On single-peaked domains and min–max rules

Abstract

We consider social choice problems where the admissible set of preferences of each agent is single-peaked. First, we show that if all the agents have the same admissible set of preferences, then every unanimous and strategy-proof social choice function (SCF) is tops-only. Next, we consider situations where different agents have different admissible sets of single-peaked preferences. We show by means of an example that unanimous and strategy-proof SCFs need not be tops-only in this situation, and consequently provide a sufficient condition on the admissible sets of preferences of the agents so that unanimity and strategy-proofness guarantee tops-onlyness. Finally, we characterize all domains on which (i) every unanimous and strategy-proof SCF is a min–max rule, and (ii) every min–max rule is strategy-proof. As an application of our result, we obtain a characterization of the unanimous and strategy-proof social choice functions on maximal single-peaked domains (Moulin in Public Choice 35(4):437–455. https://doi.org/10.1007/BF00128122, 1980; Weymark in SERIEs 2(4):529–550. https://doi.org/10.1007/s13209-011-0064-5, 2011), minimally rich single-peaked domains (Peters et al. in J Math Econ 52:123–127. https://doi.org/10.1016/j.jmateco.2014.03.008. http://www.sciencedirect.com/science/article/pii/S0304406814000470, 2014), maximal regular single-crossing domains (Saporiti in Theor Econ 4(2):127–163, 2009, J Econ Theory 154:216–228. https://doi.org/10.1016/j.jet.2014.09.006. http://www.sciencedirect.com/science/article/pii/S0022053114001276, 2009), and distance based single-peaked domains.

Appendix

1.1 Proof of Theorem 2

Proof

Let $P_{N} \in \mathcal {S}_{N}$, $i \in N$, and $P'_{i} \in \mathcal {S}$ be such that $r_{1}(P_{i}) = r_{1}(P'_{i})$. It is enough to show that $f(P_{N}) = f(P'_{i},P_{N {\setminus } i})$. Suppose not. Let $r_{1}(P_{i}) = r_{1}(P'_{i}) = x$, $f(P_{N}) = z$, and $f(P'_{i},P_{N {\setminus } i}) = y$. By strategy-proofness, $z P_{i} y$ and $y P'_{i} z$. Since $\mathcal {S}_{i}$ is single-peaked and $r_{1}(P_{i}) = r_{1}(P'_{i})$, this means either $y< x < z$ or $z< x < y$. Assume without loss of generality that $y< x < z$. Let $\bar{N} = \{j \in N \mid r_{1}(P_{j}) \ge x\}$ and let $\bar{P}_{N} \in \mathcal {S}_{N}$ be such that $\bar{P}_{j} = P_{i}$ for all $j \in \bar{N}$, and $\bar{P}_{j} = P_{j}$ for all $j \notin \bar{N}$.

Claim 1. $f(P'_{i},P_{N {\setminus } i}) = y$ implies $f(\bar{P}_{N}) = y$.

By Pareto optimality, $f(\bar{P}_{N}) \le x$. If $f(\bar{P}_{N}) \in (y,x]$ then agents in $\bar{N}$ manipulates f at $(P'_{i},P_{N {\setminus } i})$ via $\bar{P}_{N}$. On the other hand, if $f(\bar{P}_{N}) < y$, then agents in $\bar{N}$ manipulate f at $\bar{P}_{N}$ via $(P'_{i},P_{N {\setminus } i})$. This completes the proof of Claim 1.

Claim 2. $f(P_{N}) = z$ implies $f(\bar{P}_{N}) \ne y$.

Because $z \bar{P}_{j} y$ for all $j \in \bar{N}$, if $f(\bar{P}_{N}) = y$, then agents in $\bar{N}$ manipulates f at $\bar{P}_{N}$ via $P_{N}$. This completes the proof of Claim 2.

However, Claim 2 contradicts Claim 1. This completes the proof of the theorem.   $\square$

1.2 Proof of Theorem 3

Proof

We prove the theorem for the case where $\mathcal {S}_{N}$ is right-connected, the proof is the same for the case where $\mathcal {S}_{N}$ is left-connected.

