Proofs in Section 1.3 - A Test for Risk-Averse Expected Utility

Chapter 3 A Test for Risk-Averse Expected Utility

A.2 Proofs in Section 1.3

Lemma 10. Let U∗=_E_λ[u(φ)] : λ ∈ Λ∗ . For every vector u ∈ U∗, there exist vectors{uk_{: k ∈}

κ(

I

T

)} ⊆ U∗such that

uk_k< uk

for all k∈ κ(

I

_T

), and

uk_k< uk_k0

for all k6= k0∈ κ(

_I

_T

Proof. The proof relies on similar techniques as in Abreu, Dutta, and Smith (1994). Whenever possible, I will cite intermediate results in Abreu, Dutta, and Smith (1994) without reproducing their proofs. Compared to Abreu, Dutta, and Smith (1994), the main difference is that instead of imposing NEU, it is shown that a matching environment always satisfies NEU.

I will first establish that the set Φ_T satisfies what Abreu, Dutta, and Smith (1994) called the non-equivalent utilities (NEU) condition for institutions in

I

_T

. For every k ∈ κ(T ), there exists aj(k)∈

A

T

such that vj(k)(Ik)> 0. Define the assignment φ0by

φ0(I_l) =        b A_l if Il∈

T

/ 0 otherwise

For each k ∈ κ(

_I

_T

), define the assignment φk by

φk(Il) =                b A_l if Il∈

T

{a_j(k)} if Il= Ik / 0 otherwise

as well.

There are at least two elements in κ(

_I

_T

). For each pair of k 6= k0∈ κ(

_I

_T

), u_k(φ0) = u_k(φk0) = 0, while by strictness of institution preferences, u_k0(φ0) = 0 6= u_k0({a_j(k0₎}) = u_k0(φk0). Therefore, for each pair k 6= k0 ∈ κ(

I

_T

), there do not exist scalars α> 0 and β such that u_k(φ) = αuk0(φ) + β for all φ ∈ ΦT. This verifies the NEU condition in Abreu, Dutta, and Smith

(1994) for indices in κ(

_I

_T

Since co(U_T) = co({u(φ) : φ ∈ Φ_T}), Lemma 1 and Lemma 2 in Abreu, Dutta, and Smith (1994) then ensure the existence of vectors {u_bk: k ∈ κ(

_I

_T

)} ⊆ co(U_T) that satisfyu_bk_k<ub

k0 k for

all k 6= k0∈ κ(

_I

_T

). Since {u(φ) : φ ∈ Φ_T} is a finite set, for each k ∈ κ(

_I

_T

), there exists an element ukin {u(φ) : φ ∈ Φ_T} that minimizes I_k’s payoff.

For an arbitrary vector u ∈ U∗, define

uk= ε(1 − η)uk+ ηεub

k_{+ (1 − ε)u}

for each k ∈ κ(

I

_T

). Observe that if εη> 0, then uk k< u

k; for all 0< ε < 1 and 0 < η < 1,

uk∈ co(U_T); for small enough ε> 0, uk

k> uTk ; and finally, for small enough η> 0, it must be

true that uk_k< uk. Therefore, there must exist {uk: k ∈ κ(

I

T

)} ⊆ U∗such that

uk_k< uk

for all k ∈ κ(

_I

_T

), and

uk_k< uk_k0

for all k 6= k0∈ κ(

I

_T

Lemma 11. For each institution Ik∈

I

T

, there exists φ_k∈ ΦT such that uk(φ_k) = uT_k , and

max

B∈DA_k(φ_k)

Lemma 12. For each institution Ik∈

T

and every φ ∈ ΦT,

max

B∈DA_k(φ)

u_k(B) = u_k(φ)

Lemma 13. Let Φ0⊆ Φ be a subset of static assignments, and ψ :

_H

F _{→ Φ}0_{be a self-enforcing}

assignment process. Fix an institution I_k∈

_I

. If for every φ ∈ Φ0, there exists a subset of agents Aφ_{∈ D}A

k(φ) that satisfies uk(Aφ) ≥ uk, then

u_k(ψ) ≥ u_k

Proof. Suppose by contradiction that ψ is a self-enforcing assignment process, but uk(ψ)< uk.

