Multiple votes, multiple candidacies and polarization

Abstract

We use the citizen-candidate model to study the differential incentives that alternative voting rules provide for candidate entry, and their effect on policy polarization. In particular, we show that allowing voters to cast multiple votes leads to equilibria which support multiple candidate clusters. These equilibria are more polarized than those obtained under the Plurality Rule. This result differs from the one obtained in the existing literature, where the set of candidates is exogenous. Thus, our paper contributes to the scholarly literature as well as public debate on the merits of using different voting rules by highlighting the importance of endogenous candidacy.

Appendix

Proof of Lemma 1

Let $\left( e,\alpha \right)$ be a 1-position equilibrium. If two or more candidates with the same ideal policy were standing for election, one of them would be strictly better off deviating from his candidacy strategy and not stand for election since his ideal policy would still be adopted with probability one and he would save on the candidacy cost $\delta$. Hence, a single candidate stands for election.

(Necessity) Let e be a candidacy profile such that $e_{i}=1$ for potential candidate $i\in \mathcal {P}$ and $e_{j}=0$ for every $j\in \mathcal {P}$, $j\ne i$. Suppose $x_{i}=x_{L}$. If a potential candidate j with $x_{j}=x_{R}$ were to enter the race, candidates i and j would split the votes equally (since $u\left( \left| x_{L}-m\right| \right) =u\left( \left| x_{R}-m\right| \right)$) and each would be elected with probability 1/2. As a result, if Condition (2) were to fail, candidate j would want to run against candidate i and would be better off deviating from his candidacy strategy. The same argument applies if $x_{i}=x_{R}$.

(Sufficiency) If $x_{i}=x_{M}$ (i.e., Condition (1) is satisfied), then no potential candidate at $x_{L}$ or $x_{R}$ wants to enter the race since he would be defeated. If $x_{i}=x_{L}$ (resp. $x_{R}$), then Condition (2) implies no $j\in \mathcal {P}$ with $x_{j}=x_{R}$ (resp. $x_{L}$) wants to enter the race. Moreover, the concavity of $u\left( .\right)$ implies $u\left( \left| x_{L}-x_{M}\right| \right) \ge \frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}$ which, together with Condition (2), implies no candidate $j\in \mathcal {P}$ with $x_{j}=x_{M}$ wants to stand against candidate i. $\square$

Proof of Lemma 2

Let $\left( e,\alpha \right)$ be a 2-position equilibrium. With only two positions, sure losers would be better off deviating from their candidacy strategy and not run for election. Hence all candidates must be tying for first place. The proof is in four parts.

We first establish the necessity of Condition (1). Assume, by way of contradiction, that $x_{i}\in \left\{ x_{L},x_{M}\right\}$ for every candidate $i\in \mathcal {C}\left( e\right)$. Denote by $c_{L}$ (resp. $c_{M}$) the number of candidates at $x_{L}$ (resp. $x_{M}$). We proceed in four steps to establish the contradiction.

Step 1 $c_{h}\le t$ for every $h\in \left\{ L,M\right\}$. Assume, by way of contradiction, that $c_{M}>t$. To simplify notation, we denote by $V_{h}\equiv V_{h}\left( \mathcal {C},\alpha \right)$ the vote total of a candidate at $x_{h}$. Since all candidates tie for first place, it must be that $V_{L}=V_{M}$. Suppose a candidate i at $x_{M}$ were to deviate by not running for election. We let $\widetilde{\mathcal {C}}\equiv \mathcal {C}{\backslash } \left\{ i\right\}$ and denote by $\widetilde{V} _{h}\equiv V_{h}\left( \widetilde{\mathcal {C}},\alpha \right)$ the vote total of every remaining candidate at $x_{h}$ following candidate i’s deviation.²⁸ Given $c_{M}>t$, the vote total of each candidate at $x_{L}$ would remain unchanged, i.e., $\widetilde{V}_{L}=V_{L}$. At the same time, the vote total of each candidate at $x_{M}$ would strictly increase, i.e., $\widetilde{V} _{M}=\frac{c_{M}}{c_{M}-1}V_{M}>V_{M}$. It follows that $\widetilde{V}_{M}> \widetilde{V}_{L}$, and $x_{M}$ is now adopted with probability one. Candidate i is therefore strictly better off deviating since his ideal policy is adopted with higher probability and he saves on the candidacy cost; a contradiction. Hence $c_{M}\le t$. A similar argument implies $c_{L}\le t$.

Step 2 $c_{h}<s$ for every $h\in \left\{ L,M\right\}$. Given $c_{M}\le t$, $V_{M}\ge 1-F\left( \underline{x}\right)$. If $c_{M}\ge s$, then no citizen with ideal policy $x>\underline{x}$ casts a vote for a candidate at $x_{L}$. This, together with $c_{L}\le t$, implies $V_{L}=F\left( \underline{x}\right)$. Since $F\left( \underline{x}\right) <1/2$, $V_{M}>V_{L}$; a contradiction. Hence, $c_{M}<s$.

If $c_{L}\ge s$, then vote totals are equal to

$$\begin{aligned} \left\{ \begin{array}{l} V_{L}=F\left( \underline{x}\right) +\left( \frac{s-c_{M}}{c_{L}}\right) \left[ 1-F\left( \underline{x}\right) \right] \\ V_{M}=1-F\left( \underline{x}\right) . \end{array} \right. \end{aligned}$$

Moreover, $V_{L}=V_{M}$ implies $x_{M}$ is adopted with probability $\pi _{M}=\frac{c_{M}}{c_{L}+c_{M}}<1/2$. Suppose another potential candidate at $x_{M}$ were to enter the race. (We know he exists since $c_{M}<s\le c_{L}\le p$.) Vote totals would become

$$\begin{aligned} \left\{ \begin{array}{l} \widetilde{V}_{L}=F\left( \underline{x}\right) +\left( \frac{s-c_{M}-1}{c_{L} }\right) \left[ 1-F\left( \underline{x}\right) \right] <V_{L} \\ \widetilde{V}_{M}=1-F\left( \underline{x}\right) =V_{M}. \end{array} \right. \end{aligned}$$

It follows that $\widetilde{V}_{M}>\widetilde{V}_{L}$, and $x_{M}$ is adopted with probability $\widetilde{\pi }_{M}=1$. Given $\pi _{M}<1/2$, a potential candidate at $x_{M}$ is strictly better off entering the race; a contradiction. Hence $c_{L}<s$.