It is sufficient to show that $f(P_{N})=f(P'_{i},P_{N {\setminus } i})$ for all $P_{i},P'_{i} \in \mathcal {S}_{i}$ and all $P_{N {\setminus } i} \in \mathcal {S}_{N {\setminus } i}$ with $r_{1}(P_{i}) = r_{1}(P'_{i})$. Assume for contradiction, $f(P_{N}) \ne f(P'_{i},P_{N {\setminus } i})$. By strategy-proofness, we have $f(P_{N}) P_{i} f(P'_{i},P_{N {\setminus } i})$ and $f(P'_{i},P_{N {\setminus } i})P'_{i} f(P_{N})$. Assume without loss of generality, $f(P'_{i},P_{N {\setminus } i})< r_{1}(P_{i}) < f(P_{N})$. Suppose $f(P'_{i},P_{N {\setminus } i}) = y$, $r_{1}(P_{i}) = x$, and $f(P_{N}) = z$. Let $N_{1} = \{j \in N \mid r_{1}(P_{j}) \ge z\}$. By Pareto optimality, $N_{1} \ne \emptyset$. Consider $P^{1}_{N} \in \mathcal {S}_{N}$ such that $r_{1}(P^{1}_{j}) = z-1$ and $r_{2}(P^{1}_{j}) = z$ for all $j \in N_{1}$, and $P^{1}_{k} = P_{k}$ for all $k \notin N_{1}$. By group-strategy-proofness, $f(P^{1}_{N}) \in \{z,z-1\}$. By Pareto optimality, $f(P^{1}_{N}) \le z-1$. Combining, $f(P^{1}_{N}) = z-1$. Now, let $N_{2} = \{j \in N \mid r_{1}(P_{j}) = z-1\}$. Consider $P^{2}_{N} \in \mathcal {S}_{N}$ such that $r_{1}(P^{2}_{j}) = z-2$ and $r_{2}(P^{2}_{j}) = z-1$ for all $j \in N_{2}$, and $P^{2}_{k} = P_{k}$ for all $k \notin N_{2}$. Using similar logic as before, $f(P^{2}_{N}) = z-2$. Continuing in this manner, we construct a profile $\hat{P}_{N} \in \mathcal {S}_{N}$ such that $r_{1}(\hat{P}_{j}) = x$ for all $j \in N$ with $r_{1}(P_{j}) \ge x$, $\hat{P}_{k} = P_{k}$ for all $k \in N$ such that $r_{1}(\hat{P}_{k}) < x$, and $f(\hat{P}_{N}) = x$.

Now, consider the profile $\bar{P}_{N} \in \mathcal {S}_{N}$ such that $\bar{P}_{i} = P'_{i}$, $\bar{P}_{k} = P_{k}$ if $k \in N$ such that $r_{1}(P_{k}) < x$, and $r_{1}(\bar{P}_{k}) = x$ if $k \in N {\setminus } i$ such that $r_{1}(P_{k}) \ge x$. By Pareto optimality, $f(\bar{P}_{N}) \le x$. This, together with group-strategy-proofness from $(P'_{i},P_{N {\setminus } i})$ to $\bar{P}_{N}$, implies $f(\bar{P}_{N}) \ge y$. However, by group-strategy-proofness from $\bar{P}_{N}$ to $(P'_{i},P_{N {\setminus } i})$, we $f(\bar{P}_{N}) \le y$. Combining, $f(\bar{P}_{N}) = y$.

Note that $\hat{P}_{j} = \bar{P}_{j}$ for all $j \in N$ with $r_{1}(P_{j}) < x$ and $r_{1}(\hat{P}_{j}) = r_{1}(\bar{P}_{j})$ for all $j \in N$ with $r_{1}(P_{j}) \ge x$. This means the group of agents $\{j \in N \mid r_{1}(\bar{P}_{j}) = x\}$ manipulates at $\bar{P}_{N}$ via $\hat{P}_{N}$, a contradiction. This completes the proof of the theorem.   $\square$

1.3 Proof of Theorem 4

Proof

(If part) Note that a min–max rule is unanimous by definition (on any domain). We show that such a rule is strategy-proof on $\mathcal {S}_{N}$. For all $i \in N$, let $\bar{\mathcal {S}}_i$ be the maximal set of single-peaked preferences. By Weymark (2011), a min–max rule is strategy-proof on $\bar{\mathcal {S}}_N$. Since $\mathcal {S}_{i} \subseteq \bar{\mathcal {S}}_{i}$ for all $i \in N$, a min–max rule must be strategy-proof on $\mathcal {S}_N$. This completes the proof of the if part.

(Only-if part) Let $\mathcal {S}_{i}$ be a top-connected set of single-peaked preferences for all $i \in N$ and let $f: \mathcal {S}_{N} \rightarrow X$ be a unanimous and strategy-proof SCF. We show that f is a min–max rule. First, we establish a few properties of f in the following sequence of lemmas.

