Constitutionally consistent voting rules over single-peaked domains

Abstract

Constitutional consistency requires that the voting rule produce the same outcome at any vote profile as the one it produces at its induced vote profile for any given set of voting rules (or constitution) consisting of the voting rule itself. We consider this type of consistency in two voting models with single-peaked preferences, one with a finite set of alternatives and the other, when the set of alternatives is the interval [0, 1]. We show that cumulative-threshold rules are the only unanimous, anonymous and constitutionally consistent voting rules. These rules assign monotone decreasing (increasing) thresholds to each alternative and pick the minimum (maximum) alternative from the range of the vote profile that receives more cumulative votes (votes received by all the alternatives smaller (or greater) than itself) than the threshold assigned to it. This class of rules consists of the min, max and median rules. The addition of continuity leads to the characterization of k-median rules in the interval voting model.

Appendix

Proof of Theorem 1

It is easy to verify that the cumulative-threshold rules satisfy unanimity and anonymity. We show that they satisfy constitutional consistency. $\square$

Let $F^{q}$ be a cumulative threshold rule with monotone decreasing thresholds $(q_{x})_{x\in X}$. We show that $F^{q}$ satisfies constitutional consistency. Take any v and $\mathcal {A}\subseteq \mathcal {F}$ such that $F^{q}\in \mathcal {A}$ and $F^{q}(v)=x^{*}$. By the nature of single-peaked preferences for any meta vote profile$v^{\mathcal {A}} \in \mathcal {A}(\succeq )$ for all $\succeq \in \mathcal {S}(v)$,

• (i) $[v_{i} \le F^{q}(v) ] \Rightarrow [v_{i}^{\mathcal {A}}(v) \le F^{q}(v)] \text { for all } i \in N.$

• (ii) $[v_{i} \ge F^{q}(v) ] \Rightarrow [v_{i}^{\mathcal {A}}(v) \ge F^{q}(v)] \text { for all } i \in N.$

• (iii) $[v_{i} = F^{q}(v) ]\Rightarrow [v_{i}^{\mathcal {A}}(v) = F^{q}(v)]$ for all $i \in N.$

Since $F^{q}(v)=x^{*}$ and by the above arguments, $\#\{i|v_{i}^{\mathcal {A}}(v) \le F^{q}(v)\}\ge q_{x^{*}}$. We show that for all $F\in \mathcal {A}$ such that $F(v)=x< F^{q}(v)=x^{*}$ we have $\#\{i|v_{i}^{\mathcal {A}}(v) \le x\}< q_{x}$. Suppose for contradiction that there exists a voting rule $F'\in A$ such that $F'(v)=\hat{x}<F^{q}(v)$ and $\#\{i|v_{i}^{\mathcal {A}}(v) \le \hat{x}\}\ge q_{\hat{x}}$. Since $\hat{x}<x^{*}$, by definition of cumulative-threshold rules, $[\#\{i|v_{i}\le \hat{x}\}\ge q_{\hat{x}}]\Rightarrow [F^{q}(v)=\hat{x}]$. This is a contradiction to the fact that $F^{q}(v)=x^{*}\ne \hat{x}.$

Let the range of $v^{\mathcal {A}}(v)$ be $[\underline{v}^{\mathcal {A}}(v),\overline{v}^{\mathcal {A}}(v)]$. By arguments presented above,

$$\begin{aligned} F^{q}(v^{\mathcal {A}}(v))=x^{*} = \arg \min _{x\in [\underline{v}^{\mathcal {A}}(v),\overline{v}^{\mathcal {A}}(v)]} \left( \sum _{x'\le x}n_{x'} \ge q_{x}\right) . \end{aligned}$$

This implies that $F^{q}(v^{\mathcal {A}}(v))= F^{q}(v)= x^{*}$. Therefore, $F^{q}$ is constitutionally consistent.

We now show the converse. Suppose a voting rule F satisfies all the axioms in the statement of Theorem 1. We show that it is a cumulative-threshold rule with respect to monotone decreasing thresholds $\{q_{x}\}_{x\in X}$. We prove this in a series of claims.

Claim 1

For any $v\in X^{n}$ let $\underline{v}=\min _{i}\{v_{i}\}_{i=1}^{n}$ and $\overline{v}=\max _{i}\{v_{i}\}_{i=1}^{n}$. Then $F(v) \in [\underline{v} , \overline{v}]$ for all $v\in X^{n}$.

