller or larger than $\widehat{s}_{-i}(c_{M})-\widehat{s}_{-i}(c_{M+2})$. More generally, the results in Lemma 2 allow us to rank the different two-way ties for election involving candidates other than candidates $c_{M}$ and $c_{M+1}$. Most specifically, if $1\le k\le M$, define candidate $c_{k}$’s “main contender” as $c_{M+1}$ and if $M+1\le k\le K$, define $c_{k}$’s “main contender” as $c_{M}$. Then, rank the candidates according to (the inverse of) their distance, in terms of expected votes, to their main contender. As seen above, candidates $c_{M}$ and $c_{M+1}$ share the first rank in this ordering.

Consider now the candidate with the second position (either $c_{M-1}$ or $c_{M+2}$), call this candidate c(2). The next most likely pivot-type event involves a tie (or a near tie) between c(2) and its main contender. Therefore, the voter should vote for c(2) if and only if she prefers c(2) to its main contender. Remember that the vote for or against c(2)’s main contender ($c_{M}$ or $c_{M+1}$) has already been decided by the pairwise comparison between candidates $c_{M}$ and $c_{M+1}$. Indeed the event “Candidate c(2)’s main contender is caught in a tie for election with candidate c(2)” is much less likely than a tie for election between $c_{M}$ and $c_{M+1}$.

What is the next most likely pivot-type event, involving at least one candidate other than candidates c(2), $c_{M}$ and $c_{M+1}$? Denoting by c(k), for $2\le k\le K-1$ the candidate with the k’s position in the ordering defined in the previous paragraph, one may check that the next most likely pivot-type event, involving at least one candidate other than candidates c(2), $c_{M}$ and $c_{M+1}$ is a tie (or a near tie) between c(3) and its main contender. Therefore, the voter should vote for c(3) if and only if she prefers c(3) to its main contender.

The same reasoning can be generalized by considering all the candidates in turn. Thus the strategic recommendation described in Proposition 3. $\square$

Remark 12

In Sect. 6, we tackle the rule called “V-restricted Approval”, whereby a voter can only approve up to V candidates. Note that the proof above also characterizes the best responses in that case. Indeed, in that case, the voter considers all the candidates in turn, according to the priority order defined in the proof. Note that the assumption that for any pair of candidates $\left( c,c^{\prime }\right)$, $\left| \widehat{s}_{-i}(c)-\widehat{s}_{-i}(c^{\prime })\right| \ge 3$ in Proposition 11 guarantees that there is no ambiguity when defining this priority order. As long as she does not hit the vote-budget constraint ( V votes), the voter votes for a candidate if and only if her utility for this candidate is larger than her utility for its main contender.

1.4 Proof of Proposition 4

Consider a profile of strategies (ballots) from voters other than voter $i:s_{-i}=\left( s_{j}\right) _{j\ne i}.$ Let $\widehat{s}_{-i}$ denote the vector of expected scores obtained by the candidates from the votes of all the voters except voter i. Let the candidates be labelled in such a way that:

$$\begin{aligned} \widehat{s}_{-i}(c_{1})>\widehat{s}_{-i}(c_{2})>\cdots> \widehat{s}_{-i}(c_{M})>\widehat{s}_{-i}(c_{M+1})>\cdots >\widehat{s}_{-i}(c_{K}). \end{aligned}$$

• For $M=1$ (one person to be elected), the best response described in Proposition 3 prescribes (i) to identify the critical candidates ($c_{1}$ and $c_{2}$), (ii) to approve $c_{1}$ if and only if $u_{i}(c_{1})>u_{i}(c_{2})$, (iii) for $k\ge 2$, to approve $c_{k}$ if and only if $u_{i}(c_{k})>u_{i}(c_{1})$. This recommendation prescribes voting for all candidates strictly preferred to $c_{1}$ if $u_{i}(c_{1})<u_{i}(c_{2})$, and voting for all candidates weakly preferred to $c_{1}$ if $u_{i}(c_{1})>u_{i}(c_{2})$. This always produces a sincere ballot, whatever the voter’s preferences over the candidates. This property for $M=1$ was already noticed in Laslier (2009).

• For $M=K-1$, this rule always produces a sincere ballot. Indeed, if $u_{i}(c_{M})>u_{i}(c_{M+1})=u_{i}(c_{K})$: for any candidate c, she should vote for c if and only if she strictly prefers c to $c_{K}$. This always produces a sincere ballot. If $u_{i}(c_{M})<u_{i}(c_{M+1})$: the voter should vote for a candidate c if and only if she weakly prefers c to $c_{K}$. This always produces a sincere ballot.