By Corollary 1 and Theorem 3, f must satisfy Pareto property and tops-onlyness. Also, by Theorem 1, f is group strategy-proof. Our next lemma shows that f is uncompromising.

Lemma 3

The SCF f is uncompromising.

Proof

Let $P_{N} \in \mathcal {S}_{N}$, $i \in N$, and $P'_{i} \in \mathcal {S}_{i}$ be such that $r_{1}(P_{i})< f(P_{N})$ and $r_{1}(P'_{i}) \le f(P_{N})$. It is sufficient to show $f(P'_{i},P_{N {\setminus } i}) = f(P_{N})$. Suppose $r_{1}(P_{i}) = x$, $f(P_{N}) = y$, and $f(P'_{i},P_{N {\setminus } i}) = y'$.

By strategy-proofness, we must have $y' < x$. This is because, if $y' \in [x,y)$, then agent i manipulates at $P_{N}$ via $P'_{i}$. On the other hand, if $y' > y$, then by means of the fact that $r_{1}(P'_{i}) \le y$, agent i manipulates at $(P'_{i},P_{N {\setminus } i})$ via $P_{i}$.

Because $y'<x$, we assume without loss of generality that $r_1(P'_{i}) = y'$ and $\min (\tau (P'_{i},P_{N {\setminus } i} ) ) = y'$.¹³ Assume for contradiction that $y \ne y'$.

Let $T = \{j \in N \mid r_1(P_{j}) < x\}$. For $j \in T$, let $P'_{j} \in \mathcal {S}_{j}$ be such that $r_1(P'_{j})=x$.

Claim 1. $f(P_N)=y$ implies $f(P'_{T},P_{N {\setminus } T})= y.$

If T is empty, then there is nothing to show. Suppose T is non-empty. By Pareto property, $f(P'_{T},P_{N {\setminus } T}) \ge x$. If $f(P'_{T},P_{N {\setminus } T}) \in [x,y)$, then the agents in T manipulate f at $P_N$ via $(P'_{T},P_{N {\setminus } T})$. On the other hand, if $f(P'_{T},P_{N {\setminus } T}) > y$, then the agents in T manipulate f at $(P'_{T},P_{N {\setminus } T})$ via $P_N$. This completes the proof of Claim 1.

Let $T' = T \cup i$. For all $j \in T'$, let $\tilde{P}_{j} \in \mathcal {S}_j$ be such that $r_{1}(\tilde{P}_{j}) = x$.

Claim 2. $f(P_i', P_{N {\setminus } i})=y'$ implies $f(\tilde{P}_{T'},P_{N {\setminus } T'}) = x$.

Let $T''$ be the set of agents whose top-ranked alternative is $y'$ at the profile $(P'_{i},P_{N {\setminus } i})$. More formally, $T'' =i \cup \{j \in N \mid r_{1}(P_{j}) = y'\}$. Consider the profile $\bar{P}_{N} \in \mathcal {S}_{N}$ such that $r_{1}(\bar{P}_{j}) = y'+1$ for all $j \in T''$ and $\bar{P}_{j}=P_{j}$ for all other agents. As $\mathcal {S}_N$ is top-connected, we can assume $r_{2}(\bar{P}_{j}) = y'$ for all $j \in T''$. However, since $f(P_i', P_{N {\setminus } i})=y'$, by group strategy-proofness, $f(\bar{P}_{N})\in \{y',y'+1\}$ as otherwise agents in $T''$ manipulate f at $\bar{P}_N$ via $(P_i',P_{N {\setminus } i})$. Since $\min (\tau (\bar{P}_{N}))=y'+1$, by Pareto property,

$$\begin{aligned} f(\bar{P}_{N})=y'+1. \end{aligned}$$

Using similar logic, we can construct a profile $\hat{P}_{N} \in \mathcal {S}_{N}$ where $r_{1}(\hat{P}_{j}) = y'+2$ and $r_{2}(\hat{P}_{j}) = y'+1$ for all agents j with $r_{1}(\bar{P}_{j}) = y'+1$ and $\hat{P}_{j}=\bar{P}_{j}$ for all other agents, and conclude that

$$\begin{aligned} f(\hat{P}_{N}) = y'+2.\end{aligned}$$

Continuing in this manner, we move all the agents j in $T'$ to a preference $\tilde{P}_{j} \in \mathcal {S}_{j}$ with $r_{1}(\tilde{P}_{j}) = x$ and $r_2(\tilde{P}_j) = x-1$ while keeping the preferences of all other agents unchanged and conclude that

$$\begin{aligned} f(\tilde{P}_{T'},P_{N {\setminus } T'}) = x. \end{aligned}$$

This and tops-onlyness completes the proof of Claim 2.