Step 3 $c_{M}<c_{L}$. Given that $c_{h}<s$ for every $h\in \left\{ L,M\right\}$, candidates’ vote totals are given by

$$\begin{aligned} \left\{ \begin{array}{l} V_{L}=F\left( \underline{x}\right) +\frac{\min \left\{ c_{L}+c_{M}-1,s\right\} -c_{M}}{c_{L}}\left[ 1-F\left( \underline{x}\right) \right] \\ V_{M}=\frac{\min \left\{ c_{L}+c_{M}-1,s\right\} -c_{L}}{c_{M}}F\left( \underline{x}\right) +\left[ 1-F\left( \underline{x}\right) \right] . \end{array} \right. \end{aligned}$$

Simple algebra and $F\left( \underline{x}\right) <1/2$ establish that $V_{M}=V_{L}$ only if $c_{M}<c_{L}$. Observe that $c_{M}<c_{L}$ implies $x_{M}$ is adopted with probability $\pi _{M}=\frac{c_{M}}{c_{M}+c_{L}}<1/2$.

Step 4 Suppose another potential candidate at $x_{M}$ were to enter the race. (We know he exists since $c_{M}<c_{L}\le p$.) There are two possibilities.

One possibility is $\left( c_{L}+c_{M}\right) \le s$, in which case every citizen casts $\left( c_{L}+c_{M}\right)$ votes instead of $\left( c_{L}+c_{M}-1\right)$. In this case the vote totals become

$$\begin{aligned} \left\{ \begin{array}{l} \widetilde{V}_{L}=V_{L}=F\left( \underline{x}\right) +\left( \frac{c_{L}-1}{ c_{L}}\right) \left[ 1-F\left( \underline{x}\right) \right] \\ \widetilde{V}_{M}=\frac{c_{M}}{c_{M}+1}F\left( \underline{x}\right) +\left[ 1-F\left( \underline{x}\right) \right] >\frac{c_{M}-1}{c_{M}}F\left( \underline{x}\right) +\left[ 1-F\left( \underline{x}\right) \right] =V_{M}. \end{array} \right. \end{aligned}$$

Thus, $\widetilde{V}_{M}>\widetilde{V}_{L}$ and $\widetilde{\pi }_{M}=1$.

The other possibility is $\left( c_{L}+c_{M}\right) >s$, in which case every citizen will cast exactly s votes before and after the entry of another candidate at $x_{M}$. In this case the vote totals become

$$\begin{aligned} \left\{ \begin{array}{l} \widetilde{V}_{L}=F\left( \underline{x}\right) +\left( \frac{s-c_{M}-1}{c_{L} }\right) \left[ 1-F\left( \underline{x}\right) \right] <F\left( \underline{x} \right) +\left( \frac{s-c_{M}}{c_{L}}\right) \left[ 1-F\left( \underline{x} \right) \right] =V_{L} \\ \widetilde{V}_{M}=\frac{s-c_{L}}{c_{M}+1}F\left( \underline{x}\right) +\left[ 1-F\left( \underline{x}\right) \right] <\frac{s-c_{L}}{c_{M}}F\left( \underline{x}\right) +\left[ 1-F\left( \underline{x}\right) \right] =V_{M}. \end{array} \right. \end{aligned}$$

Simple algebra establishes that $V_{L}=V_{M}$ implies $\widetilde{V}_{M}> \widetilde{V}_{L}$ and, therefore, $\widetilde{\pi }_{M}=1$.

In both cases, $\widetilde{\pi }_{M}=1$ and $\pi _{M}<1/2$ imply another potential candidate at $x_{M}$ wants to enter the race; a contradiction. Hence, it cannot be that $x_{i}\in \left\{ x_{L},x_{M}\right\}$ for every $i\in \mathcal {C}\left( e\right)$. A similar argument applies to $x_{i}\in \left\{ x_{M},x_{R}\right\}$ for every $i\in \mathcal {C}\left( e\right)$.

Secondly, we establish the necessity of Condition (2). That $c_{h}\le p$ for every $h\in \left\{ L,R\right\}$ is obvious. That $c_{h}\le t$ is shown as in Step 1 above. We now establish $c_{h}\ge \min \left\{ s,p\right\}$. W.l.o.g. suppose $c_{R}\ge c_{L}$, which implies that $x_{L}$ is adopted with probability $\pi _{L}=\frac{c_{L}}{c_{L}+c_{R}}\le \frac{1}{2}$. Assume by way of contradiction that $c_{L}<\min \left\{ s,p\right\}$. Proceeding as in Step 4 above (replacing M by R and making a minor adjustment if $c_{R}\ge s$), we can establish that another potential candidate at $x_{L}$ (whom we know to exist since $c_{L}<p$) would be better off entering the race; a contradiction. Hence $c_{R}\ge c_{L}\ge \min \left\{ s,p\right\}$.

Given the discussion above, it is obvious that when $s=t$, we have $c_{L}=c_{R}$. Now consider the voting rules with $s<t$, and suppose it were that $c_{L}<c_{R}$. Given the discussion above, it must be that $s\le c_{L}<c_{R}\le t$. Vote totals are $V_{L\ }=V_{R}=1/2$. Note that if a candidate at $x_{R}$ were to deviate by not standing for election, vote totals of the remaining candidates would be left unchanged,²⁹ and the probability that $x_{R}$ is adopted would drop from $\pi _{R}=\frac{c_{R}}{c_{L}+c_{R}}$ to $\widetilde{\pi }_{R}= \frac{c_{R}-1}{c_{L}+c_{R}-1}$. Thus, a candidate at $x_{R}$ does not want to deviate only if

$$\begin{aligned} \delta \le -\frac{c_{L}}{c\left( c-1\right) }u\left( \left| x_{L}-x_{R}\right| \right) , \end{aligned}$$

(1)

where $c=c_{L}+c_{R}$. If another potential candidate at $x_{L}$ were to enter the race, vote totals would be left unchanged (given $c_{L}<t$, $c_{R}>s$ and the definition of sincere voting) and the probability that $x_{L}$ is adopted would increase from $\pi _{L}=\frac{c_{L}}{c_{L}+c_{R}}$ to $\widetilde{\pi }_{L}=\frac{c_{L}+1}{c_{L}+c_{R}+1}$. Thus, another potential candidate at $x_{L}$ does not want to deviate by entering the race only if

$$\begin{aligned} \delta >-\frac{c_{R}}{c\left( c+1\right) }u\left( \left| x_{L}-x_{R}\right| \right) . \end{aligned}$$

(2)

Simple algebra shows that (1) and (2) cannot hold simultaneously, which gives us a contradiction. Hence $c_{L}=c_{R}$.