I will show that institution I_k has a feasible profitable deviation plan from ψ, so ψ must not be self-enforcing. Consider the following deviation plan d_k0 from ψ: for every ex post history h∈

_H

F_{, define}

d_k0(h) = Aψ(h)_.

By assumption, ψ(h) ∈ Φ0for all h ∈

_H

F, so d_k0 is well-defined and feasible. To see that d_k0 is profitable, observe that at every ex post history h, institution Ik’s stage-game payoff from the

manipulated static assignment [ψ(h), (Ik, dk0(h))] is uk(Aψ(h)) ≥ uk> uk(ψ). Since this is true for

every ex post history h, Ik’s total discounted payoff from the deviation plan satisfies

u_k([ψ, (Ik, dk0)])> uk(ψ).

The deviation plan d0_kis both feasible and profitable for institution Ik, which is a contra-

diction to the self-enforcement of ψ. So u_k(ψ) ≥ u_k.

Proof of Theorem 2. Let {φ

Lemma 11. Fix λ0∈ Λ∗and define u0_{= E}_λ0[u(φ)]. By Lemma 10, there exist vectors {uk: k ∈ κ(

I

T

)} ⊆ U∗, such that

uk_k< u0_k

for every k ∈ κ(

I

_T

), and

uk_k< uk_k0

for all k, k0∈ κ(

I

_T

), k 6= k0. For each k ∈ κ(

_I

_T

), let λk∈ Λ∗be the distribution over Φ_T that give rise to the payoff vectors uk.

By Lemma 9, it suffices to consider the matching process represented by the automaton (Θ, p0, f , γ), where

• Θ =

θ(e, φ) : e ∈ κ(

_I

_T

) ∪ {0}, φ ∈ Φ_T ∪ θ(k,t) : k ∈ κ(

I

_T

), 0 ≤ t < L is the set of all possible states;

• p0_{is the initial distribution over states, which satisfies p}0_{(θ(0, φ)) = λ}0_{(φ) for all φ ∈ Φ}

• f : Θ → Φ is the output function, where f (θ(e, φ)) = φ and f (θ(k,t)) = φ

• γ : Θ × Φ → ∆(Θ) is the transition function. For states {θ(k,t)|0 ≤ t < L − 1}, γ is defined as γ θ(k,t), φ0 =                θ(k0, 0) if φ06= φ k; φ 0_{= [φ} k0, (Ik0, B)] for some k 0_{∈ κ(}

_I

_T

₎ and B ∈ DA_k0(φ k0) θ(k,t + 1) otherwise

For states θ(k, L − 1), the transition is defined as γ θ(k, L − 1), φ0 =                θ(k0, 0) if φ06= φ k; φ 0_{= [φ} k0, (Ik0, B)] for some k 0_{∈ κ(}

_I

_T

₎ and B ∈ DA_k0(φ k0) pk otherwise

where pk is the distribution over states that satisfies pk(θ(k, φ)) = λk_{(φ) for all k ∈ κ(}

_I

_T

₎

and φ ∈ Φ_T.

For states θ(e, φ), the transition is

γ θ(e, φ), φ0 =       

θ(k0, 0) if φ06= φ; φ0= [φ, (Ik0, B)] for some k0∈ κ(

I

T

) and B ∈ DA_k0(φ) pe otherwise

Note that owing to the identifiability of deviating institution, for any θ ∈ Θ and assignment φ06= f (θ) which can result from an institution’s deviation, we can uniquely identify the institution, so the transition above is well-defined. Any φ06= f (θ) that cannot possibly result from a deviation by an institution is ignored by the transition.

The assignment process represented by the above automaton randomizes over Φ_T accord- ing to λ0in every period. It remains to check that it is self-enforcing, or equivalently,

1. every agent’s payoff is greater than or equal to 0 in all automaton states, 2. no institution has any profitable one-shot deviation in any automaton state,

3. for every institution, the stage-game payoffs across all automaton states are bounded. Point 3 above follows because the set of states Θ is finite. In every state θ ∈ Θ, the recommended assignment f (θ) ∈ Φ_T is individually rational for all agents, so point 1 follows as well. It remains to verify point 2.