Now, we complete the proof of the lemma. Consider the profiles $(P'_{T},P_{N {\setminus } T})$ and $(\tilde{P}_{T'},P_{N {\setminus } T'})$. Note that for an agent j, if $r_{1}(P_{j}) > x$, then her preference is the same in both the profiles $(P'_{T},P_{N {\setminus } T})$ and $(\tilde{P}_{T'},P_{N {\setminus } T'})$. Moreover, for an agent j, if $r_{1}(P_{j}) \le x$, then her top-ranked alternative is x in both the profiles. Therefore, the top-alternatives of each agent in these two profiles are the same. However, since $f(P'_{T},P_{N {\setminus } T}) \ne f(\tilde{P}_{T'},P_{N {\setminus } T'})$, Claim 1 and 2 contradict tops-onlyness of f. This completes the proof of the lemma.   $\square$

The following lemma establishes that f is a min–max rule.

Lemma 4

The SCF f is a min–max rule.

Proof

For all $S \subseteq N$, let $(P^{a}_{S},P^{b}_{N {\setminus } S}) \in \mathcal {S}_{N}$ be such that $r_{1}(P^{a}_{i}) = a$ for all $i \in S$ and $r_{1}(P^{b}_{i}) = b$ for all $i \in N {\setminus } S$. Define $\beta _{S} = f(P^{a}_{S},P^{b}_{N {\setminus } S})$ for all $S \subseteq N$. Clearly, $\beta _{S} \in X$ for all $S \subseteq N$. By unanimity, $\beta _{\varnothing } = b$ and $\beta _{N} = a$. Also, by strategy-proofness, $\beta _{S} \le \beta _{T}$ for all $T \subseteq S$.

Take $P_{N} \in \mathcal {S}_{N}$. We show $f(P_{N}) = \min \nolimits _{S \subseteq N}\{\max \nolimits _{i \in S}\{r_{1}(P_{i}),\beta _{S}\}\}$. Suppose $S_{1} = \{i \in N \mid r_{1}(P_{i}) < f(P_{N})\}$, $S_{2} = \{i \in N \mid f(P_{N}) < r_{1}(P_{i})\}$, and $S_{3} = \{i \in N \mid r_{1}(P_{i}) = f(P_{N})\}$. By strategy-proofness and uncompromisingness, $\beta _{S_{1} \cup S_{3}} \le f(P_{N}) \le \beta _{S_{1}}$. Consider the expression $\min \nolimits _{S \subseteq N}\{\max \nolimits _{i \in S} \{r_{1}(P_{i}), \beta _{S}\}\}$. Take $S \subseteq S_{1}$. Then, by Condition (iii) in Definition 9, $\beta _{S_{1}} \le \beta _{S}$. Since $r_{1}(P_{i}) < f(P_{N})$ for all $i \in S$ and $f(P_{N}) \le \beta _{S_{1}} \le \beta _{S}$, we have $\max \nolimits _{i \in S} \{r_{1}(P_{i}), \beta _{S}\} = \beta _{S}$. Clearly, for all $S \subseteq N$ such that $S \cap S_{2} \ne \varnothing$, we have $\max \nolimits _{i \in S} \{r_{1}(P_{i}), \beta _{S}\} > f(P_{N})$. Consider $S \subseteq N$ such that $S \cap S_{2} = \varnothing$ and $S \cap S_{3} \ne \varnothing$. Then, $S \subseteq S_{1} \cup S_{3}$, and hence $\beta _{S_{1} \cup S_{3}} \le \beta _{S}$. Therefore, $\max \nolimits _{i \in S} \{r_{1}(P_{i}), \beta _{S}\} = \max \{f(P_{N}), \beta _{S}\} \ge \max \{f(P_{N}), \beta _{S_{1} \cup S_{3}}\}$. Since $\beta _{S_{1} \cup S_{3}} \le f(P_{N})$, we have $\max \{f(P_{N}), \beta _{S_{1} \cup S_{3}}\} = f(P_{N})$. Combining all these, we have $\min \nolimits \nolimits _{S \subseteq N}\{\max \nolimits _{i \in S} \{r_{1}(P_{i}), \beta _{S}\}\} = \min \{f(P_{N}),\beta _{S_{1}}\}$. Because $f(P_{N}) \le \beta _{S_{1}}$, we have $\min \{f(P_{N}),\beta _{S_{1}}\} = f(P_{N})$. This completes the proof of the lemma.   $\square$