Suppose $c_{L}=c_{R}=\min \left\{ s,p\right\}$. If a candidate, say at $x_{L}$, were to deviate by not running for election, then every citizen preferring $x_{L}$ would have to cast more votes for the candidates at $x_{R}$ than the number of votes every citizen preferring $x_{R}$ would have to cast for the candidates at $x_{L}$. As a result, vote totals would be such that $\widetilde{V}_{R}>\widetilde{V}_{L}$, and the probability that $x_{L}$ would be adopted would drop from $\pi _{L}=1/2$ to $\widetilde{\pi }_{L\ }=0$. Thus, a candidate at $x_{L}$ does not want to deviate only if $\delta \le -\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}$.

Suppose $c_{L}=c_{R}>s$. If a candidate, say at $x_{L}$, were to deviate by not running for election, then again, as discussed, the vote totals would be left unchanged, i.e., we would have $\widetilde{V}_{L}=\widetilde{V}_{R}=1/2$ and the probability with which $x_{L}$ would be adopted would drop from $\pi _{L\ }=1/2$ to $\widetilde{\pi }_{L}=\frac{c_{L}-1}{c_{L}+c_{R}-1}$. Thus, a candidate at $x_{L}$ does not want to deviate only if $\delta \le -\frac{ u\left( \left| x_{L}-x_{R}\right| \right) }{2\left( c-1\right) }$.

Suppose $c_{L}=c_{R}<\min \left\{ t,p\right\}$. If another potential candidate, say at $x_{L}$, were to enter the race (we know he exists since $c_{L}<p$), vote totals would be left unchanged, i.e., we would have $\widetilde{V}_{L}=\widetilde{V}_{R}=1/2$ and the probability with which $x_{L}$ would be adopted would increase from $\pi _{L}=1/2$ to $\widetilde{ \pi }_{L}=\frac{c_{L}+1}{c_{L}+c_{R}+1}$. Thus, a candidate at $x_{L}$ does not want to deviate only if $\delta >-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2\left( c+1\right) }$.

Thirdly, we establish the necessity of Condition (3). For $\left( e,\alpha \right)$ to be an equilibrium, it must be that no potential candidate at $x_{M}$ wants to stand for election. Suppose a potential candidate i at $x_{M}$ were to enter the race, i.e., $\widetilde{e}_{i}=1$ for some $i\in \mathcal {P}$ with $x_{i}=x_{M}$ and $\widetilde{e}_{j}=e_{j}$ for all $j\in \mathcal {P}$, $j\ne i$. Given Condition (2) above, there are three cases to consider.

Case 1 $c_{L}=c_{R}=t$. Let the voting profile $\alpha \left( \mathcal {C}\left( \widetilde{e}\right) \right)$ be as followed: for every citizen n with ideal policy $x_{n}<\underline{x}$, $\alpha _{j}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ for every candidate j with $x_{j}=x_{L}$; for every citizen n with ideal policy $x_{n}\in \left( \underline{x},1/2\right)$, $\alpha _{i}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ and $\alpha _{j}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ for $\left( t-1\right)$ of the candidates with $x_{j}=x_{L}$; for every citizen n with ideal policy $x_{n}\in \left( 1/2,\overline{x}\right)$, $\alpha _{i}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ and $\alpha _{j}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ for $\left( t-1\right)$ of the candidates with $x_{j}=x_{R}$; and, finally, for every citizen n with ideal policy $x_{n}>\overline{x}$, $\alpha _{j}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ for every candidate j with $x_{j}=x_{R}$. Observe that this vote profile maximizes the vote totals of candidates at $x_{L}$ and $x_{R}$, and minimizes the vote total of the candidate at $x_{M}$. Candidates’ vote totals are given by

$$\begin{aligned} \left\{ \begin{array}{l} \widetilde{V}_{L}=F\left( \underline{x}\right) +\frac{t-1}{t}\left[ \frac{1}{ 2}-F\left( \underline{x}\right) \right] \\ \widetilde{V}_{M}=F\left( \overline{x}\right) -F\left( \underline{x}\right) \\ \widetilde{V}_{R}=\frac{t-1}{t}\left[ F\left( \overline{x}\right) -\frac{1}{2 }\right] +\left[ 1-F\left( \overline{x}\right) \right] . \end{array} \right. \end{aligned}$$

Suppose that $F\left( \overline{x}\right) \ge 1-F\left( \underline{x} \right)$, which implies $\widetilde{V}_{L}\ge \widetilde{V}_{R}$. (A similar argument holds for $F\left( \overline{x}\right) <1-F\left( \underline{x}\right)$.) There are four sub-cases to consider.

Case 1(i) If $F\left( \overline{x}\right) >\frac{1}{2} +\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1}{2 }\right]$, then we have $\widetilde{V}_{M}>\widetilde{V}_{L}$, and candidate i at $x_{M}$ is elected outright. For him to not be willing to enter the race, it must be that the candidacy cost $\delta$ exceeds his expected utility gain $\left[ 0-u\left( \left| x_{L}-x_{M}\right| \right) \right]$.

Case 1(ii) If $F\left( \overline{x}\right) =\frac{1}{2 }+\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1}{ 2}\right]$ and $F\left( \overline{x}\right) >1-F\left( \underline{x}\right)$, then we have $\widetilde{V}_{M}=\widetilde{V}_{L}>\widetilde{V}_{R}$, and candidate i at $x_{M}$ ties with the t candidates at $x_{L}$. For him to not be willing to enter the race, it must be that the candidacy cost exceeds his expected utility gain, which is equal to

$$\begin{aligned} \left[ \frac{t}{t+1}u\left( \left| x_{L}-x_{M}\right| \right) -u\left( \left| x_{L}-x_{M}\right| \right) \right] . \end{aligned}$$

Case 1(iii) If $F\left( \overline{x}\right) =\frac{1}{ 2}+\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1 }{2}\right]$ and $F\left( \overline{x}\right) =1-F\left( \underline{x} \right)$, then we have $\widetilde{V}_{M}=\widetilde{V}_{L}=\widetilde{V} _{R}$, and candidate i at $x_{M}$ ties with the 2t candidates at $x_{L}$ and $x_{R}$. For him to not be willing to enter the race, it must be that the candidacy cost exceeds his expected utility gain, which is equal to $\left[ \frac{2t}{2t+1}u\left( \left| x_{L}-x_{M}\right| \right) -u\left( \left| x_{L}-x_{M}\right| \right) \right]$.

