The possibility of Arrovian social choice with the process of nomination

Abstract

In this paper, we introduce an Arrovian social choice framework with the process of nomination. We consider a two-stage social choice procedure in which some alternatives are first nominated by aggregating the opinions of nominators, and then the society makes a collective choice from the nominated alternatives by aggregating the preferences of voters. Each nominator’s opinion is a positive, negative, or neutral view as to whether each alternative deserves to be eligible for collective decision making. If a voter is a nominator, his preference space is restricted by his opinion as follows: he always prefers positive alternatives to neutral ones and neutral alternatives to negative ones, according to his opinion. When each nominating voter has such a preference space, we first characterize Arrow-consistent preference domains at the second stage of the social choice framework. Second, we find a resolution of Arrow’s impossibility theorem when at least one nominating voter exists.

Appendix

We provide the proofs of some of our results in the Appendix. The next three lemmas are used to prove Theorem 1.

Lemma 1

Given an opinion profile $J\in \mathscr {J}$ with $\#f(J)\ge 3$, suppose that there exists an extreme trichotomist i on f(J) with $\{x\}=T_{i}^{J}(f(J))$. If there exist $y\in f(J)\backslash \{x\}$ and a voter $j\in \mathcal {V}\backslash \{i\}$ such that $y\notin T_{i}^{J}(f(J))$ and $y\in T_{j}^{J}(f(J))$, then there exists an SCC with f(J) satisfying WP, I, and ND.

Proof

For all $\mathbf R \in \mathcal {D}^{J}$, construct an SCC with f(J) as follows:

$$ C_{f}(\mathbf R ,f(J))=B(f(J),R_{j})\cup \{x\}. $$

It is easy to see that $C_{f}$ is well-defined and satisfies WP and I. We now show that $C_{f}$ satisfies ND. Since $y\in T_{j}^{J}(f(J))$, there exists $\mathbf R \in \mathcal {D}^{J}$ with $yP_{j}x$. However, we have $x\in C_{f}(\mathbf R ,f(J))$ by construction. Therefore, j is not a dictator on f(J). Since we have $\{x\}=T_{i}^{J}(f(J))$, $y\notin T_{i}^{J}(f(J))$, and $y\in T_{j}^{J}(f(J))$, there exists $\mathbf R '\in \mathcal {D}^{J}$ with $xP'_{i}y$ and $y\in B(f(J),R'_{j})$. By construction, we have $y\in C_{f}(\mathbf R ',f(J))$, which implies that i is not a dictator on f(J). It is easy to show that no other voter is a dictator on f(J). $\square$

Lemma 2

Given an opinion profile $J\in \mathscr {J}$ with $\#f(J)\ge 3$, suppose that there exists a trichotomist i on f(J) who is not an extreme trichotomist on f(J). If there exist $x\in f(J)$ and a voter $j\in \mathcal {V}\backslash \{i\}$ such that $x\notin T_{i}^{J}(f(J))$ and $x\in T_{j}^{J}(f(J))$, then there exists an SCC with f(J) satisfying WP, I, and ND.

Proof

For all $\mathbf R \in \mathcal {D}^{J}$, construct an SCC with f(J) as follows:

$$ C_{f}(\mathbf R ,f(J))=B(T_{i}^{J}(f(J)),R_{j}). $$

It is easy to see that $C_{f}$ is well-defined and satisfies WP and I. It is possible to show that it satisfies ND by a similar proof to Lemma 1. $\square$

Lemma 3

Given an opinion profile $J\in \mathscr {J}$ with $\#f(J)\ge 3$, suppose the case where every voter i is a trichotomist on f(J) such that the restriction of $\mathcal {D}_{j}^{J}$ to $T_{j}^{J}(f(J))$ are the same for all voters $j\in \mathcal {V}$.

• (1) If $\# T_{j}^{J}(f(J))\ge 3$, then there exists an SCC with f(J) satisfying WP and I if and only if there exists a dictator on f(J).

• (2) If $\# T_{j}^{J}(f(J))=2$, then there exists an SCC with f(J) satisfying WP, I, and ND.

• (3) If $\# T_{j}^{J}(f(J))=1$, then there exists an SCC with f(J) satisfying WP if and only if there exists a dictator on f(J).

