A one-shot proof of Arrow’s theorem and the Gibbard–Satterthwaite theorem

Abstract

This paper provides a simple and transparent proof of a new social choice impossibility theorem. The Gibbard–Satterthwaite theorem and Arrow’s impossibility theorem are straightforward corollaries.

1 Introduction

One of the impossibility theorems introduced by Yu (2013) can help prove both the Gibbard–Satterthwaite theorem (Gibbard 1973; Satterthwaite 1975) and Arrow’s impossibility theorem (Arrow 1963) succinctly. In this paper, we offer a direct proof of this theorem, which resembles Yu (2012) and employs the “pivotal voter” technique devised and perfected by Barberá (1980), Geanakoplos (2005), and Reny (2001). We comment on different approaches of establishing classical theorems in the last section.

2 The setup

The following terminology follows Yu (2013) closely.

A set of individuals $\mathcal N =\{1,\ldots ,N\}$ with $N\ge 2$, each have some preferences over $M\ge 3$ alternatives $A=\{a_1,\ldots ,a_M\}$. Throughout the paper, we label individuals with $n$ and number alternatives with $i,j,k$. Let $\mathcal P$ be the set of all possible relations on $M$ that are complete, asymmetric, and transitive, i.e., strict preference relations.¹ A preference profile is an ordered list $\vec {\succ }=(\succ _1,\ldots ,\succ _N)$ with $\succ _n\in \mathcal P$ for every $1\le n\le N$. All possible preference profiles form the collection $\mathcal P ^\mathcal N$.

Definition 1

(SCF) A social choice function $F{:}\,\, \mathcal P ^\mathcal N \rightarrow A$ assigns to each $\vec {\succ }\in \mathcal P ^\mathcal N$ a choice $F(\vec {\succ })\in A$.

Definition 2

(SPF) A social preference function $R{:}\,\, \mathcal P ^\mathcal N \rightarrow \mathcal P$ assigns to each $\vec {\succ }\in \mathcal P ^\mathcal N$ a preference relation $R(\vec {\succ })\in \mathcal P$.

Several definitions facilitate communication.

Definition 3

For profile $\vec {\succ }$, $a_i$ dominates $a_j$ if $a_i\succ _n a_j$ for every $n$.

Definition 4

Two profiles $\vec {\succ }$ and $\vec {\succ }^{\prime }$ agree on $\{a_i, a_j\}$ if $a_i\succ _n a_j$ is necessary and sufficient for $a_i\succ ^{\prime }_n a_j$ for every $n$.

Definition 5

$A^{\prime }\subset A$ is at the top of $\vec {\succ }$ if $a_i\succ _n a_j$ for every $n$, every $a_i\in A^{\prime }$, and every $a_j\in A\backslash A^{\prime }$. If in addition $A^{\prime }=\{a_i,a_j\}$, we call $\vec {\succ }$ a $\{i,j\}$-runoff.

Definition 6

A runoff generating function $T_{ij}{:}\,\,\mathcal P ^\mathcal N \rightarrow \mathcal P ^\mathcal N$ brings $A^{\prime }=\{a_i,a_j\}$ to the top, keeping internal rankings of $A^{\prime }$ and $A\backslash A^{\prime }$ intact.

3 The main theorem

Definition 7

(D) A SCF $F$ is dictatorial if there exists a social choice dictator $n$ such that $F(\vec {\succ })\succ _na_i$ for every $a_i\ne F(\vec {\succ })$ and every $\vec {\succ }$.

Definition 8

(O) A SCF $F$ is onto if $F(\mathcal P ^\mathcal N )=A$.

Definition 9

(WP) A SCF $F$ is weakly Paretian if $a_i$ dominating $a_j$ implies $F(\vec {\succ })\ne a_j$ for every $i,j$ and every $\vec {\succ }$.

Definition 10

(CM) A SCF $F$ is Condorcet monotonic if whenever $\vec {\succ }$ and $\{i,j\}$-runoff $\vec {\succ }^{\prime }$ agree on $\{a_i,a_j\},\, F(\vec {\succ })=a_i$ implies $F(\vec {\succ }^{\prime })=a_i$.

This condition requires every social choice to be a generalized Condorcet winner in that it has to win any runoff it enters that keeps intact its rankings against the opponent. Losing a $\{i,j\}$-runoff prevents $a_i$ from winning any profile that agrees with it on $\{a_i,a_j\}$, no matter what rule—for example, simple majority, super majority, or weighted majority—determines the runoff outcome. In particular, if $F(\vec {\succ })=a_i,\, F(T_{ij}(\vec {\succ }))=a_i$ by (CM). We often apply the contrapositive statement.

Lemma 1

(O) and (CM) imply (WP).

