Weak independence and the Pareto principle

Abstract

In this paper, the independence of irrelevant alternatives and the Pareto principle are simultaneously weakened in the Arrovian framework of social choice. Moreover, we also relax transitivity of social preferences. We show that impossibility remains under weaker versions of Arrow’s original conditions. Our results complement the recent work by Coban and Sanver (Soc Choice Welf 43(4):953–961, 2014).

Appendix: Proof of Proposition 9

Let us introduce some definitions. A set $A \subseteq N$ is said to be decisive over (x, y) for f if, for all ${\mathbf {R}} \in {\mathcal {R}}^N$,

$$\begin{aligned} (x,y)\in \bigcap _{i \in A}P(R_i) \Rightarrow (x,y) \in P(f({\mathbf {R}})). \end{aligned}$$

A decisive set A over (x, y) for f is minimal if there exists no proper subset $A'$ of A that is decisive over (x, y) for f: for every proper subset $A'$ of A, there exists ${\mathbf {R}}\in {\mathcal {R}}^N$ such that

$$\begin{aligned} (x,y) \in \bigcap _{i \in A'}P(R_i)\quad \hbox {and}\quad (y,x) \in f({\mathbf {R}}). \end{aligned}$$

In the presence of PR, it is the case that for all $j \in A$, $(y,x) \in f({\mathbf {R}})$ when $(y,x) \in \bigcap _{i \in A {\setminus } \{j\}}P(R_i)$ and $(x,y) \in \bigcap _{i \in \{ j \} \cup (N {\setminus } A) }P(R_i)$.

A coalition $A \subseteq N$ is said to be almost weakly decisive over (x, y) for f if, for all ${\mathbf {R}} \in {\mathcal {R}}^N$,

$$\begin{aligned} \left[ (x,y) \in \bigcap _{i \in A}P(R_i) \hbox { and } (y,x) \in \bigcap _{i \in N {\setminus } A}P(R_i)\right] \Rightarrow (x,y) \in f({\mathbf {R}}). \end{aligned}$$

Let $\mathcal {W}^*_f(x,y)$ be the family of coalitions almost weakly decisive over (x, y) for f.

The proof of Proposition 9 proceeds in a similar manner to that of Theorem 3 of Mas-Colell and Sonnenschein (1972).

Lemma 2

Suppose that $\# X \ge 3$. If a Paretian QSWF f satisfies WI and PR,  there exists $v \in N$ such that $\{ v \} \in {\mathcal {W}}^*(x,y)$ for some pair (x, y) of alternatives.

Proof

Suppose that $\{ i \} \notin \mathcal {W}^*(a,b)$ for all $a,b \in X$ and for all $i \in N$. Let $M \subseteq V$ be a minimal set decisive over (y, z) for f. If M is a singleton, the claim is immediately follows. In the rest of the proof, we suppose that $\# M \ge 2$. Take any $v,v' \in M$ ($v \ne v'$). Let ${\mathbf {R}}^1 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{ (x,y),(y,z) \} \subseteq P(R^1_v),\\&\{ (y,z),(z,x) \} \subseteq P(R^1_i)\quad \hbox {for all } i \in M {\setminus } \{ v \}, \\&\{ (z,x),(x,y) \} \subseteq P(R^1_i)\quad \hbox {for all } i \in N {\setminus } M. \end{aligned}$$

Since M is decisive over (y, z) for f, we have $(y,z) \in P(f({\mathbf {R}}^1))$. Since $\{ v \} \notin \mathcal {W}^*(x,z)$, $(z,x) \in P(f({\mathbf {R}}^1))$. Quasi-transitivity implies that $(y,x) \in P(f({\mathbf {R}}^1))$.

Let ${\mathbf {R}}^2 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&(y,z)\in P(R^2_v)\quad \hbox {and}\ (z,y)\in P(R^2_{v'}),\\&(y,z)\in P(R^2_i)\quad \hbox {for all } i \in M {\setminus } \{ v,v' \},\\&(z,y)\in P(R^2_i)\quad \hbox {for all } i \in N {\setminus } M. \end{aligned}$$

Since M is a minimal decisive set over (y, z), it follows that $(z,y) \in f({\mathbf {R}}^2)$.

Let ${\mathbf {R}}^3 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&(y,z)\in I(R^3_v)\quad \hbox {and}\quad (z,y)\in P(R^3_{v'}),\\&(y,z)\in P(R^3_i)\quad \hbox {for all } i \in M {\setminus } \{ v,v' \},\\&(z,y)\in P(R^3_i)\quad \hbox {for all } i \in N {\setminus } M. \end{aligned}$$

Comparing ${\mathbf {R}}^3$ with ${\mathbf {R}}^2$, PR implies that $(z,y) \in P(f({\mathbf {R}}^3))$.