• Whenever $M\ge 2$ and $K\ge M+2$, there exist preferences for voter i such that strategic voting entails casting a non-sincere ballot. Suppose that voter i has preferences over the candidates such that:

  $$\begin{aligned} u_{i}(c_{M})>u_{i}(c_{K})>u_{i}(c_{1})>u_{i}(c_{M+1}), \end{aligned}$$

  which is possible whenever $M\ge 2$ and $K\ge M+2$. The voter should approve the expected winners ($c_{1},c_{2},\ldots ,c_{M}$) if and only if she prefers them to the strongest expected loser $c_{M+1}$: given her preferences, this implies in particular voting for $c_{1}$. She should approve the expected losers ($c_{M+1},c_{M+2},\ldots ,c_{K}$) if and only if she prefers them to the weakest expected winner c: given her preferences, this implies in particular not voting for $c_{K}$. One concludes that such a voter should approve $c_{1}$ but not $c_{K}$, although she prefers $c_{K}$ to $c_{1}$. This results in a non-sincere ballot. $\square$

1.5 Proof of Proposition 10

In the single-peaked case, the majority tournament is transitive. By Proposition 9, there is an equilibrium where the M top candidates according to the majority tournament are elected (Point 1 in Proposition 10).

By Remark 6, we also know that at any equilibrium, the Condorcet winner is elected. Let us now show that the elected committee forms a segment in the ordered set of candidates.

Consider an equilibrium (by Proposition 9, we know that (at least) one equilibrium exists). Denote by $x_{M}$ (resp. $x_{M+1}$) the position of the weakest winner in the ordered set of candidates (resp., the position of the strongest loser in the ordered set of candidates). Without loss of generality, assume that in the ordered set of candidates, $x_{M}<x_{M+1}$.

Consider first a candidate (assuming there is one), whose position in the set of ordered candidates is x, such that $x_{M}<x<x_{M+1}$. Let us show that necessarily, x is an expected winner (with a slight abuse of language, we will use in the sequel the same notation to denote a candidate, and its position in the ordered set of candidates). Indeed, assume by contradiction that x is an expected loser. By Proposition 5 Point 3, it must by the case that

$$\begin{aligned} N(x,x_{M})<N(x_{M+1},x_{M}). \end{aligned}$$

Since preferences are single peaked and $x_{M}<x<x_{M+1}$, it must be the case that

$$\begin{aligned} N(x,x_{M})>N(x_{M+1},x_{M}), \end{aligned}$$

yielding a contradiction. Therefore, any candidate located between $x_{M}$ and $x_{M+1}$ (if any) must be an expected winner.

Consider now a candidate, say x, (assuming there is one) such that $x<x_{M}$. Let us show that necessarily, x is an expected loser. Indeed, assume by contradiction that x is an expected winner. By Proposition 5 Point 2, it must by the case that

$$\begin{aligned} N(x,x_{M+1})>N(x_{M},x_{M+1}). \end{aligned}$$

Since preferences are single peaked and $x<x_{M} < x_{M+1}$, it must be the case that

$$\begin{aligned} N(x,x_{M+1})<N(x_{M},x_{M+1}), \end{aligned}$$

yielding a contradiction. Therefore, any candidate located on the left-hand side of $x_{M}$ (if any) must be an expected loser.

Consider last a candidate, say x, (assuming there is one) such that $x>x_{M+1}$. Let us show that necessarily, x is an expected loser. Indeed, assume by contradiction that x is an expected winner. By Proposition 5, the scores of $x_{M+1}$, $x_M$ and x are such that:

$$\begin{aligned} N(x_{M+1}, x_M)< N(x_{M},x_{M+1}) < N(x,x_{M+1}). \end{aligned}$$

Now, by single-peakedness, $x_M< x_{M+1} < x$ implies:

$$\begin{aligned} N(x,x_{M+1})< N(x,x_{M}) < N(x_{M+1}, x_M), \end{aligned}$$

yielding a contradiction. Therefore, any candidate located on the right-hand side of $x_{M+1}$ (if any) must be an expected loser.

