Incomplete information, proportional representation and strategic voting

Abstract

We introduce incomplete information to a multiparty election under proportional representation: each voter knows her preferences and votes strategically to maximize her payoffs, but is uncertain about the number and the preferences of the other voters. Parties are assumed to be purely office motivated and, hence, the resulting governments are always minimum winning. In this framework, we prove (a) generic existence of equilibria where only two parties receive a positive fraction of the votes and therefore lead to single-party governments and (b) generic inexistence of equilibria that lead to coalition governments. That is, contrary to common wisdom, a proportional rule is found not to promote sincere voting and to be favorable towards single-party governments. The existence of two-party equilibria that lead to single-party governments is robust to parties having ideological concerns.

Appendix (Proofs)

Lemma 1

If party J obtains the majority of the vote (that is $V_{J}>\frac{1}{2}k$) then by forming a single-party government $G=J$ where $W_{J}^{G}=1$ it obtains payoff $U_{J}(J)=1$. For any other coalition government $G\ne J$ then it is straightforward that $U_{J}(G)<U_{J}(J)$ since it holds that $W_{J}^{G}<1.$ If no party has a majority of votes then party J maximizes its utility by forming a coalition government with Q if and only if $V_{Q}\le V_{R}.$ ¹⁶ It is obvious that if all parties tie then any coalition is minimum winning. If at least one party’s number of votes is different from the ones of the other parties then (a) in case all number of votes are different the two bigger parties want to form a coalition with the smallest one and the smallest one wants to form a coalition with the second smallest one (b) if two parties tie and receive less votes than the other party then they wish to form a coalition with each other and (c) if two parties tie and receive more votes than the other party then both want to form a coalition with the smallest party while the smallest party is indifferent. In all three cases only a minimal winning coalition is stable, winning and vote-share consistent. $\square$

Proposition 1

A BNE consists of strategies and beliefs. We will prove that if the expected vote share of A is zero and the expected votes shares of B and C are strictly positive then a $t_{ABC}$ will vote for B (equivalently one can show that a $t_{ACB}$ voter will vote for C). Consider the first pivotal event $\hat{V}_{B}=\hat{V}_{C}-1$ as presented in Table 1, where $\hat{V}_{B}$ and $\hat{V}_{C}$ denote the number of votes that B and C receive by the rest of the population. If $\hat{V}_{B}=0$ then the voter is strictly better off voting A. If $\hat{V}_{B}=1$ then the voter is indifferent between voting A and B. If $\hat{V}_{B}>1$ then the voter is better off voting B for a large enough number of voters. Notice that $prob(\hat{V}_{B}=0|\hat{V}_{B}=\hat{V}_{C}-1)\rightarrow 0$, $prob(\hat{V}_{B}=1|\hat{V}_{B}=\hat{V}_{C}-1)\rightarrow 0$ and $prob(\hat{V} _{B}>1|\hat{V}_{B}=\hat{V}_{C}-1)\rightarrow 1$ as $n\rightarrow \infty$ and that utility differences are bounded from above and below (conditional on $\hat{V}_{B}=\hat{V}_{C}-1$ our voter’s choice will affect her utility by at most $\frac{1}{2}$ independently of how many voters vote). Therefore, conditional on $\hat{V}_{B}=\hat{V}_{C}-1$ the voter decides which party to support based on the most probable event in large elections (i.e. $\hat{V} _{B}>1$) and votes B (i.e. the expected utility difference by voting B rather than A is equal to $\frac{1}{2}$). Exactly the same argument applies for the second pivotal event where $\hat{V}_{B}=\hat{V}_{C}$. Consider now the third pivotal event $\hat{V}_{B}=\hat{V}_{C}+1$. If $\hat{V} _{C}=0$ then the voter is strictly better off voting A. If $\hat{V}_{C}=1$ then the voter is indifferent between voting A and B. If $\hat{V}_{B}>1$ then for a large enough number of voters the voter is essentially indifferent between voting A and B. Given the same argument regarding pivotal probabilities, conditional on $\hat{V}_{B}=\hat{V}_{C}+1$ the expected utility difference by voting B rather than A is negative but converges to zero as $n\rightarrow \infty$. Notice now that all three pivotal events are equiprobable (Myerson 2000). Given the positive and non-degenerate utility difference from the first two events a $t_{ABC}$ will vote for B (equivalently a $t_{ACB}$ voter will vote for C). Hence, beliefs $\{v_{A}=0,v_{B}=p(t_{BAC})+p(t_{BCA})+p(t_{ABC}),v_{C}=p(t_{CAB})+p(t_{CBA})+p(t_{ACB})\}$ and the strategy profile that is consistent with these beliefs form a BNE. $\square$

