Triple-consistent social choice and the majority rule

Abstract

We define generalized (preference) domains $\mathcal{D}$ as subsets of the hypercube {−1,1}^D, where each of the D coordinates relates to a yes-no issue. Given a finite set of n individuals, a profile assigns each individual to an element of $\mathcal{D}$. We prove that, for any domain $\mathcal{D}$, the outcome of issue-wise majority voting φ _m belongs to $\mathcal{D}$ at any profile where φ _m is well-defined if and only if this is true when φ _m is applied to any profile involving only 3 elements of $\mathcal{D}$. We call this property triple-consistency. We characterize the class of anonymous issue-wise voting rules that are triple-consistent, and give several interpretations of the result, each being related to a specific collective choice problem.

Appendix: Proof of Theorem 3

The sufficiency part follows from Theorem 2 together with Propositions 1 and 2. We prove the necessary part through several intermediate lemmas. Let φ be an anonymous and independent aggregation rule. Independence together with anonymity implies that for any domain $\mathcal{D}$, for any issue d∈{1,…,D}, for any n≥3 and for any $(\mathcal{D},n)$-profile $\pi \in \mathcal{D}_{\varphi }^{n}$, [φ(π)]^d only depends on n ₁(π∣ _d ). Consider the set $\mathcal{D}_{\varphi }^{3}$ of all admissible profiles which involve exactly 3 individuals. Pick any $\pi \in \mathcal{D}_{\varphi }^{3}$ together with an issue d. Then, there exists a _d ∈{−1,1} such that for all issues d,

Case 1::

$[n_{a_{d}}(\pi \mid _{d})=2\mbox{ and }n_{-a_{d}}(\pi \mid _{d})=1\Rightarrow [ \varphi (\pi )]^{d}=a_{d}]$ (property M ⁺);

Case 2::

$[n_{a_{d}}(\pi \mid _{d})=2\mbox{ and }n_{-a_{d}}(\pi \mid _{d})=1\Rightarrow [ \varphi (\pi )]^{d}=-a_{d}]$ (property M ⁻);

Case 3::

$[n_{-a_{d}}(\pi \mid _{d})\in \{1,2\}\Rightarrow [ \varphi (\pi )]^{d}=a_{d}]$ (property M ⁼).

Suppose that for some domain $\mathcal{D}$, φ violates the majority criterion at some issue d. Thus there exist h>h′>0, d∈{1,…,D}, a _d ∈{−1,1} and $\pi \in \mathcal{D}_{\varphi }^{h+h^{\prime }}$ such that $n_{a_{d}}(\pi \mid _{d})=h$, $n_{-a_{d}}(\pi \mid _{d})=h^{\prime }$ and [φ(π)]^d=−a _d . We denote by h _φ the smallest integer h for which the above situation holds for φ.

Lemma 1

If φ≠φ _m is triple-consistent, then h _φ =2.

Proof

Suppose that for some domain $\widetilde{\mathcal{D}}$ there exists an admissible $(\widetilde{\mathcal{D}},h_{\varphi }+h^{\prime })$-profile $\widetilde{\pi }$ such that $n_{a_{d}}(\widetilde{\pi }\mid _{d})=h_{\varphi }\geq 3$, $n_{-a_{d}}(\widetilde{\pi }\mid _{d})=h^{\prime }$ and $[\varphi (\widetilde{\pi })]^{d}=-a_{d}$ for some issue d. Write a=a _d . Define $\mathcal{D}=\{x,y,z\}\in \{-1,1\}^{3}$, where x=(−a,a,−a), y=(a,−a,a), and z=(a,a,a). Consider the $(\mathcal{D},3)$-profile π=(x,y,z). Since n _a (π∣ _d )<h _φ for all d, then φ(π)=z, and thus $\mathcal{D}$ is 3-stable under φ. Consider the $(\mathcal{D},h_{\varphi }+h^{\prime })$-profile $\pi ^{\prime }=(\overset{h^{\prime }}{\overbrace{x,\ldots,x}},\overset{h^{\prime }}{\overbrace{y,\ldots,y}},\overset{h_{\varphi }-h^{\prime }}{\overbrace{z,\ldots,z}})$. Since for all issues d, $n_{a}(\pi ^{\prime }\mid _{d})=h_{\varphi }=n_{a}(\widetilde{\pi }\mid _{d})$ and $n_{-a}(\pi ^{\prime }\mid _{d}) =h^{\prime }=n_{-a}(\widetilde{\pi }\mid _{d})$, then $\pi ^{\prime }\in \mathcal{D}_{\varphi }^{h_{\varphi }+h^{\prime }}$. Moreover, [φ(π′)]^d=−a. It follows from independence that $\varphi (\pi ^{\prime })=(-a,-a,-a)\notin \mathcal{D}$, and therefore φ is not triple-consistent. Thus, h _φ ≤2. Finally, since h _φ >h′>0, then h _φ =2. □

An immediate consequence of Lemma 1 is:

Lemma 2

If φ is triple-consistent and such that property M ⁺ holds, then for all $\mathcal{D}\in \varDelta$, for all n≥3, and for all non-unanimous profiles $\pi \in \mathcal{D}^{n}$, φ(π)=φ _m (π).

