Arrowian Social Equilibrium: Indecisiveness, Influence and Rational Social Choices under Majority Rule

A.1 Proof of Proposition 4.1

Consider any preference profile ≻ = ( ≻ 1 , … , ≻ n ) ∈ B n for which θ ≻ ( S ) > θ ̲ ( S ) = | S | n n 2 − 1 | S | − 1 , for all S ∈ X with |S| > 2. For any such profile, we show below that an ASEM exists by explicitly constructing one.

Step 1: Showing that MR candidates exist for any issue.

For any issue S ∈ X with |S| > 2, first note that the set of potential MR winners under an ASEM is given by:

This is because in any ASEM, for any such S, c _i (S) ∈ m(S, ≻ _i ), and hence for an alternative to be chosen by a majority, a necessary condition is that it is in a majority of individual maximal sets. We first establish that if θ ≻ ( S ) > θ ̲ ( S ) , then M(S) ≠ ∅, for any such S. To do so assume otherwise; suppose M(S) =∅, for some such S. For any x ∈ S, let I _S (x) = {i ∈ I: x ∈ m(S; ≻ _i )}. Under our supposition | I S ( x ) | ≤ [ n 2 ] , for all x ∈ S. It should be straightforward to verify that:

This implies that:

i.e., θ ≻ ( S ) ≤ θ ̲ ( S ) , which brings us to our desired contradiction.

Step 2: Construction of ASEM choice profile ( c 0 , c 1 , … , c n ) ∈ C n + 1

In this step, we explicitly construct a choice profile ( c 0 , c 1 , … , c n ) ∈ C n + 1 , and show that it constitutes an ASEM. To that end, first, observe the following: if S , T ∈ X such that S ⊆ T, then for any i ∈ I, m(T; ≻ _i ) ∩ S ⊆ m(S; ≻ _i ). This is because if for any x ∈ S, ∄y ∈ T, such that y ≻ _i x, then clearly ∄y ∈ S ⊆ T, such that y ≻ _i x. In turn, this implies that M(T) ∩ S ⊆ M(S).

For notational convenience, in the construction below, write X = {x ₁, …, x _k }. Since, from Step 1, M(S) ≠ ∅ for all S ∈ X , |S| > 2, we can construct c 0 : X → X recursively as follows:

Step 2.1: Wlog, let x ₁ ∈ M(X).

Let c ₀(S) = x ₁, for all S ∈ X such that x ₁ ∈ S.

Step 2.2: Wlog, let x ₂ ∈ M(X \{x ₁}). Let c ₀(S) = x ₂, for all S ∈ X , such that, x ₂ ∈ S and x ₁∉S.

Step 2.k-2: Wlog, let x _k−2 ∈ M(X\{x ₁, x ₂, …, x _k−3}) = M({x _k , x _k−1, x _k−2}).

Let c ₀(S) = x _k−2, for all S ∈ X such that x _k−2 ∈ S and x ₁, …, x _k−3 ∉ S.

Observe that in this manner, we have covered all issues except for the issue {x _k , x _k−1}. Since n is odd, even if all individuals have complete preferences over x _k and x _k−1, we have, M({x _k , x _k−1}) ≠ ∅. Wlog say, x _k−1 ∈ M({x _k , x _k−1}). Let c ₀({x _k , x _k−1}) = x _k−1. It is straightforward to see that c ₀ constructed thus is a choice function. It also directly follows from the construction that c ₀ satisfies WARP. Let ≻₀ ⊆ X × X be defined by x ≻₀ y if c ₀({x, y}) = x. Since, c ₀ satisfies WARP, it follows that ≻₀ is a strict preference ranking and it rationalizes c ₀, i.e., for all S ∈ X , c ₀(S) = m(S; ≻₀). Finally, note that in this construction, drawing on the fact that if T′ ⊆ T, then M(T) ∩ T′ ⊆ M(T′), we have that for all S ∈ X , if c ₀(S) = x, then x ∈ M(S).

Next, we construct c ₁, c ₂, …, c _n as follows. For any S ∈ X and i ∈ I, let:

Since ≻₀ is a strict preference ranking, the second stage is always decisive and, hence, any such c _i is a well defined choice function (specifically, an RSM).

All that, therefore, remains to be shown to establish that the collection (c ₀, c ₁, …, c _n ) is an ASEM is that for all S ∈ X , c ₀(S) = f _mr(c ₁(S), …, c _n (S)). Consider any S ∈ X . Let c ₀(S) = x. Then it follows that x ≻₀ y, for all y ∈ S\{x}. Accordingly, for any i ∈ I, if x ∈ m(S; ≻ _i ), then c _i (S) = x. This is, of course, true if m(S; ≻ _i ) = {x}. Even otherwise, this is so because x is the best alternative in m(S; ≻ _i ) according to ≻₀. Further, we know that x ∈ M(S). That is,

A.2 Proof of Proposition 4.2

The proof for the case n even is exactly along the same lines as n odd. The only difference emerges in Step 2, following the (k − 2)-th step when we arrive at the issue {x _k , x _k−1}. Now, it is possible that all individuals have complete preferences over these two alternatives and exactly half prefer x _k to x _k−1 and the other half x _k−1 to x _k . In that case M({x _k , x _k−1}) = ∅. This is precisely where the role of the additional condition comes in. According to it, there exists some i ∈ I such that neither x _k ≻ _i x _k−1, nor x _k−1 ≻ _i x _k . This implies M({x _k , x _k−1}) ≠ ∅ and the rest of the argument follows as above.