Proofs of Proposition 2 and Theorem 1

We first provide explicit formulas for the equilibrium welfare of each agent under SD and T T C. To do so, we introduce some additional notation. For each k ∈ N , let N^k ≡ {1, . . . , k − 1}. Then under SD, N^k is the set agents with higher priority than agent k. Similarly, under T T C, N^k is the set of agents whose endowments have common values higher than the common value of agent k’s endowment. For each

pair l, k ∈ {1, . . . , m} with l ≤ k, let

Lemma 2 below allows us to interpret these products in terms of agents’ options sets:

under either SD or T T C, Q(l, k) represents the probability that a_l is the object with the highest common value in agent k’s option set.

Lemma 2. For each pair l, k ∈ {1, . . . , n} with l ≤ k, Q(l, k) is the probability that (i) a_l has the highest common value in agent k’s option set under SD and (ii) a_l has the highest common value in agent k’s option set under T T C.

Proof. Let l, k ∈ {1, . . . , n} with l < k. After verifying the probabilities for l < k, the formula for l = k follows by subtraction. Let σ^≺, σ^µ ∈ A^N be equilibria under SD and T T C respectively and let P^≺, P^µ ∈ P^N be preferences determined by σ^≺ and σ^µ. That is, P^≺ and P^µ are random variables that depend on the realizations of ε_σ^≺ and ε_σ^µ respectively.

Without loss of generality, suppose that for each i ∈ N , σ^≺_i = a_i+1 and σ^µ_i = a_i

where a_m+1 = a₁ if necessary.

(i) Priority rule. According to σ^≺, agents in N^kreceive at least k − 1 of the objects {a₁, . . . , a_k}. Thus, agent k’s option set contains exactly one of these objects and the cases are mutually exclusive. Now suppose al∈ Ok(P_−k^≺, SD). Then no agent in N^k receives a_l. We verify the probabilities by induction on l.

Case 1.1: l = 1. Then for each i ∈ N^k, agent i receives a_i+1. Therefore, a_i+1 P_i^≺ a₁ which occurs with probability p_1(i+1). Since these events are independent, the probability that a₁ ∈ O_k(P_−k^≺ , SD) is p₁₂· p₁₃· · · p_1k = Q(1, k).

Case 1.2: l = 2. Then agent 1 receives a₁ and for each i ∈ N^k\{1}, agent i re-ceives a_i+1. Therefore, a₁ P₁^≺ a₂ which occurs with probability 1−p₁₂ = 1−Q(1, 2) = Q(2, 2). Also, for each i ∈ N^k\{1}, a_i+1 P_i^≺ a₂ which occurs with probability p_2(i+1). Since these events are independent, the probability that a₂ ∈ O_k(P_−k^≺, SD) is

(1 − p₁₂)p₂₃· p₂₄· · · p_2k = Q(2, 2)P (2, k) = Q(2, k).

Case 1.3: l ≥ 3. Suppose the claim is true for each i < l. Then the agents in N^k collectively receive {a₁, . . . , a_l−1, a_l+1, . . . , a_k}. Moreover, since each agent i always receives an object among {a₁, . . . , a_i+1}, this implies that the agents N^l collectively receive {a₁, . . . , a_l−1}. Therefore, {a₁, . . . , a_l−1} ∩ O_l(P_−l^≺, SD) = ∅. By hypothesis, for each i < l, Q(i, l) is the probability that a_i ∈ O_l(P_−l^≺, SD), so 1 − Q(i, l) is the probability a_i 6∈ O_l(P_−l^≺, SD). By mutual exclusivity, 1 − Pl−1

i=1Q(i, l) is the probability that {a₁, . . . , a_l−1} ∩ Ol(P_−l^≺, SD) = ∅. Next, for each i ∈ N^k\(N^l∪ {l}), agent i receives a_i+1. Therefore, a_i+1 P_i^≺ al which occurs with probability p_l(i+1).

Since these events are independent, the probability that a_l∈ O_k(P_−k^≺, SD) is

1 −

l−1

X

i=1

Q(i, l)

!

p_l(l+1)· p_l(l+2)· · · p_lk = Q(l, l) · P (l, k) = Q(l, k).