Proof

Let $a_i$ dominate $a_j$ in $\vec {\succ }$. (O) ensures that $F(\vec {\succ }^{\prime })=a_i$ for some $\vec {\succ }^{\prime }$. By (CM), the choice remains $a_i$ for $T_{ik}(\vec {\succ }^{\prime })$, where $k\notin \{i,j\}$, and further for $T_{ij}(T_{ik}(\vec {\succ }^{\prime }))$. Since $a_j$ loses the $\{i,j\}$-runoff $T_{ij}(T_{ik}(\vec {\succ }^{\prime }))$ that agrees with $\vec {\succ }$ on $\{a_i,a_j\}$, the choice for $\vec {\succ }$ cannot be $a_j$. $\square$

Theorem 1

If a SCF $F$ satisfies (O) and (CM), then it is dictatorial.

Proof

Pick any $\vec {\succ }^{\prime }$ with $\{a_i\}$ and $\{a_i, a_j\}$ at the top. Swap the positions of $\{a_i, a_j\}$ sequentially from 1 to $N$. By (WP), the choice is either $a_i$ or $a_j$, starting with $a_i$ and ending with $a_j$. The $(i,j)$-pivotal voter $n_{ij}$ is the first whose swap makes a difference. By (CM), any two $(i,j)$-runoffs that agree on $\{a_i, a_j\}$ select the same alternative, so $n_{ij}$ is independent of which $\vec {\succ }^{\prime }$ (with $\{a_i\}$ and $\{a_i, a_j\}$ at the top) we start with.

	1	$\ldots$	$n_{ij}-1$	$n_{ij}$	$n_{ij}+1$	$\ldots$	$N$
$\vec {\succ }^{\prime \prime }$	$a_j$	$\ldots$	$a_j$	$a_i$	$a_i$	$\ldots$	$a_i$
	$\square$	$\ldots$	$\square$
	$a_i$	$\ldots$	$a_i$	$a_j$	$a_j$	$\ldots$	$a_j$
	$\square$	$\ldots$	$\square$	$a_k$	$a_k$	$\ldots$	$a_k$
$\vec {\succ }^{\prime \prime \prime }$	$\square$	$\ldots$	$\square$
	$a_j$	$\ldots$	$a_j$	$a_j$	$a_i$	$\ldots$	$a_i$
	$\square$	$\ldots$	$\square$		$\square$	$\ldots$	$\square$
	$a_i$	$\ldots$	$a_i$	$a_i$	$a_j$	$\ldots$	$a_j$
	$\square$	$\ldots$	$\square$	$a_k$	$\square$	$\ldots$	$\square$

Consider depicted profiles $\vec {\succ }^{\prime \prime }$ and $\vec {\succ }^{\prime \prime \prime }$ with $\{a_i,a_j,a_k\}$ at the top, where columns correspond to voters and squares mark possible positions of $a_k$. The definition of $n_{ij}$ informs us that $F(T_{ij}(\vec {\succ }^{\prime \prime }))=a_i$ and $F(T_{ij}(\vec {\succ }^{\prime \prime \prime }))=a_j$, so $F(\vec {\succ }^{\prime \prime })\ne a_j$ and $F(\vec {\succ }^{\prime \prime \prime })\ne a_i$ by (CM). For $\vec {\succ }^{\prime \prime }$, the choice is $a_i$, for (WP) rules out $a_k$ and others. Hence, $F(T_{ik}(\vec {\succ }^{\prime \prime }))=a_i$, implying that in defining $n_{ik}$, no swap before $n_{ij}$ makes a difference, i.e., $n_{ik}\ge n_{ij}$. But $j$ and $k$ are arbitrary, so $n_{ij}=n_{ik}$, i.e., $n_{i-}$ refers to the same individual.

Moreover, $T_{ik}(\vec {\succ }^{\prime \prime })$ and $\vec {\succ }^{\prime \prime \prime }$ agree on $\{a_i, a_k\}$, so $F(\vec {\succ }^{\prime \prime \prime })$ is not $a_k$ due to its loss to $a_i$ in $\{i,k\}$-runoff $T_{ik}(\vec {\succ }^{\prime \prime })$. We are left with $F(\vec {\succ }^{\prime \prime \prime })=a_j$, so $F(T_{jk}(\vec {\succ }^{\prime \prime \prime }))=a_j$, demanding by (CM) that

$$\begin{aligned} a_j\succ _{n_{ij}} a_k\text { implies } F(\vec {\succ })\ne a_k. \end{aligned}$$

(*)

In defining $n_{jk}$, ($*$) says that no swap before $n_{ij}$ makes a difference, so $n_{jk}\ge n_{ij}$ or $n_{j-}\ge n_{i-}$. But $i$ and $j$ are arbitrary, confirming $n_{j-}=n_{i-}$. The single pivotal voter can eliminate any alternative except her favorite by ($*$). $\square$

4 Classical impossibility theorems

Given Theorem 1, we can prove the Gibbard–Satterthwaite theorem through a simple lemma. Let $(\succ ^{\prime }_n,\vec {\succ }_{-n})$ represent the new profile constructed by replacing the preferences of individual $n$ in $\vec {\succ }$ with $\succ ^{\prime }_n\in \mathcal P$.