For every state θ, I use U (θ) = (U1(θ), . . . ,UK(θ)) to denote the discounted expected

payoff profile for the institutions in state θ. By construction,

U(θ(e, φ)) = (1 − δ)u(φ) + δue, for all e ∈ κ(

I

_T

) ∪ {0},

and

U(θ(k,t)) = (1 − δL−t)u(φ

k) + δ

L−t_uk_{, for all k ∈ κ(}

_I

_T

_{), 0 ≤ t ≤ L − 1.}

In addition, since ul(φ) = ul( bAl) for every Il∈

T

and all φ ∈ ΦT, the above equalities simplify to

U_l(θ(e, φ)) = Ul(θ(k,t)) = ul( bAl) (A.3)

for every Il∈

T

I now verify no institution has profitable one shot deviations in any automaton states. For statesθ(e, φ) : e ∈ κ(

I

_T

) ∪ {0}, φ ∈ Φ_T : there are three cases to consider.

Case 1: I_k0∈

T

. By A.3, the continuation value of I_k0is identical across states. By Lemma 12, no feasible deviation for I_k0 can improve its stage-game payoff. I_k0 does not have any profitable one shot deviation.

Case 2: Ik0∈

I

T

, k06= e. Choose a number Z > sup_{{φ∈Φ,k∈κ(}_I_\_T_)}u_k(φ). For any assignment φ ∈ Φ_T, the manipulated assignment resulting from a feasible deviation must still be an assignment in Φ. So Z is larger than the payoff any institution can obtain in any feasible deviation from an assignment in Φ_T.

Consider a one-shot deviation (I_k0, B) by institution I_k0. Without deviation, I_k0 has value (1 − δ)uk0(φ) + δuek0. After deviation, Ik0 yields less than

(1 − δ)Z + δUk0(θ(e, 0)) = (1 − δ)Z + δ(1 − δL)uTk0+ δ L+1_uk0

There is no profitable one-shot deviation for Ik0 if (1 − δ)uk0(φ) + δuek0 ≥ (1 − δ)Z + δ(1 − δ L_)uT k0+ δ L+1_uk0 k0

As δ → 1, the LHS converges to ue_k0 while the RHS converges to uk 0

k0. By construction, uek0> uk 0 k0. It

follows that such deviations are not profitable for δ high enough.

Case 3: I_k0∈

I

T

, k0= e. Without deviation, I_k0 has value (1 − δ)u_k0(φ) + δuk0

k0. After deviation,

I_k0 yields less than

(1 − δ)Z + δUk0(θ(k0, 0)) = (1 − δ)Z + δ(1 − δL)uTk0+ δL+1uk 0 k0.

There is no profitable one-shot deviation for Ik0 if

(1 − δ)uk0(φ) + δuk 0 k0≥ (1 − δ)Z + δ(1 − δ L_)uT k0+ δL+1uk 0 k0.

The inequality is equivalent to

Z− u_k0(φ) ≤ δ(1 +. . . + δL−1)[uk 0 k0− uTk0] By construction, uk_k00− uT_k0 > 0. Choose L large enough so that L(uk

k0− uTk0)> Z − uk0(φ). As δ → 1,

the LHS remains unchanged while the RHS converges to L(uk_k00− uT_k0), so such deviations are not profitable for δ high enough.

For statesθ(k,t) : k ∈ κ(

I

_T

), 0 ≤ t ≤ L − 1 : there are three cases to consider.

Case 1: I_k0∈

T

Case 2: Ik0∈

I

T

, k06= k. Without deviation, institution I_k0 has payoff

(1 − δL−t)u_k0(φ k) + δ

L−t_uk k0

With any deviation, Ik0 has payoff less than

(1 − δ)Z + δ(1 − δL)uT_k0+ δL+1uk 0 k0

There is no profitable one-shot deviation for I_k0 if

(1 − δL−t)u_k0(φ k) + δ L−t_uk k0≥ (1 − δ)Z + δ(1 − δ L_)uT k0+ δ L+1_uk0 k0

Observe that as δ → 1, the LHS converges to uk_k0 for all t such that 0 ≤ t ≤ L, while the RHS converges to uk_k00. By construction uk_k0> uk

k0. So the above inequality holds for sufficiently high δ. Case 3: Ik0∈

I

T

, k0= k. Without deviation, institution I_k0 has payoff

(1 − δL−t)uT_k0+ δL−tuk 0 k0.