Finally, since agent k’s option set under SD always includes exactly one of {a₁, . . . , a_k}, the probability that a_k ∈ O_k(P_−k^≺ , SD) is 1 −Pk−1

i=1 Q(i, k) = Q(k, k).

(ii) Top trading cycles rule. According to σ^µ, each agent i reports preferences that agree with P₀ on A\{µ_i}. Thus, for each pair i, j ∈ N with i < j, O_j(P_−j^µ , T T C) ⊆ O_i(P_−i^µ, T T C). Moreover, if agent i receives a_j, then agent j receives a_i and for each h ∈ N with i < h < j, agent h receives µ_h = a_h. In particular, the events a_i ∈ O_j(P_−j^µ , T T C) and a_h ∈ O_j(P_−j^µ , T T C) are mutually exclusive. Now suppose a_l ∈ O_k(P_−k^µ , T T C). Then no agent in N^k receives a_l. We verify the probabilities by induction on l.

Case 2.1: l = 1. Then ak P₁^µ a₁. This occurs with probability p_1k. Furthermore, {a₁, . . . , a_k−1}∩O₁(P₋₁^µ , T T C) = {a₁}. For each i ∈ N^k\{1}, a_i 6∈ O₁(P₋₁^µ , T T C) im-plies a_i P_i^µ a₁ which occurs with probability p_1i. Since these events are independent, the probability that a₁ ∈ O_k(P_−k^µ , T T C) is p_1k· p₁₂· p₁₃· · · p_1(k−1) = Q(1, k).

Case 2.2: l = 2. Then a_k P₂^µ a₂. This occurs with probability p_2k. Furthermore, {a₁, . . . , a_k−1} ∩ O₂(P₋₂^µ , T T C) = {a₂}. For each i ∈ N^k\{2}, a_i 6∈ O₂(P₋₂^µ , T T C) implies a_i P_i^µa₂. This occurs for i = 1 with probability 1 − p₁₂= 1 − Q(1, 2) and for i > 2 with probability p_2i. Since these events are independent, the probability that a₂ ∈ O_k(P_−k^µ , T T C) is p_2k(1 − p₁₂) · p₂₃· · · p_2(k−1) = Q(2, k).

Case 2.3: l ≥ 3. Suppose the claim is true for each i < l. Then a_k P_l^µ a_l. This

occurs with probability p_lk. Furthermore, {a₁, . . . , a_k−1} ∩ O_l(P_−l^µ, T T C) = {a_l}. By

Finally, since agent k’s option set under T T C always includes a_k, the probabil-ity that a_k is the object in O_k(P_−k^µ , T T C) with the highest common value is then Ok(P_−k^µ , T T C) is 1 −Pk−1

i=1 Q(i, k) = Q(k, k).

We now provide formulas for the each agent’s equilibrium utility. For ease of notation, we adopt the conventions that a_m+1 ≡ a₁ and for each k ∈ N , p_k(m+1) ≡ 0

Proof. Let σ ∈ A^N be an equilibrium under SD, P ∈ P^N be preferences determined by σ, and k ∈ N . Without loss of generality, suppose σ_k = a_k+1. We interpret the expressions in the formula as expected utilities conditional on the realization of agent k’s option set.

According to SD, exactly one of {a₁, . . . , a_k} is in O_k(P−k, SD) so these events are mutually exclusive and exhaustive. By Lemma 2, for each l ∈ {1, . . . , k}, a_l ∈ O_k(P−k, SD) with probability Q(l, k). In this case, agent k reports preferences with either al or a_k+1 at the top of Ok(P−k, SD) and receives that object. If ε_k(k+1) <

v_l − v_k+1, then a_l P_k a_k+1 and agent k receives a_l. This occurs with probability 1−p_l(k+1)and yields a conditional expected utility of v_l. If instead ε_k(k+1) > v_l−v_k+1, then a_k+1 P_k a_l and agent k receives a_k+1. This occurs with probability p_l(k+1) and yields a conditional expected utility of v_k+1+ η_(k+1)l. Thus

Uk σ, SD

al∈ Ok(P−k, SD)
= (1 − p_l(k+1))vl+ p_l(k+1)(v_k+1+ η_(k+1)l).