Case 1(iv) If $F\left( \overline{x}\right) <\frac{1}{2 }+\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1}{ 2}\right]$, then we have $\widetilde{V}_{M}<\widetilde{V}_{L}$, and candidate i at $x_{M}$ is not elected and does not want to enter the race.

These four sub-cases exhaust all possibilities.

Case 2 $s\le c_{L}=c_{R}<t$. Let the voting profile $\alpha \left( \widetilde{e}\right)$ be as follows: for every citizen n with ideal policy $x_{n}<\underline{x}$, $\alpha _{j}^{n}\left( \mathcal {C} \left( \widetilde{e}\right) \right) =1$ for every candidate j with $x_{j}=x_{L}$; for every citizen n with ideal policy $x_{n}\in \left( \underline{x},1/2\right)$, $\alpha _{i}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ and $\alpha _{j}^{n}\left( \mathcal {C} \left( \widetilde{e}\right) \right) =1$ for every candidate j with $x_{j}=x_{L}$; for every citizen n with ideal policy $x_{n}\in \left( 1/2, \overline{x}\right)$, $\alpha _{i}^{n}\left( \mathcal {C}\left( \widetilde{e} \right) \right) =1$ and $\alpha _{j}^{n}\left( \mathcal {C}\left( \widetilde{e }\right) \right) =1$ for every candidate j with $x_{j}=x_{R}$; and, finally, for every citizen n with ideal policy $x_{n}>\overline{x}$, $\alpha _{j}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =1$ for every candidate j with $x_{j}=x_{R}$. Otherwise, $\alpha _{k}^{n}\left( \mathcal {C}\left( \widetilde{e}\right) \right) =0$ for every other candidate k. Observe that this vote profile maximizes the vote totals of candidates at $x_{L}$ and $x_{R}$, and minimizes the vote total of the candidate at $x_{M}$. Candidates’ vote totals are given by

$$\begin{aligned} \left\{ \begin{array}{l} \widetilde{V}_{L}=\widetilde{V}_{R}=1/2 \\ \widetilde{V}_{M}=F\left( \overline{x}\right) -F\left( \underline{x}\right) . \end{array} \right. \end{aligned}$$

Proceeding as in Case 1 above, we obtain the three conditions in the statement.

Case 3 $c_{L}=c_{R}=p<s$. In any voting profile, every citizen casts a vote for the candidate at $x_{M}$. At the same time, every citizen n with ideal policy $x_{n}<1/2$ (resp. $x_{n}>1/2$) does not vote for at least one of the candidates at $x_{R}$ (resp. $x_{L}$). As a result, vote totals are such that $\widetilde{V}_{M}=1>\max \left\{ \widetilde{V}_{L},\widetilde{V}_{R}\right\}$, and candidate i at $x_{M}$ is elected outright. For him to not be willing to enter the race, it must be that the candidacy cost $\delta$ exceeds his expected utility gain, which is equal to $\left[ 0-u\left( \left| x_{L}-x_{M}\right| \right) \right]$.

Finally, it is easy to check that together the conditions in Lemma 2 are sufficient for a 2-position equilibrium to exist. $\square$

Proof of Lemma 3

Consider an election held under an $\left( s,t\right)$-rule where $s=t$. Let $\left( e,\alpha \right)$ be a 3-position equilibrium. We denote by $c_{h}$ the number of candidates at $x_{h}$ for $h=L,M,R$. We start by establishing that $\max \left\{ c_{L}+c_{M},c_{M}+c_{R}\right\} >s=t$. Suppose the contrary. Given that $c_{L}+c_{M}\le s$, every citizen n with ideal policy $x_{n}<1/2$ casts a vote for each of the candidates at $x_{M}$ and does not vote for at least one of the candidates at $x_{R}$. Likewise, $c_{M}+c_{R}\le s$ implies that every citizen n with ideal policy $x_{n}>1/2$ casts a vote for each of the candidates at $x_{M}$ and does not vote for at least one of the candidates at $x_{L}$. Vote totals are therefore such that $V_{M}=1>\max \left\{ V_{L},V_{R}\right\}$; $x_{M}$ is thus adopted with probability $\pi _{M}=1$. A candidate at $x_{L}$ would be better off not running for election since it would still be the case that $\widetilde{V}_{M}>\max \left\{ \widetilde{V}_{L},\widetilde{V}_{R}\right\}$ and $\widetilde{\pi }_{M}=1$, but he would save on the candidacy cost. This contradicts that $\left( e,\alpha \right)$ is an equilibrium.

We then establish that $\min \left\{ c_{L}+c_{M},c_{M}+c_{R}\right\} \le s=t$. Suppose the contrary. We start by observing that all candidates must then be tying for first place. Indeed, if $\pi _{M}=0$, a candidate at $x_{M}$ would get his least-preferred policy ($x_{L}$ or $x_{R}$) adopted; he would therefore be better off not running since he would save on the candidacy cost and could not get a worse policy adopted. Likewise, if $\pi _{L}=0$, a candidate at $x_{L}$ would be better off not running for election. This is because $c_{L}+c_{M}>t$ implies that neither of the votes he would have received would go to the candidates at $x_{R}$. Vote totals would then be such that $\widetilde{V}_{L}\ge V_{L}$, $\widetilde{V} _{M}>V_{M}$ and $\widetilde{V}_{R}=V_{R}$, the inequality is strict because $c_{L}\le t$ (more than t candidates at a position would split votes). This, together with $\pi _{M}>0$, implies $\widetilde{\pi }_{R}=0$, and either $x_{L}$ or $x_{M}$ would be adopted. As a result, a candidate at $x_{L}$ would want to deviate and not run for election since he would get a weakly preferred policy adopted and would save on the candidacy cost. Hence, it must be that $\pi _{L}>0$. Likewise, it must be that $\pi _{R}>0$.

