Penn Institute for Economic Research
Department of Economics
University of Pennsylvania
3718 Locust Walk
Philadelphia, PA 19104-6297
pier@econ.upenn.edu
http://economics.sas.upenn.edu/pier
PIER Working Paper 16-007
Efficient Mechanisms with Information Acquisition
By
Yunan Li
http://ssrn.com/abstract=2757692
Efficient Mechanisms with Information Acquisition
Yunan Li∗
University of Pennsylvania
February 15, 2016
Latest Version
Abstract
This paper studies the design of ex ante efficient mechanisms in situations where
a single item is for sale, and agents have positively interdependent values and can
covertly acquire information at a cost before participating in a mechanism. I find
that when interdependency is low and/or the number of agents is large, the ex post
efficient mechanism is also ex ante efficient. In cases of high interdependency and/or a
small number of agents, ex ante efficient mechanisms discourage agents from acquiring
excessive information by introducing randomization to the ex post efficient allocation
rule in areas where the information’s precision increases most rapidly.
Keywords: Auctions, Mechanism Design, Information Acquisition, Efficiency
JEL Classification Numbers: C70, D44, D82, D86
∗
Author’s addresses: Yunan Li, Economics Department, University of Pennsylvania, Philadelphia, PA,
USA, yunanli@sas.upenn.edu. I am grateful to Rakesh Vohra, Steven Matthews, Mallesh Pai, Aislinn Bohren,
Hanming Fang, Jason Hartline, Sangmok Lee, George Mailath, Andrew Postlewaite and the participants in
seminars at University of Pennsylvania, 3rd summer school of the Econometric Society and 11th World
Congress of the Econometric Society for extremely helpful discussions. All remaining errors are my responsibility.
1
1
Introduction
Most literature on mechanism design assumes that the amount of information possessed
by agents is exogenous. In many important settings, however, this assumption does not apply.
For example, in auctions for offshore oil and gas leases in the U.S., companies use seismic
surveys to collect information about the tracts offered for sale before participating in the
auctions. Another example is the sale of financial or business assets, in which buyers perform
due diligence to investigate the quality and compatibility of the assets before submitting
offers. In other words, in these settings the information held by agents is endogenous.
Moreover, it is costly to acquire information. In the example of U.S. auctions for offshore oil
and gas leases (see Haile et al. (2010)), companies can choose to conduct two-dimensional
(2-D) or three dimensional (3-D) seismic surveys. 3-D surveys can produce more accurate
information, and thus were used in 80% of wells drilled in the Gulf of Mexico by 1996.
However, this number was only 5% in 1989 when 3-D surveys were more expensive than 2-D
surveys.1 Similarly, the legal and accounting costs of performing due diligence often amount
to millions of dollars in the sale of a business asset (see Quint and Hendricks (2013) and
Bergemann et al. (2009)).
Furthermore, the incentives for agents to acquire information depend on the design of
a mechanism. This can be seen from earlier studies that compare first price auctions with
second price auctions in terms of the incentives they provide for agents to collect information
in advance (among them Matthews (1984a), Stegeman (1996) and Persico (2000)). More recently, Bergemann and Välimäki (2002) consider the socially optimal information acquisition
in the context of general mechanism design. They focus on mechanisms that implement the
ex post efficient allocations given acquired private information, and find that ex ante efficient
information acquisition can be achieved if agents have independent private values. However,
if agents’ values are interdependent, then ex post efficient mechanisms will result in socially
1
For instance, it costs $1 million to examine a 50 square mile 3-D seismic survey in 1990, while this
number was less than $100, 000 in 2000.
2
sub-optimal information acquisition. In a follow-up paper, Bergemann et al. (2009) study
the equilibrium level of information acquisition when agents face binary information decisions, and find that ex post efficient mechanisms result in excessive information acquisition
in equilibrium. In summary, there is a conflict between the provision of ex ante efficient
incentives to acquire information and the ex post efficient use of information. The question
regarding how to design an ex ante efficient mechanism to balance the two trade-offs remains
open.
This paper studies the design of ex ante efficient mechanisms in the sale of a single object
when agents’ values are positively interdependent. The true value of the object to each agent
is initially unknown. Before participating in a mechanism agents can simultaneously and
independently decide how much information to acquire. Information acquisition is costly
and the information choice of each agent is his private information. In this paper, I assume
that the information structures are supermodular ordered. Ganuza and Penalva (2010) first
introduce the notion of “supermodular precision”, and Shi (2012) later uses it when studying
revenue-maximizing mechanisms with endogenous information.
In the main body of the paper, I focus on symmetric mechanisms and symmetric equilibria
in which agents acquire the same amount of information before participating in a mechanism.
Firstly, I show that the social planner never withholds the item in an ex ante efficient
mechanism. Intuitively, whenever the social planner withholds the object, she can also
allocate it randomly. By doing so, the social surplus from the allocation increases while an
agent’s ex ante incentive to acquire information remains unaffected. Though intuitive, the
proof of the result is non-trivial because of the presence of the non-standard information
acquisition constraint. In addition, this result facilitates the analysis by allowing me to work
with the interim allocation rule directly.