Definition 11

(SP) A SCF $F$ is strategy-proof if $F(\succ ^{\prime }_n,\vec {\succ }_{-n})\ne F(\vec {\succ })$ implies $F(\vec {\succ })\succ _n F(\succ ^{\prime }_n,\vec {\succ }_{-n})$ for every $n$, every $\vec {\succ }$, and every $\succ ^{\prime }_n$.

This condition demands that an individual is worse off misreporting her preferences whenever a misreport can influence the choice, so truth-telling is optimal.

Lemma 2

(SP) imply (CM).

Proof

Let $\vec {\succ }$ and $\{i,j\}$-runoff $\vec {\succ }^{\prime }$ agree on $\{a_i,a_j\}$. Suppose that $F(\vec {\succ })=a_i$ and $F(\succ ^{\prime }_1,\vec {\succ }_{-1})=a_k$ for $k\ne i$. By (SP), $a_i\succ _1 a_k$. Define $\vec {\succ }^{\prime \prime }=(\succ ^{\prime }_1,\vec {\succ }_{-1})$. Given $\vec {\succ }=(\succ _1,\vec {\succ }^{\prime \prime }_{-1})$, (SP) entails $a_k\succ ^{\prime }_1 a_i$. Since $\vec {\succ }^{\prime }$ is a $\{i,j\}$-runoff, $k=j$, contradicting the agreement of $\vec {\succ }$ and $\vec {\succ }^{\prime }$ on $\{a_i,a_j\}$. So $F(\succ ^{\prime }_1,\vec {\succ }_{-1})$ has to remain $a_i$. Likewise, the process of updating $\vec {\succ }$ to $\vec {\succ }^{\prime }$ one by one keeps the choice unaltered. $\square$

Theorem 2

(Gibbard–Satterthwaite) If a SCF $F$ satisfies (O) and (SP), then it is dictatorial.

Proving Arrow’s theorem is equally simple.

Definition 12

(AD) A SPF $R$ is Arrow dictatorial if there exists a social preference dictator $n$ such that $a_i\succ _n a_j$ implies $a_iR(\vec {\succ })a_j$ for every $i,j$ and every $\vec {\succ }$.

Definition 13

(AU) A SPF $R$ is Arrow unanimous if $a_i\succ _n a_j$ for every $n$ implies $a_iR(\vec {\succ })a_j$ for every $i,j$ and every $\vec {\succ }$.

Definition 14

(AIIA) A SPF $R$ is Arrow independent of irrelevant alternatives if whenever $\vec {\succ }$ and $\vec {\succ }^{\prime }$ agree on $\{a_i, a_j\},\, a_iR(\vec {\succ })a_j$ implies $a_iR(\vec {\succ }^{\prime })a_j$.

Theorem 3

(Arrow’s) If a SPF $R$ satisfies (AU) and (AIIA), then it is Arrow dictatorial.

Proof

Given $R$, we can define a SCF $F^R$ that selects the alternative ranked highest by $R$. $F^R$ obviously satisfies (O), because by (AU), $F^R(\vec {\succ })=a_i$ if only $\{a_i\}$ is at the top. To see (CM), let $\vec {\succ }$ and $\{i,j\}$-runoff $\vec {\succ }^{\prime }$ agree on $\{a_i,a_j\}$. When $F^R(\vec {\succ })=a_i$, $a_iR(\vec {\succ })a_j$, so by (AIIA), $a_iR(\vec {\succ }^{\prime })a_j$. But $a_iR(\vec {\succ }^{\prime })a_k$ for $k\notin \{i,j\}$ by (AU), so $F^R(\vec {\succ }^{\prime })$ has to be $a_i$.

Theorem 1 thus presents a social choice dictator $n$ of $F^R$. She is a social preference dictator too. If $a_i\succ _n a_j$, individual $n$ dictates $F^R(T_{ij}(\vec {\succ }))=a_i$, so $a_iR(T_{ij}(\vec {\succ }))a_j$ and by (AIIA) $a_iR(\vec {\succ })a_j$. $\square$

5 Concluding remarks

The simplicity is no coincidence. Our main theorem employs assumptions that are almost logically minimal (Yu 2013). Traditionally, people prove the Gibbard–Satterthwaite theorem as a corollary of the Muller–Satterthwaite theorem (Muller and Satterthwaite 1977). But the “monotonicity” condition assumed is more restrictive than (CM), making it a weaker theorem than ours. Reny (2001) has to circumvent the difficulty by giving parallel proofs of Arrow’s theorem and the Muller–Satterthwaite theorem. Mas-Colell et al. (1995) prove Arrow’s theorem first, but as a step toward the Muller–Satterthwaite theorem, the construction of $R^F$, a SPF derived from a SCF, is necessarily more involved than that of $F^R$. Building on these earlier works,² our framework offers an alternative way of unifying and teaching classical theorems.