Proof

Suppose for contradiction that $F(v) \notin [\underline{v},\overline{v}]$. Without loss of generality assume that $F(v)=x>\overline{v}$. Let $\mathcal {A}=\{F,F_{1}\}$ be such that $F(v)=x$ and $F_{1}(v)=\overline{v}$. Then by definition of meta sets for all $v^{\mathcal {A}} \in \mathcal {A}(\succeq )$ for all $\succeq \in \mathcal {S}(v)$,

$$\begin{aligned}{}[v_{i}\le \overline{v}] \Leftrightarrow [\mathcal {A}(\succeq _{i}) =\{F_{1}\}] . \end{aligned}$$

$\square$

This implies that $v^{\mathcal {A}}=(F_{1},\ldots , F_{1})$ is a meta-profile of v at $\mathcal {A}$ where each voter ‘votes’ for $F_{1}$ i.e. $v_{i}^{\mathcal {A}}=F_{1}$ for all $i\in N$. By constitutional consistency,

$$\begin{aligned} F(\overline{v}^{\mathcal {A}}(v))=F(v)=x. \end{aligned}$$

By unanimity, $F(\overline{v}^{\mathcal {A}}(v))= F(F(v),\ldots , F(v))=F(\overline{v}, \ldots , \overline{v})=\overline{v}$. Since $x>\overline{v}$ this is a contradiction. Therefore, $F(v) \in [\underline{v} , \overline{v}]$ for all $v\in X^{n}$.

Let $\underline{x}$ and $\overline{x}$ be the minimum and the maximum alternatives according in X to the ordering $\le$. For any $x\in X\backslash \{\bar{x}\}$ let $x_{+1}$ denote the alternative that succeeds x in the ordering $\le$. Similarly, for all $x \in X\backslash \{\underline{x}\}$ let $x_{-1}$ be the alternative that precedes x in the ordering $\le$. Due to anonymity, we can concentrate only on the number of votes received by an alternative. Let $v=(x^{n_{x}},(x_{+1})^{n-n_{x}})$ be a vote profile where $n_{x}$ voters vote for x and the remaining vote for $x_{+1}$.

We now define thresholds$q=\{q_{x}\}_{x\in X}$ such that $q_{x} \in \{1,\ldots ,n\}$ for each $x\in X$ as follows. Let $q_{x}$ for all $x\in X\backslash \{\overline{x}\}$ be such that,

$$\begin{aligned} q_{x}\in \text {arg}\min \limits _{n_{x}\in N} \left( F(x^{n_{x}},(x_{+1})^{n-n_{x}})=x\right) . \end{aligned}$$

Therefore, $q_{x}$ is the minimum number of votes required for x to be chosen when voters only vote for x and $x_{+1}$ for any $x\in X\backslash \{\overline{x}\}$. By unanimity we know that there exists such a $q_{x}$ since $F(x^{n_{x}},(x_{+1})^{n-n_{x}})=x$ when $n_{x}=n$. We set $q_{\overline{x}}=q_{\overline{x}{-1}}$.

Claim 2

Suppose $x\in X$ and $q_{x}$ is its threshold. Let $v'=(x^{n_{x}},\hat{x}^{n-n_{x}})\in X^{n}$ be a vote profile. Then

$$\begin{aligned}{}[\hat{x} \ge x \text { and } n_{x}\ge q_{x}] \Rightarrow [F(v')=x]. \end{aligned}$$

We prove this using the results of Border and Jordan (1983) and Moulin (1980). By strategy-proofness any unilateral deviation does not change the outcome. By iteratively applying this step to voters with reported peaks greater than x we can show that $F(v')=x$ for any $v'$ which satisfies the conditions in the statement of the claim.

Claim 3

We show that the thresholds are monotonic decreasing i.e. $q_{x}\ge q_{x_{+1}}$ for all $x\in X$.

Proof

We show this by contradiction. Suppose there exists $x, x_{+1}\in X$ such that $q_{x}<q_{x_{+1}}$. By definition $q_{\bar{x}}=q_{\bar{x}_{-1}}$. Therefore, an implication of our assumption is that $x_{+1}<\bar{x}$. $\square$

Since $x_{+1}< \overline{x}$ we can always find an alternative $(x_{+2})\in X$ such that $x<x_{+1}<x_{+2}\le \bar{x}$. Consider the profile $v=((x_{+1})^{q_{x}},(x_{+2})^{n-q_{x}})$. We argue that $F(v)\notin \{x_{+1},x_{+2}\}$ thus contradicting Claim 1.

Suppose $F(v)= x_{+1}$. By definition of threshold for x and the fact that,

$F((x_{+1})^{q_{x}},(x_{+2})^{n-q_{x}})=x_{+1}$ we have $q_{x_{+1}}\le q_{x}$. This is a contradiction to our assumption that $q_{x_{+1}}>q_{x}$. Therefore, $F(v)\ne x_{+1}$.