Secondly, I show that it is socially optimal for agents to acquire no more information
than when the ex post efficient mechanism is used. This is consistent with Bergemann and
Välimäki (2002) and Bergemann et al. (2009). For any given information choice satisfying the
3
above condition, I fully characterize mechanisms that implement this choice and maximize
ex ante social surplus. An ex ante efficient mechanism discourages agents from excessive
information acquisition by sometimes randomly allocating the item. Specifically, an ex ante
efficient interim allocation rule randomizes in areas in which the accuracy of an agent’s
posterior estimate can be significantly improved if an additional piece of information is
acquired. In special cases in which the improvements are the same for all possible posterior
estimates, i.e., the information structures are uniformly supermodular ordered, it is optimal
to allocate the object uniformly at random with some probability.
To illustrate this result, consider a simple example where an agent’s type can take four
possible values: v1L < v1H < v2L < v2H . Initially, each agent only knows whether v ∈
{v1L , v1H } or v ∈ {v2L , v2H }. Agents have access to the following information technology.
By incurring a cost c > 0, an agent with type v1H (or v1L ) can observe a high (or low) signal
with probability 1/2 + p and a low (or high) signal with probability 1/2 − p, and an agent
with type v2H (or v2L ) can observe a high (or low) signal with probability 1/2 + εp and a
low (or high) signal with probability 1/2 − εp, where p ∈ (0, 1/2) and ε ∈ [0, 1]. If ε = 1,
then the signal is uniformly informative over the entire support. If ε < 1, then agents receive
more precise information over {v1L , v1H } than {v2L , v2H }. Intuitively, allocating the object
randomly over {v1L , v1H } can more effectively lower the value of information to agents as
well as their incentives to collect information. This is demonstrated in the extreme case of
ε = 0 in which randomization over {v2L , v2H } has no impact on the value of information.
This intuition carries over to cases in which agents have a continuum of types and access to
a continuum of information choices.
Methodologically, when characterizing optimal mechanisms, I use an approach first proposed by Reid (1968) and later introduced into the mechanism design problem by Mierendorff
(2009). My proof, however, is not a straightforward modification of Mierendorff (2009). In
Mierendorff (2009), the interim allocation rule is discontinuous at one known point. In my
model, the interim allocation rule could be discontinuous at most countably many unknown
4
points.
I also study how the socially optimal information choice is affected by model primitives
such as the interdependency of agents’ values and the number of agents. In general, it is
difficult to solve the optimal information choice analytically. Hence, I restrict attention to
the special case in which the information structures are uniformly supermodular ordered.
Under this assumption, I show that the optimal level of information gathering decreases
as the interdependency of agents’ values increases, and gathering no information is optimal
in the case of pure common value. Furthermore, when the ex post efficient mechanism is
used, the amount of information acquired by each agent diminishes as the number of agents
increases to infinity. As a result, the ex post efficient mechanism is also ex ante efficient for
a large number of agents.
Lastly, I study ex ante efficient mechanisms without imposing symmetry restrictions. As
in the symmetric case, the social planner never withholds the item in an ex ante efficient
mechanism. Since characterizing optimal mechanisms in the general setting is extremely
hard, I restrict attention to the special case in which the information structures are uniformly
supermodular ordered. Under this assumption, I provide conditions under which the socially
optimal information choices are the same for all agents and there exists a symmetric ex ante
efficient mechanism.
This paper is related to the literature of mechanism design with endogenous information. Earlier papers focus on the comparison between first- and second-price auctions. For
example, Matthews (1984a) considers a first-price auction with pure common values, and examines how an increase in the number of agents affects the information acquisition. Stegeman
(1996) finds that both auctions lead to identical and, more importantly, efficient incentives
for information acquisition when agents’ values are private and independent. In contrast,
Persico (2000) finds that first-price auctions provide stronger incentive for agents to acquire
information than second-price auctions do when their values are affiliated. The two most
closely related papers are Bergemann and Välimäki (2002) and Bergemann et al. (2009).
5
Both study the efficiency of information acquisition by agents when the ex post efficient
mechanism is used. Instead, I focus on the design of ex ante efficient mechanisms.
This paper is also related to papers that study the revenue-maximizing mechanisms
with endogenous information acquisition. The most closely related paper is Shi (2012) who
considers the sale of a single asset when buyers have independent private values and who,
before the auction, can simultaneously and independently decide how much information to
acquire. He finds that the optimal reserve price is always below the standard monopoly price
to encourage information acquisition. Several other papers consider the case where the seller
can control the timing of information acquisition (see, for example, Levin and Smith (1994),
Ye (2004), and Crémer et al. (2009)).
The rest of the paper is organized as follows. Section 2 presents the model. Section 3
contains the main results. More specifically, Section 3.2 characterizes optimal symmetric
mechanisms for each given information choice and Section 3.3 studies the socially optimal
information choice. Section 4 examines ex ante efficient mechanisms without imposing symmetry restrictions. Section 5 concludes. All omitted proofs are relegated to appendix.