Without loss of generality, suppose $c_{L}\le c_{R}$. Since all candidates tie for first place, $x_{h}$ (for $h=L,M,R$) is adopted with probability $\pi _{h}=\frac{ c_{h}}{c_{L}+c_{M}+c_{R}}$. Since $c_{L}\le c_{R}$, we then have $\pi _{L}\le \pi _{R}$. If a candidate at $x_{L}$ were to deviate by not running for election, $c_{L}+c_{M}>t$ and $c_{L}\le t$ imply again that vote totals would be such that $\widetilde{V}_{L}\ge V_{L}$, $\widetilde{V}_{M}>V_{M}$ and $\widetilde{V}_{R}=V_{R}$. This, together with $\pi _{M}>0$, implies $\widetilde{\pi }_{R}=0$. Given the concavity of $u\left( .\right)$, we have

$$\begin{aligned} \sum _{h\in \left\{ L,M,R\right\} }\widetilde{\pi }_{h}u^{L}\left( x_{h}\right) \ge \sum _{h\in \left\{ L,M,R\right\} }\pi _{h}u^{L}\left( x_{h}\right) . \end{aligned}$$

A candidate at $x_{L}$ would therefore be better off not running since it would increase his expected utility over policy outcomes and he would save on the candidacy cost. Hence, it must be that $\min \left\{ c_{L}+c_{M},c_{M}+c_{R}\right\} \le s=t$.

We are now ready to establish that $\left( e,\alpha \right)$ cannot be an equilibrium. This is obvious for $\left( s,t\right)$-rules where $t\ge 2p$ since $\max \left\{ c_{L}+c_{M},c_{M}+c_{R}\right\} >t$ cannot hold. From now on, consider $\left( s,t\right)$-rules where $t<2p$. Without loss of generality, suppose $c_{L}+c_{M}>t\ge c_{M}+c_{R}$. Vote totals are equal to

$$\begin{aligned} \left\{ \begin{array}{l} V_{L}=F\left( \underline{x}\right) +\left( \frac{t-c_{M}}{c_{L}}\right) \left[ \frac{1}{2}-F\left( \underline{x}\right) \right] +\left( \frac{ t-c_{M}-c_{R}}{c_{L}}\right) \frac{1}{2} \\ V_{M}=\left( \frac{t-c_{L}}{c_{M}}\right) F\left( \underline{x}\right) + \left[ 1-F\left( \underline{x}\right) \right] \\ V_{R}=\frac{1}{2}. \end{array} \right. \end{aligned}$$

Since $F\left( \underline{x}\right) <1/2$, it must be the case that $V_{M}>1/2=V_{R}$, and $x_{R}$ is adopted with probability $\pi _{R}=0$. By the same argument as above, $\pi _{M}>0$ and $\pi _{L}>0$ (the latter since $c_{L}+c_{M}>t$), which requires $V_{L}=V_{M}$. Simple algebra shows that $V_{L}=V_{M}$ only if $c_{L}>c_{M}$ and $t>\left( c_{M}+c_{R}\right)$.

If another potential candidate at $x_{R}$ (we know he exists since $c_{R}<c_{L}\le p$) were to enter the race, vote totals would be such that $\widetilde{V}_{L}<V_{L}$, $\widetilde{V}_{M}=V_{M}$ and $\widetilde{V} _{R}=V_{R}$. Since $V_{L}=V_{M}>V_{R}$, $x_{M}$ would be implemented with probability $\widetilde{\pi }_{M}=1$. Thus, no other potential candidate at $x_{R}$ wants to enter the race only if

$$\begin{aligned} \delta >\frac{c_{L}}{c_{L}+c_{M}}\left[ u\left( \left| x_{R}-x_{M}\right| \right) -u\left( \left| x_{R}-x_{L}\right| \right) \right] . \end{aligned}$$

(3)

If a candidate at $x_{R}$ were to deviate by not running for election, vote totals would be such that $\widetilde{V}_{L}>V_{L}$, $\widetilde{V} _{M}=V_{M}$ and $\widetilde{V}_{R}=V_{R}$. Since $V_{L}=V_{M}>V_{R}$, $x_{L}$ would be implemented with probability $\widetilde{\pi }_{L}=1$. Thus, a candidate at $x_{R}$ does not want to deviate by not running only if

$$\begin{aligned} \delta \le \frac{c_{M}}{c_{L}+c_{M}}\left[ u\left( \left| x_{R}-x_{M}\right| \right) -u\left( \left| x_{R}-x_{L}\right| \right) \right] . \end{aligned}$$

(4)

Taken together, (3) and (4) imply $c_{M}>c_{L}$, which contradicts our earlier finding that $c_{L}>c_{M}$. $\square$

The following lemma will be useful to prove Propositions 1 and 2. It establishes that under an $\left( s,t\right)$-rule, we have that: (1) every policy which is supported by a 1-position equilibrium is strictly less polarized than any policy which is supported by a 2-position equilibrium; and (2) the set of policies which can be supported by 1- or 2-position equilibria is an interval centered around the median m.

Lemma 4

Let the election be held under an $\left( s,t\right)$-rule.

• (1) For every $x\in X$ supported by a 1-position equilibrium and every $y\in X$ supported by a 2-position equilibrium, we have that $\left| x-m\right| <\left| y-m\right|$.

• (2) The set of policies which can be supported by 1- or 2-position equilibria is an interval centered around m.

Proof of Lemma 4

(1) We start by establishing that every 1-position equilibrium is strictly less polarized than any 2-position equilibrium. Pick a 1-position equilibrium $\left( e,\alpha \right)$. We know from Lemma 1 that a single candidate i runs unopposed. One possibility is $x_{i}=x_{M}$, in which case the result follows directly from $x_{j}\in \left\{ x_{L},x_{R}\right\}$ for every candidate j in a 2-position equilibrium (by Condition (1) of Lemma 2). The other possibility is that $x_{i}\in \left\{ x_{L},x_{R}\right\}$, in which case Condition (2) of Lemma 1 implies $\delta >-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2 }$. Pick a 2-position equilibrium $\left( \widehat{e},\widehat{\alpha } \right)$. We know from Condition (1) of Lemma 2 that $x_{j}\in \left\{ \widehat{x}_{L},\widehat{x}_{R}\right\}$ for every candidate $j\in \mathcal { C}\left( \widehat{e}\right)$. Moreover, Condition (2) of Lemma 2 implies $- \frac{u\left( \left| \widehat{x}_{L}-\widehat{x}_{R}\right| \right) }{2}\ge \delta$.³⁰ We then have $u\left( \left| x_{L}-x_{R}\right| \right) >u\left( \left| \widehat{x}_{L}-\widehat{x}_{R}\right| \right)$. Since $u\left( .\right)$ is a strictly decreasing function, the latter inequality implies $\left| \widehat{x}_{L}-m\right| >\left| x_{L}-m\right|$. Hence, the 1-position equilibrium $\left( e,\alpha \right)$ is strictly less polarized than the 2-position equilibrium $\left( \widehat{e},\widehat{\alpha }\right)$.