We generalize below Lemma 2 to any admissible profile, unanimous or not. An intermediate step is:

Lemma 3

If φ is triple-consistent and unanimous in restriction to 3-program profiles, then φ is unanimous.

Proof

Define $\mathcal{D}=\{x,y,z,t\}\subset \{-1,1\}^{3}$, where x=(a,a,a), y=(−a,−a,a), z=(a,−a,a), and t=(−a,a,a). Since $\mathcal{D}$ contains all programs with a as third coordinate, and since φ is unanimous in restriction to $\mathcal{D}_{\varphi }^{3}$, then $\mathcal{D}$ is 3-stable under φ. Thus, triple-consistency ensures that $\mathcal{D}$ is stable under φ. Suppose that $n_{a}(\widetilde{\pi }\mid _{d})=n$ and $[\varphi (\widetilde{\pi })]^{d}=-a$ for some $\widetilde{\mathcal{D}}\in \varDelta$, some n≥3 and some $\widetilde{\pi }\in \widetilde{\mathcal{D}}_{\varphi }^{n}$. It follows from anonymity together with independence that [φ(π)]^d=−a for any $\pi \in \mathcal{D}_{\varphi }^{n}$ such that n _a (π∣ _d )=n. Thus $\mathcal{D}$ is not stable under φ, in contradiction with triple-consistency. □

Lemma 4

If φ is triple-consistent and satisfies property M ⁺, then φ=φ _m .

Proof

From Lemma 2, together with Lemma 3, it suffices to prove that φ is unanimous at all profiles $\pi \in \mathcal{D}_{\varphi }^{3}$. Suppose the contrary. It follows from anonymity and independence that there exists a∈{1,−1} such that for all $\pi \in \mathcal{D}_{\varphi }^{3}$ and all d, n _a (π∣ _d )=3 and [φ(π)]^d=−a (Property A). Define $\mathcal{D}=\{x,y,z,t,-x,-y, -z,-t\}\subset \{-1,1\}^{4}$, where x=(a,a,a,−a), y=(a,a,−a,a), z=(a,−a,a,a), and t=(−a,a,a,a). From M ⁺, together with Property A, we get that φ(x,y,z)=t, φ(x,y,t)=z, φ(x,z,t)=y, and φ(y,z,t)=x. Furthermore, for any w≠w′≠w′′∈{x,y,z,t}, we have φ(w,w,w′)=−w′=φ(−w,−w,w′), φ(w,−w,w′)=w′, φ(w,−w,−w′)=−w′, $\varphi (w,w^{\prime },w^{\prime \prime })=\varphi (-w,-w^{\prime },w^{\prime \prime })\in \mathcal{D}$, and $\varphi (-w,-w^{\prime }, -w^{\prime \prime })=-\varphi (w,w^{\prime },w^{\prime \prime })\in \mathcal{D}$. Thus $\mathcal{D}$ is 3-stable under φ. Finally, we get from Lemma 2 that $\varphi (x,y,z,t)=(a,a,a,a)\notin \mathcal{D}$. Thus, φ is not triple-consistent, a contradiction. □

We now turn to aggregation rules sharing property M ⁼.

Lemma 5

If φ is triple-consistent and such that property M ⁼ holds, then φ is almost constant.

Proof

Consider the domain $\mathcal{D}^{\ast }\subset \{-1,1\}^{h+1}$ defined in Table 1, where issue-wise positions appear in columns and programs in rows.