When deviating from φ

k0, by Lemma 11, Ik’s stage-game payoff is at most uTk . So Ik0’s discounted

expected payoff from deviation is at most

(1 − δ)uT_k0 + δ(1 − δL)uT_k0+ δL+1uk 0 k0= (1 − δ L+1_)uT k0+ δ L+1_uk0 k0

Institution Ik0 has no profitable deviation if

(1 − δL−t)uT_k0 + δL−tuk 0 k0≥ (1 − δ L+1_)uT k0+ δ L+1_uk0 k0, (A.4)

uk_k00≥ uT_k0,

which is true by construction. So Ik0 has no profitable one-shot deviation.

We have verified that there is no profitable one-shot deviation in any states of the automaton. This completes the proof.

Proof of Theorem 1. To prove the result, it is sufficient to show that if ψ is a self-enforcing assignment process, then

1. ψ(h) ∈ Φ_A at every ex post history h ∈

_H

2. ψ(Ik|h) = bAkfor every Ik∈

T

at every ex post history h ∈

H

F; and

3. u_k(ψ) ≥ uT_k for every I_k∈

_I

_T

. I now establish these claims in order.

1. This follows from the definition of self-enforcing matching process.

2. If

_T

= /0, there is nothing to prove. Suppose

T

= {(Ik1, bAk1), . . . , (IkG, bAkG)}. The proof proceeds by induction.

First I establish that in every self-enforcing assignment process ψ, ψ(Ik1|h) = bAk1 for all h∈

_H

F_{. Suppose by contradiction that ψ(I}

k1|eh) 6= bAk1 for some self-enforcing assignment process ψ at some eh∈

_H

F_.

By the construction of

_T

, uk1(B) ≤ uk1( bAk1) for all B ⊆

A

. By the strictness of institutions’ preferences, uk1(ψ(Ik1|eh))< uk1( bAk1). In addition, since uk1( bAk1) is the highest possible stage-game payoff for Ik1, uk1(ψ) ≤ uk1( bAk1) for every assignment process ψ. Together

these imply

u_k₁(ψ|

h) = (1 − δ)uk1(ψ(Ik1|eh)) + δuk1(ψ|eh,ψ(eh))

< (1 − δ)uk1( bAk1) + δuk1( bAk1) = uk1( bAk1)

(A.5)

Since ψ is a self-enforcing assignment process, ψ|

ehmust also be self-enforcing, so ψ|eh(h) ∈

Φ_A at every h ∈

H

F. In addition, for every φ ∈ Φ_A, by the construction of

T

and the strictness of agents’ preferences, vj(Ik1)> vj(φ) for every aj∈ bAk1\φ(Ik1). So bAk1∈ DAk1(φ) for all φ ∈ Φ_A. By Lemma 13,

u_k₁(ψ|

h) ≥ uk1( bAk1) (A.6)

Inequalities (A.5) and (A.6) cannot be true at the same time, a contradiction. So ψ(Ik1|h) = b

A_k₁ for all h ∈

_H

Suppose it has been shown that in every self-enforcing assignment process ψ, ψ(Iki|h) = bAki for i = 1, . . . , g − 1 at every ex post history h ∈

H

F_{. I show that this implies that in every}

self-enforcing assignment process ψ, ψ(Ikg|h) = bAkg at every ex post history h ∈

H

F_.