The proof of the only-if part of Theorem 4 follows from Lemmas 3 and 4.   $\square$

1.4 Proof of Theorem 5

Proof

The proof of the if part follows from Theorem 4. We proceed to prove the only-if part. Let $\mathcal {D}_{N}$ be a min–max domain. We show that $\mathcal {D}_{i}$ is top-connected single-peaked for all $i \in N$. We show this in two steps: in Step 1 we show that $\mathcal {D}_i$ is single-peaked for all $i \in N$, and in Step 2, we show that $\mathcal {D}_i$ is top-connected for all $i\in N$.

Step 1. Suppose that $\mathcal {D}_{i}$ is not single-peaked for some $i \in N$. Then, there is $Q \in \mathcal {D}_{i}$ and $x,y \in X$ such that $x< y < r_{1}(Q)$ and xQy. Consider the min–max rule $f^{\beta }$ with respect to $(\beta _{S})_{S \subseteq N}$ such that $\beta _{S} = x$ for all $\emptyset \subsetneq S \subsetneq N$. Take $P_{N} \in {\mathcal {D}}_N$ such that $P_{i} = Q$ and $r_{1}(P_{j}) = y$ for all $j \in N {\setminus } i$. By the definition of $f^{\beta }$, $f^{\beta }(P_{N}) = y$. Now, take $P'_{i} \in \mathcal {D}_{i}$ with $r_{1}(P'_{i}) = x$. Again, by the definition of $f^{\beta }$, $f^{\beta }(P'_{i},P_{N {\setminus } i}) = x$. This means agent i manipulates at $P_{N}$ via $P'_{i}$, which is a contradiction to the assumption that $\mathcal {D}_{N}$ is a min–max domain. This completes Step 1.

Step 2. In this step, we show that $\mathcal {D}_{i}$ satisfies top-connectedness for all $i \in N$. Assume for contradiction that $\mathcal {D}_{i}$ is not top-connected for some $i \in N$. By definition, $\mathcal {D}_{i}$ is regular. Since $\mathcal {D}_{i}$ is single-peaked, for all $P \in \mathcal {D}_{i}$, $r_{1}(P)=a ~(\text{ or } b)$ implies $r_{2}(P)=a+1 ~ (\text{ or } b-1)$. Again, because $\mathcal {D}_{i}$ is single-peaked, for all $P \in \mathcal {D}_{i}$ and all $x \in X {\setminus } \{a,b\}$, $r_{1}(P)=x$ implies $r_{2}(P) \in \{x-1,x+1\}$. Since $\mathcal {D}_{i}$ violates top-connectedness, assume without loss of generality that there exists $x \in X {\setminus } \{a,b\}$ such that for all $P \in \mathcal {D}_{i}$, $r_{1}(P) = x$ implies $r_{2}(P) = x-1$. Consider the following SCF:¹⁴

$$\begin{aligned}f(P_{N}) = \left\{ \begin{array}{ll} x &{}\quad \text{ if } r_1(P_{i}) = x \quad \text{ and } \quad x P_{j} (x-1) \quad \text{ for } \text{ all } \,\,j \in N {\setminus } i, \\ x-1 &{}\quad \text{ if } r_1(P_{i}) = x \quad \text{ and } \quad (x-1) P_{j} x \quad \text{ for } \text{ some } \,\,j \in N {\setminus } i , \\ r_1(P_{i}) &{}\quad \text{ otherwise }. \end{array} \right. \end{aligned}$$

It is left to the reader to verify that f is unanimous and strategy-proof. We show that f is not uncompromising, which in turn means that f is not a min–max rule. Let $P_{N} \in \mathcal {D}_{N}$ be such that $r_{1}(P_{i}) = x$ and $r_{1}(P_{j}) = x-1$ for some $j \ne i$, and let $P'_{i} \in \mathcal {D}_{i}$ be such that $r_{1}(P'_{i}) = x+1$. Then, by the definition of f, $f(P_{N}) = x-1$ and $f(P'_{i},P_{N {\setminus } i}) = x+1$. Therefore, because $f(P_N)=x-1$ and $x-1 \le r_1(P_i) \le r_1(P_i')$, the fact that $f(P'_{i},P_{N {\setminus } i}) = x+1$ is a violation of uncompromisingness. This completes Step 2 and the proof of the only-if part.   $\square$