(2) We continue by establishing that the set of policies supported by 1- or 2-position equilibria is an interval centered around m. We know from Lemma 1 that $x_{M}=m$ is supported by a 1-position equilibrium. Moreover, given the symmetry of $x_{L}$ and $x_{R}$ around the median $m=1/2$, we can restrict attention to left positions $x\in \left[ 0,\frac{1}{2}\right)$. Pick any $x\in \left[ 0,\frac{1}{2}\right)$, and suppose it is supported by an equilibrium (for the platform configuration $\left\{ x,\frac{1}{2} ,1-x\right\}$). Pick any $x^{\prime }\in \left( x,\frac{1}{2}\right)$. We must show that $x^{\prime }$ can be supported by an equilibrium (for the platform configuration $\left\{ x^{\prime },\frac{1}{2},1-x^{\prime }\right\}$). There are two cases to consider.

Case 1 x is supported by a 1-position equilibrium. Condition (2) of Lemma 1 implies that $\delta >-\frac{u\left( \left| x-\left( 1-x\right) \right| \right) }{2}$. Moreover, $\left| x^{\prime }-\frac{1}{2}\right| <\left| x-\frac{1}{2}\right|$ and $u\left( \cdot \right)$ strictly decreasing, imply $u\left( \left| x^{\prime }-\left( 1-x^{\prime }\right) \right| \right) >u\left( \left| x-\left( 1-x\right) \right| \right)$. Hence $\delta >-\frac{ u\left( \left| x^{\prime }-\left( 1-x^{\prime }\right) \right| \right) }{2}$; $x^{\prime }$ is therefore supported by a 1-position equilibrium.

Case 2 x is supported by a 2-position equilibrium. There are two possibilities. One possibility is that $\delta >-\frac{u\left( \left| x^{\prime }-\left( 1-x^{\prime }\right) \right| \right) }{2}$, in which case $x^{\prime }$ is supported by a 1-position equilibrium (for the platform configuration $\left\{ x^{\prime },\frac{1}{2},1-x^{\prime }\right\}$). Another possibility is that $\delta \le -\frac{u\left( \left| x^{\prime }-\left( 1-x^{\prime }\right) \right| \right) }{2}$. In this case, there is a pair $c_{L}^{\prime }$ and $c_{R}^{\prime }$ that satisfies Condition (2) of Lemma 2 for $x^{\prime }$. To establish that $x^{\prime }$ is supported by a 2-position equilibrium, it is therefore sufficient to show that Condition (3) of Lemma 2 is satisfied as well. By assumption, Condition (3) of Lemma 2 is satisfied for x. Moreover, $u\left( \left| x^{\prime }-\left( 1-x^{\prime }\right) \right| \right) >u\left( \left| x-\left( 1-x\right) \right| \right)$ and Condition (2) of Lemma 2 imply that $c_{L}^{\prime }=c_{R}^{\prime }\le c_{L}=c_{R}$. There are three subcases to consider.

Case 2(i) Condition 3(c) is satisfied for x. Condition 3(c) is then satisfied for $x^{\prime }$ as well.

Case 2(ii) Condition 3(b) is satisfied for x. Condition 3(b) applies for $x^{\prime }$ as well. Let $\underline{x}^{\prime }\equiv \frac{x^{\prime }+x_{M}}{2}$ and $\overline{x}^{\prime }\equiv \frac{ x_{M}+\left( 1-x^{\prime }\right) }{2}$. Since $x<x^{\prime }<\frac{1}{2} =x_{M}$, we have $\underline{x}<\underline{x}^{\prime }$ and $\overline{x}> \overline{x}^{\prime }$, implying $F\left( \underline{x}\right) <F\left( \underline{x}^{\prime }\right)$ and $F\left( \overline{x}\right) >F\left( \overline{x}^{\prime }\right)$. This, together with $c_{L}^{\prime }=c_{R}^{\prime }\le c_{L}=c_{R}$, implies that Condition 3(b) is satisfied for $x^{\prime }$.

Case 2(iii) Condition 3(a) is satisfied for x. Suppose $F\left( \overline{x}\right) \ge 1-F\left( \underline{x}\right)$. (A similar argument holds for $F\left( \overline{x}\right) <1-F\left( \underline{x}\right)$.) There are three possibilities to consider.

One possibility is that $c_{L}^{\prime }=c_{R}^{\prime }=t$ and $F\left( \overline{x}^{\prime }\right) \ge 1-F\left( \underline{x}^{\prime }\right)$. Since $F\left( \underline{x}\right) <F\left( \underline{x}^{\prime }\right)$, $F\left( \overline{x}\right) >F\left( \overline{x}^{\prime }\right)$ and $u\left( \left| x^{\prime }-x_{M}\right| \right) >u\left( \left| x-x_{M}\right| \right)$, Condition 3(a) is necessarily satisfied for $x^{\prime }$.

The second possibility is that $c_{L}^{\prime }=c_{R}^{\prime }=t$ and $F\left( \overline{x}^{\prime }\right) <1-F\left( \underline{x}^{\prime }\right)$. If $F\left( \overline{x}\right) \le \frac{1}{2}+\frac{1}{t} \left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1}{2}\right]$, then simple algebra establishes that $F\left( \overline{x}^{\prime }\right) < \frac{1}{2}+\frac{t}{1+t}F\left( \underline{x}^{\prime }\right)$.³¹ If $F\left( \overline{x} \right) >\frac{1}{2}+\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x }\right) -\frac{1}{2}\right]$, then $\delta >-u\left( \left| x-x_{M}\right| \right)$ which, together with $u\left( \left| x^{\prime }-x_{M}\right| \right) >u\left( \left| x-x_{M}\right| \right)$, implies $\delta >-u\left( \left| x^{\prime }-x_{M}\right| \right)$. In any case, Condition 3(a) is satisfied for $x^{\prime }$.

The third possibility is that $s\le c_{L}^{\prime }=c_{R}^{\prime }<t$. If $F\left( \overline{x}\right) \le \frac{1}{2}+\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1}{2}\right]$, then simple algebra establishes that $F\left( \overline{x}^{\prime }\right) <\frac{1}{2} +F\left( \underline{x}^{\prime }\right)$.³² If $F\left( \overline{x}\right) >\frac{1}{2}+\frac{1}{t} \left[ \left( t+1\right) F\left( \underline{x}\right) -\frac{1}{2}\right]$, then $\delta >-u\left( \left| x-x_{M}\right| \right)$ which, together with $u\left( \left| x^{\prime }-x_{M}\right| \right) >u\left( \left| x-x_{M}\right| \right)$, implies $\delta >-u\left( \left| x^{\prime }-x_{M}\right| \right)$. In any case, Condition 3(b) is satisfied for $x^{\prime }$.