These remarks show that the set of expected winners forms a segment (in the set of ordered candidates) and conclude the proof of Point 2

To conclude the proof of the proposition, note that, since the Condorcet winner belongs to the set expected winners, and the set of winners is a segment, at most M distinct sets of winners can be supported at equilibrium. $\square$

Comment on Proposition 10: An example with single-peaked preferences where M distinct sets of winners can be elected at equilibrium. Proposition 10 states that if voters have single-peaked preferences, there are at most M distinct set of elected candidates which can be supported at equilibrium. We provide below a simple example showing that this maximum number can be reached.

Assume that there is a continuum of voters, with bliss points uniformly distributed on the interval $[-1,+1]$.

A voter with bliss point $x^{*}$ evaluates a candidate at position x with the utility function $u(x,x^{*})=-\left| x-x^{*}\right|$.

Assume $M\$candidates are to be elected among $K=2M$ candidates. Candidates are labeled $l_1,\ldots ,l_M,x_1,\ldots ,x_M$ and located as follows:

$$\begin{aligned} -1= & {} l_{M}<l_{M-1}<\cdots<l_{2}<l_{1}<-0.5; \\ 0.4= & {} x_{1}<x_{2}<\cdots<x_{M-1}<x_{M}<+0.5. \end{aligned}$$

The M top candidates according to the majority tournament are $x_{1}$, $x_{2}$, ..., $x_{M}$ (ranked in that order from the Condorcet winner, $x_{1}$, to the $M^{\text {th}}$ candidate in the majority tournament, $x_{M}$). The remaining candidates are $l_{1}$, $l_{2}$, ..., $l_{M}$, ranked in that order from the $\left( M+1\right) ^{\text {th}}$ candidate in the majority tournament, $l_{1}$, to the Condorcet loser, $l_{M}$.

Necessary and sufficient condition for the set of winners to be the set

$$\begin{aligned} \left\{ l_{M-k},l_{M-k-1},\ldots ,l_{2},l_{l},x_{1},x_{2},\ldots ,x_{k-1},x_{k}\right\} , \end{aligned}$$

with $x_{k}$ as the weakest winner and $l_{M-k+1}$ as the strongest loser are (see Proposition 5):

$$\begin{aligned} \text {Condition 1}\text {: }&N(x_{k},l_{M-k+1})>1/2, \\ \text {Condition 2a}\text {: }&N(l_{j},l_{M-k+1})>N(x_{k},l_{M-k+1})\text { for }j<M-k+1 \\ \text {Condition 2b}\text {: }&N(x_{j},l_{M-k+1})>N(x_{k},l_{M-k+1})\text { for }j<k\text { (if }k\ge 2\text {)} \\ \text {Condition 3a}\text {: }&N(l_{j},x_{k})<N(l_{M-k+1},x_{k})\text { for }j>M-k+1 \\ \text {Condition 3b}\text {: }&N(x_{j},x_{k})<N(l_{M-k+1},x_{k})\text { for }j>k \end{aligned}$$

It is straightforward to check that Conditions 1, 2a, 2b, 3a are satisfied. Condition 3b is satisfied if and only if:

$$\begin{aligned} N(x_{k+1},x_{k})<N(l_{M-k+1},x_{k}). \end{aligned}$$

Note that

$$\begin{aligned} N(l_{M-k+1},x_{k})>N(l_{M},x_{1}). \end{aligned}$$

The voters who prefer $l_{M}=-1$ to $x_1=.4$ are located between $-1$ and $-.3$, a segment of length .7, so that $N(l_{M},x_{1})=35~\%$. Likewise:

$$\begin{aligned} N(x_{k+1},x_{k})<N(x_{2},x_{1})<30~\%. \end{aligned}$$

Therefore $N(x_{k+1},x_{k})<N(l_{M-k+1},x_{k})$, and Condition 3b is satisfied.

This shows that, in this example, for any $k=1,2,\ldots ,M$, the set of candidates $\left\{ l_{M-k},x_{M-k-1},\ldots ,l_{2},l_{1},x_{1},x_{2},\ldots ,x_{k-1},x_{k}\right\}$ can be elected at equilibrium. There can be M distinct sets of winners at equilibrium, including one where the set of elected candidates is $\left\{ l_{M-1},l_{M-2},\ldots ,l_{1},x_{1}\right\}$. Note that, in this equilibrium, among the M elected candidates, only the Condorcet winner belongs to the set of the M top candidates according to the majority tournament.