Suppose $F(v)=x_{+2}$. Consider the constitution $\mathcal {A}=\{F,F_{1}\}$ such that $F_{1}(v)=x$. By similar arguments as those made above there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that,

$$\begin{aligned} {[}v_{i}= & {} x_{+1}] \Leftrightarrow [\mathcal {A}(\succeq _{i}) =\{F_{1}\}]. \\ {[}v_{i}= & {} x_{+2}] \Leftrightarrow [\mathcal {A}(\succeq _{i}) =\{F\}]. \end{aligned}$$

By definition of $q_{x}$ we can construct a meta profile of v, $v^{\mathcal {A}}= (F_{1}^{q_{x}},F^{n-q_{x}})$. By constitutional consistency,

$$\begin{aligned} F(v^{\mathcal {A}}(v))=F(F_{1}(v)^{q_{x}},F(v)^{n-q_{x}}) =F(x^{q_{x}},(x_{+2})^{n-q_{x}})=x_{+2}. \end{aligned}$$

By Claim 2, $F(x^{q_{x}},(x_{+2})^{n-q_{x}}) =x$. This is a contradiction. By the above arguments, $F(v) \notin \{x_{+1},x_{+2}\}$. This is a contradiction to Claim 1. Therefore, $q_{x}\ge q_{x_{+1}}$ for all $x\in X$.

Claim 4

Suppose $v\in X^{n}$ is a vote profile. Let $x^{*} = \text {arg}\min \nolimits _{x'\in [\underline{v},\overline{v}]} \sum _{x'\le x}n_{x'} \ge q_{x}$. Then $F(v) =x^{*}$.

Proof

By definition of thresholds $q=\{q_{x}\}_{x\in X}$ there exists $x^{*}\in [\underline{v},\overline{v}]$ such that $x^{*}= \text {arg}\min \nolimits _{x\in [\underline{v},\overline{v}]} \sum _{x'\le x}n_{x'} \ge q_{x}$. We show that $F(v)= x^{*}$.

Suppose for contradiction that $F(v)=\hat{x}\ne x^{*}$ for some $x\in [\underline{v},\overline{v}]$. Let $\bar{n}_{x}=\sum _{x'\le x}n_{x}$ for any $x\in X$ be the cumulative vote for x. We consider two separate cases.

Case 1  Suppose $\hat{x}$ such that $\bar{n}_{\hat{x}} < q_{\hat{x}}$. There are two sub-cases.

Case 1(a)   Suppose $\hat{x}< x^{*}$. Consider the constitution $\mathcal {A}=\{F,F_{1}\}$ such that $F(v)=\hat{x}$ and $F_{1}(v)=\hat{x}_{+1}$. Note that there exists such an alternative $x_{+1}\in [\underline{v},\overline{v}]$ since $\hat{x}<x^{*}$. By definition of meta sets there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that,

$$\begin{aligned} {[}v_{i}\le \hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F\}]. \\ {[}v_{i} \ge \hat{x}_{+1}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F_{1}\}]. \end{aligned}$$

Using similar arguments as the ones used above $v^{\mathcal {A}}= (F^{\bar{n}_{\hat{x}}},F_{1}^{n-\bar{n}_{\hat{x}}})$ is a meta profile of v and the corresponding dual vote profile is $v^{\mathcal {A}}(v)=(F(v)^{\bar{n}_{\hat{x}}},F_{1} (v)^{n-\bar{n}_{\hat{x}}})=(\hat{x}^{\bar{n}_{\hat{x}}}, (\hat{x}_{+1})^{n-\bar{n}_{\hat{x}}})$. Therefore by constitutional consistency,

$$\begin{aligned} F\left( \hat{x}^{\bar{n}_{\hat{x}}}, (\hat{x}_{+1})^{n -\bar{n}_{\hat{x}}}\right) =F(v)=\hat{x}. \end{aligned}$$

By our assumption $\hat{x}\in [\underline{v},\overline{v}]$ is such that $\bar{n}_{\hat{x}} <q_{\hat{x}}$. Therefore, the cumulative vote for $\hat{x}$ does not exceed its prescribed threshold. By definition of thresholds, $q_{x}$ is such that $F(\hat{x}^{\bar{n}_{\hat{x}}},(\hat{x}_{+1})^{n -\bar{n}_{\hat{x}}}) =\hat{x}_{+1}.$ This is a contradiction.