2
Model
There are n agents, indexed by i ∈ {1, · · · , n}, who compete for a single object. Each
agent i has a payoff-relevant type θi , which is unknown ex ante. The true value of the object
to agent i is
vi (θ) ≡ θi + γ
X
θj ,
j6=i
where θ ≡ (θ1 , . . . , θn ) and γ ∈ [0, 1] is a measure of interdependence.2 Each agent i has a
quasi-linear utility. His payoff is qi vi (θ) − ti if he gets the object with probability qi ∈ [0, 1]
2
Under this specification, the range of possible valuations for agent i depends on the number of agents.
An alternative normalized specification is given by
γ X
vi (θ) ≡ (1 − γ)θi +
θj .
n−1
j6=i
6
and pays ti ∈ R.
Initially, each agent i only knows that θi ’s are independently distributed with a common
prior distribution F and support [θ, θ] ⊂ R+ . F has a positive and continuous density
f . Each agent i can covertly acquire a costly signal xi about his type θi by selecting a
joint distribution of (xi , θi ) from a family of joint distributions G(xi , θi |αi ), indexed by αi ∈
A ≡ [α, α] ⊂ R. For each αi ∈ A, we also refer the joint distribution G(·, ·|αi ) as an
information structure. Let g denote the density function associated with G. For all αi ∈ A,
R
G(·, ·|αi ) admits the same marginal distribution of θi , i.e., R g(xi , θi |αi )dxi = f (θi ) for all
θi ∈ [θ, θ]. Throughout the paper, I assume that E[θ|x, α] is strictly increasing in x for all
α ∈ A, i.e., a higher signal results in a higher conditional expectation.3 A signal with a
higher αi is more precise, in a sense which I formally define below. Let C(αi ) denote the
cost of acquiring a signal with precision αi . As is standard in the literature, I assume that
C is non-negative, strictly increasing, twice continuously differentiable and strictly convex.
Furthermore, C(α) = C 0 (α) = 0.
2.1
Information Order
Let G(x|α) ≡
R
G(x|θ, α)dF (θ) denote the marginal distribution of x given α. I define
a new signal by applying the probability integral transformation on the original signal: s ≡
G(x|α). The transformed signal is uniformly distributed on [0, 1].4 Clearly, the transformed
signal has the same informational content as the original signal. Furthermore, since any
two transformed signals have the same marginal distribution, their realizations are directly
comparable. Therefore, I will henceforth work with the transformed signal si directly. Let
With this specification, the range of possible valuations for agent i does not depend on the number of agents,
but now the single-crossing condition does. Most of my results hold for either specifications.
3
For each α ∈ A, let G(θ|x, α) denote the conditional distribution of θ given x. Then one sufficient
condition for this is to assume that G(θ|x, α) have the monotone likelihood ratio property.
4
s is uniform on [0, 1] only if G(x) is continuous and strictly increasing. This can be assumed without
loss of generality. If G has a discontinuity at z, where P(x̃ = z) = p, x can be transformed into x∗ , which
has a continuous and strictly increasing distribution function using the following construction proposed in
Lehmann (1988): x∗ = x for x < z, x∗ = x + pU if x = z, where U is uniform on (0, 1), and x∗ = x + p for
x > z.
7
w(s, α) ≡ E[θ|s, α] be the conditional expectation of θ given s and α. Then w(s, α) is strictly
increasing in s. Let H(w|α) ≡ P(w(s, α) ≤ w) denote the distribution of w(s, α) with density
h(w|α). I assume that H(w|α) and h(w|α) are twice continuously differentiable. Throughout
the paper, I impose the following assumption on information structures:
Assumption 1 The information structures are supermodular ordered, i.e.,
−
Hα (w|α)
is strictly increasing in w on [w(0, α), w(1, α)] for all α.
h(w|α)
To better understand Assumption 1, note that wα (s, α) = −Hα (w(s, α)|α)/h(w(s, α)|α)
which is strictly increasing in s. Thus, w(s, α) is supermodular in (s, α). Furthermore, I
show in the appendix that Assumption 1 implies the following property of w(s, α):
Lemma 1 Suppose the information structures are supermodular. Then for all s, s0 ∈ (0, 1)
and s > s0 ,
w(s, α) − w(s0 , η) > w(s0 , α) − w(s, η).
(1)
Intuitively, if s contains little information about θ, then w(s, α) does not vary much as s
changes and its distribution concentrates around E[θ]. As s becomes more informative about
θ, w(s, α) varies more as s changes and its distribution becomes more dispersed. Formally,
if α > α0 then w(s, α) is strictly larger than w(s, α0 ) in the dispersive order.5 Ganuza and
Penalva (2010) first develop an information order called “supermodular precision” based on
this notion of dispersion. Shi (2012) also imposes this assumption for some of his results.
For some results of the paper, I strengthen Assumption 1 to that the information structures are uniformly supermodular ordered. The meaning of “uniformly” will become clear
through the characterization of optimal mechanisms in Section 3.2.
5
(See Ganuza and Penalva (2010)) Let Y and Z be two real-valued random variables with distributions
F and G, respectively. We say Y is greater than Z in the dispersive order if for all q, p ∈ (0, 1) and q > p,
F −1 (q) − F −1 (p) ≥ G−1 (q) − G−1 (p).