Proposition 2

Consider that there exists a BNE that leads to a coalition government with a non-degenerate probability. Since we focus on pure strategy equilibria, the beliefs which are part of such a BNE should be consistent with $0<v_{J}<v_{Q}<v_{R}<1/2$ for any generic distribution of players’ types. Let us assume without loss of generality that $0<v_{A}<v_{B}<v_{C}<1/2$. In such case the probability with which the minimal winning coalition $\{A,B\}$ will be formed after the election converges to one as n becomes arbitrarily large. Let us study the problem that a $t_{CAB}$ voter faces. Her vote may increase the weight of a party in a coalition government (such a change in the weight is bounded from below by $\frac{1}{k+1}$) or/and it may determine the coalition government that will be formed. Conditional on the outcome being such that a vote cannot affect the coalition government that will be formed, a $t_{CAB}$ voter obviously prefers to vote for party A since the coalition government $\{A,B\}$ will be formed with infinitely larger probability than any other coalition government. Now, conditional on her vote being pivotal for the determination of the exact coalition (or single-party) government a $t_{CAB}$ voter could be better off by not voting for party A. Since parties are purely office motivated, all such pivotal events are a subset of the cases in which the difference between the votes of (at least) two parties is not more than one vote plus the case in which a party lacks one vote to form a single-party government. Moreover, the probability of any such pivotal event must not exceed the probability of the most probable tie which is given by $\frac{ e^{-n(\sqrt{\hat{\psi }}-\sqrt{\hat{\omega }})^{2}}}{2\sqrt{\pi n\sqrt{\hat{ \psi }\hat{\omega }}}}$ (see Myerson 2000) where $(\hat{\psi },\hat{\omega } )=\arg \min \{(\sqrt{\psi }-\sqrt{\omega })^{2}$ such that $\psi$ and $\omega$ take distinct values from the set $\{v_{A},v_{B},v_{C},1-v_{A},1-v_{B},1-v_{C}\}\}$.

Then, to establish that a $t_{CAB}$ voter would prefer to vote for A it is enough to show that

$$\begin{aligned}&\lim _{n\rightarrow +\infty }\frac{E\left( \frac{1}{k+1}|\frac{k}{2}-1>\hat{V} _{C}>\max \{\hat{V}_{A},\hat{V}_{B}\}+1\right) \times prob\left( \frac{k}{2}-1>\hat{V} _{C}>\max \{\hat{V}_{A},\hat{V}_{B}\}+1\right) }{\frac{e^{-n\left( \sqrt{\hat{\psi }}- \sqrt{\hat{\omega }}\right) ^{2}}}{2\sqrt{\pi n\sqrt{\hat{\psi }\hat{\omega }}}}} =+\infty . \end{aligned}$$

Given that $E(\frac{1}{k+1}|\frac{k}{2}-1>\hat{V}_{C}>\max \{\hat{V}_{A}, \hat{V}_{B}\}+1)>\frac{1}{2}E(\frac{1}{k+1}|k\ge m)=\frac{1}{2}\frac{\Gamma (1+m)-\Gamma (1+m,n)}{n\Gamma (1+m)}$ where $\Gamma (z)$ is the Euler gamma function, $\Gamma (a,z)$ is the incomplete gamma function and m a sufficiently large integer, we can establish the above relationship if the following is true:

$$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{prob\left( \frac{k}{2}-1>\hat{V}_{C}>\max \left\{ \hat{V}_{A},\hat{V}_{B}\right\} +1\right) }{\frac{2\Gamma (1+m)}{\Gamma (1+m)-\Gamma (1+m,n)}\frac{\sqrt{n}e^{-n\left( \sqrt{\hat{\psi }}-\sqrt{\hat{\omega }}\right) ^{2}}}{2 \sqrt{\pi \sqrt{\hat{\psi }\hat{\omega }}}}}=+\infty . \end{aligned}$$