Case 1(b)  Suppose $\hat{x}> x^{*}$. Consider the constitution $\mathcal {A}=\{F,F_{1}\}$ such that $F(v)=\hat{x}$ and $F_{1}(v)=x^{*}$. By similar arguments as the ones used above there exists a preference profile $\succeq \in \mathcal {S}(v)$,

$$\begin{aligned}{}[v_{i} \le x^{*}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F_{1}\}] . \\ {[}x^{*}<v_{i}<\hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F,F_{1}\}]. \\ {[}v_{i}\ge \hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F\}] . \end{aligned}$$

Therefore $v^{\mathcal {A}}= (F_{1}^{\bar{n}_{x^{*}}},F^{n -\bar{n}_{x^{*}}})$ is a meta profile of v at $\mathcal {A}$ and $v^{\mathcal {A}}(v)=(x^{*(\bar{n}_{x^{*}})},\hat{x}^{n -\bar{n}_{x^{*}}})$ is a dual profile of v. By constitutional consistency,

$$\begin{aligned} F(x^{*(\bar{n}_{x^{*}})},\hat{x}^{n-\bar{n}_{x^{*}}}) =F(v) = \hat{x}. \end{aligned}$$

Note that $\bar{n}_{x^{*}} < q_{x^{*}}$. By Claim 2 and by definition of threshold, $q_{x}$ is such that $F(x^{*(\bar{n}_{x^{*}})}, \hat{x}^{n-\bar{n}_{x^{*}}}) =x^{*}.$ This is a contradiction.

Case 2  Suppose $F(v) = \hat{v}\ne x^{*}$ such that $\text {arg}\min \nolimits _{x\in [\underline{v},\overline{v}]} \sum _{x'\le x}n_{x'} \ge q_{\hat{x}}$. As assumed above $x^{*} = \text {arg}\min \nolimits _{x\in [\underline{v},\overline{v}]} \sum _{x'\le x}n_{x'} \ge q_{x}$. Therefore, $x^{*}<\hat{x}$. We can use similar arguments as the ones made in Case 1(b) to show that $F(v)=\hat{x}\ne x^{*}$ will lead to a contradiction. This completes the proof. $\square$

Proof of Theorem 2

Necessity can be proved using the same arguments as the ones used in the proof of Theorem 1. We prove the converse. Suppose a voting rule F satisfies all the axioms in the statement of Theorem 2. We show that it is a cumulative-threshold rule(lsc) with respect to some $q_{x}:X\rightarrow N$ such that $q_{x}$ is a monotone-decreasing and lower semi-continuous threshold function for all $x\in X$. We prove this part in separate claims. $\square$

Claim 5

For any $v\in X^{n}$ let $\underline{v}=\min _{i}\{v_{i}\}_{i=1}^{n}$ and $\overline{v}=\max _{i}\{v_{i}\}_{i=1}^{n}$. Then $\underline{v} \le F(v)\le \overline{v}$ for all $v\in X^{n}$.

Proof

Suppose for contradiction that $F(v) \notin [\underline{v},\overline{v}]$. Without loss of generality assume that $F(v)=x>\overline{v}$. Let $\mathcal {A}=\{F,F_{1}\}$ be such that $F(v)=x$ and $F_{1}(v)=\overline{v}$. By definition of meta sets and the fact that there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that,

$$\begin{aligned} {[}v_{i}\le \overline{v}] \Leftrightarrow [\mathcal {A}(\succeq _{i}) =\{F_{1}\}] . \end{aligned}$$

$\square$

This implies that $(\overline{v}^{n})$ is a meta-profile of v at $\mathcal {A}$. By constitutional consistency,

$$\begin{aligned} F(\overline{v}^{n})=x. \end{aligned}$$

By unanimity$F(\overline{v}^{n})= \overline{v}\ne x$. This is a contradiction.

By anonymity we can consider profiles as follows. Let $v=(x^{n_{x}},(x+\varepsilon )^{n-n_{x}})$ be a profile such that $n_{x}$ voters vote for x and the remaining vote for $x+\varepsilon$ for some $\varepsilon \in (0, 1-x].$

We define $q_{x}:[0,1]\rightarrow N$ for all $x\in X$ as follows. For every $x\in [0,1)$,

$$\begin{aligned} q_{x} \in \text {arg}\min \limits _{k\in N} F(x^{k}, (x+\varepsilon _{x})^{n-k})=x \quad \text { for some } \varepsilon _{x} \in (0, 1-x] \end{aligned}$$

Therefore, $q_{x}$ is the least number of votes for x to be selected when every other voter is voting for some alternative strictly greater than x. Note that such a $q_{x}$ exists for any given $x\in X$ since by unanimity$F(x^{k}, (x+\varepsilon _{x})^{n-k})=x$ if $k=n$. We set $q_{1}=1$.

