Rank-dominant strategy and sincere voting

Abstract

This study considers a voting rule wherein each player sincerely votes when he/she has no information about the preferences of the other players. We introduce the concept of rank-dominant strategies to discuss the situation where a player is completely ignorant in the preferences of the other players and decision theoretic justification of the concept. We show that under the plurality voting rule with the equal probability random tie-breaking, sincere voting is always the rank-dominant strategy of each voter. We also discuss other scoring rules and show that sincere voting may not be a rank-dominant strategy of a voter even with the equal probability random tie-breaking.

Appendix

1.1 Proof of Fact 1

Suppose that there exists a rank-dominant strategy of i denoted by $m_{i}$ . We show the first result. Since $m_{i}$ is a rank-dominant strategy of i , $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) \ge U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$ for all $s\in \left\{ 1,\ldots ,\theta _{i}\right\}$ and all $m_{i}^{\prime }\in M_{i}$. Then $m_{i}$ is not rank-dominated by any strategy in $M_{i}$.

We show the second result. If $m_{i}^{\prime }\in M_{i}$ is also a rank-dominant strategy of i, then both $U_{i}^{f}\left( m_{i}^{\prime }, \mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right) \ge U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)$ and $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) \ge U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$ for all $s\in \left\{ 1,\ldots ,\theta _{i}\right\}$. Therefore, $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) =U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$ for all $s\in \left\{ 1,\ldots ,\theta _{i}\right\}$.

We show the third result. If $m_{i}^{\prime }\in M_{i}$ is not a rank-dominant strategy of i, then there is $m_{i}^{\prime \prime }$ such that $U_{i}^{f}\left( m_{i}^{\prime \prime },\mathrm {m}_{-i}^{t}\left( m_{i}^{\prime \prime }\right) \right) >U_{i}^{f}\left( m_{i}^{\prime }, \mathrm {m}_{-i}^{t}\left( m_{i}^{\prime }\right) \right)$ for some $t\in \left\{ 1,\ldots ,\theta _{i}\right\}$. Since $m_{i}$ is a rank-dominant strategy, then $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{t}\left( m_{i}\right) \right) \ge U_{i}^{f}\left( m_{i}^{\prime \prime },\mathrm {m} _{-i}^{t}\left( m_{i}^{\prime \prime }\right) \right) >U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{t}\left( m_{i}^{\prime }\right) \right)$. Moreover, since $m_{i}$ a rank-dominant strategy, $U_{i}^{f}\left( m_{i}, \mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) \ge U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$ for all $s\in \left\{ 1,\ldots ,\theta _{i}\right\}$. Therefore, $m_{i}$ rank-dominates $m_{i}^{\prime }$.

1.2 Proof of Lemma 1

We show the necessary part (the only-if-part). Suppose that $m_{i}^{\prime }$ is rank-dominated by $m_{i}$. First, we show (1). Toward a contradiction, suppose that (1) is not satisfied for $x\ge 0$ or (1) is satisfied in equality for any x. In the latter case, the fact that (1) is satisfied in equality for any x contradicts that $m_{i}^{\prime }$ is not rank-dominated by $m_{i}$.

Thus, we consider the former case. Then there is $x\ge 0$ such that $\left| M_{-i}^{f}\left( m_{i}^{\prime },x\right) \right| >\left| M_{-i}^{f}\left( m_{i},x\right) \right| \ge 0$. Then $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right) \ge x$ for some $s=1,\ldots ,\theta _{i}$. If $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{\theta _{i}}\left( m_{i}^{\prime }\right) \right) \ge x$, then $x>U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{\theta _{i}}\left( m_{i}\right) \right)$. Then $U_{i}^{f}\left( m_{i}^{\prime }, \mathrm {m}_{-i}^{\theta _{i}}\left( m_{i}^{\prime }\right) \right) \ge x>U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{\theta _{i}}\left( m_{i}\right) \right)$ contradicts that $m_{i}^{\prime }$ is rank-dominated by $m_{i}$. Thus, there is $s=1,\ldots ,\theta _{i}-1$ such that $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right) \ge x>U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s+1}\left( m_{i}^{\prime }\right) \right)$. Then $x>U_{i}^{f}\left( m_{i},\mathrm {m} _{-i}^{s}\left( m_{i}\right) \right)$ and $U_{i}^{f}\left( m_{i}^{\prime }, \mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right) >U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)$. This also contradicts that $m_{i}^{\prime }$ is rank-dominated by $m_{i}$. Therefore, we have (1) for all $x\ge 0$.