Suppose by contradiction that ψ(Ikg|eh) 6= bAkg for some self-enforcing assignment process ψ and some eh∈

H

Since by the inductive hypothesis ψ(Iki|h) = bAkifor all 1 ≤ i ≤ g − 1 and all h ∈

H

F_{, it must}

be that ψ(Ikg|h) ⊆

A

\ ∪ g−1

i=1 Ab_k_iat all h ∈

H

F. From the construction of

T

, u_k_g(ψ(I_k_g|h)) ≤ u_k_g( bA_k_g) for all h ∈

_H

F, it follows that at every ex post history h ∈

H

u_k_g(ψ|h) ≤ bAkg

u_k_g(ψ(Ikg|eh))< ukg( bAkg). Together these imply

u_k_g(ψ|

h) = (1 − δ)ukg(ψ(Ikg|eh)) + δukg(ψ|eh,ψ(eh))

< (1 − δ)ukg( bAkg) + δukg( bAkg) = ukg( bAkg)

(A.7)

Since ψ is a self-enforcing assignment process, ψ|

h must also be self-enforcing. Let

Φg= {φ ∈ Φ_A : φ(Iki) = bAki for i = 1, . . . , g − 1}. By the inductive hypothesis, ψ|eh(h) ∈ Φ g

at every h ∈

H

F. In addition, for every φ ∈ Φg, by the construction of

T

and the strictness of agents’ preferences, vj(Ikg)> vj(φ) for every aj∈ bAkg\φ(Ikg). So bAkg ∈ D

A kg(φ) for all φ ∈ Φg. By Lemma 13, u_k₁(ψ| e h) ≥ uk1( bAk1) (A.8) Inequalities (A.7) and (A.8) cannot be true at the same time, a contradiction. So ψ(Ikg|h) =

A_k_g for all h ∈

_H

F, completing the induction argument.

3. Let ψ be a self-enforcing assignment process. ψ(h) ∈ Φ_T for all ex post histories h ∈

_H

F. For every institution Ik∈

I

T

, from the definition of uT_k , there exists Aφ∈ DA_k(φ) for every

φ ∈ Φ_T such that

u_k(Aφ_{) ≥ u}T k .

By Lemma 13, uk(ψ) ≥ uT_k for every Ik∈

I

T

Proof of Corollary 1. Suppose all agents share a common ordinal ranking I_k₁ I_k₂ . . . IkK over the institutions. For each institution Ikg, 1 ≤ g ≤ K, define

A_k_g = arg max

B⊆A\∪g−1_i=1Ab_ki

Then

_T

= {(Ik1, bAk1), . . . , (IkK, bAkK)} is the top coalition sequence of the matching market. Let φ_T be the static assignment that satisfies φ_T(Ik) = bAkfor k = 1, . . . , K.

Claim 1. φ_T(I_k) = bA_k is the unique static stable matching.

The proof is by induction. It is clear that in every static stable assignment, Ik1 must be matched to bA_k₁, otherwise bA_k₁ is a feasible profitable deviation for Ik1. Suppose Ik1, . . . , Ikg−1 is matched to bA_k₁, . . . , bA_k_g−1, respectively, in every static assignment. Let φ be an arbitrary static stable assignment. I prove that φ(Ikg) = bAkg.

Suppose by contradiction that φ(Ikg) 6= bAkg, then by the inductive hypothesis, ψ(aj) ∈

I

\{I_k₁, . . . , Ikg−1} for every aj ∈ bAkg\φ(Ikg). By the definition of top coalition sequence and the strictness of agents’ preferences, vj(Ikg)> vj(φ) for every aj∈ bAkg\φ(Ikg), so bAkg ∈ D

kg(φ). Similarly, by the inductive hypothesis, φ(I_k_g) ⊆

_A

\ ∪g−1_i=1 Ab_k_i, so u_k_g( bA_k_g)> u_k_g(φ) by the definition of a top coalition sequence and the strictness of institutions’ preferences. This implies that bA_k_g is a profitable and feasible deviation, a contradiction to φ being a stable assignment. So φ(Ikg) = bAkg. This completes the inductive step.

Claim 2. There is a unique self-enforcing assignment process ψ_T, where ψ_T(h) = φ_T for all h∈

_H

F_.

Since

_I

⊆

_T

, Φ_T = {φ : φ(Ik) = bAk∀k = 1, . . . , K}. So φT is the unique element of the

set Φ_T. The claim then follows from Theorem 1.

In document Essays in Microeconomic Theory (Page 105-116)