These three subcases exhaust all possibilities. In each case, Condition (3) of Lemma 2 is satisfied. Since we have already shown that the other two conditions of Lemma 2 are satisfied, we have established that $x^{\prime }$ is supported by a 2-position equilibrium. $\square$

Proof of Proposition 1

At the risk of a slight abuse of notation, we shall call the equilibrium set the set of policies which are supported by equilibria for the corresponding platform configurations. We establish the result by comparing the equilibrium set under an $\left( s,t\right)$-rule where $s=t>1$ with the equilibrium set under the Plurality Rule.

We know from Lemma 3 that there are no 3-position equilibria under either of the two rules. Hence, all equilibria are 1- and 2-position equilibria. Moreover, we know from Lemma 4 that every 1-position equilibrium is strictly less polarized than any 2-position equilibrium and that the set of policies supported by 1- or 2-position equilibria is an interval centered around m.

We start by considering an $\left( s,t\right)$-rule where $1<s=t\le p$. We know from Lemma 1 that the sets of 1-position equilibria are equivalent under the Plurality Rule and the $\left( s,t\right)$-rule. Given Lemma 4 and the non-existence of 3-position equilibria, the result stated in Proposition 1 is then established by showing that the set of 2-position equilibria under the $\left( s,t\right)$-rule is a superset of the set of 2-position equilibria under the Plurality Rule. This is trivial if there are no 2-position equilibria under the Plurality Rule. So suppose $\left( e,\alpha \right)$ is a 2-position equilibrium under the Plurality Rule. Denote the two positions by $x_{L}$ and $x_{R}$. Condition (2) of Lemma 2 implies $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$ and $c_{L}=c_{R}=1$. Moreover, Condition 3(a) must hold with $t=1$. To prove the result, it is sufficient to construct an equivalent equilibrium under the $\left( s,t\right)$-rule. Let the number of candidates at each position be $\widehat{c}_{L}=\widehat{c}_{R}=s$($=t$) and $\widehat{c}_{M}=0$. Thus, Condition (1) of Lemma 2 is satisfied. Given $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$, Condition (2) of Lemma 2 is satisfied as well. Finally, given $\widehat{c}_{L}=\widehat{c} _{R}=t$, Condition 3(a) applies. Since $t>1$, the fact that this subcondition is satisfied for the Plurality Rule implies it is satisfied for the $\left( s,t\right)$-rule as well. Finally, $c_{L}=c_{R}$ and $\widehat{ c}_{L}=\widehat{c}_{R}$ imply that $x_{L}$ and $x_{R}$ are each adopted with probability 1/2 under either of the two rules. Hence, we have constructed an equivalent 2-position equilibrium.

We now consider an $\left( s,t\right)$-rule where $s=t>p$. We show that the set of 2-position equilibria under the $\left( s,t\right)$-rule is a subset of the set of 2-position equilibria under the Plurality Rule. This is trivial if there are no 2-position equilibria under the $\left( s,t\right)$-rule. So suppose $\left( e,\alpha \right)$ is a 2-position equilibrium under the $\left( s,t\right)$-rule. Denote the positions by $x_{L}$ and $x_{R}$. Condition (2) of Lemma 2 implies $c_{L}=c_{R}=p$ and $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$. Moreover, Condition 3(c) must be satisfied, implying $\delta >-u\left( \left| x_{L}-x_{M}\right| \right)$. We now construct an equivalent equilibrium under the Plurality Rule. Let the number of candidates at each position be $\widehat{c}_{L}=\widehat{c}_{R}=1$ and $\widehat{c}_{M}=0$. Thus, Condition (1) of Lemma 2 is satisfied. Since $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$, Condition (2) is satisfied as well. Finally, Condition 3(a) applies. Since $\delta >-u\left( \left| x_{L}-x_{M}\right| \right)$, this condition is satisfied as well. Hence, we have constructed an equivalent 2-position equilibrium. $\square$

Proof of Proposition 2

(1) Consider the $\left( s,t\right)$- and $\left( t,t\right)$-rules where $s<t$. We start with a pair of observations. First, we know from Lemma 1 that 1-position equilibria are equivalent under both rules. Second, we know from Lemma 4 that under both rules, every 1-position equilibrium is strictly less polarized than any 2-position equilibrium and the set of policies supported by 1- or 2-position equilibria is an interval centered around m.

We now establish that the set of 2-position equilibria is more (or equally) polarized under the $\left( s,t\right)$-rule than under the $\left( t,t\right)$-rule. Given our observations above, it is sufficient to show that the set of policies which are supported by 2-position equilibria under the $\left( s,t\right)$-rule is a superset of the corresponding set under the $\left( t,t\right)$-rule. Let $\left( e^{\prime },\alpha ^{\prime }\right)$ be a 2-position equilibrium under the $\left( t,t\right)$-rule. Denote by $x_{L}$ and $x_{R}$ the two positions. We now construct an equivalent equilibrium under the $\left( s,t\right)$-rule. There are three cases to consider.

Case 1 $s<t\le p$. Condition (2) of Lemma 2 implies $c_{L}^{\prime }=c_{R}^{\prime }=t$ and $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$. Moreover, Condition 3(a) must be satisfied.

Suppose now that the election is held under the $\left( s,t\right)$-rule. Let there be $c_{L}=c_{R}\in \left\{ s,s+1,\ldots ,t\right\}$ candidates at $x_{L}$ and at $x_{R}$ such that Condition (2) of Lemma 2 is satisfied. (This is possible since $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$.) Also, let $c_{M}=0$, which implies Condition (1) of Lemma 2 is satisfied as well. It remains to show that Condition (3) is satisfied. This is obvious if $c_{L}=c_{R}=t$ since the condition is already satisfied under the $\left( t,t\right)$-rule. In case, $s\le c_{L}=c_{R}<t$, Condition 3(b) applies. Observe that $F\left( \underline{x}\right) >\frac{1 }{t}\left[ \left( t+1\right) F\left( \underline{x}\right) -1/2\right]$ and $F\left( \underline{x}\right) >\frac{t}{1+t}F\left( \underline{x}\right)$. This, together with the fact that Condition 3(a) is satisfied for $c_{L}^{\prime }=c_{R}^{\prime }=t$, implies Condition 3(b) is satisfied for $s\le c_{L}=c_{R}<t$. Hence, all three conditions of Lemma 2 are satisfied.