Second, we show that the strict inequality of (1) is satisfied for some x. Since $m_{i}^{\prime }$ is rank-dominated by $m_{i}$, we can let $s=1,\ldots ,\theta _{i}$ be such that $U_{i}^{f}\left( m_{i},\mathrm {m} _{-i}^{s}\left( m_{i}\right) \right) >U_{i}^{f}\left( m_{i}^{\prime }, \mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$. Then

$$\begin{aligned} \left| M_{-i}^{f}\left( m_{i},x\right) \right| \ge s>\left| M_{-i}^{f}\left( m_{i}^{\prime },x\right) \right| \end{aligned}$$

for any $x\in \left( U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m} _{-i}^{s}\left( m_{i}^{\prime }\right) \right) ,U_{i}^{f}\left( m_{i}, \mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) \right)$. Hence, the strict inequality of (1) is satisfied for some $x\ge 0$.

Next, show the sufficiency part (the if-part). Suppose that (1) is satisfied for any $x\ge 0$ and the strict inequality of (1) is satisfied for some $x\ge 0$. First, toward a contradiction, suppose that $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) <U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$ for some $s=1,\ldots ,\theta _{i}$. Let $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right) =x$ . Then

$$\begin{aligned} \left| M_{-i}^{f}\left( m_{i}^{\prime },x\right) \right| =s>\left| M_{-i}^{f}\left( m_{i},x\right) \right| , \end{aligned}$$

contradicting that (1) is satisfied for any $x\ge 0$. Therefore, $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) \ge U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)$ for all $s=1,\ldots ,\theta _{i}$.

Finally, let x be such that $\left| M_{-i}^{f}\left( m_{i},x\right) \right| >\left| M_{-i}^{f}\left( m_{i}^{\prime },x\right) \right|$ and $\left| M_{-i}^{f}\left( m_{i},x\right) \right| =t$ . Then we have $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{t}\left( m_{i}\right) \right) >U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m} _{-i}^{t}\left( m_{i}^{\prime }\right) \right)$. Therefore, $m_{i}^{\prime }$ is rank-dominated by $m_{i}$.

1.3 Proof of Proposition 1

First, we show the second result, because the first result is straightforward from this. By Lemma 1,

$$\begin{aligned} \left| \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right| \ge \left| \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right| \text { for all }\delta =1,\ldots ,\Delta \end{aligned}$$

and the strict inequality holds for some $\delta =1,\cdots ,\Delta$. By Assumption 1,

$$\begin{aligned} v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right) \ge v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right) \text { for all }\delta =1,\ldots ,\Delta , \end{aligned}$$

(5)

and the strict inequality holds for some $\delta =1,\ldots ,\Delta$. Therefore, we have $V_{i}^{f}\left( m_{i}\right) >V_{i}^{f}\left( m_{i}^{\prime }\right)$.

By the second result of Proposition 1 and the third result of Fact 1, we have the first result of Proposition 1.

Next, we show the third result. Suppose that $m_{i}^{\prime }$ is not rank-dominated by $m_{i}$ and $\mu _{i}^{f}\left( m_{i}\right) \ne \mu _{i}^{f}\left( m_{i}^{\prime }\right)$. Then there is $s\in \{1,\ldots ,\theta _{i}\}$, such that $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) <U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m} _{-i}^{s}\left( m_{i}^{\prime }\right) \right)$. We choose the largest integer $s^{*}$ satisfying $U_{i}^{f}\left( m_{i},\mathrm {m} _{-i}^{s^{*}}\left( m_{i}\right) \right) <U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s^{*}}\left( m_{i}^{\prime }\right) \right)$ and let $x_{i}^{\delta }=U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s^{*}}\left( m_{i}^{\prime }\right) \right)$.

First, suppose $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{1}\left( m_{i}\right) \right) <x_{i}^{\delta }$ and $s^{*}=\theta _{i}$. Then

$$\begin{aligned} U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)\le & {} x_{i}^{\delta +1}\text { for all }s\in \{1,\ldots ,\theta _{i}\}, \end{aligned}$$

(6)

$$\begin{aligned} U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)\ge & {} x_{i}^{\delta }\text { for all }s\in \{1,\ldots ,\theta _{i}\}. \end{aligned}$$

(7)

Therefore, we have

$$\begin{aligned} V_{i}^{f}\left( m_{i}\right) \le x_{i}^{\delta +1}<x_{i}^{\delta }\le V_{i}^{f}\left( m_{i}^{\prime }\right) . \end{aligned}$$

Second, suppose $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{1}\left( m_{i}\right) \right) <x_{i}^{\delta }$ and $s^{*}<\theta _{i}$. Then

$$\begin{aligned} U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)\ge & {} x_{i}^{\delta }\text { for all }s\in \{1,\ldots ,s^{*}\}, \end{aligned}$$