8
Assumption 2 The information structures are uniformly supermodular ordered, i.e., for
all α and all w ∈ [w(0, α), w(1, α)] there exists a positive function b : A → R++ such that
−
w−µ
Hα (w|α)
=
.
h(w|α)
b(α)
Uniformly supermodular-ordered information structures include the following two commonly used information technologies in the literature:6
Example 1 (The Linear Experiments) Let θi ’s be independently distributed with a common prior F and mean µ. Agent i can obtain a costly signal xi about θi at a cost. Consider
the following information structure, which is called “truth-or-noise” in Lewis and Sappington
(1994), Johnson and Myatt (2006) and Shi (2012). With probability αi ∈ [0, 1], xi is equal
to his true type θi , and with probability 1 − αi , xi is an independent draw from F . Since the
marginal distribution of xi is F , the transformed signal is si ≡ F (xi ). Then the posterior
estimate of an agent who chooses αi and receives si is w(si , αi ) = αi F −1 (si ) + (1 − αi )µ.
Therefore we have
−
wi − µ
Hαi (wi |αi )
=
.
h(wi |αi )
αi
It is clear that the information structures are uniformly supermodular ordered.
Example 2 (Normal Experiments) Let θi ’s be independently distributed with a normal
iid
iid
distribution: θi ∼ N (µ, 1/β). Agent i can observe a costly signal xi = θi + εi , where εi ∼
N (0, 1/αi ). Since the marginal distribution of xi is also normal, i.e., xi ∼ N (µ, (β+αi )/βαi ),
√
√
the transformed signal is si ≡ Φ βαi (xi − µ)/ β + αi . Then the posterior estimate of
an agent who chooses αi and receives si is
√
αi Φ−1 (si )
w(si , αi ) = µ + p
.
β(α + βi )
6
See, for example, Ganuza and Penalva (2010) and Shi (2012).
9
Therefore we have
−
Hαi (wi |αi )
β(wi − µ)
=
.
h(wi |αi )
2αi (αi + β)
It is clear that the information structures are uniformly supermodular ordered.
2.2
Timing
The game proceeds in the following way: The social planner announces a mechanism.
After observing the mechanism, each agent i simultaneously chooses an information structure
αi and observes the realized signal si . Both αi and si are agent i’s private information. Then
each agent decides whether to participate in the mechanism. Each participating agent reports
his private information. Based on their reports, an allocation and payments are implemented
according to the announced mechanism.
I assume that the payoff structure, the timing of the game, the information structures
and the prior distribution F are common knowledge. The solution concept is Bayesian Nash
equilibrium.
2.3
Mechanisms
The private information of agent i is two-dimensional, including the choice of information
structure αi and the realized signal si . However, similar to Biais et al. (2000), Szalay
(2009) and Shi (2012), the usual difficulties inherent in multi-dimensional mechanism design
problem do not arise here, since the posterior estimate, w(si , αi ), summarizes all the private
information needed to compute agent i’s valuation of the object:
Eθ [vi (θ)|αi , si ] = w(si , αi ) + γ
X
E[θj ].
j6=i
Furthermore, the seller cannot screen agents with the same posterior estimate but different
choices of information structures. Hence, one can appeal to the Revelation Principle and
10
focus on direct mechanisms in which agents report their posterior estimates directly. For
ease of notation, I use wi to denote wi (si , αi ) and w ≡ (w1 , . . . , wn ). A direct mechanism
is a pair (q, t), where, given the reported vector of posterior estimates w, qi (w) denotes
the probability agent i is allocated the object, and ti (w) denotes agent i’s payment. Define
agent i’s interim allocation rule as
Qi (wi ) ≡ Ew−i [qi (wi , w−i )|α−i ],
(2)
where α−i are his opponents’ information structures. Then the interim utility of agent i who
has a posterior estimate wi and reports wi0 is
"
Ui (wi , wi0 ) ≡ wi Qi (wi0 ) + Ew−i γ
X
j6=i
#
θj qi (wi0 , w−i ) − ti (wi0 , w−i ) α−i .
Note that Qi (wi ) and Ui (wi , wi0 ) also depend on α−i , and I suppress the dependence for ease
of notation.
I now describe the constraints that the social planner’s mechanism must satisfy. First,
the mechanism must satisfy the individual rationality constraint (IR), i.e.,
Ui (wi ) ≡ Ui (wi , wi ) ≥ 0, ∀wi ∈ [θ, θ],
so that the agents are willing to participate in the mechanism. Second the mechanism must
satisfy the Bayesian incentive compatibility constraint (IC), i.e.,
Ui (wi ) ≥ Ui (wi , wi ), ∀wi , wi0 ∈ [θ, θ],
so that truth-telling is a Bayesian Nash equilibrium. By the standard argument (see, for
example, Myerson (1981)), the (IC) constraint holds if and only if
Qi (wi ) is non-decreasing in wi ,
11
(MON)
and Ui (wi ) is absolutely continuous and satisfies the following envelope condition
wi
Z
Ui (wi ) = Ui (θ) +
Qi (w̃i )dw̃i .
(3)
θ
Since the social planner’s goal is to maximize social surplus and transfers between agents
and the social planner do not affect social surplus, we can guarantee that the (IR) constraint
is satisfied by making sufficiently large lump sum transfers to agents. Hence, we can safely
ignore the (IR) constraint for the remainder of the paper.