Let ${\mathbf {R}}^4 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{ (x,y),(y,z) \} \subseteq I(R^4_v),\\&\{ (z,y),(y,x) \} \subseteq P(R^4_{v'}),\\&\{ (y,x),(x,z) \} \subseteq P(R^4_i)\quad \hbox {for all } i \in M {\setminus } \{ v,v' \},\\&\{ (x,z),(z,y) \} \subseteq P(R^4_i)\quad \hbox {for all } i \in N {\setminus } M. \end{aligned}$$

Since ${\mathbf {R}}^4|_{\{ y,z \}}={\mathbf {R}}^3|_{\{ y,z \}}$, WI implies that $(z,y) \in f({\mathbf {R}}^4)$. Comparing ${\mathbf {R}}^4$ with ${\mathbf {R}}^1$, WI and PR implies that $(y,x) \in P(f({\mathbf {R}}^4))$. Quasi-transitivity implies that $(z,x) \in f({\mathbf {R}}^4)$.

Let ${\mathbf {R}}^5 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&(z,x) \in P(R^5_v)\quad \hbox {and}\ (z,x) \in P(R^5_{v'}),\\&(x,z) \in P(R^5_i)\quad \hbox {for all } i\in M {\setminus } \{ v,v' \}, \\&(x,z) \in P(R^5_i)\quad \hbox {for all } i \in N {\setminus } M. \end{aligned}$$

Comparing ${\mathbf {R}}^5$ with ${\mathbf {R}}^4$, PR implies that $(z,x) \in P(f({\mathbf {R}}^5))$. In the light of PR, $\{v,v' \}$ is decisive over (z, x) for f.

Let ${\mathbf {R}}^6 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{(y,z),(z,x)\} \subseteq P(R^6_v),\\&\{(z,x),(x,y)\} \subseteq P(R^6_{v'}),\\&\{(x,y,(y,z)\} \subseteq P(R^6_i)\quad \hbox {for all } i \in N {\setminus } \{ v,v' \}. \end{aligned}$$

Since $\{ v,v'\}$ is decisive over (z, x) for f, we have $(z,x) \in P(f({\mathbf {R}}^6))$. Since $\{ v \} \notin \mathcal {W}^*(y,x)$, it follows that $(x,y) \in P(f({\mathbf {R}}^6))$. Quasi-transitivity implies that $(z,y) \in P(f({\mathbf {R}}^6))$. This implies that $\{ v' \} \in \mathcal {W}^*(z,y)$. The proof is complete. $\square$

Lemma 3

Suppose that $\# X \ge 3,$ $\# N \ge 4,$ and a Paretian QSWF f satisfies WI and PR. Let $x,y \in X$. If $\{ v \} \in {\mathcal {W}}^* (x,y),$ then $\{ v \} \in {\mathcal {W}}^* (x,z) \hbox { for all } z \in X$.

Proof

Suppose that $\{ v \} \in \mathcal {W}^* (x,y)$. Since $\# N \ge 4$, there exists a partition $(A_1,A_2,A_3)$ of $N {\setminus } \{ v \}$. Let ${\mathbf {R}}^1 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{(x,y), (y,z) \} \subseteq P(R^1_v),\\&[ (x,y) \in I(R^1_i) \wedge (y,z) \in P(R^1_i)]\quad \hbox {for all } i \in A_1,\\&\{(y,z),(z,x) \} \subseteq P(R^1_i)\quad \hbox {for all } i \in A_2,\\&\{(y,z),(z,x) \} \subseteq P(R^1_i)\quad \hbox {for all } i \in A_3. \end{aligned}$$

Since $\{ v \} \in \mathcal {W}^* (x,y)$, PR implies that $(x,y) \in P(f({\mathbf {R}}^1))$. Since f is Paretian, we have $(y,z) \in P(f({\mathbf {R}}^1))$. By quasi-transitivity, we have $(x,z) \in P(f({\mathbf {R}}^1))$.

Let ${\mathbf {R}}^2 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{(y,x), (x,z) \} \subseteq P(R^2_v),\\&\{(y,x), (x,z) \} \subseteq P(R^2_i)\quad \hbox {for all } i \in A_1,\\&\{(z,y),(y,x) \} \subseteq P(R^2_i)\quad \hbox {for all } i \in A_2,\\&[(x,z) \in I(R_i) \wedge (y,x) \in P(R^2_i)]\quad \hbox {for all } i \in A_3. \end{aligned}$$

Comparing ${\mathbf {R}}^2$ with ${\mathbf {R}}^1$, PR implies that $(x,z) \in P(f({\mathbf {R}}^2))$. Since f is Paretian, $(y,x) \in P(f({\mathbf {R}}^2))$. Thus, $(y,z) \in P(f({\mathbf {R}}^2))$.