(8)

$$\begin{aligned} U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)\ge & {} x_{i}^{\Delta }\text { for all }s\in \{s^{*}+1,\ldots ,\theta _{i}\}, \end{aligned}$$

(9)

and (6) are satisfied. In this case, let

$$\begin{aligned} v\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right) =1-\epsilon \text {.} \end{aligned}$$

Then

$$\begin{aligned} V_{i}^{f}\left( m_{i}\right)\le & {} x_{i}^{\delta +1}, \\ V_{i}^{f}\left( m_{i}^{\prime }\right)\ge & {} \left( 1-\epsilon \right) x_{i}^{\delta }+\epsilon x_{i}^{\Delta }. \end{aligned}$$

Therefore, if $\epsilon$ is sufficiently small, then $V_{i}^{f}\left( m_{i}^{\prime }\right) >V_{i}^{f}\left( m_{i}\right)$.

Third, suppose $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{1}\left( m_{i}\right) \right) \ge x_{i}^{\delta }$ and $s^{*}=\theta _{i}$. Then

$$\begin{aligned} U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)\le & {} x_{i}^{1}\text { for all }s\in \{1,\ldots ,s^{*}-1\}, \end{aligned}$$

(10)

$$\begin{aligned} U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)\le & {} x_{i}^{\delta +1}\text { for all }s\in \{s^{*},\ldots ,\theta _{i}\}, \end{aligned}$$

(11)

and (7) are satisfied. In this case, let

$$\begin{aligned} v\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right) =\epsilon \text {.} \end{aligned}$$

Then

$$\begin{aligned} V_{i}^{f}\left( m_{i}\right)\le & {} \epsilon x_{i}^{1}+\left( 1-\epsilon \right) x_{i}^{\delta +1}, \\ V_{i}^{f}\left( m_{i}^{\prime }\right)\ge & {} x_{i}^{\delta }\text {.} \end{aligned}$$

Therefore, if $\epsilon$ is sufficiently small, then $V_{i}^{f}\left( m_{i}^{\prime }\right) >V_{i}^{f}\left( m_{i}\right)$.

Finally, suppose $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{1}\left( m_{i}\right) \right) \ge x_{i}^{\delta }$ and $s^{*}<\theta _{i}$. Then (8), (9), (10) and (11) are satisfied. In this case, since $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s^{*}}\left( m_{i}\right) \right) <U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m} _{-i}^{s^{*}}\left( m_{i}^{\prime }\right) \right) =x_{i}^{\delta }$ and $U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{1}\left( m_{i}\right) \right) \ge x_{i}^{\delta }$,

$$\begin{aligned} \emptyset \ne \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \subsetneq \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \text {.} \end{aligned}$$

Therefore, we can let

$$\begin{aligned} v\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right) =\epsilon \text { and }v\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right) =1-\epsilon \end{aligned}$$

where $\epsilon \in \left( 0,1/2\right)$. Then

$$\begin{aligned} V_{i}^{f}\left( m_{i}\right)\le & {} \epsilon x_{i}^{1}+\left( 1-\epsilon \right) x_{i}^{\delta +1} \\ V_{i}^{f}\left( m_{i}^{\prime }\right)\ge & {} \left( 1-\epsilon \right) x_{i}^{\delta }+\epsilon x_{i}^{\Delta } \end{aligned}$$

We have

$$\begin{aligned} V_{i}^{f}\left( m_{i}^{\prime }\right) -V_{i}^{f}\left( m_{i}\right) \ge -\left( x_{i}^{1}-x_{i}^{\Delta }\right) \epsilon +\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) \left( 1-\epsilon \right) \text {.} \end{aligned}$$

Therefore, if $\epsilon$ is sufficiently small, then the right-hand side is positive and thus $V_{i}^{f}\left( m_{i}^{\prime }\right) >V_{i}^{f}\left( m_{i}\right)$.

1.4 Proof of Proposition 2

First, we show the second result, because we use this to show the first result. Suppose that $m_{i}$ is rank-dominated by $m_{i}^{\prime }$. Then there is $t\in \{1,\ldots ,\theta _{i}\}$ such that $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{t}\left( m_{i}^{\prime }\right) \right) >U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{t}\left( m_{i}\right) \right)$. Let x be such that $\psi _{i}^{x}=t$. If $m_{i}\notin M_{i}^{x-1}\left( \psi _{i}\right)$, then the proof is finished. If $m_{i}\in M_{i}^{x-1}\left( \psi _{i}\right)$, then $m_{i}^{\prime }\in M_{i}^{x-1}\left( \psi _{i}\right)$. This is because $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m} _{-i}^{s}\left( m_{i}^{\prime }\right) \right) \ge U_{i}^{f}\left( m_{i}, \mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)$ for all $s\in \{1,\ldots ,\theta _{i}\}$. Then $m_{i}\notin M_{i}^{x}\left( \psi _{i}\right)$ and therefore $m_{i}\notin M_{i}^{*}\left( \psi _{i}\right)$.