Lastly, since the information structure chosen by an agent is unobservable, the mechanism
must also satisfy the information acquisition constraint (IA): no agent stands to gain by
deviating from the equilibrium choice αi∗ , i.e.,
!
"
αi∗ ∈ argmax Ew qi (w) wi + γ
αi
X
E[θj ]
j6=i
#
− ti (w) αi , αj = αj∗ ∀j 6= i − C(αi ).
Using the envelope condition (3) and the fact that the support of wi is [w(0, αi ), w(1, αi )],
agent i’s payoff from choosing αi can be rewritten as
"
!
Ew qi (w) wi + γ
X
E[θj ]
j6=i
Z
#
∗
− ti (w) αi , αj = αj ∀j 6= i − C(αi )
w(1,αi )
[1 − H(wi |αi )] Qi (wi )dwi − C(αi ).
= Ui (w(0, αi )) +
w(0,αi )
The social planner chooses mechanism (q, t) and a vector of recommendations of information structures α∗ to maximize social surplus,
"
max E
α∗ ,(q,t)
X
i
(wi + γ
X
j6=i
#
X
wj )qi (w) αi = αi∗ ∀i −
C(αi∗ ),
i
12
subject to the (MON), (IA) and the feasibility constraint (F), i.e.,
0 ≤ qi (w) ≤ 1,
X
qi (w) ≤ 1, ∀w.
(F)
i
We say a mechanism (q, t) is ex ante efficient or optimal if there exists α∗ such that α∗ and
(q, t) solve the social planner’s problem.
3
Efficient Mechanisms
In this section, I restrict attention to mechanisms that treat all agents symmetrically7 as
well as symmetric equilibria in which all agents acquire the same information structure, i.e.,
αi∗ = α∗ for all i. The symmetric assumption significantly simplifies the analysis. However,
this assumption may result in a loss of generality. Section 4 presents a study of ex ante
efficient mechanisms without imposing symmetry restrictions, and provides conditions under
which the socially optimal information choices are the same for all agents and there exists a
symmetric ex ante efficient mechanism. Note that when q is symmetric, the corresponding
interim allocation rule Qi is independent of i. From hereon, I drop the subscript i from Q,
w and α whenever the meaning is clear.
I start solving the social planner’s problem by replacing the (IA) constraint by a one-sided
first-order necessary condition. In an earlier paper, Bergemann and Välimäki (2002) show
that if the social planner adopts the ex post efficient mechanism, then agents tend to acquire
more information than the socially desired level. This suggests that an ex ante efficient
mechanism would sacrifice some allocation efficiency to discourage agents from acquiring
information. Thus, I hypothesize that to ensure the (IA) constraint holds in an ex ante
efficient mechanism, it suffices to ensure that no agent has the incentive to choose αi > α∗ :
7
The formal definition of symmetric mechanisms can be found in the appendix.
13
for all αi > αi∗ ,
Ui (w(0, αi∗ ))
Z
w(1,α∗i )
+
w(0,α∗i )
Z
[1 − H(wi |αi∗ )]Qi (wi )dwi − C(αi∗ )
w(1,αi )
[1 − H(wi |αi )]Qi (wi )dwi − C(αi ).
≥ Ui (w(0, αi )) +
w(0,αi )
This implies the following one-sided first-order condition:
Hαi (wi |αi∗ )
∗
E −
Qi (wi ) αi = αi ≤ C 0 (αi∗ ).
h(wi |αi∗ )
(IA0 )
The left-hand side of the above inequality is agent i’s marginal benefit from acquiring αi∗ ,
and the right-hand side is the marginal cost. In Lemma 7 in the appendix, I show that
for any non-decreasing Qi , an agent’s marginal benefit from acquiring information is nonnegative. Subsequently, I consider the relaxed problem of the social planner by replacing
the (IA) constraint by the (IA0 ) constraint. I later show that (IA0 ) holds with equality when
α∗ is chosen optimally. The first-order approach is valid if the second-order condition of the
agents’ optimization problem is satisfied. Appendix A.3 provides sufficient conditions that
ensure the first-order approach is valid.
The (IA0 ) constraint is easier to work with than the original (IA) constraint. However,
it is still nonstandard and prevents me from solving the social planner’s problem directly as
in Myerson (1981). To overcome this difficulty, I focus on reduced form auctions. Formally,
Definition 1 We say that q implements Q : [θ, θ] → [0, 1] and Q is the reduced form of q
if q satisfies (2) and (F) for all w ∈ [θ, θ]. Q is implementable if q exists implementing Q.
One important prior result I use in this paper is the necessary and sufficient condition of
Maskin and Riley (1984) and Matthews (1984b) which characterizes the interim allocation
rules implementable by symmetric mechanisms. By Theorem 1 in Matthews (1984b), any
14
non-decreasing function Q : [θ, θ] → [0, 1] is implementable if and only if it satisfies
Z
θ
Y (w) ≡
H(z|α∗ )n−1 − Q(z) h(z|α∗ )dz ≥ 0, ∀w ∈ [θ, θ].
(F0 )
w
Thus, given (MON), we can replace (F) by (F0 ). Note that the support of w is W ≡
[w(0, α∗ ), w(1, α∗ )] ⊂ [θ, θ]. Therefore, (F0 ) imposes no restriction on Q in [θ, θ]\W .