Claim 6

Pick any $x\in X$ and let $q_{x}$ be as defined above. Let $v= (x^{q_{x}}, (x+\varepsilon _{x})^{n-q_{x}})$. Let $v'=(x^{n_{x}},\hat{x}^{n-n_{x}})$ be another profile. Then

$$\begin{aligned} {[}\hat{x} \ge x \text { and } n_{x}\ge q_{x}]\Rightarrow [F(v')=x]. \end{aligned}$$

We again prove this using the results of Border and Jordan (1983) and Moulin (1980). By strategy-proofness any unilateral deviation does not change the outcome. By iteratively applying this step to voters with reported peaks greater than x we can show that $F(v')=x$ for any $v'$ which satisfies the conditions in the statement of the claim.

Claim 7

We show that the thresholds are monotonic decreasing i.e. $[x>\hat{x}]\Rightarrow [q_{\hat{x}}\le q_{x}]$ for all $x,\hat{x}\in X$.

Proof

We show this by contradiction. Suppose there exists $x, \hat{x}\in X$ such that $\hat{x}>x$ and $q_{\hat{x}}>q_{x}$. By definition $q(1)=1$. Therefore, an implication of our assumption is that $x,\hat{x}<1$. Pick any $\tilde{x}\in (\hat{x},1]$. $\square$

Consider the profile $v=(\hat{x}^{q_{x}},\tilde{x}^{n-q_{x}})$. We argue that $F(v)\notin [\hat{x},\tilde{x}]$ thus contradicting Claim 5.

Suppose $F(v)= \hat{x}$. By definition of threshold for x, $F(\hat{x}^{q_{x}},\tilde{x}^{n-q_{x}})=\hat{x}$ implies that $q(\hat{x})=q_{x}$. This is a contradiction to our assumption that $q_{\tilde{x}}>q_{x}$. Therefore, $F(v)\ne \hat{x}$.

Suppose $F(v)=x_{1}\in (\hat{x},\tilde{x}]$. Consider the constitution $\mathcal {A}=\{F,F_{1}\}$ such that $F_{1}(v)=x$. By similar arguments as the ones made above there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that,

$$\begin{aligned} {[}v_{i}= \hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F_{1}\}]. \\ {[}v_{i}= \tilde{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F\}]. \end{aligned}$$

By definition of $q_{x}$ we can construct a meta profile of v such that $v^{\mathcal {A}}= (F_{1}^{q_{x}},F^{n-q_{x}})$. By constitutional consistency,

$$\begin{aligned} F(v^{\mathcal {A}}(v))=F(F_{1}(v)^{q_{x}},F(v)^{n-q_{x}}) =F\left( x^{q_{x}},x_{1}^{n-q_{x}}\right) =x_{1}. \end{aligned}$$

By Claim 6, $F(x^{q_{x}},\hat{x}^{n-q_{x}}) =x$. This is a contradiction to the above equation. Therefore, $F(v)\ne x_{1}$. By the above arguments, $F(v) \notin [\hat{x},\tilde{x}]$. This is a contradiction to Claim 6. Therefore, $\hat{x} > x$ implies $q_{\hat{x}}\le q_{x}$.

Claim 8

Suppose $v\in X^{n}$ is a vote profile. Let $x^{*} \in \text {arg}\min \nolimits _{x\in [\underline{v},\overline{v}]} \sum _{x'\le x}n_{x} \ge q_{x}$ such that $x^{*}\in [\underline{v},\overline{v}]$. Then $F(v) =x^{*}$.

Proof

By Claim 5, we have $F(v)\in [\underline{v},\overline{v}]$. Suppose for contradiction that $F(v)=\hat{x}\ne x^{*}$ for some $x\in X$. Let $\bar{n}_{x}=\sum _{x'\le x}n_{x}$ for any $x\in X$ be the cumulative vote for x. We consider separate cases.

First we show that F is only defined if the thresholds $q_{x}$ are lower semi-continuous. Suppose for contradiction that they are not. Then as constructed in Example 7, F would not be well-defined. This would lead to a contradiction since F is assumed to be a well-defined function. Therefore, $q_{x}$ is lower semi-continuous.

Case 1  Suppose $\hat{x}$ such that $\sum _{x\le \hat{x}}n_{x} < q_{\hat{x}}$. There are two sub-cases.

Case 1(a)  Suppose $\hat{x}<x^{*}$. Consider the following set $\mathcal {A}=\{F,F_{1}\}$ such that $F(v)=\hat{x}$ and $F_{1}(v)=\hat{x}+\varepsilon _{\hat{x}}$. By definition of meta sets there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that,

$$\begin{aligned} {[}v_{i}\le \hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F\}]. \\ {[}v_{i} \ge \hat{x}+\varepsilon _{\hat{x}}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F,F_{1}\}] . \\ \end{aligned}$$

Using similar arguments as the ones used above $v^{\mathcal {A}}= (F^{\bar{n}_{\hat{x}}},F_{1}^{n-\bar{n}_{\hat{x}}})$ is a meta profile of v at $\mathcal {A}$ and $v^{\mathcal {\mathcal {A}}}(v) =(\hat{x}^{\bar{n}_{\hat{x}}},(\hat{x}+\varepsilon _{x})^{n -\bar{n}_{\hat{x}}})$ is the corresponding dual profile. Therefore by constitutional consistency,

$$\begin{aligned} F(\hat{x}^{\bar{n}_{\hat{x}}},(\hat{x} +\varepsilon _{\hat{x}})^{n-\bar{n}_{\hat{x}}}) =\hat{x}. \end{aligned}$$

By assumption $\bar{n}_{\hat{x}} < q_{\hat{x}}$. By definition of $q_{\hat{x}}$, $F(\hat{x}^{\bar{n}_{\hat{x}}},(\hat{x} +\varepsilon _{x})^{n-\bar{n}_{\hat{x}}}) \ne \hat{x}.$ This is a contradiction.

Case 1(b)  Suppose $\hat{x}> x^{*}$. Let $\mathcal {A}=\{F,F_{1}\}$ such that $F(v)=\hat{x}$ and $F_{1}(v)=x^{*}$. By using similar arguments as the ones used above there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that,

$$\begin{aligned} {[}v_{i} \le x^{*}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F_{1}\}]. \\ {[}x^{*}<v_{i}<\hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F,F_{1}\}] . \\ {[}v_{i}\ge \hat{x}]\Leftrightarrow & {} [\mathcal {A}(\succeq _{i}) =\{F\}]. \end{aligned}$$

Therefore $v^{\mathcal {A}}= (F_{1}^{\bar{n}_{x^{*}}}, F^{n-\bar{n}_{x^{*}}})$ is a meta profile of v at $\mathcal {A}$ and $v^{\mathcal {A}}(v)=(x^{*(\bar{n}_{x^{*}})}, \hat{x}^{(n-\bar{n}_{x^{*}})})$. By constitutional consistency,

$$\begin{aligned} F\left( x^{*(\bar{n}_{x^{*}})},\hat{x}^{n-\bar{n}_{x^{*}}}\right) =\hat{x}. \end{aligned}$$

Note that $\bar{n}_{x^{*}} \ge q_{x^{*}}$. By claim 7 and by definition of $q_{x}\hbox {s}$, $F(x^{*(\bar{n}_{x^{*}})}, \hat{x}^{n-\bar{n}_{x^{*}}}) =x^{*}.$ This is a contradiction.

Case 2  Suppose $F(v)=\hat{x}$ such that $\sum _{x'\le x}n_{\hat{x}} \ge q_{\hat{x}}$ but $\hat{x}\ne x^{*}$. As assumed above $x^{*} = \text {arg}\min \nolimits _{x\in [\underline{v},\overline{v}]} \sum _{x'\le x}n_{x} \ge q_{x}$. Therefore, $x^{*}<\hat{x}$. We can use similar arguments as the ones made in Case 1(b) to show that $F(v)=\hat{x}\ne x^{*}$ will lead to a contradiction. This completes the proof. $\square$

Proof of Theorem 3

It is easy to verify that k-median rules satisfy continuity in addition to all the other axioms. By proof of Theorem 2 we know that F is a cumulative-threshold rul(lsc) such that the thresholds are monotonic decreasing and lower semi-continuous. $\square$

Claim 9

We show that $q_{x}=q_{x'}$ for all $x,x'\in [0,1]$ such that $x\ne x'$.

Proof

We first show that $q_{x}:X\rightarrow N$ is continuous in x if and only if F(v) is continuous in $v_{i}= x_{i}$ for all $i\in N$. Suppose $\delta \in \mathbb {R}$ such that $|x-x'|<\delta$. By definition of the thresholds, $|q_{x}-q_{x'}|= |(\arg \min _{k\in N}F(x^{k},(x+\varepsilon _{x})^{n-k}))- (\arg \min _{k\in N}(F(x'^{(k)},(x'+\varepsilon _{x'})^{n-k}))|$.

If F is not continuous at x then $|(\arg \min _{k\in N}F(x^{k},(x+\varepsilon _{x})^{n-k}))- (\arg \min _{k\in N}(F(x'^{(k)},(x'+\varepsilon _{x'})^{n-k}))|\ne 0$ as $x'\rightarrow x.$ Moreover, due to the fact that the thresholds are integers, an implication of the above statement is that $|(\arg \min _{k\in N}F(x^{k},(x+\varepsilon _{x})^{n-k}))- (\arg \min _{k\in N}(F(x'^{(k)},(x'+\varepsilon _{x'})^{n-k}))| \ge 1.$ Therefore, for any $\varepsilon <1$ there does not exist $\delta$ such that $|x-x'|<\delta$ and $|q_{x}-q_{x'}|<\varepsilon$. Hence, $q_{x}$continuous if F is continuous.

It is easy to verify that if q is continuous then so is F. Therefore, $q_{x}$ is continuous if and only if F is continuous. Since the thresholds are integer valued and the domain [0, 1] is connected, $q_{x}$ must be connected. Therefore, $q_{x}$ is constant. $\square$

Independence of the axioms.

We now show for each axiom there exists a voting rule which satisfies all the other axioms.

• Constitutional consistency  The m-plurality rule satisfies all the other axioms except constitutional consistency. Its violation of the axiom has been shown in the section on constitutional consistency.

• Unanimity  The following version of the generalized median rule satisfies all axioms except unanimity. There exists $\alpha _{1},\ldots \alpha _{n+1}$ such that $F^{gm} =median(v_{1},\ldots ,v_{n},\alpha _{1},\ldots ,\alpha _{n+1})$. To see how this violates unanimity suppose $\alpha _{t}=a\in X$ for all $t\in \{1,\ldots ,n+1\}$. Then $F^{gm}(b^{n},a^{n+1})=median(b^{n}, a^{n+1}) =a$. Similar arguments as the ones used to show that cumulative-threshold rules satisfy constitutional consistency can be made to show that $F^{gm}$ also satisfies constitutional consistency. It is clearly anonymous.

• Anonymity  The dictator rule satisfies all the axioms except anonymity. The argument for it satisfying constitutional consistency was provided in the section on the latter axiom. It is easy to see that it satisfies unanimity.

Claim 10

Suppose F satisfies constitutional consistency and unanimity. Then F is strategy-proof.

Proof

We prove this by contradiction. Suppose F satisfies constitutional consistency and unanimity but is not strategy-proof. Suppose v is the truthful vote profile of peaks and let $F(v)=x$. By Claim 1 of the proof of Theorem 1 we know by constitutional consistency and unanimity that $F(v) \in \{\underline{v},\overline{v}\}$.

By our assumption there exists a strategy $v_{i1}'$ for some $i_{1}\in N$ such that for $v'=(v_{i1}',v_{-i})$, we have $F(v_{i1}',v_{-i}) \succ _{i} F(v)$. W.l.o.g. assume that $v_{i1}>F(v)=x$. Without loss of generality we can write the votes in ascending order according to $\le$. Therefore, we can assume that $v_{1}\le v_{2}\cdots \le v_{n}$. Let $v_{k}=x$ for some $k\in N$.¹⁸ There are two cases:

Case 1  Suppose $v_{i1}'>F(v)=x$. Then by single-peakedness, $F(v')=x'>x$. Consider the following set of rules $\mathcal {A}_{1}=\{F,F',F_{1},F_{2},\ldots ,F_{k-1}\}$ such that $F'(v)=x'$ and $F_{i}(v)=v_{i}$ for all $i\in \{1,\ldots ,k-1\}$. By single-peakedness there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that (i) $[v_{i}=F_{i}(v)] \Leftrightarrow [v^{\mathcal {A}_{1}}_{i}=F_{i}]$ for all $i\in \{1,\ldots ,k-1\}$, (ii) $[v_{k}=x]\Leftrightarrow [v_{k}^{\mathcal {A}_{1}}=F]$ and (iii) $[v_{i} > x]\Leftrightarrow [v^{\mathcal {A}_{1}}_{i}=F']$ for all $i\in \{k+1,\ldots ,n\}$. Therefore, by constitutional consistency,

$$\begin{aligned} F(v^{\mathcal {A}_{1}}(v))= & {} F(F_{1}(v),F_{2}(v),\ldots ,F(v),F'(v),\ldots F'(v))\\= & {} F\left( v_{1},v_{2},\ldots ,v_{k-1},x,\underbrace{x',\ldots ,x'}_{n-k}\right) =F(v). \end{aligned}$$

Again, consider the following set of rules $\mathcal {A}_{2}=\{F,F_{1},F_{2},\ldots ,F_{k}\}$ such that $F_{i}(v')=v_{i}'$ for all $i\in \{1,\ldots ,k\}$. By single-peakedness there exists a preference profile $\succeq \in \mathcal {S}(v')$ such that (i) $[v_{i}'=F_{i}(v')] \Leftrightarrow [v'^{\mathcal {A}_{2}}_{i}=F_{i}]$ for all $i\in \{1,\ldots ,k\}$ and (ii) $[v_{i}' > x]\Leftrightarrow [v'^{\mathcal {A}_{2}}_{i}=F]$ for all $i\in \{k+1,\ldots ,n\}$. Therefore, by constitutional consistency,

$$\begin{aligned} F(v'^{\mathcal {A}_{2}}(v'))= & {} F (F_{1}(v'),F_{2}(v'),\ldots ,F(v'),\ldots F(v')) \\= & {} F\left( v'_{1},v'_{2},\ldots ,v_{k}',\underbrace{x',\ldots ,x'}_{n-k}\right) =F(v'). \end{aligned}$$