We show the first result. Let $RDS_{i}$ be the set of rank-dominant strategies of i. Suppose $RDS_{i}\ne \emptyset$. First, we show $M_{i}^{*}\left( \psi _{i}\right) \subseteq RDS_{i}$. By the third result of Fact 1 and the second result of Proposition 2, if $m_{i}^{\prime }\notin RDS_{i}$, then $m_{i}^{\prime }\notin M_{i}^{*}\left( \psi _{i}\right)$ . Therefore, $M_{i}^{*}\left( \psi _{i}\right) \subseteq RDS_{i}$. Second, we show $RDS_{i}\subseteq M_{i}^{*}\left( \psi _{i}\right)$. Since $M_{i}^{*}\left( \psi _{i}\right) \ne \emptyset$ and any non-rank-dominant strategy is not in $M_{i}^{*}\left( \psi _{i}\right)$ , a rank-dominant strategy is in $M_{i}^{*}\left( \psi _{i}\right)$. By the second result of Fact 1, any rank-dominant strategy is in $M_{i}^{*}\left( \psi _{i}\right)$; that is, $RDS_{i}\subseteq M_{i}^{*}\left( \psi _{i}\right)$. By the two facts, we have $RDS_{i}=M_{i}^{*}\left( \psi _{i}\right)$.

1.5 Proof of remark 1

Suppose that player i has a Choquet expected utility function given by ( 3) and (4) is satisfied. Moreover, suppose that $m_{i}^{\prime }$ is a prudent strategy and $m_{i}$ is not. We show if $z_{i}$ is sufficiently large, then $V_{i}^{f}\left( m_{i}^{\prime }\right) >V_{i}^{f}\left( m_{i}\right)$. There exists $s^{*}\in \left\{ 1,\ldots ,\theta _{i}\right\}$ such that

$$\begin{aligned} U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s^{*}}\left( m_{i}^{\prime }\right) \right)> & {} U_{i}^{f}\left( m_{i},\mathrm {m} _{-i}^{s^{*}}\left( m_{i}\right) \right) \text { and} \nonumber \\ U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)= & {} U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right) \text { for all }s\in \left\{ s^{*}+1,\ldots ,\theta _{i}\right\} \text {.} \end{aligned}$$

(12)

Let $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s^{*}}\left( m_{i}^{\prime }\right) \right) =x^{d^{\prime }}$ and $U_{i}^{f}\left( m_{i}, \mathrm {m}_{-i}^{s^{*}}\left( m_{i}\right) \right) =x^{d}$ where $d^{\prime }<d$. Then $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m} _{-i}^{s^{*}+1}\left( m_{i}^{\prime }\right) \right) \le x^{d}$. Let $s^{**}$ be the largest integer such that $U_{i}^{f}\left( m_{i}, \mathrm {m}_{-i}^{s^{**}}\left( m_{i}\right) \right) =x^{d}$. By (12),

$$\begin{aligned} \sum _{\delta =d+1}^{\Delta }\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right) =\sum _{\delta =d+1}^{\Delta }\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right) . \end{aligned}$$

(13)

Moreover, since

$$\begin{aligned} U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s}\left( m_{i}\right) \right)\le & {} x_{i}^{1}\text { for }s\in \left\{ 1,\ldots ,s^{*}-1\right\} , \\ U_{i}^{f}\left( m_{i},\mathrm {m}_{-i}^{s^{*}}\left( m_{i}\right) \right)= & {} x^{d}\le x^{d^{\prime }+1}, \\ U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right)\ge & {} x_{i}^{d^{\prime }}\text { for }s\in \left\{ 1,\ldots ,s^{*}-1\right\} , \end{aligned}$$

we have

$$\begin{aligned} \sum _{\delta =1}^{d^{\prime }-1}\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right)\ge & {} x_{i}^{d^{\prime }}\left( \frac{ s^{*}-1}{\theta _{i}}\right) ^{z_{i}} \end{aligned}$$

(14)

$$\begin{aligned} \sum _{\delta =1}^{d-1}\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right)\le & {} x_{i}^{1}\left( \frac{s^{*}-1}{\theta _{i}}\right) ^{z_{i}}. \end{aligned}$$

(15)

First, suppose $s^{*}=s^{**}$. Then since $U_{i}^{f}\left( m_{i}^{\prime },\mathrm {m}_{-i}^{s}\left( m_{i}^{\prime }\right) \right) \ne x^{d}$ for any s,

$$\begin{aligned} \sum _{\delta =d^{\prime }}^{d}\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right)= & {} x_{i}^{d^{\prime }}\left( \frac{s^{*}}{ \theta _{i}}\right) ^{z_{i}}-x_{i}^{d+1}\left( \frac{s^{*}}{\theta _{i}} \right) ^{z_{i}} \end{aligned}$$

(16)

$$\begin{aligned} \left( x_{i}^{d}-x_{i}^{d+1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right)= & {} x_{i}^{d}\left( \frac{s^{*}}{ \theta _{i}}\right) ^{z_{i}}-x_{i}^{d+1}\left( \frac{s^{*}}{\theta _{i}} \right) ^{z_{i}}\text { }. \end{aligned}$$

(17)

By (13), (14), (15), (16) and (17), we have

$$\begin{aligned}&V_{i}^{f}\left( m_{i}^{\prime }\right) -V_{i}^{f}\left( m_{i}\right) \ge -\left( x_{i}^{1}-x_{i}^{d^{\prime }}\right) \left( \frac{s^{*}-1}{ \theta _{i}}\right) ^{z_{i}}+\left( x_{i}^{d^{\prime }}-x_{i}^{d}\right) \left( \frac{s^{*}}{\theta _{i}}\right) ^{z_{i}} \nonumber \\&\qquad =\left( \frac{s^{*}}{\theta _{i}}\right) ^{z_{i}}\left[ -\left( x_{i}^{1}-x_{i}^{d^{\prime }}\right) \left( \frac{s^{*}-1}{s^{*}} \right) ^{^{z_{i}}}+\left( x_{i}^{d^{\prime }}-x_{i}^{d}\right) \right] . \end{aligned}$$

(18)

Second, $s^{**}>s^{*}$. Then

$$\begin{aligned}&\sum _{\delta =d^{\prime }}^{d}\left( x_{i}^{\delta }-x_{i}^{\delta +1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}^{\prime }\right) \right) =\left( x_{i}^{d^{\prime }}-x_{i}^{d}\right) \left( \frac{s^{*}}{\theta _{i}}\right) ^{z_{i}}+\left( x_{i}^{d}-x_{i}^{d+1}\right) \left( \frac{s^{**}}{\theta _{i}} \right) ^{z_{i}} \end{aligned}$$

(19)

$$\begin{aligned}&\left( x_{i}^{d}-x_{i}^{d+1}\right) v_{i}\left( \bigcup \limits _{j=1}^{\delta }E_{i}^{j}\left( m_{i}\right) \right) =\left( x_{i}^{d}-x_{i}^{d+1}\right) \left( \frac{s^{**}}{\theta _{i}}\right) ^{z_{i}}. \end{aligned}$$

(20)

By (13), (14), (15), (17), (19) and (20), we also have (18). Therefore, if $z_{i}$ is sufficiently large, then $V_{i}^{f}\left( m_{i}^{\prime }\right) >V_{i}^{f}\left( m_{i}\right)$.

1.6 Proof of Theorem 1

Suppose that a deterministic outcome function f satisfies neutrality and elementary monotonicity. First, let $m_{i}$ be a strict sincere strategy of i. We show that $\left| M_{-i}^{f}\left( m_{i},x\right) \right| \ge \left| M_{-i}^{f}\left( m_{i}^{\prime },x\right) \right|$ is satisfied for all $m_{i}^{\prime }\in M_{i}$ and for all $x\ge 0$. A similar fact is shown by Majumdar and Sen (2004) in the proof of their Theorem 3.1.

For notational simplicity, we let

$$\begin{aligned} \left| \left\{ m_{-i}\in M_{-i}\text { }\left| \text { }f_{a}\left( m_{i},m_{-i}\right) =1\right. \right\} \right| =\xi \left( m_{i},a\right) . \end{aligned}$$

First, we show

$$\begin{aligned} \xi \left( m_{i},m_{i}^{k}\right) =\xi \left( m_{i}^{\prime },m_{i}^{\prime k}\right) \text { for any }m_{i},m_{i}^{\prime }\in M_{i}, \end{aligned}$$

(21)