Now all the constraints are expressed as functions of the interim allocation rule Q, while
the objective function, i.e., the expected social surplus, is not. The expected social surplus is
the summation of the payoffs of all agents. In the case of independent private value (γ = 0),
agent i’s valuation of the object and his winning probability are independent conditional
on his private information. Thus, each agent’s expected payoff is completely determined by
Q. This is not true in general when γ > 0, as other agents’ private information affects his
winning probability and his value. However, this is true for a socially optimal mechanism,
which never withholds the object:
Theorem 1 Suppose α∗ and (q, t) solve the relaxed problem of the social planner, then
X
qi (w) = 1 for almost all w ∈ W n .
(4)
i
Using Theorem 1 and Law of iterated expectations, we can rewrite the social planner’s
objective function as a function of Q:
"
E
X
i
(wi + γ
X
j6=i
#
X
wj )qi (w) αi = α∗ ∀i =
E [(1 − γ)wi Q(wi )| αi = α∗ ] + nγE[θ].
i
Since the second term, nγE[θ], is a constant, we ignore it from hereon. To summarize, the
social planner’s relaxed problem, denoted by (P 0 ), can be rewritten as:
max
E [(1 − γ)wQ(w)| α∗ ] − C(α∗ ),
∗
α ,Q
15
subject to
Z
θ
Y (w) =
[H(z|α∗ )n−1 − Q(z)]h(z|α∗ )dz ≥ 0, ∀w ∈ [θ, θ].
(F0 )
Q(w) is non-decreasing in w,
∗
Hα (w|α∗ )
α ≤ C 0 (α∗ ).
Q(w)
E −
h(w|α∗ )
(MON)
w
(IA0 )
In addition to being instrumental in solving the social planner’s problem, the result of
Theorem 1 has some inherent economic interest. Obviously, the efficient mechanism never
withholds the object when information is exogenous. This is not obvious when information
is endogenous, since, by withholding the object occasionally, the social planner can discourage agents from information acquisition, which may improve efficiency ex ante. However,
intuitively, whenever the social planner withholds the object, she can also allocate it randomly. By doing so, the social surplus from the allocation increases while an agent’s ex ante
incentive to acquire information remains unaffected.
Though intuitive, the proof of Theorem 1 is non-trivial. This is because the resulting
mechanism, by simply randomizing the object whenever it is withheld, is likely to violate the
P
(IC) and/or (IA0 ) constraints. To illustrate this, let A be a set of types such that i qi (w) < 1
whenever w ∈ An , and suppose there exist (w, w) ∩ A = ∅ such that inf A < w < w < sup A.
P
If we simply redefine q such that it remains unchanged outside An and i qi (w) = 1
whenever w ∈ An , then the resulting Q remains unchanged for w ∈ (w, w) but increases
for all w ∈ A. If we allocate the object too often to agents whose types are in [θ, w] ∩ A,
the resulting Q will no longer be non-decreasing and therefore violate the (IC) constraint.
If we allocate the object too often to agents whose types are in [w, θ] ∩ A, this will increase
an agent’s marginal benefit from acquiring information and lead to a violation of the (IA0 )
constraint. Thus to ensure the new q generating a higher social surplus while satisfying all
the constraints, one needs to adjust q not only for w inside An , but also for w outside An .
Section 3.1 contains the proof of this result.
16
3.1
Proof of Theorem 1
I prove Theorem 1 by proving the following two lemmas. Observe first that if αi = α∗
for all i, then (4) is violated if and only if Y (w(0, α∗ )) > 0. Then
Lemma 2 Suppose αi = α∗ for all i. Let Q be any interim allocation rule satisfying (F0 ),
(MON), (IA0 ) and Y (w(0, α∗ )) > 0, then there exists Q̂ satisfying (F0 ), (MON) and (IA0 )
such that
Q̂(w) ≥ Q(w), ∀w ∈ W,
(5)
and ” > ” holds for a set of w with positive measure.
The intuition behind the proof of Lemma 2 can be illustrated by Figure 1. If Q satisfies
the assumptions of Lemma 2, then one can construct a Q̂ by increasing Q at the lower end
of its domain as in Figure 1. Clearly, the resulting Q̂ is non-decreasing and, furthermore,
implementable if the change is sufficiently small. It remains to verify that Q̂ also satisfies
the (IA0 ) constraint. Intuitively, agents have fewer incentives to acquire information if the
outcome is less sensitive to a change in their private information, which is the case when
they face a less steep allocation rule. I show in Lemma 7 that if Q̂ is less steep than Q in
the sense that it differs from Q by a non-increasing function (as in Figure 1), then for any
α, Q̂ gives agents a smaller marginal benefit of acquiring information. Hence, Q̂ satisfies the
(IA0 ) constraint as Q.
The gap between Lemma 2 and Theorem 1 is that social surplus is not just a function
of Q when γ > 0. We need to show that, for any ex-post allocation rule q implementing Q,
we can find a q̂ that implements Q̂ and yields higher social surplus. This is the result of
Lemma 3.