Case 2 $s\le p<t$. Condition (2) of Lemma 2 implies $c_{L}^{\prime }=c_{R}^{\prime }=p$ and $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$. Also, Condition 3(c) must be satisfied, which implies $\delta >-u\left( \left| x_{L}-x_{M}\right| \right)$.

Suppose now that the election is held under the $\left( s,t\right)$-rule. Proceeding in the same fashion as in Case 1 above, let $c_{M}=0$ and $c_{L}=c_{R}\in \left\{ s,\ldots ,p\right\}$ such that Conditions (1) and (2) of Lemma 2 are satisfied. (Again, this is possible since $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$.) Since $s\le c_{L}=c_{R}\le p<t$, Condition 3(b) applies. Observe that this condition is necessarily satisfied since $\delta >-u\left( \left| x_{L}-x_{M}\right| \right)$.

Case 3 $p<s<t$. As in Case 2 above, we have $c_{L}^{\prime }=c_{R}^{\prime }=p$ and $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta >-u\left( \left| x_{L}-x_{M}\right| \right)$.

Suppose now that the election is held under the $\left( s,t\right)$-rule. Let $c_{M}=0$ and $c_{L}=c_{R}=p$; Conditions $\left( 1\right)$ and $\left( 2\right)$ of Lemma 2 are therefore satisfied. Moreover, since $c_{L}=c_{R}=p<s$, Condition 3(c) applies, and is satisfied since $\delta >-u\left( \left| x_{L}-x_{M}\right| \right)$.

These three cases exhaust all possibilities. In each case we have constructed a 2-position equilibrium under the $\left( s,t\right)$-rule which is equivalent to $\left( e^{\prime },\alpha ^{\prime }\right)$ since $c_{L}^{\prime }=c_{R}^{\prime }$ and $c_{L}=c_{R}$ imply $x_{L}$ and $x_{R}$ are each adopted with probability 1/2 in both equilibria. Hence, we have shown that any 2-position equilibrium under the $\left( t,t\right)$-rule has an equivalent equilibrium under the $\left( s,t\right)$-rule. However, the reverse is not true.³³ Hence, the set of 2-position equilibria is more polarized under the $\left( s,t\right)$-rule than under the $\left( t,t\right)$-rule.

Finally, we know from Lemma 3 that there are no 3-position equilibrium under the $\left( t,t\right)$-rule. 3-position equilibria may however exist under the $\left( s,t\right)$-rule (e.g., see Example 1 in the online appendix). Since the set of policies supported by 1- or 2-position equilibria is an interval centered around m (by Lemma 4), the only policies that can be supported only by 3-position equilibria are more polarized than policies supported by 1- and 2-position equilibria.

We have thus established that an $\left( s,t\right)$-rule where $s<t$ can support more polarization than the $\left( t,t\right)$-rule.

(2) We now prove that an $\left( s,t\right)$-rule where $s<t$ supports more (or as much) polarization than the $\left( s,s\right)$-rule. For the same reason as in part (1) of the proof, we only need to show that the set of policies which are supported by 2-position equilibria under the $\left( s,t\right)$-rule is a superset of the corresponding set under the $\left( s,s\right)$-rule. Let $\left( e^{\prime },\alpha ^{\prime }\right)$ be a 2-position equilibrium under the $\left( s,s\right)$-rule. Denote by $x_{L}$ and $x_{R}$ the two positions. We now construct an equivalent equilibrium under the $\left( s,t\right)$-rule. There are two cases to consider.

Case 1 $s\le p$. Condition (2) of Lemma 2 implies $c_{L}^{\prime }=c_{R}^{\prime }=s$ and $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta$. Moreover, Condition 3(a) must be satisfied.

Suppose the election is held under the $\left( s,t\right)$-rule. Let there be $c_{L}=c_{R}\in \left\{ s,\ldots ,\min \left\{ t,p\right\} \right\}$ candidates at $x_{L}$ and at $x_{R}$ such that Condition (2) of Lemma 2 is satisfied. Also, let $c_{M}=0$, which implies Condition (1) of Lemma 2 is satisfied. It remains to show that Condition (3) is satisfied. One possibility to consider is $c_{L}=c_{R}=t$, in which case Condition 3(a) applies. Since $\frac{1}{t}\left[ \left( t+1\right) F\left( \underline{x} \right) -1/2\right]$ and $\frac{t}{1+t}F\left( \underline{x}\right)$ are both strictly increasing in t and $t>s$, the fact that Condition 3(a) is satisfied for $c_{L}^{\prime }=c_{R}^{\prime }=s$ implies it is satisfied for $c_{L}=c_{R}=t$ as well. The other possibility is $s\le c_{L}=c_{R}<t$, in which case Condition 3(b) applies. Since $F\left( \underline{x}\right) > \frac{1}{t^{\prime }}\left[ \left( t^{\prime }+1\right) F\left( \underline{x} \right) -1/2\right]$ and $F\left( \underline{x}\right) >\frac{s}{1+s} F\left( \underline{x}\right)$, the fact that Condition 3(a) is satisfied for $c_{L}^{\prime }=c_{R}^{\prime }=s$ implies Condition 3(b) is satisfied for $s\le c_{L}=c_{R}<t$. Hence, all three conditions of Lemma 2 are satisfied.

Case 2 $p<s$. Condition (2) of Lemma 2 implies $c_{L}^{\prime }=c_{R}^{\prime }=p$ and $-\frac{u\left( \left| x_{L}-x_{R}\right| \right) }{2}\ge \delta >-u\left( \left| x_{L}-x_{R}\right| \right)$.

Suppose the election is held under the $\left( s,t\right)$-rule. Let $c_{M}=0$ and $c_{L}=c_{R}=p$; Conditions (1) and (2) of Lemma 2 are therefore satisfied. Moreover, since $c_{L}=c_{R}=p<s$, condition 3(c) applies, and is satisfied since $\delta >-u\left( \left| x_{L}-x_{M}\right| \right)$.

These two cases exhaust all possibilities. In each case we have constructed a 2-position equilibrium under the $\left( s,t\right)$-rule which is equivalent to $\left( e^{\prime },\alpha ^{\prime }\right)$ since $c_{L}^{\prime }=c_{R}^{\prime }$ and $c_{L}=c_{R}$ imply $x_{L}$ and $x_{R}$ are each adopted with probability 1/2 in both equilibria. By the same argument as in part (1) of the proof we have then established the result.   $\square$