But note that $v_{i}'^{\mathcal {A}_{2}}(v')=v_{i}^{\mathcal {A}_{1}} (v)$ for all $i\in \{1,\ldots ,k\}$ and $v_{i}'^{\mathcal {A}_{2}}(v') =v^{\mathcal {A}_{1}}(v)=x'$ for all $i\in \{k+1,\ldots n\}$. Therefore, $v'^{\mathcal {A}_{2}}(v') =v^{\mathcal {A}_{1}}(v)$ and $F(v')=F(v'^{\mathcal {A}_{2}}(v'))=F(v^{\mathcal {A}_{1}}(v))=F(v)$. This a contradiction.

Case 2  Suppose $v_{i1}'\le F(v)=x$. Then by single-peakedness, $F(v')=x'<x$. Consider the following set of rules $\mathcal {A}_{1}=\{F,F',F_{k+1},F_{k+2},\ldots ,F_{n}\}$ such that $F'(v)=x'$ and $F_{i}(v)=v_{i}$ for all $i\in \{k+1,\ldots ,n\}$. By single-peakedness there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that (i) $[v_{i}= F_{i}(v)] \Leftrightarrow [v^{\mathcal {A}_{1}}_{i}=F_{i}]$ for all $i\in \{k+1,\ldots ,n\}$, (ii) $[v_{k}=x]\Leftrightarrow [v^{\mathcal {A}_{1}}_{i}=F]$ and (iii) $[v_{i} <x]\Leftrightarrow [v^{\mathcal {A}_{1}}_{i}=F']$ for all $i\in \{1,\ldots ,k-1\}$. Therefore, by constitutional consistency,

$$\begin{aligned} F(v^{\mathcal {A}_{1}}(v))= & {} F(F'(v),F'(v),\ldots ,F(v), F_{k+1}(v),\ldots F_{n}(v))\\= & {} F\left( \underbrace{x',\ldots ,x'}_{k-1 \;times},x, v_{k+1},\ldots ,v_{n} \right) =F(v). \end{aligned}$$

Again, consider the following set of rules $\mathcal {A}_{2}=\{F,F_{k},F_{k+1},\ldots ,F_{n}\}$ such that $F_{i}(v)=v'_{i}$ for all $i\in \{k,\ldots ,n\}$. By single-peakedness there exists a preference profile $\succeq \in \mathcal {S}(v)$ such that (i) $[v_{i}'= F_{i}(v')] \Leftrightarrow [v'^{\mathcal {A}_{2}}=F_{i}]$ for all $i\in \{k,\ldots ,n\}$ and (ii) $[v'_{i} <x]\Leftrightarrow [v'^{\mathcal {A}_{2}}_{i}=F]$ for all $i\in \{1,\ldots ,k-1\}$. Therefore, by constitutional consistency,

$$\begin{aligned} F(v'^{\mathcal {A}_{2}}(v'))= & {} F(F(v'),F(v'),\ldots ,F_{k}(v'),\ldots F_{n}(v')) \\= & {} F\left( \underbrace{x',\ldots ,x'}_{k-1 \;times}, v'_{k},\ldots ,v'_{n}\right) =F(v'). \end{aligned}$$

But note that $v_{i}'^{\mathcal {A}_{2}}(v')=v_{i}^{\mathcal {A}_{1}} (v)$ for all $i\in \{k,\ldots ,n\}$ and $v_{i}'^{\mathcal {A}_{2}}(v') =v^{\mathcal {A}_{1}}_{i}(v)=x'$ for all $i\in \{1,\ldots ,k-1\}$. Therefore, $v'^{\mathcal {A}_{2}}(v') =v^{\mathcal {A}_{1}}(v)$ and $F(v')=F(v'^{\mathcal {A}_{2}}(v'))=F(v^{\mathcal {A}_{1}}(v))=F(v)$. This a contradiction.

There is an additional case where $F(v)=x$ and $\not \exists i\in N$ such that $v_{i}=x$. W.l.o.g. assume that $v_{\bar{i}}<x<v_{\bar{i} +1}$ for some $\bar{i}\in N$. If $v_{i1}>x$ then repeat the above arguments in Case 1 with $k=\bar{i}$. If $v_{i1}\le x$ then repeat the above arguments in Case 2 with $k=\bar{i}+1$. $\square$