Lemma 3 Suppose two implementable allocation rules Q and Q̂ satisfy (5). Let q be an
ex-post allocation rule that implements Q, then there exists an ex-post allocation rule q̂ that
17
Figure 1: Lemma 2
Q
Q̂
wi
0
implements Q̂ and satisfies
"
E
X
i
(wi + γ
X
j6=i
#
"
#
X
X
wj )q̂i (w) αi = α∗ ∀i > E
wj )qi (w) αi = α∗ ∀i .
(wi + γ
i
j6=i
The proof of Lemma 3 relies on the following technical lemma. Let m denote the probability measure on W corresponding to H(wi |α∗ ), then
Lemma 4 Let Q : W → [0, 1] be implementable and ρ : W n → [0, 1] be a symmetric
measurable function, then there exists a symmetric ex post allocation rule q implementing
P
Q such that i qi (w) ≥ ρ(w) for almost all w ∈ W n if and only if for each measurable set
A ⊂ W , the following inequality is satisfied
Z
1
Q(wi )dm(wi ) ≥
n
A
Z
ρ(w)dmn (w).
(6)
An
It is clear that inequality (6) is necessary. Suppose there exists an allocation rule q
P
such that i qi (w) ≥ ρ(w) for almost all w ∈ W n implementing Q. For any measurable set
R
A ⊂ W , the probability of an agent whose type is in A getting the object, n A Q(wi )dm(wi ),
must exceed the probability of some agent getting the object when all agents’ types are in
R P
R
A, An i qi (w)dmn (w), which by assumption is greater than An ρ(w)dmn (w). A similar
argument to that in the proof of Proposition 3.1 in Border (1991) proves that inequality (6)
18
is also sufficient for such a q to exist and implement Q. With Lemma 4 in hand, it is easy
to prove Lemma 3.
Proof of Lemma 3. Consider two implementable allocation rules Q and Q̂ satisfying (5).
Let q be a symmetric ex-post allocation rule that implements Q. Define ρ : W n → [0, 1] by
P
ρ(w) ≡ i qi (w). Then
Z
Z
1
Q̂(wi )dm(wi ) ≥
Q(wi )dm(wi ) ≥
n
A
A
Z
ρ(w)dmn (w).
An
By Lemma 4 there exists an allocation rule q̂ implementing Q̂ such that
P
n
i qi (w) for almost all w ∈ W . Hence,
P
q̂i (w) ≥ ρ(w) =
#
E
wj )q̂i (w) αi = α∗ ∀i
(wi + γ
i
j6=i
"
!
!
#
i
X h
X
X
=
E (1 − γ)wi Q̂(wi ) αi = α∗ + E
γ
wi
q̂i (w) αi = α∗ ∀i
i
i
i
!
"
!
#
X
X
X
>
E [(1 − γ)wi Q(wi )| αi = α∗ ] + E
γ
wi
qi (w) αi = α∗ ∀i
i
i
"
# i
X
X
=E
wj )qi (w) αi = α∗ ∀i .
(wi + γ
"
X
i
X
j6=i
This completes the proof.
3.2
Efficient Mechanisms for Fixed α∗
I solve (P 0 ) in two steps. In this section, I solve the following sub-problem for each
α∗ ∈ A, denoted by (P 0 -α∗ ):
V (α∗ ) ≡ max E [ wQ(w)| α∗ ] subject to (F0 ), (MON) and (IA0 ).
Q
In Section 3.3, I will solve α∗ ∈ argmaxα∈A (1 − γ)V (α) − C(α).
19
Fix α∗ . If α∗ is such that
Hα (w|α∗ )
0
∗
∗ n−1 ∗
E −
H(w|α
)
α ≤ C (α ),
h(w|α∗ )
(7)
then clearly the ex post efficient mechanism, i.e., Q(w) = H(w|α∗ )n−1 for all w, solves
(P 0 -α∗ ). Thus, in the remainder of this section, I assume α∗ is such that (7) is violated.
I first present an informal argument to derive the optimal solution to (P 0 -α∗ ). If we ignore
the monotonicity constraint (MON), then we can use the following Lagrangian relaxation to
get an intuition of the optimal solution:
L ≡
Z
w(1,α∗ )
ϕλX (H(w|α∗ ))Q(w)h(w|α∗ )dw + λX C 0 (α∗ ),
(8)
w(0,α∗ )
where λX > 0 is the Lagrangian multiplier associated with (IA0 ) and ϕλX (t, α∗ ) is defined by
ϕλX (t, α∗ ) ≡ H −1 (t|α∗ ) + λX
Hα (H −1 (t|α∗ )|α∗ )
, t ∈ [0, 1].
h(H −1 (t|α∗ )|α∗ )
Note that since H(·|α) is strictly increasing, ϕλX (·, α∗ ) is strictly increasing (or decreasing)
if and only if ϕλX (H(·|α∗ ), α∗ ) is strictly increasing (or decreasing). If ϕλX (H(·|α∗ ), α∗ ) is
strictly increasing, then the allocation rule Q that maximizes L is non-decreasing and, more
importantly, ex post efficient: Q(w) = H(w|α∗ )n−1 for all w. However, this contradicts the
assumption that condition (7) is violated. Hence, the Lagrangian multiplier λX must be
such that ϕλX (H(·|α∗ ), α∗ ) is not strictly increasing.