Table 1 Domain $\mathcal{D}^{\ast }$

Then, we define $\mathcal{D}=\{1,-1\}^{h+1}-\{(-x_{h+2})\}$. Suppose that there exist a∈{−1,1} and $h\neq h^{\prime } \in \mathbb{N}$ such that h+h′>3 and, for some d and some $\pi \in \mathcal{D}_{\varphi }^{h+h^{\prime }}$, n _a (π∣ _d )=h′, n _−a (π∣ _d )=h, and [φ(π)]^d=−a. Pick up any triple $\{x,y,z\}\subset \mathcal{D}$ such that profile $(x,y,z) \in \mathcal{D}_{\varphi }^{3}$. From the definition of $\mathcal{D}$, we have that for all d, there exists w∈{x,y,z} such that w ^d=−a. Hence, property M ⁼ implies that φ(x,y,z)≠(−x _h+2). Therefore $\varphi (x,y,z)\in \mathcal{D}$, and $\mathcal{D}$ is 3-stable under φ. Finally, define the $(\mathcal{D},h+h^{\prime })$-profile $\pi ^{\ast }=(x_{1},x_{2},\ldots,x_{h+1},\overset{h^{\prime }-1}{\overbrace{x_{h+2},\ldots,x_{h+2}}})$ that contains only programs in $\mathcal{D}^{\ast }\subset \mathcal{D}$. Since ∀d∈{1,…,h+1}, then n _a (π ^∗∣ _d )=h′ and n _−a (π ^∗∣ _d )=h. Since n _a (π ^∗∣ _d )=n _a (π∣ _d )=h′ and n _−a (π ^∗∣ _d )=n _−a (π∣ _d )=h, then $\pi ^{\ast }\in \mathcal{D}_{\varphi }^{h+h^{\prime }}$. It follows from anonymity together with independence that $\varphi (\pi ^{\ast })=-x_{h+2}\notin \mathcal{D}$. Thus, $\mathcal{D}$ is not stable under φ, in contradiction with triple-consistency. Therefore, we have shown that $\forall \mathcal{D}\in \varDelta$, ∀n≥3, $\forall \pi \in \mathcal{D}_{\varphi }^{n}$, ∀a∈{−1,1}, 0<n _a (π∣ _d )<n⇒[φ(π)]^d=a, and thus φ is almost constant. □

It remains to consider property M ⁻.

Lemma 6

If φ is triple-consistent and such that property M ⁻ holds, then $\varphi (\pi )=\varphi _{m}^{-}(\pi )$ for all $\mathcal{D}\in \varDelta$ and all $\pi \in \mathcal{D}_{\varphi }^{3}$ .

Proof

It suffices to prove that ∀a∈{1,−1}, $\forall \pi \in \mathcal{D}_{\varphi }^{3}$, ∀d, n _a (π∣ _d )=3⇒[φ(π)]^d=−a. Suppose the contrary, and consider again the domain $\mathcal{D}$ defined in the proof of Lemma 4, where $\forall w\in \mathcal{D}$, $-w\in \mathcal{D}$. Moreover, for any non-unanimous profile $(w,w^{\prime },w'')\in \mathcal{D}^{3}$, one has from property M ⁻ that φ(w,w′,w′′)=−φ _m (w,w′,w′′). Since $\mathcal{D}$ is 3-stable under φ _m , then $\mathcal{D}$ is 3-stable under φ. Now consider profile $\widetilde{\pi }=(x,y,z,t)$. Observe that ∀d, $n_{a}(\widetilde{\pi }\mid _{d})=3$ and $n_{-a}(\widetilde{\pi }\mid _{d})=1$. From independence together with anonymity, we get that $\varphi (\widetilde{\pi })=\{(a,a,a,a), (-a,-a,-a,-a)\}$. Thus $\varphi (\widetilde{\pi })\notin \mathcal{D}$, therefore φ is not triple-consistent, a contradiction. Finally, independence together with anonymity ensures that $\forall \overline{\mathcal{D}} \in \varDelta$, $\forall \overline{\pi }\in \overline{\mathcal{D}}_{\varphi }^{3}$, $n_{a}(\overline{\pi }\mid _{d})=3\Rightarrow [ \varphi (\overline{\pi })]^{d}=-a$, and the proof is complete. □

Lemma 7

Let φ be triple-consistent and such that property M ⁻ holds. Let $h>h^{\prime }\in \mathbb{N}$ with h+h′ odd. Consider any domain $\mathcal{D}$. Then for any $\pi \in \mathcal{D}_{\varphi }^{h+h^{\prime }}$, for any a∈{−1,1} and for any issue d,

• (1) if [n _a (π∣ _d )=h and n _−a (π∣ _d )=h′⇒[φ(π)]^d=−a], then [n _−a (π∣ _d )=n⇒[φ(π)]^d=a];

• (2) if [n _a (π∣ _d )=h and n _−a (π∣ _d )=h′⇒[φ(π)]^d=a], then [n _−a (π∣ _d )=n⇒[φ(π)]^d=−a].

Proof

From Lemma 6, $\varphi =\varphi _{m}^{-}$ in restriction to 3-program profiles. Define $\mathcal{D}=\{x,y,z,-x,-y,-z\}\subset \{-1,1\}^{3}$ by x=(−a,a,a), y=(−a,a,−a) and z=(−a,−a,a). Using property M ⁻, it is easily checked that $\mathcal{D}$ is 3-stable under φ. Consider the $(\mathcal{D},h+h^{\prime })$-profile $\pi =(\overset{h-h^{\prime }}{\overbrace{x,\ldots,x}},\overset{h^{\prime }}{\overbrace{y,\ldots,y}},\overset{h^{\prime }}{\overbrace{z,\ldots,z}})$. Then n _a (π∣ _d )=h and n _−a (π∣ _d )=h′ for d=2,3. Suppose that [φ(π)]²=−a. From independence and anonymity, we get [φ(π)]³=−a. Moreover, triple-consistency requires φ(π)=(a,−a,−a). Since n _−a (π∣₁)=h+h′, then we must have [φ(π)]¹=a, and assertion (1) follows from independence and anonymity. Similarly, if [φ(π)]²=a, then [φ(π)]³=a, while triple-consistency requires φ(π)=(−a,a,a). Since n _−a (π∣₁)=h+h′, then we must have [φ(π)]¹=−a, and assertion (2) follows from independence and anonymity. □