Pure office motivation and $0<v_{A}<v_{B}<v_{C}<1/2$ suggest that $prob( \frac{k}{2}-1>\hat{V}_{C}>\max \{\hat{V}_{A},\hat{V}_{B}\}+1)\rightarrow 1$ as $n\rightarrow +\infty$, m being positive suggests that $\lim _{n\rightarrow +\infty }\frac{2\Gamma (1+m)}{\Gamma (1+m)-\Gamma (1+m,n)} =2$ and $(\sqrt{\hat{\psi }}-\sqrt{\hat{\omega }})^{2}>0$ suggests that $\lim _{n\rightarrow +\infty }\frac{\sqrt{n}e^{-n(\sqrt{\hat{\psi }}-\sqrt{\hat{ \omega }})^{2}}}{2\sqrt{\pi \sqrt{\hat{\psi }\hat{\omega }}}}=0$. Hence, the required relationship holds.

This proves that when a strategy profile generates expected vote shares $0<v_{A}<v_{B}<v_{C}<1/2$, then a $t_{CAB}$ voter is strictly better off by voting for A than for C. Similarly one can show that a $t_{CBA}$ voter is strictly better off by voting for B than for C and that C is not a best response to such expectations for any voter type. That is, beliefs which are consistent with $0<v_{A}<v_{B}<v_{C}<1/2$ cannot be part of a BNE since they lead to $\hat{V}_{C}=0$ and, hence, to a voting behavior which is inconsistent with them. $\square$

Lemma 2

Similar to Lemma 1 if a party J obtains the majority of the votes then by performing a single-party government $G=J$ where $W_{J}^{G}=1$ it maximizes its payoff $U_{J}(J)=1$. To show that under policy motives the government coalition may be non-minimal let $\lambda _{A}=\lambda _{B}=\lambda _{C}=\lambda <1$ and $\tilde{V}_{A}=\tilde{V} _{C}+\varepsilon =\tilde{V}_{B}+2\varepsilon$ where $\tilde{V} _J=V_J/(V_A+V_B+V_C)$ denotes party’s J vote share. Assume that the policy preferences of party A are compatible with those of a voter of type $t_{ABC}$ and the policy preferences of party B are compatible with those of a voter of type $t_{BAC}.$ Then due to both office seeking incentives and ideology proximity party A strictly prefers to form a coalition with B than with C. Now, B strictly prefers to form a coalition with A rather than with C if and only if $U_{B}(\{A,B\})>U_{B}(\{B,C\}).$ The latter holds if and only if:

$$\begin{aligned}&\lambda \frac{V_{B}}{V_{A}+V_{B}}+(1-\lambda )\left[ \frac{V_{B}}{V_{A}+V_{B}} u_{t_{BAC}}(B)+\frac{V_{A}}{V_{A}+V_{B}}u_{t_{BAC}}(A)\right]>\lambda \frac{V_{B}}{ V_{B}+V_{C}}\\&\qquad +\;(1-\lambda )\left[ \frac{V_{B}}{V_{B}+V_{C}}u_{t_{BAC}}(B)+\frac{V_{C} }{V_{B}+V_{C}}u_{t_{BAC}}(C)\right] \\&\quad \Longrightarrow \lambda \frac{V_{B}}{V_{A}+V_{B}}+(1-\lambda )\left[ \frac{V_{B}}{V_{A}+V_{B}} \right]>\lambda \frac{V_{B}}{V_{B}+V_{C}}\\&\qquad +\;(1-\lambda )\left[ \frac{V_{B}}{V_{B}+V_{C}}- \frac{V_{C}}{V_{B}+V_{C}}\right] \\&\quad \Longrightarrow \frac{V_{B}}{V_{A}+V_{B}}>\frac{V_{B}}{V_{B}+V_{C}}-(1-\lambda )\frac{V_{C} }{V_{B}+V_{C}}\\&\quad \Longrightarrow \frac{\tilde{V}_{B}}{2\tilde{V}_{B}+2\varepsilon }>\frac{\tilde{V}_{B}}{2 \tilde{V}_{B}+\varepsilon }-(1-\lambda )\frac{\tilde{V}_{B}+\varepsilon }{2\tilde{V}_{B}+\varepsilon }\\&\quad \Longrightarrow \frac{\tilde{V}_{B}}{2\tilde{V}_{B}+2\varepsilon }>\frac{\tilde{V}_{B}- \tilde{V}_{B}-\varepsilon +\lambda \tilde{V}_{B}+\lambda \varepsilon }{2 \tilde{V}_{B}+\varepsilon }\\&\quad \Longrightarrow \frac{\tilde{V}_{B}}{2\tilde{V}_{B}+2\varepsilon }>\frac{\lambda \tilde{V} _{B}+\varepsilon (1-\lambda )}{2\tilde{V}_{B}+\varepsilon } \end{aligned}$$

For small values of $\varepsilon$ (that is for $\varepsilon \rightarrow 0$) the above becomes $\frac{1}{2}>\frac{\lambda }{2}$ and holds for any $\lambda \in [0,1).$ $\square$

Lemma 3

Let $V_{A}=V_{B}+\eta >V_{C}=1,$ where $\eta \in \{-1,0,1\}$. Therefore, $\frac{k}{2}-1\le V_{A}\le \frac{k}{2}+1$. Assume that $\eta =1$ (the equivalent arguments holds for $\eta =-1$). In this case $V_{A}= \frac{k}{2}$ and the only winning coalition in which C participates is $\{A,C\}.$ Thus, we are interested in finding for which values of k the $\{A,C\}$ coalition is also stable. Notice that $\{A,C\}$ is stable if and only if A prefers to form a coalition with C than with B. It is obvious that $\frac{V_{A}}{V_{A}+V_{C}}>\frac{V_{A}}{V_{A}+V_{B}}$, that is, if A is a purely office-motivated party it would prefer to form a coalition with C. To establish our result we need to find the condition that guarantees that $\frac{V_{A}}{V_{A}+V_{C}}u_{t_{ABC}}(A)+\frac{V_{C}}{ V_{A}+V_{C}}u_{t_{ABC}}(C)>\frac{V_{A}}{V_{A}+V_{B}}u_{t_{ABC}}(A)+\frac{ V_{B}}{V_{A}+V_{B}}u_{t_{ABC}}(B)$ as well. The latter inequality implies that even if party A is purely ideology motivated and even if it dislikes the policy of party C the most (party A has preferences compatible with a $t_{ABC}$ voter), it would still prefer to form a coalition with C rather than with B. We know that $\frac{V_{A}}{V_{A}+V_{C}}u_{t_{ABC}}(A)+ \frac{V_{C}}{V_{A}+V_{C}}u_{t_{ABC}}(C)=\frac{V_{A}-V_{C}}{V_{A}+V_{C}}$ and that $\frac{V_{A}}{V_{A}+V_{B}}u_{t_{ABC}}(A)+\frac{V_{B}}{V_{A}+V_{B}} u_{t_{ABC}}(B)=\frac{V_{A}}{V_{A}+V_{B}}.$ We notice that $\frac{V_{A}-V_{C} }{V_{A}+V_{C}}=\frac{\frac{k}{2}-1}{\frac{k}{2}+1}$ and that $\frac{V_{A}}{ V_{A}+V_{B}}=\frac{\frac{k}{2}}{k-1}.$ Therefore, the required inequality holds if $\frac{\frac{k}{2}-1}{\frac{k}{2}+1}>\frac{\frac{k}{2}}{k-1}$ and since k is always a natural number, the later inequality holds if and only if $k>7.$ We replicate this exercise for $\eta =0$ and we also find that $k>7$ is a sufficient condition for the existence of a stable coalition in which C participates. The only difference of the $\eta =0$ case compared to the cases in which $\eta \in \{-1,1\}$ is that there are two winning coalitions in which C participates; $\{A,C\}$ and $\{B,C\}.$ Since both A and B received exactly the same votes, party C will choose partner according to its ideological preferences (or it will choose a partner randomly if it is purely office-motivated) and $k>7$ will guarantee that A (B) strictly prefers to be in a coalition with C rather than with B ( A ). $\square$

Proposition 3

A BNE consists of strategies and beliefs. We will prove that if the expected vote share of A is zero and the expected vote shares of the other two parties are positive (formally, $v_{A}=0$, $v_{B}>0$ and $v_{C}>0$ such that $v_{A}+v_{B}+v_{C}=1$) then a $t_{ABC}$ will vote for B (equivalently one can show that a $t_{ACB}$ voter will vote for C) . That is, we will show that beliefs $\{v_{A}=0,v_{B}=p(t_{BAC})+p(t_{BCA})+p(t_{ABC}),v_{C}=p(t_{CAB})+p(t_{CBA})+p(t_{ACB})\}$ and the strategy profile which is consistent with these beliefs form a BNE. We know that $A^{\prime }s$ ideological preferences should be given by one of the strict orders $\{A\succ B\succ C\}$ or $\{A\succ C\succ B\}$ (assume without loss of generality that $A^{\prime }s$ preferences are given by the first linear order). For large polities ($n\rightarrow \infty$) the probability that there exists a positive measure of voters with preferences that are compatible with $A^{\prime }s$ ideological preferences converges to one. Let us investigate the problem that a voter i with such preferences faces in this case. If she expects that $v_{A}=0$, $v_{B}>0$ and $v_{C}>0$ then the only possibility that her vote can define the outcome is when $\hat{V}_{B}=\hat{V}_{C}+\eta$ where $\eta \in \{-1,0,1\}$. Notice that a) in such eventualities voting for C is always a strictly dominated strategy and b) $\eta \in \{-1,1\}$ implies that k is even and $\eta =0$ implies that k is odd. We detail below the possible pivotal eventualities.

(I) If $\eta =-1$ then voter i gets utility $-\frac{1}{2}$ if she votes for B (because B and C will form a coalition in which they will have equal weights). (a) If $k>7$ and our voter votes for A then both coalitions $\{A,C\}$ and $\{B,C\}$ are stable. In this case the coalition $\{A,C\}$ is formed with probability $0<<\xi<<1$ and the coalition $\{B,C\}$ is formed with probability $1-\xi .$ Voter i gets utility $-\frac{\hat{V} _{C}-1}{1+\hat{V}_{C}}$ if she votes for A and $\{A,C\}$ is formed (because the condition of Lemma 3 is satisfied and A and C can form a coalition in which they will have weights $\frac{1}{1+\hat{V}_{C}}$ and $\frac{\hat{V}_{C}}{1+\hat{V}_{C}}$ respectively) and utility strictly less than $-\frac{1}{2}$ if she votes for A and $\{B,C\}$ is formed. (b) If $k\le 7$ voter i gets utility of, at most, equal to 0 when the voter votes for A (and government $\{A,C\}$ is formed if $\hat{V} _{A}=0,\hat{V}_{B}=0, \hat{V}_{C}=1$).