The stated formula follows by taking an expectation over the realization of the option set.

Lemma 4. Let σ ∈ A^N be an equilibrium under T T C. Then for each k ∈ N ,

U_k(σ, T T C) =

k−1

X

l=1

Q(l, k) [(1 − p_lk)v_l+ p_lk(v_k+ η_kl)]

+

n

X

l=k+1

Q(k, l − 1) − Q(k, l)
[(1 − p_kl)(v_k+ η_kl) + p_klv_l]

+ Q(k, n)(1 − p_k(n+1))(vk+ η_k(n+1)) + p_k(n+1)v_n+1 .

Proof. Let σ ∈ A^N be an equilibrium under T T C, P ∈ P^N be preferences de-termined by σ, and k ∈ N . Without loss of generality, suppose that σ_k = a_k. We interpret the expressions in the formula as expected utilities conditional on

the realization of agent k’s option set. Suppose O_k(P−k, T T C)\{a_k} 6= ∅ and let a_l ∈ O_k(P−k, T T C)\{a_k} be the object with the highest common value.

Case 1: l < k. Then agent k reports preferences with either al or ak at the top of O_k(P_−k, T T C) and receives that object. If ε_kk < v_l− v_k, then a_l P_k a_k and agent k receives a_l. This occurs with probability 1 − p_lk and yields a conditional expected utility of v_l. If instead ε_kk > v_l − v_k, then a_k P_k a_l and agent k receives a_k. This occurs with probability p_lk and yields a conditional expected utility of v_k+ η_kl. Then

Uk σ, T T C

al ∈ Ok(P−k, T T C)
= (1 − plk)vl+ plk(vk+ ηkl).

By Lemma 2, this occurs with probability Q(l, k). Taking expectations yields the first summation in the stated formula.

Case 2: k < l ≤ n. Then agent k reports preferences with either ak or al at the top of O_k(P_−k, T T C) and receives that object. If ε_kk > v_l− v_k, then a_k P_k a_l and agent k receives a_k. This occurs with probability 1 − p_kl and yields a conditional expected utility of v_k + η_kl. If instead ε_kk < v_l − v_k, then a_l P_k a_k and agent k receives a_l. This occurs with probability p_kl and yields a conditional expected utility of v_l. Then

U_k σ, T T C

O_k(P_−k, T T C) ∩ {a₁, . . . , a_l} = {a_k, a_l}
= (1 − pkl)v_l+ p_kl(v_k+ η_kl).

We now compute the probability that O_k(P−k, T T C) ∩ {a₁, . . . , a_l} = {a_k, a_l}.

This requires that {a₁, . . . , a_k−1} ∩ O_k(P−k, T T C) = ∅. By Lemma 2, this occurs with probability Q(k, k) and this depends only on the preferences of agents in N^k.

Now consider i ∈ N with k < i. Then also {a₁, . . . , a_k−1} ∩ O_i(P−i, T T C) = ∅. Since each such agent i reports preferences that agree with P₀ on A\{µ_i}, agent i reports preferences with either a_k or a_i at the top of A\{a₁, . . . , a_k−1}. If ε_ii < v_k− v_i, then ak Pi ai and ai ∈ Ok(P−k, T T C). This occurs with probability 1 − pki. If instead ε_ii> v_k− v_i, then a_i P_i a_k and a_i 6∈ O_k(P_−k, T T C). This occurs with probability p_ki. Conditional on the preferences of agents in N^k, for each pair i, j ∈ N with k < i < j, the events a_i ∈ O_k(P−k, T T C) and a_j ∈ O_k(P−k, T T C) are independent. Combining results,

P r Ok(P−k, T T C) ∩ {a₁, . . . , al} = {ak, al}

= Q(k, k) · p_k(k+1)· p_k(k+2)· · · p_k(l−1)· (1 − p_kl)

= Q(k, l − 1) − Q(k, l).