Proof

Let $C_{f}:\mathcal {D}^{J}\twoheadrightarrow f(J)$ satisfy WP and I. Let $Z=T_{j}^{J}(f(J))$ for all $j\in \mathcal {V}$, and $\mathbf R |_{Z}$ and $\mathcal {D}^{J}|_{Z}$ denote the restrictions of $\mathbf R$ and $\mathcal {D}^{J}$ to Z, respectively. Define $C_{f}^{*}:\mathcal {D}^{J}|_{Z}\twoheadrightarrow Z$ by setting $C_{f}^{*}(\mathbf R |_{Z},Z)=C_{f}(\mathbf R ,f(J))$ for all $\mathbf R \in \mathcal {D}^{J}$. Because $C_{f}$ satisfies WP, the alternatives in $f(J)\backslash Z$ are never chosen from f(J) for all $\mathbf R \in \mathcal {D}^{J}$. Therefore, $C^{*}_{f}$ is well-defined.

(1) Suppose $\# Z\ge 3$. It is easy to show that for all $j\in \mathcal {V}$, $\mathcal {D}_{j}^{J}|_{Z}=\mathscr {R}|_{Z}$ and $C^{*}_{f}$ satisfies WP and I. Since $C_{f}^{*}$ can be seen as a standard social choice correspondence with a fixed agenda, it follows from Denicolò (1985, 1993) that $C_{f}^{*}$ is dictatorial, which implies that $C_{f}$ is also dictatorial. The converse is obvious.

(2) Suppose $\# Z=2$. Let $Z=\{x,y\}$. Consider the following $C_{f}^{*}$; $x\in C_{f}^{*}(\mathbf R |_{Z},Z)$ if and only if $\# \{j\in \mathcal {V}|xR_{j}y\}\ge \# \{j\in \mathcal {V}|yR_{j}x\}$. It is clear that $C_{f}^{*}$ satisfies WP, I, and ND, which implies $C_{f}$ satisfies WP, I, and ND.

(3) Suppose $\# Z=1$. Let $Z=\{x\}$. Since $C_{f}$ satisfies WP, we have $C_{f}(\mathbf R ,f(J))=\{x\}$ for all $\mathbf R \in \mathcal {D}^{J}$, which implies that $C_{f}$ is dictatorial. The converse is obvious. $\square$

Proof of Theorem 1

Suppose that either condition 1 or condition 2 fails. If condition 1 fails, then the restriction of every voter’s preference space $\mathcal {D}_{j}^{J}$ to f(J) is unrestricted. By Denicolò (1985, 1993), there exists no SCC with f(J) satisfying WP, I, and ND. If condition 2 fails, then $\# T_{j}^{J}(f(J))\ne 2$ for all $j\in \mathcal {V}$ and every voter is a trichotomist on f(J) such that the restriction of $\mathcal {D}_{j}^{J}$ to $T_{j}^{J}(f(J))$ are the same for all $j\in \mathcal {V}$. By Lemma 3, there exists no SCC with f(J) satisfying WP, I, and ND. Therefore, if there exists an SCC with f(J) satisfying WP, I, and ND, then conditions 1 and 2 hold.

Conversely, suppose that conditions 1 and 2 hold. Then, there exists a trichotomist $i\in \mathcal {M}\cap \mathcal {V}$ on f(J). If i is an extreme trichotomist on f(J), then by Lemma 1, there exists an SCC with f(J) satisfying WP, I, and ND. Suppose that i is not an extreme trichotomist on f(J). Then, if $\# T_{i}^{J}(f(J))=2$, then we have two possible cases; (i) for all voters $j\in \mathcal {V}$, the restriction of $\mathcal {D}_{j}^{J}$ to $T_{j}^{J}(f(J))$ are the same and $\# T_{j}^{J}(f(J))=2$, and (ii) there exist $x\in f(J)$ and $j\in \mathcal {V}\backslash \{i\}$ such that either $x\notin T_{i}^{J}(f(J))$ and $x\in T_{j}^{J}(f(J))$ or $x\in T_{i}^{J}(f(J))$ and $x\notin T_{j}^{J}(f(J))$. For Case (i), there exists an SCC with f(J) satisfying WP, I, and ND by Lemma 3. For Case (ii), there exists an SCC with f(J) satisfying WP, I, and ND by Lemmas 1 and 2.