(II) If $\eta =0$ then voter i gets utility 0 if she votes for B (because B will form a single-party government) and if $k>7$ voter i gets utility $\frac{1}{1+\hat{V}_{B}}$ if she votes for A. This is because the condition of Lemma 3 is satisfied and, thus, B will prefer to form a coalition with A rather than with C. Due to ideological proximity with B , A will prefer to form a coalition with$\ B$ and, thus A and B will form a coalition in which they will have weights $\frac{1}{1+\hat{V}_{B} }$ and $\frac{\hat{V}_{B}}{1+\hat{V}_{B}}$ respectively. If $k\le 7$ voter i gets utility of, at most, equal to 1 by voting A (and a single-party government $\{A\}$ is formed if $\hat{V}_{A}=0,\hat{V}_{B}=0, \hat{V}_{C}=0$).

(III) If $\eta =1$ then voter i gets utility 0 if she votes for B (because B will form a single-party government) and if $k>7$ voter i gets utility $\frac{1}{1+\hat{V}_{B}}$ if she votes for A. This is because the condition of Lemma 3 is satisfied and, thus, B will prefer to form a coalition with A rather than with C. Due to ideological proximity with B , A will prefer to form a coalition with$\ B$ and, thus A and B will form a coalition in which they will have weights $\frac{1}{1+\hat{V}_{B} }$ and $\frac{\hat{V}_{B}}{1+\hat{V}_{B}}$ respectively. If $k\le 7$ voter i gets utility of, at most, equal to $\frac{1}{2}$ by voting A (and government $\{A,B\}$ is formed if $\hat{V}_{A}=0,\hat{V}_{B}=1, \hat{V} _{C}=0$).

Formally, voter i will vote for her second ranked candidate B if the following holds:

$$\begin{aligned} prob(\hat{V}_{B}= & {} \hat{V}_{C}-1\,and\,k>7)\\&\times \left[ -\frac{1}{2}-\xi \left( -E\left( \frac{\hat{V}_{C}-1}{1+\hat{V}_{C}}|\hat{V}_{B}=\hat{V}_{C}-1\,and \,k>7\right) \right) -(1-\xi )\left( -\frac{1}{2}\right) \right] \\&+\;prob\left( \hat{V}_{B} =\hat{V}_{C}-1\,and\,k\le 7\right) \left[ -\frac{1}{2}-0\right] \\&+\;prob\left( \hat{V}_{B} =\hat{V}_{C}\,and\,k>7\right) \left[ 0-E\left( \frac{1}{1+\hat{V }_{B}}|\hat{V}_{B}=\hat{V}_{C}\,and\,k>7\right) \right] \\&+\;prob\left( \hat{V}_{B} =\hat{V}_{C}\,and\,k\le 7\right) [0-1]\\&+\;prob\left( \hat{V}_{B} =\hat{V}_{C}+1\,and\,k>7\right) \\&\times \left[ 0-E\left( \frac{1}{1+ \hat{V}_{B}}|\hat{V}_{B}=\hat{V}_{C}+1\,and\,k>7\right) \right] \\&+\;prob\left( \hat{V}_{B} =\hat{V}_{C}+1\,and\,k\le 7\right) \left[ 0-\frac{1}{2} \right] >0 \end{aligned}$$

We observe that:

$$\begin{aligned} prob(\hat{V}_{B}=\hat{V}_{C}-1\,and\,k\le 7)= & {} prob(\hat{V}_{B}=\hat{V} _{C}-1|k\le 7)\times prob(k\le 7),\\ prob(\hat{V}_{B}=\hat{V}_{C}\,and\,k\le 7)= & {} prob(\hat{V}_{B}=\hat{V} _{C}|k\le 7)\times prob(k\le 7)\, \hbox {and}\\ prob(\hat{V}_{B}=\hat{V}_{C}+1\,and\,k\le 7)= & {} prob(\hat{V}_{B}=\hat{V} _{C}+1|k\le 7)\times prob(k\le 7). \end{aligned}$$

It is obvious that all $prob(\hat{V}_{B}=\hat{V}_{C}-1|k\le 7),prob(\hat{V} _{B}=\hat{V}_{C}|k\le 7)$ and $prob(\hat{V}_{B}=\hat{V}_{C}+1|k\le 7)$ take a positive value significantly higher than 0. On the other side we have that $prob(k\le 7)=\overset{7}{\sum \nolimits _{m=0}}\frac{n^{m}}{m!} e^{-n}$.

Notice that $prob(\hat{V}_{B}=\hat{V}_{C}-1)=$ $prob(\hat{V}_{B}= \hat{V}_{C}-1$ and $k>7)+prob(\hat{V}_{B}=\hat{V}_{C}-1$ and $k\le 7)$ and $prob(\hat{V}_{B}=\hat{V}_{C}-1$ and $k\le 7)=prob(\hat{V}_{B}=\hat{V} _{C}-1|k\le 7)\times prob(k\le 7).$ That is, $1=$ $\frac{prob(\hat{V}_{B}= \hat{V}_{C}-1\,and\,k>7)}{prob(\hat{V}_{B}=\hat{V}_{C}-1)}+prob( \hat{V}_{B}=\hat{V}_{C}-1|k\le 7)\times \frac{prob(k\le 7)}{prob(\hat{V} _{B}=\hat{V}_{C}-1)}.$

By Myerson (2000), we know that $prob(\hat{V}_{B}=\hat{V}_{C}-1)\simeq \frac{ e^{n(2\sqrt{v_{B}v_{C}}-v_{B}-v_{C})}}{2\sqrt{\pi n\sqrt{v_{B}v_{C}}}}$ for $n\rightarrow \infty .$ So it must be the case that for $n\rightarrow \infty ,$ $prob(k\le 7)$ becomes infinitely smaller than $prob(\hat{V}_{B}=\hat{V} _{C}-1).$ This is due to the fact that $\overset{7}{\sum \nolimits _{m=0}}\frac{ n^{m}}{m!}e^{-n}<\frac{e^{n(2\sqrt{v_{B}v_{C}}-v_{B}-v_{C})}}{2\sqrt{\pi n \sqrt{v_{B}v_{C}}}}$ is equivalent to $\frac{2\sqrt{\pi n\sqrt{v_{B}v_{C}}} \overset{7}{\sum \nolimits _{m=0}}\frac{n^{m}}{m!}}{e^{n(2\sqrt{v_{B}v_{C}} -v_{B}-v_{C})+n}}<1$ and $2\sqrt{\pi n\sqrt{v_{B}v_{C}}}\overset{7}{ \sum \nolimits _{m=0}}\frac{n^{m}}{m!}$ increases in a polynomial rate while $e^{n(2\sqrt{v_{B}v_{C}}-v_{B}-v_{C})+n}$ increases in an exponential rate ($n(2\sqrt{v_{B}v_{C}}-v_{B}-v_{C})+n$ is always positive for any $v_{B}>0$ and $v_{C}>0$). That is, $\frac{2\sqrt{\pi n\sqrt{v_{B}v_{C}}}\overset{7}{ \sum \nolimits _{m=0}}\frac{n^{m}}{m!}}{e^{n(2\sqrt{v_{B}v_{C}}-v_{B}-v_{C})+n}} \rightarrow 0$ and therefore $\frac{prob(k\le 7)}{prob(\hat{V}_{B}=\hat{V} _{C}-1)}\rightarrow 0.$ This implies that for $n\rightarrow \infty$ we have $1=$ $\frac{prob(\hat{V}_{B}=\hat{V}_{C}-1\,and\,k>7)}{prob(\hat{ V}_{B}=\hat{V}_{C}-1)}+prob(\hat{V}_{B}=\hat{V}_{C}-1|k\le 7)\times 0$ which is equivalent to $\frac{prob(\hat{V}_{B}=\hat{V}_{C}-1\,and\,k>7)}{prob(\hat{V}_{B}=\hat{V}_{C}-1)}\rightarrow 1$ and to $prob(\hat{V} _{B}=\hat{V}_{C}-1$ and $k>7)\rightarrow prob(\hat{V}_{B}=\hat{V}_{C}-1).$