Taking expectations yields the second summation in the stated formula.

Case 3: n < l. Then l = n + 1. Agent k reports preferences with one of a_k and a_n+1 at the top of Ok(P−k, T T C) and receives that object. If εkk > v_n+1 − vk, then a_k P_k a_n+1 and agent k receives a_k. This occurs with probability 1 − p_k(n+1) and yields a conditional expected utility of v_k+ η_k(n+1). If instead ε_kk< v_n+1− v_k, then a_n+1 P_k a_k and agent k receives a_l. This occurs with probability p_k(n+1) and yields a conditional expected utility of v_n+1. Then

Uk σ, T T C

Ok(P−k, T T C) ∩ {a₁, . . . , al} = {ak, a_n+1}

= (1 − p_k(n+1))(v_k+ η_k(n+1)) + p_k(n+1)v_n+1.

We now compute P r(O_k(P−k, T T C) ∩ {a₁, . . . , a_l} = {a_k, a_l}). If n < m, then Case 3 occurs with the remaining probability after considering the events of Cases 1 and 2,

1 −

k−1

X

h=1

Q(h, k) −

n

X

l=k+1

Q(k, l − 1) − Q(k, l) = Q(k, k) − Q(k, k) − Q(k, n)

= Q(k, n).

This yields final term in the stated formula.

If n = m, then Case 3 does not occur. Instead, still with probability Q(k, n), O_k(P_−k, T T C)\{a_k} = ∅. Then O_k(P_−k, T T C) = {a_k} and agent k receives a_k regardless of his preferences. This yields a conditional expected utility of v_k. Given our conventions η_k(m+1) = 0 and p_k(m+1) = 0, the expression in Case 3 simplifies to v_k and the stated formula applies.

For reference, Corollaries 2 and 3 simplify the utility formulas from Lemmas 3 and 4 for the special case of m = n = 2.

Corollary 2. Let σ^≺∈ A^N be an equilibrium under SD. If n = m = 2, then

U₁(σ^≺, SD) = (1 − p₁₂)v₁+ p₁₂(v₂+ η₂₁) and U₂(σ^≺, SD) = (1 − p₁₂)v₂+ p₁₂v₁.

Corollary 3. Let σ^µ∈ A^N be an equilibrium under T T C. If n = m = 2, then

U₁(σ^µ, T T C) = (1 − p₁₂)(1 − p₁₂)(v₁+ η₁₂) + p₁₂v₂ + p₁₂v₁ and

U₂(σ^µ, T T C) = p₁₂p₁₂(v₂+ η₂₁) + (1 − p₁₂)v₁ + (1 − p₁₂)v₂.

A.4.1 Proof of Proposition 2

We argue by inspecting the formulas derived in Lemmas 3 and 4. Let n, m ∈ N with n ≤ m and let σ^≺, σ^µ∈ A^N be equilibria under SD and T T C respectively. We proceed by induction on the number of agents and objects. Let U^n,m(ϕ) be the sum of the agents’ ex-ante equilibrium utilities under ϕ. Let U₀(k) ≡ Pk

i=1v_i. So U₀(n) is the utilitarian welfare without learning.

Step 1: n = m = 2. By Corollary 2,

U^2,2(SD) = U₁^2,2(σ^≺, SD) + U₂^2,2(σ^≺, SD)

= [(1 − p₁₂)v₁+ p₁₂(v₂+ η₂₁)] + [(1 − p₁₂)v₂+ p₁₂v₁]

= U₀(n) + p₁₂η₂₁.

By Lemma 1(2), (1 − p₁₂)η₁₂ = p₁₂η₂₁. Now by Corollary 3,

U^2,2(T T C) = U₁^2,2(σ^µ, T T C) + U₂^2,2(σ^µ, T T C)

= (1 − p₁₂)(1 − p₁₂)(v₁+ η₁₂) + p₁₂v₂ + p₁₂v₁

+ p₁₂p₁₂(v₂+ η₂₁) + (1 − p₁₂)v₁ + (1 − p₁₂)v₂

= U₀(n) + (1 − p₁₂)(1 − p₁₂)η₁₂+ p₁₂p₁₂η₂₁

= U₀(n) + (1 − p₁₂)p₁₂η₂₁+ p₁₂p₁₂η₂₁

= U₀(n) + p₁₂η₂₁.

Therefore, U^2,2(SD) = U^2,2(T T C).

Step 2: Increasing the number of objects. Suppose U^n,n(SD) = U^n,n(T T C).

Under either rule, objects {a_n+2, a_n+3, . . . , a_m} are never allocated and so all agents are indifferent to their presence. Thus, it suffices to consider m = n + 1. Since our hypothesis applies to all problems with n agents and n objects, we may suppose the new object has the lowest common value among all objects, namely v_n+1.

Under SD, only agent n ever receives a_n+1, and the difference in utilitarian welfare between U^n,n(SD) and U^n,n+1(SD) is the difference in agent n’s ex-ante expected utility. Then by Lemma 3,

Under T T C, the arrival of the new object increases each agent’s ex-ante expected utility. For each k ∈ N , the difference is the last term in the expression for Uk in Lemma 4:

U_k^n,n+1(σ^µ, T T C) − U_k^n,n(σ^µ, T T C)

= Q(k, n)(1 − p_k(n+1))(v_k+ η_k(n+1)) + p_k(n+1)v_n+1 − Q(k, n)vk

= Q(k, n)(1 − p_k(n+1))η_k(n+1)+ p_k(n+1)(v_n+1− v_k).

By Lemma 1(2), (1 − p_k(n+1))η_k(n+1)l = p_k(n+1)η_(n+1)k. Substituting and summing

Step 3: Increasing the number of agents. Suppose U^n−1,n(SD) = U^n−1,n(T T C). Since our hypothesis applies to all problems with n − 1 agents and n objects, we may assume that the new agent has the lowest priority under SD and is endowed with an under T T C.

Under SD, the arrival of agent n has no effect on the allocations or ante ex-pected utilities of the original agents. Therefore, the difference in utilitarian welfare is the ex-ante expected utility of agent n. Also, agent n has no meaningful investiga-tion decision. This is reflected by our conveninvestiga-tions p_l(m+1) = 0 and η_(m+1)l = 0. Then by Lemma 3,

=

n

X

l=1

Q(l, n)v_l.

Under T T C, the arrival of the new agent decreases each original agent’s ex-ante expected utility. The difference arises because a_n may now be unavailable. In terms of Lemma 4, for each k ∈ N , the comparison between a_k and a_n moves from the final term into the second summation. The difference is

U_k^n,n(σ^µ,T T C) − U_k^n−1,n(σ^µ, T T C)

Now agent n’s ex-ante expected utility is

U_n^n,n(σ^µ, T T C) =

n−1

X

l=1

Q(l, n)(1 − pln)v_l+ p_ln(v_n+ η_nl) + Q(n, n)vn.

Combining results, profile under T T C Lorenz dominates the equilibrium utility profile under SD when there are n agents. For each n ∈ N and each k ∈ N with k ≤ n, let Ukⁿ(SD) be the equilibrium utility under SD of the agent with kth highest priority when there are n agents. Similarly, let U_kⁿ(T T C) be the equilibrium utility under T T C of the agent whose endowment has the kth highest priority when there are n agents. Explicit formulas for theses expressions are computed in Lemmas 3 and 4.

By Proposition 2, each rule yields the same total utility. Moreover, the proof of Proposition 2 shows that (i) the utility of a given agent under SD is independent of the number of agents with lower priority; and (ii) the utility of a given agent under T T C is decreasing in the number of agents with lower priority. That is, for each n, k ∈ N with k ≤ m,

Comparing the utilities of the less well off agents,

n

<

We omit the straightforward argument that priority rules are efficient.⁴⁰ There are two cases to consider in showing that TTC rules are not efficient when m ≥ 3.

Case 1: 2 = n and m = 3. We show by example that T T C is not ex-ante efficient.

40See Bade (2015) for a proof that can be translated to our setting.

In document Learning matters: reappraising object allocation rules when agents strategically investigate (Page 51-66)