Next, if $\# T_{i}^{J}(f(J))\ne 2$, then there exist $x\in f(J)$ and $j\in \mathcal {V}\backslash \{i\}$ such that either $x\notin T_{i}^{J}(f(J))$ and $x\in T_{j}^{J}(f(J))$ or $x\in T_{i}^{J}(f(J))$ and $x\notin T_{j}^{J}(f(J))$. By Lemmas 1 and 2, there exists an SCC with f(J) satisfying WP, I, and ND. $\square$

Proof of Theorem 5

Consider an SCR with f satisfying the hypotheses of Theorem 5 except for ND. Consider the completely positive opinion profile $J\in \mathscr {J}$. Since we have $f(J)=X$ by construction of f and the preference domain $\mathcal {D}^{J}$ is unrestricted, there exists a dictator d on f(J) by Theorem 1.

Since f is a subrule of the unanimous rule, we have $f(J')\subseteq T_{j}^{J'}(X)$ for all $J'\in \mathscr {J}$ and all $j\in \mathcal {V}$. We now show that d is a dictator on $f(J')$. We distinguish two cases; (i) $\#f(J')\ge 3$ and (ii) $\#f(J')<3$.

Case (i) Since the SCC with $f(J')$ violates condition 1 of Theorem 1, there exists a dictator $d'$ on $f(J')$. We now show $d=d'$. Consider $x,y\in f(J')$ and $\mathbf R \in \mathcal {D}^{J'}$ such that $\{x\}=B(X,R_{d})$ and $\{y\}=B(X,R_{d'})$. Since $\mathcal {D}^{J}$ is unrestricted, we have $\mathbf R \in \mathcal {D}^{J}\cap \mathcal {D}^{J'}$. Then, we have $C_{f}(\mathbf R ,f(J))=\{x\}$ by the dictatorship of d on f(J), and also we have $C_{f}(\mathbf R ,f(J'))=\{y\}$ by the dictatorship of $d'$ on $f(J')$. Since we have $f(J')\subseteq f(J)$ and $C_{f}(\mathbf R ,f(J))\cap f(J')\ne \emptyset$, it follows from FCA that $C_{f}(\mathbf R ,f(J))\cap f(J')=C_{f}(\mathbf R ,f(J'))$, which is a contradiction. Therefore, we have $d=d'$.

Case (ii) Suppose to the contrary that d is not a dictator on $f(J')$. Then, there exist $x,y\in f(J')$ and $\mathbf R \in \mathcal {D}^{J'}$ such that $xP_{d}y$ and $y\in C_{f}(\mathbf R ,f(J'))$, which implies that $f(J')=\{x,y\}$. Let $\mathbf R '\in \mathcal {D}^{J'}$ be a preference profile such that $\mathbf R '|_{\{x,y\}}=\mathbf R |_{\{x,y\}}$ and for all $z\notin f(J')$, $xP'_{j}z$ and $yP'_{j}z$ for all $j\in \mathcal {V}$. If we have $y\notin C_{f}(\mathbf R ',f(J'))$, then we have $x\in C_{f}(\mathbf R ',f(J'))$. However, it follows from I that $y\notin C_{f}(\mathbf R ,f(J'))$, which is a contradiction. Therefore, we obtain $y\in C_{f}(\mathbf R ',f(J'))$. Since $\mathcal {D}^{J}$ is unrestricted, we have $\mathbf R '\in \mathcal {D}^{J}\cap \mathcal {D}^{J'}$. Since we have $f(J')\subseteq f(J)$ and it follows from WP that $C_{f}(\mathbf R ',f(J))\subseteq \{x,y\}$, we have $C_{f}(\mathbf R ',f(J))\cap f(J')\ne \emptyset$. By FCA, we have $C_{f}(\mathbf R ',f(J))\cap f(J')=C_{f}(\mathbf R ',f(J'))$. Therefore, we have $y\in C_{f}(\mathbf R ',f(J))$, which is a contradiction to the dictatorship of d on f(J). Thus, the dictator on f(J) is also a dictator on $f(J')$. $\square$