Lemma 8

Let φ be triple-consistent and such that property M ⁻ holds. For any domain $\mathcal{D}$, for any $n\in \mathbb{N}$ and any $\pi \in \mathcal{D}_{\varphi _{m}}^{n}$, $\varphi (\pi )\in \{\varphi _{m}(\pi ), \varphi _{m}^{-}(\pi )\}$.

Proof

Consider any domain $\mathcal{D}$. Suppose first that there exists a∈{1,−1} together with $\{h,h^{\prime },k,k^{\prime }\}\subset \mathbb{N}$ with h>h′, k>k′, and n=h+h′=k+k′, such that, for two $(\mathcal{D},n)$-profiles $\pi ,\pi ^{\prime }\in \mathcal{D}_{\varphi }^{n}$, one has for some issue d:

• (1) n _a (π∣ _d )=h, n _−a (π∣ _d )=h′, and [φ(π)]^d=a;

• (2) n _a (π′∣ _d )=k, n _−a (π′∣ _d )=k′, and [φ(′π)]^d=−a.

It follows from Lemma 7 that for any issue d′, [n _−a (π∣ _d′)=n⇒[φ(π)]^d=−a] and [n _−a (π′∣ _d′)=n⇒[φ(π)]^d=a], which clearly contradicts hat φ is independent and anonymous. Thus, we have proven that:

• (3) ∀a∈{1,−1}, ∀d, either [n _a (π∣ _d )>n _−a (π∣ _d ) and [φ(π)]^d=a], or [n _a (π∣ _d )>n _−a (π∣ _d ) and [φ(π)]^d=−a].

Next, suppose that there exist $\mathcal{D}\in \varDelta$, $n \in \mathbb{N}$ and a∈{1,−1} such that for some $\pi \in \mathcal{D}_{\varphi }^{n}$ and some d, we have

• (4) [n _a (π∣ _d )>n _−a (π∣ _d )⇒[φ(π)]^d=−a] and [n _−a (π∣ _d )>n _a (π∣ _d )⇒[φ(π)]^d=−a].

Consider $\mathcal{D}^{\ast }=\{x,y\}\subset \{-1,1\}^{2}$, with x=(a,−a) and y=(−a,a). It follows from Lemma 6 that φ(x,x,y)=y and φ(x,y,y)=x, and thus $\mathcal{D}^{\ast }$ is 3-stable under φ. Then define the $(\mathcal{D}^{\ast },t+t^{\prime })$-profile $\pi =(\overset{t}{\overbrace{x,\ldots,x}},\overset{t^{\prime }}{\overbrace{y,\ldots,y})}$, where t>t′. One gets from independence together with (4) that $\varphi (\pi )=(-a,-a)\notin \mathcal{D}^{\ast }$, therefore φ is not triple-consistent, a contradiction. Thus, we have either [n _a (π∣ _d )>n _−a (π∣ _d )⇒[φ(π)]^d=−a] or [n _a (π∣ _d )>n _−a (π∣ _d )⇒[φ(π)]^d=−a]. Using anonymity together with independence, this proves that for all $\mathcal{D}\in \varDelta$, for all d, for all $n \in \mathbb{N}$ and all a∈{1,−1}, and for all $\pi \in \mathcal{D}_{\varphi _{m}}^{n}$, either

• (5) [n _a (π∣ _d )>n _−a (π∣ _d )⇒[φ(π)]^d=a] and [n _−a (π∣ _d )>n _a (π∣ _d )⇒[φ(π)]^d=−a]

or

• (6) [n _a (π∣ _d )>n _−a (π∣ _d )⇒[φ(π)]^d=−a] and [n _−a (π∣ _d )>n _a (π∣ _d )⇒[φ(π)]^d=a].

Since (5) defines φ _m while (6) defines $\varphi _{m}^{-}$, the proof is complete. □

The proof of Theorem 3 is now complete. Indeed, consider any anonymous, independent and triple-consistent aggregation rule φ. If M ⁺ holds, then Lemma 4 implies that φ=φ _m . If M ⁼ holds, then, from Lemma 5, φ is almost constant. If M ⁻ holds, Lemma 8 imply that $\varphi \in \varPhi _{m}^{+-}$.