Let ${\mathbf {R}}^3 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{(x,y), (y,z) \} \subseteq P(R^3_v),\\&\{(y,z), (z,x) \} \subseteq P(R^3_i)\quad \hbox {for all } i \in A_1,\\&\{ (x,y),(y,z) \} \subseteq I(R^3_i)\quad \hbox {for all } i \in A_2,\\&\{(y,z), (z,x) \} \subseteq P(R^3_i) \quad \hbox {for all } i \in A_3. \end{aligned}$$

Since $\{ v \} \in {\mathcal {W}}^* (x,y)$, PR implies that $(x,y) \in P(f({\mathbf {R}}^3))$. Comparing ${\mathbf {R}}^3$ with ${\mathbf {R}}^2$, WI and PR imply that $(y,z) \in P(f({\mathbf {R}}^3))$. Quasi-transitivity implies that $(x,z) \in P(f({\mathbf {R}}^3))$.

Let ${\mathbf {R}}^4 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{(y,x), (x,z) \} \subseteq P(R^4_v),\\&\{(z,y), (y,x) \} \subseteq P(R^4_i)\quad \hbox {for all } i \in A_1,\\&\{ (y,x),(x,z) \} \subseteq P(R^4_i)\quad \hbox {for all } i \in A_2,\\&\{(z,y), (y,x) \} \subseteq P(R^4_i)\quad \hbox {for all } i \in A_3. \end{aligned}$$

Comparing ${\mathbf {R}}^4$ with ${\mathbf {R}}^3$, WI and PR imply that $(x,z) \in P(f({\mathbf {R}}^4))$. Since f is Paretian, $(y,x) \in P(f({\mathbf {R}}^4))$. Quasi-transitivity implies that $(y,z) \in P(f({\mathbf {R}}^4))$.

Let ${\mathbf {R}}^5 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&\{(x,y), (y,z) \} \subseteq P(R^5_v),\\&[(z,x) \in P(R^5_i) \wedge (x,y) \in I(R^5_i)] \quad \hbox {for all } i \in A_1,\\&\{ (y,z),(z,x) \} \subseteq P(R^5_i) \quad \hbox {for all } i \in A_2,\\&[(y,z) \in I(R^5_i) \wedge (z,x) \in P(R^5_i)]\quad \hbox {for all } i \in A_3. \end{aligned}$$

Since $\{ v \} \in \mathcal {W}^* (x,y)$, PR implies that $(x,y) \in P(f({\mathbf {R}}^5))$. Comparing ${\mathbf {R}}^5$ with ${\mathbf {R}}^4$, WI and PR implies that $(y,z) \in P(f({\mathbf {R}}^5))$. By quasi-transitivity, we have $(x,z) \in P(f({\mathbf {R}}^5))$. From WI, if $(x,z) \in P(R^5_v)$ and $(z,x) \in P(R_i)$ for all $i \in N {\setminus } \{ v \}$, we have $(x,z) \in f({\mathbf {R}})$. Thus, it follows that $\{ v \} \in {\mathcal {W}}^* (x,z)$. $\square$

The proof of the following lemma is similar to Lemma 3. We omit it.

Lemma 4

Suppose that $\# X \ge 3,$ $\# N \ge 4,$ and a Paretian QSWF f satisfies WI and PR. Let $x,y \in X$. If $\{ v \} \in \mathcal {W}^* (x,y),$ then $\{ v \} \in {\mathcal {W}}^* (z,y) \hbox { for all } z \in X$.

Proof of Proposition 9

By Lemma 2, there exists $v \in N$ such that $\{ v \} \in \mathcal {W}^*(x,y)$ for some (x, y). Lemmas 3 and 4 implies that $\{ v \} \in \mathcal {W}^* (w,z) \hbox { for all } w,z \in X$. PR implies that $\{ v \}$ is a vetoer for f.

Now, we prove the uniqueness. Suppose that there exists two vetoers, v and $v'$. Let ${\mathbf {R}}^1 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&(x,y) \in P(R^1_v)\quad \hbox {and}\quad (y,x) \in P(R^1_{v'}),\\&(x,y) \in I(R^1_i)\quad \hbox {for all } i \in N {\setminus } \{ v,v' \}. \end{aligned}$$

Since both v and $v'$ are vetoers, it follows that $(x,y) \in f({\mathbf {R}}^1)$ and $(y,x) \in f({\mathbf {R}}^1)$. Let ${\mathbf {R}}^2 \in {\mathcal {R}}^N$ such that

$$\begin{aligned}&(x,y) \in P(R^2_v)\quad \hbox {and}\quad (y,x) \in P(R^2_{v'}),\\&(x,y) \in P(R^2_i)\quad \hbox {for all } i \in N {\setminus } \{ v,v' \}. \end{aligned}$$

By PR, we have $(x,y) \in P(f({\mathbf {R}}^2))$. This contradicts the fact that $v'$ is a vetoer. $\square$