Suppose ϕλX (H(·|α∗ ), α∗ ) is non-decreasing but not strictly increasing. Let Q|(w,w) denote
the restriction of Q to an interval (w, w). Intuitively, if ϕλX (H(·|α∗ ), α∗ ) is strictly increasing
on an interval (w, w), then it is optimal to choose Q|(w,w) = H(·|α∗ )n−1 so that the allocation
rule is ex post efficient on (w, w); if ϕλX (H(·|α∗ ), α∗ ) is constant on an maximally chosen
interval (w, w), then any non-decreasing function Q|(w,w) that satisfies H(w|α∗ )n−1 ≤ Q(w) ≤
Rw
H(w|α∗ )n−1 for all w ∈ (w, w) and w [H(w|α∗ )n−1 − Q(w)]h(w|α∗ )dw = 0 is optimal. More
20
formally, any non-decreasing implementable Q maximizes L subject to (F0 ) and (MON) if
and only if Y (w(0, α∗ )) = 0 and Y (·) > 0 on (w, w) implies ϕλX (H(·|α∗ ), α∗ ) is constant
on this interval. Suppose, in addition, that Q satisfies (IA0 ) with equality, then Q solves
(P 0 -α∗ ). This can be further illustrated by the following example.
Example 3 (The Linear Experiments) Consider the information structures in Example
1. Recall that in this example we have
−
wi − µ
Hαi (wi |αi )
=
.
h(wi |αi )
αi
Hence,
ϕλX (H(w|α∗ ), α∗ ) = w − λX
w−µ
.
α∗
Suppose α∗ is such that condition (7) is violated. Then it must be that λX ≥ α∗ since
otherwise ϕλX (·, α∗ ) is strictly increasing. If λX > α∗ , then ϕλX (·, α∗ ) is strictly decreasing. Hence, the Q that maximizes L and satisfies the monotonicity constraint is constant.
However, in this case (IA0 ) constraint does not bind, a contradiction to λX > 0. Hence, in
optimum λX = α∗ and ϕλX (·, α∗ ) = µ, which is constant. Hence, any feasible monotonic
allocation rule Q satisfying condition (4) maximizes L . If Q also satisfies (IA0 ), then it
solves (P 0 -α∗ ). In particular, there exists ξ ∈ [0, 1] such that
Q̂(w) = ξH(w|α∗ )n−1 + (1 − ξ)
1
n
solves (P 0 -α∗ ).
More generally, the arguments in Example 3 can be readily extended to prove Proposition
1.
Proposition 1 Suppose the first-order approach is valid, and Assumption 2 holds. Suppose,
in addition, that α∗ is such that (7) is violated. Let λX = b(α∗ ) and Q be a non-decreasing
21
implementable allocation rule. Then Q solves (P 0 -α∗ ) if and only if Y (w(0, α∗ )) = 0 and Q
satisfies (IA0 ) with equality.
If Assumption 2 does not apply, then ϕλX (·, α∗ ) is not monotonic. In this case, an optimal
solution can be obtained by ironing ϕλX (·, α∗ ) in the following procedure first introduced by
Myerson (1981). For each t ∈ [0, 1], define
J
λX
∗
Z
t
(t, α ) ≡
ϕλX (τ, α∗ )dτ.
0
Let J
λX
J
denote the convex hull of J λX , defined by
λX
Since J
(t, α∗ ) ≡ min {βJ(t1 , α∗ ) + (1 − β)J(t2 , α∗ )|t1 , t2 ∈ [0, 1], βt1 + (1 − β)t2 = t} .
λX
(·, α∗ ) is convex, it is continuously differentiable virtually everywhere. Define
ϕλX (·, α∗ ) as follows. For each t ∈ (0, 1) such that ∂J
∂J
λX
λX
(t, α∗ )/∂t exists, let ϕλX (t, α∗ ) ≡
(t, α∗ )/∂t. Then extend ϕλX (·, α∗ ) to [0, 1] by right continuity. Then ϕλX (·, α∗ ) is
non-decreasing. Note also that if J λX (H(w|α∗ ), α∗ ) < J
λX
(H(w|α∗ ), α∗ ) for some w then
ϕλX (H(·|α∗ ), α∗ ) is constant in a neighborhood of w. By a standard argument (see, e.g.,
Myerson (1981) and Toikka (2011)), an optimal Q must satisfy the following pooling property:
J λX (H(w|α∗ ), α∗ ) < J
λX
(H(w|α∗ ), α∗ ) for all w ∈ (w, w) implies that Q is constant on
(w, w). Therefore any non-decreasing implementable Q maximizes L subject to (F0 ) and
(MON) if and only if (i) Y (w(0, α∗ )) = 0; (ii) Q satisfies the pooling property; and (iii)
Y (·) > 0 on (w, w) implies ϕλX (H(·|α∗ ), α∗ ) is constant on this interval. Suppose, in addition,
that Q satisfies (IA0 ) with equality, then Q solves (P 0 -α∗ ).
It remains to determine the multiplier λX . In order to do so I first define two extreme
allocation rules satisfying conditions (i)-(iii) given above. