A Proofs of the claims in Section 1

Proof of Theorem 1.1 For every θ, we have Σi=1Ifij(θ)=1. Therefore, for k=1,…,K , we have

Applying the Envelope Theorem (specifically, applying Corollary 4 in Milgrom and Segal (2002) to the function defined in eq. (3) in Section 1), we get

If assumption (a) of Theorem 1.1 holds, then clearly,Vk(α,θ)=0 , for k=1,⋯,K .

Under assumptions (b) of Theorem 1.1, for any k, eq. (23) becomes

for some constants C1,⋯,CJ . Hence by eq. (22), we get

□

Proof of Corollary 1.1 Let α∈intΔJ . First, note that Ni(α,θ0)=N(α) . Moreover, the assumptions on u(xi,⋅) imply that l(α, a) is a strictly concave function of A . Therefore, N(α)={a∗} for some a∗∈A . Since M(α,θ0)=[N(α,θ0)]I , this implies that

Therefore, for every j and every z∈M(α,θ0) , u(xj,ai)=u(xj,a∗)=andCj . The conclusion of the corollary follows from applying Theorem 1.1. □

Proof of Corollary 1.2 Under either assumption (a) or (b), we have ∇θV(α,θ0)=0 by Theorem 1.1 (∇θV is the gradient of V with respect to θ).^[12] Since V(α,θ0)=0 and V(θˆ,α) for some θˆ∈Θ , we have

and therefore, V cannot be concave on Θ . Had V been concave on Θ , we would have had V(α,θˆ)−V(α,θ0)≤∇Vθ(α,θ0)⋅[θˆ−θ0] for all θ∈Θ .□

Proof of Corollary 1.3 Let G(α,θ)=V(α,θ)−C(θ) . Then, our assumption on C(θ)—with either assumption (a) or (b) of Theorem 1—imply that ∇θG(α,θ0)<0 on some neighborhood of θ0 . Using the first order Taylor approximation of G at θ0 , and using the facts that G(α,θ0)=0 and Θ=[θ0,θ1]K , we conclude that there exists a neighborhood W(θ0) of θ0 such that G(α, θ') < 0 for all θ′∈W(θ0)∩Θ .□

Proof of Theorem 1.2. Fix α. The assumptions of the theorem imply that there is a unique boundary (not in the interior of A ) solution a∗ to the problem maxa∈Al(α,a) . This, in turn, implies that

The definition of li(α,θ0,a) implies that

Assumption (ii) of the theorem and assumption A1 imply that the vector-valued function ∇ali(α,θ,a) is continuous in θ for all i. Hence, there exists a neighborhood Wˆ(θ0) of θ0 such that for all θ′∈Wˆ(θ0)∩Θ , we have

for every i. Assumption (i) of the theorem implies that there exists a neighborhood W˜(θ0) of θ0 such that for all θ′∈W˜(θ0)∩Θ , the function li(α,θ′,⋅) is strictly concave. Therefore, letting W(θ0)=W˜(θ0)∩Wˆ(θ0) , a∗ is still the unique solution of maxa∈Ali(α,θ′,a) , and hence Ni(α,θ′)={a∗} for all i and for all θ′∈W(θ0)∩Θ . This implies that

on W(θ0)∩Θ . Part (b) of Theorem 1.1 now implies that Vk(α,θ′)=0 on all of W(θ0)∩Θ .

B Proofs of the claims in Section 2

Proof of Theorem 2.1 Fix α∈intΔ2 . The non-concavity of f and g imply the existence of θ', θ′′ and θ˜=λθ′+(1−λ)θ′′ , with λ ∈ (0, 1), such that

and

If V(α,θ˜)=0 , then by the second half of assumption (c), we must have that either V(α, θ') > 0 or V(α, θ′′) > 0, and in this case it is clear that

Now assume V(α,θ˜)/=0 . For every z ∈ Z, the expression for h in eq. (2) can be re-expressed as

where

and

and the value function is given by

Define Z+={z∈A2|m1(z)>0andm2(z)>0}∪{z∈A2|m2(z)=0andm2(z)>0}∪{z∈A2|m1(z)>0andm2(z)=0} .

In other words, Z+ is a subset of Z where m1(z) and m2(z) are both positive with at least one of them being strictly positive.

Define Z−={z∈A2|m1(z)≤0andm2(z)≤0} .

Because the payoffs are reciprocal with respect to the states, the subset of Z where mi(z)>0 and mj(z)<0 for i/=j is empty. Hence, we have

First we note that

By our assumptions on the range of f and g, we have

Therefore, for every z∈Z− , we have

and eq. (28) holds.

Assumption A4 implies that the supremum of h(⋅,θ˜) will be attained on Z. eq. (28), V(α,θ˜)>0 , and Z=Z−∪Z+ together imply

Therefore, there exists zˆ∈Z+ such that

Assume, without loss of generality, that m1(zˆ)>0 and m2(zˆ)>0 . Then B1 and B2 imply

and

Adding eqs. (30) and (31), and then adding C(zˆ) to both sides of the resulting inequality gives us

and

and therefore

and V(α, · ) is not concave on Θ . □

Proof of Theorem 2.2 Fix α. When f = g, eq. (26) becomes

where

and

Let θ', θ′′ and λθ′+(1−λ)θ′′ satisfy

The same argument we used in the proof of Theorem 2.1 allows us to limit our attention to the case where V(θ˜)>0 . This last inequality implies that

for some zˆ∈Z+ . Equation (35) now implies the existence of zˆ∈Z+ such that

Therefore, we have

□

Proof Corollary 2.1 Let θ' and θ′′ be two points in Θˆ . Let θ˜=λθ′+(1−λ)θ′′ for some λ ∈ (0, 1). The convexity of Θˆ and the convexity of f on Θˆ imply that θ˜∈Θˆ and

Furthermore, V(α,θ˜)>0 , which implies that there exists zˆ∈Z with m(zˆ)≥0 such that

From eq. (36), we get

and hence

□

Proof of the claim in Example 2.1 First, since f(θ) is continuous, V(α, · ) is the supremum of continuous functions in θ, and therefore it also is continuous in θ. Define θˆ=inf{θ∈[0,θ1]|V(θ)>0} . The fact the f is weakly increasing implies that V(α, · ) is weakly increasing on (θˆ,θ1] . To see this, let θ' and θ′′ be two points in (θˆ,θ1] such that θ′′ > θ'. Then, using the same argument we used in the proof of Corollary 2.1, we have V(α,θ′)=h(α,zˆ,θ′)−v∗ for some zˆ with m(zˆ)≥0 , and hence

where C and m are given by eqs. (33) and (34). Therefore,

This, combined with the fact that V is flat at zero on [0,θˆ] and the continuity of V(α, · ), implies that V(α, · ) is weakly increasing on all of [0,θ1] .□

C Proofs of the claims in Section 3

Recall that basic assumptions in Section 3 on Θ , f and g are as follows:

Θ is a set with a partial order ≥ and corresponding strict partial order >. The functions f and g are increasing in the sense that for any θ′′,θ' in Θ , θ′′ > θ' implies f(θ′′) > f(θ') and g(θ′′) > g(θ'). The range of f and g is contained in the interval [1/2,1] , and therefore, for any θ∈Θ , we have f(θ)>1−f(θ) and g(θ′′)>1−g(θ) . A matrix is stochastic if it has non-negative entries and every column adds up to one. We will need the following two useful facts about matrices. First, if the sum of each column in a matrix A is one, then so is the sum of each column in A−1 . Note this does not imply that the inverse of a stochastic matrix is a stochastic matrix, since the inverse might have negative entries. Second, if the sum of each column in the matrices A and B is one, then so is the sum of each column in the product AB.

Proof of the claim in Example 3.1.

Given the assigned values of f, g and h at θ' and θ′′, we can compute [S(θ′′)]−1 , the inverse of matrix S(θ′′), and multiply it by the matrix S(θ') to obtain the matrix M=S(θ′)[S(θ′′)]−1 . Computing M, we find

Therefore, the only matrix that satisfies the matrix equation S(θ') = MS(θ′′) is not stochastic since it has non-negative entries. Therefore, by Blackwell’s theorem, there must be some DM who strictly prefers S(θ') to S(θ′′). □

Proof of Theorem 3.1.

First, we show θ′′≥θ′⇒V(α,θ′′)≥V(α,θ′) for any agent (u,A) and any α∈intΔJ .

Let θ′′ > θ'. Let M=S(θ′)[S(θ′′]−1 (since f(θ′′)>1/2>1−f(θ′′) and g(θ′′)>1/2>1−g(θ′′) , S(θ′′) is indeed invertible). Then, we have

and brute force computations give us

The two facts about matrices that we mentioned in the beginning of Appendix C imply that the sum of each column in M is 1. Hence, to show that M is stochastic, we only need to show that the entries of M are non-negative. Since we have an explicit expression for every mij , this can be done entry-by-entry. For example, note that f(θ′′)+g(θ′′)−1>0 since f(θ′′)>f(θ′)≥1/2 and g(θ′′)>g(θ′)≥1/2 . Moreover, g(θ′′)>1−g(θ′) and f(θ′)≥1−f(θ′′) . Therefore,

Similarly, we can show that the rest of entries are non-negative. In fact, it is also possible to check that m11+m21=m12+m22=1 by brute force, rather than by relying on our previous argument. Hence, M is a stochastic matrix, and by Blackwell’s theorem (See Blackwell 1951, or for a simpler exposition Crémer 1981), we have V(α, θ′′, α) ≥V(α, θ ') for every DM (A,u) and every α∈intΔJ .

Second, we assume that V(α,θ′′)≥V(α,θ′) for any agent (A,u) and α∈intΔJ , and we show that θ′′≥θ .

Under our current assumption, Blackwell’s theorem implies the existence of a stochastic matrix M such that

Assume that θ′′ < θ'. By C3, we must have

which implies

(otherwise, we have f(θ') < f(θ′′), which contradicts θ′′ < θ' and the monotonicity of f). Similarly, C3 implies

which implies

Otherwise, we have g(θ') < g(θ′′) contradicting θ′′ < θ' and the monotonicity of g. Given eqs. (38) and (39), it is clear that assumption θ′′ < θ' cannot be true, and we must have

□

Proof of Theorem 3.2 Assume S(θ) is given by eq. (19) with I = J > 2.

First, assume θ′′ > θ'. Define

As in the proof of Theorem 3.1, the sum of each column in M is 1. Straightforward (but tedious) computations give the entries of M:

Given that θ′′ > θ' and that f(θ′′)>f(θ)>1J−1 , eq. (40) implies that each mij is non-negative, and hence M is stochastic. Applying Blackwell’s theorem yields the desired conclusion.

Second, note that the range of f is contained in the interval [1/J,1] , and therefore, for any θ∈Θ , we have f(θ)>1−f(θ)J−1 . Following the same argument in the second part of the proof of Theorem 3.1, we can show that if V(θ′′)≥V(θ′) for any agent (u,A) , then we must have θ′′≥θ′ .□