With the same logic we can demonstrate that $prob(\hat{V}_{B}=\hat{V}_{C}$ and $k>7)\rightarrow prob(\hat{V}_{B}=\hat{V}_{C})$ and that $prob(\hat{V} _{B}=\hat{V}_{C}+1$ and $k>7)\rightarrow prob(\hat{V}_{B}=\hat{V}_{C}+1).$

By this last observation and the offset theorem of Myerson (2000) we have that $\frac{prob(\hat{V}_{B}=\hat{V}_{C}\,and\,k>7)}{prob(\hat{V} _{B}=\hat{V}_{C}-1\,and\,k>7)}\rightarrow \sqrt{\frac{v_{B}}{ v_{C}}}$ and $\frac{prob(\hat{V}_{B}=\hat{V}_{C}+1\,and\,k>7)}{ prob(\hat{V}_{B}=\hat{V}_{C}-1\,and\,k>7)}\rightarrow \frac{v_{B} }{v_{C}}.$

All these imply that if we divide our relevant inequality with $prob(\hat{V} _{B}=\hat{V}_{C}-1$ and $k>7)$ we should get:

$$\begin{aligned}&\xi \left[ -\frac{1}{2}-\left( -E\left( \frac{\hat{V}_{C}-1}{1+\hat{V}_{C}}|\hat{V} _{B}=\hat{V}_{C}-1\,and\,k>7\right) \right) \right] \\&\quad +\,prob\left( \hat{V}_{B} =\hat{V}_{C}-1|k\le 7\right) \frac{prob(k\le 7)}{prob\left( \hat{V} _{B}=\hat{V}_{C}-1\,and\,k>7\right) }\left[ -\frac{1}{2}-0\right] \\&\quad +\,\sqrt{\frac{v_{B}}{v_{C}}}\left[ 0-E\left( \frac{1}{1+\hat{V}_{B}}|\hat{V}_{B}=\hat{V} _{C}\,and\,k>7\right) \right] \\&\quad +\,prob\left( \hat{V}_{B} =\hat{V}_{C}|k\le 7\right) \frac{prob(k\le 7)}{prob\left( \hat{V} _{B}=\hat{V}_{C}-1\,and\,k>7\right) }[0-1] \\&\quad +\,\frac{v_{B}}{v_{C}}\left[ 0-E\left( \frac{1}{1+\hat{V}_{B}}|\hat{V}_{B}=\hat{V}_{C}+1 \,and\,k>7\right) \right] \\&\quad +\,prob\left( \hat{V}_{B} =\hat{V}_{C}|k\le 7\right) \frac{prob\left( k\le 7\right) }{prob\left( \hat{V} _{B}=\hat{V}_{C}-1\,and\,k>7\right) }\left[ 0-\frac{1}{2}\right] >0 \end{aligned}$$

Since

$$\begin{aligned}&E\left( \frac{\hat{V}_{C}-1}{1+\hat{V}_{C}}|\hat{V}_{B}=\hat{V}_{C}-1\,and\, k>7\right) \\&\quad =1-E\left( \frac{2}{1+\hat{V}_{C}}|\hat{V}_{B}=\hat{V}_{C}-1\,and\, k>7\right) \rightarrow 1 \end{aligned}$$

and

$$\begin{aligned} E\left( \frac{1}{1+\hat{V}_{B}}|\hat{V}_{B}=\hat{V}_{C}\, and\, k>7\right) \rightarrow E\left( \frac{1}{1+\hat{V}_{B}}|\hat{V}_{B}=\hat{V}_{C}+1\, and\, k>7\right) \rightarrow 0 \end{aligned}$$

the above inequality becomes $\frac{\xi }{2}>0$ and always holds. That is, when $v_{A}=0$, $v_{B}>0$ and $v_{C}>0$ then a $t_{ABC}$ will vote for B (equivalently a $t_{ACB}$ voter will vote for C). Therefore, beliefs $\{v_{A}=0,v_{B}=p(t_{BAC})+p(t_{BCA})+p(t_{ABC}),v_{C}=p(t_{CAB})+p(t_{CBA})+p(t_{ACB})\}$ and the strategy profile which is consistent with these beliefs form a BNE. $\square$