Fix $i\in N$, $m_{i}\in M_{i}$ and $k=1,\ldots ,\left| A\right|$. To consider the nontrivial cases only, we assume that there is $m_{-i}\in M_{-i}$ such that $f_{m_{i}^{k}}\left( m\right) =1$. Let $\sigma :A\rightarrow A$ be a permutation of A such that $\sigma \left( m_{i}^{k}\right) =m_{i}^{\prime k}$ for all $k=1,\ldots ,\left| A\right| -1$. By neutrality, $f_{m_{i}^{k}}\left( m\right) =f_{m_{i}^{\prime k}}\left( m_{i}^{\prime },m_{-i}^{\sigma }\right) =1$. Hence, if there is $m_{-i}\in M_{-i}$ such that $f_{m_{i}^{k}}\left( m\right) =1$, then there is $m_{-i}^{\prime }\in M_{-i}$ such that $f_{m_{i}^{\prime k}}\left( m_{i}^{\prime },m_{-i}^{\prime }\right) =1$. Hence, we have $\xi \left( m_{i},m_{i}^{k}\right) \le \xi \left( m_{i}^{\prime },m_{i}^{\prime k}\right)$. Since we can similarly obtain $\xi \left( m_{i},m_{i}^{k}\right) \ge \xi \left( m_{i}^{\prime },m_{i}^{\prime k}\right)$, we have (21).

Second, we show

$$\begin{aligned} \xi \left( m_{i},m_{i}^{k}\right) \ge \xi \left( m_{i},m_{i}^{k+1}\right) \text { for all }k=1,\ldots ,\left| A\right| -1. \end{aligned}$$

(22)

Fix $i\in N$, $m_{i}\in M_{i}$ and $k=1,\ldots ,\left| A\right| -1$. Let $m_{i}^{k}=a$, $m_{i}^{k+1}=b$ and $m_{i}^{\prime }$ be such that $m_{i}^{\prime k}=b$, $m_{i}^{\prime k+1}=a$ and $m_{i}^{l}=m_{i}^{\prime l}$ for all $l\ne k,k+1$. That is, $m_{i}$ is an elementary a-improvement of $m_{i}^{\prime }$. To consider the nontrivial cases only, we assume that there is $m_{-i}\in M_{-i}$ such that $f_{a}\left( m_{i}^{\prime },m_{-i}\right) =1$. Then by elementary monotonicity, $f_{a}\left( m_{i},m_{-i}\right) =1$ is satisfied. This implies that $\xi \left( m_{i},a\right) \ge \xi \left( m_{i}^{\prime },a\right)$. By (21), $\xi \left( m_{i}^{\prime },a\right) =\xi \left( m_{i},b\right)$ and, therefore, $\xi \left( m_{i},a\right) \ge \xi \left( m_{i},b\right)$. We have (22).

Moreover, (22) indicates

$$\begin{aligned} \xi \left( m_{i},m_{i}^{1}\right) \ge \xi \left( m_{i},m_{i}^{2}\right) \ge \cdots \ge \xi \left( m_{i},m_{i}^{\left| A\right| }\right) . \end{aligned}$$

(23)

Now, let $m_{i}$ be a strict sincere strategy and $m_{i}^{\prime }\in M_{i}\setminus \left\{ m_{i}\right\}$. By (21), for all $K=1,2,\ldots ,\left| A\right|$,

$$\begin{aligned} \sum \limits _{k=1}^{K}\xi \left( m_{i},m_{i}^{k}\right) =\sum \limits _{k=1}^{K}\xi \left( m_{i}^{\prime },m_{i}^{\prime k}\right) \text {.} \end{aligned}$$

Moreover, by (23),

$$\begin{aligned} \sum \limits _{k=1}^{K}\xi \left( m_{i}^{\prime },m_{i}^{\prime k}\right) =\xi \left( m_{i}^{\prime },m_{i}^{\prime 1}\right) +\xi \left( m_{i}^{\prime },m_{i}^{\prime 2}\right) +\cdots +\xi \left( m_{i}^{\prime },m_{i}^{\prime K}\right) \ge \sum \limits _{k=1}^{K}\xi \left( m_{i}^{\prime },m_{i}^{k}\right) , \end{aligned}$$

because the left-hand side is the sum of the elements of the largest to the Kth largest. Hence, we have

$$\begin{aligned} \sum \limits _{k=1}^{K}\xi \left( m_{i},m_{i}^{k}\right) \ge \sum \limits _{k=1}^{K}\xi \left( m_{i}^{\prime },m_{i}^{k}\right) \text {.} \end{aligned}$$

Since $m_{i}$ is a strict sincere strategy, we have (1) is satisfied for all $m_{i}^{\prime }\in M_{i}$ and for all $x\ge 0$. By Lemma 1 and Fact 1, a strict sincere strategy is a rank-dominant strategy of i.

Finally, let $m_{i}$ be a strict sincere strategy of i and $m_{i}^{\prime }$ be a sincere strategy of i. Since $f\left( m_{i}^{\prime },m_{-i}\right) =f\left( m_{i},m_{-i}\right)$ for all $m_{-i}\in M_{-i}$, $\mu _{i}^{f}\left( m_{i}\right) =\mu _{i}^{f}\left( m_{i}^{\prime }\right)$ . Therefore, if $m_{i}$ is a rank-dominant strategy of i, then $m_{i}^{\prime }$ is also a rank-dominant strategy of i. Since a strict sincere strategy is a rank-dominant strategy of i, any sincere strategy of i is a rank-dominant strategy of i.

1.7 Proof of Theorem 2

First, we show the following claim.

Claim 1

Suppose that f is the plurality voting with the equal probability tie-breaking rule. Then for any $\left( \left| A\right| ,\left| N\right| \right)$, $\gamma \left( a_{i}^{*}\right)$ is a sincere strategy of i but $\gamma \left( a\right)$ is not sincere strategy for any $a\in A\setminus \left\{ a_{i}^{*}\right\}$.

Proof of Claim 1

Let $m_{i}$ be a strict sincere strategy of i. Since $w^{1}=1$ and $w^{2}=\cdots =w^{\left| A\right| }=0$, $S\left( m_{i},m_{-i}\right) =S\left( \gamma \left( a_{i}^{*}\right) ,m_{-i}\right)$ for any $m_{-i}\in M_{-i}$. Since f is the plurality voting with the equal probability tie-breaking rule, $f_{a}\left( m_{i},m_{-i}\right) =f_{a}\left( \gamma \left( a_{i}^{*}\right) ,m_{-i}\right)$ for $a\in A$ and all $m_{-i}\in M_{-i}$. Therefore, $\gamma \left( a_{i}^{*}\right)$ is a sincere strategy of i.

Let $a\in A\setminus \left\{ a_{i}^{*}\right\}$. First, suppose that n is odd. Let $\hat{m}_{-i}$ be such that

$$\begin{aligned} s\left( \left( \gamma \left( a_{i}^{*}\right) ,\hat{m}_{-i}\right) ,a_{i}^{*}\right) =\frac{n+1}{2}\text { and }s\left( \left( \gamma \left( a_{i}^{*}\right) ,\hat{m}_{-i}\right) ,a\right) =\frac{n-1}{2}. \end{aligned}$$

(24)

In this case, $S(\gamma \left( a_{i}^{*}\right) ,\hat{m} _{-i})=\{a_{i}^{*}\}$ and $S(\gamma \left( a\right) ,\hat{m}_{-i})=\{a\}$ . Hence, $f_{a_{i}^{*}}\left( \gamma \left( a_{i}^{*}\right) ,\hat{m} _{-i}\right) \ne f_{a_{i}^{*}}\left( \gamma \left( a\right) ,\hat{m} _{-i}\right)$. Second, suppose that n is even. Let $\tilde{m}_{-i}$ be such that

$$\begin{aligned} s\left( \left( \gamma \left( a_{i}^{*}\right) ,\tilde{m}_{-i}\right) ,a_{i}^{*}\right) =\text { }s\left( \left( \gamma \left( a_{i}^{*}\right) ,\tilde{m}_{-i}\right) ,a_{i}\right) =\frac{n}{2}. \end{aligned}$$

(25)

Then $S(\gamma \left( a_{i}^{*}\right) ,\tilde{m}_{-i})=\{a_{i}^{*},a\}$ and $S(\gamma \left( a\right) ,\tilde{m}_{-i})=\{a\}$. Hence, $f_{a_{i}^{*}}\left( \gamma \left( a_{i}^{*}\right) ,\tilde{m} _{-i}\right) \ne f_{a_{i}^{*}}\left( \gamma \left( a\right) ,\tilde{m} _{-i}\right)$. $\square$

Let $a\in A\setminus \left\{ a_{i}^{*}\right\}$. By Claim 1, to show Theorem 2, we show that $\gamma \left( a_{i}^{*}\right)$ is a rank-dominant strategy and $\gamma \left( a\right)$ is not for any $a\in A\setminus \left\{ a_{i}^{*}\right\}$; that is,

$$\begin{aligned} U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\mathrm {m} _{-i}^{s}\left( \gamma \left( a_{i}^{*}\right) \right) \right) \ge U_{i}^{f}\left( \gamma \left( a\right) ,\mathrm {m}_{-i}^{s}\left( \gamma \left( a\right) \right) \right) \end{aligned}$$

for all $s\in \left\{ 1,\cdots ,\theta _{i}\right\}$ and

$$\begin{aligned} U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\mathrm {m} _{-i}^{t}\left( \gamma \left( a_{i}^{*}\right) \right) \right) >U_{i}^{f}\left( \gamma \left( a\right) ,\mathrm {m}_{-i}^{t}\left( \gamma \left( a\right) \right) \right) \end{aligned}$$

for some $t\in \left\{ 1,\ldots ,\theta _{i}\right\}$.

We introduce the concept of pivotal strategies. For given f, $m_{i}$ is said to be a pivotal strategy of i for $m_{-i}$ if $m_{i}^{1}\in S(m)$ and there exists $z\in M_{i}\setminus \{m_{i}\}$ such that $S(m)\ne S(z,m_{-i})$.

First, suppose that $m_{-i}$ for which neither $\gamma \left( a_{i}^{*}\right)$ nor $\gamma \left( a\right)$ is a pivotal strategy of i. Then $U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,m_{-i}\right) =U_{i}^{f}\left( \gamma \left( a\right) ,m_{-i}\right)$.

Second, let

$$\begin{aligned} M_{-i}^{P}(c)=\left\{ m_{-i}\left| \text { }\gamma \left( c\right) \text { is a pivotal strategy of }i\text { for }m_{-i}\right. \right\} \subseteq M_{-i}. \end{aligned}$$

Then $\left| M_{-i}^{P}(a_{i}^{*})\right| =\left| M_{-i}^{P}(a)\right|$. Let $m_{-i}\in M_{-i}^{P}(a_{i}^{*})$. We define $m_{-i}^{\prime }$ such that the elements $a_{i}^{*}$ in $m_{-i}$ are replaced by a and the elements a in $m_{-i}$ are replaced by $a_{i}^{*}$. Then $m_{-i}^{\prime }\in M_{-i}^{P}(a)$ and $S(\gamma \left( a_{i}^{*}\right) ,m_{-i})\setminus \{a_{i}^{*}\}=S(\gamma \left( a\right) ,m_{-i}^{\prime })\setminus \{a\}$. If $a_{i}^{*},a\in S(\gamma \left( a_{i}^{*}\right) ,m_{-i})$, then $S(\gamma \left( a_{i}^{*}\right) ,m_{-i})=S(\gamma \left( a\right) ,m_{-i}^{\prime })$ and thus $U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,m_{-i}\right) =U_{i}^{f}\left( \gamma \left( a\right) ,m_{-i}^{\prime }\right)$. On the other hand, if $a_{i}^{*}\in S(\gamma \left( a_{i}^{*}\right) ,m_{-i})$ and $a\notin S(\gamma \left( a_{i}^{*}\right) ,m_{-i})$, then $a\in S(\gamma \left( a\right) ,m_{-i}^{\prime })$, $a_{i}^{*}\notin S(\gamma \left( a\right) ,m_{-i}^{\prime })$ and thus $U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,m_{-i}\right) >U_{i}^{f}\left( \gamma \left( a\right) ,m_{-i}^{\prime }\right)$. These two facts are because of the equal probability tie-breaking rule. Since $\left| M_{-i}^{P}(a_{i}^{*})\right| =\left| M_{-i}^{P}(a)\right|$, we have $U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\mathrm {m} _{-i}^{s}\left( a_{i}^{*}\right) \right) \ge U_{i}^{f}\left( \gamma \left( a\right) ,\mathrm {m}_{-i}^{s}\left( a\right) \right)$ for all $s\in \left\{ 1,\cdots ,\theta _{i}\right\}$.

Finally, we show that

$$\begin{aligned} U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\mathrm {m} _{-i}^{t}\left( a_{i}^{*}\right) \right) >U_{i}^{f}\left( \gamma \left( a\right) ,\mathrm {m}_{-i}^{t}\left( a\right) \right) \end{aligned}$$

for some $t\in \left\{ 1,\ldots ,\theta _{i}\right\}$. First, suppose that n is odd. Let $\hat{m}_{-i}$ satisfy (24). In this case, $U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\hat{m}_{-i}\right) >U_{i}^{f}\left( \gamma \left( a\right) ,\hat{m}_{-i}\right)$. Second, suppose that n is even. Let $\tilde{m}_{-i}$ satisfy (25). We have

$$\begin{aligned} U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\tilde{m}_{-i}\right) = \frac{u_{i}\left( a_{i}^{*}\right) +u_{i}\left( a\right) }{2}>u\left( a\right) =U_{i}^{f}\left( \gamma \left( a\right) ,\tilde{m}_{-i}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} U_{i}^{f}\left( \gamma \left( a_{i}^{*}\right) ,\mathrm {m} _{-i}^{t}\left( a_{i}^{*}\right) \right) >U_{i}^{f}\left( \gamma \left( a\right) ,\mathrm {m}_{-i}^{t}\left( a\right) \right) \end{aligned}$$

for some $t\in \left\{ 1,\ldots ,\theta _{i}\right\}$.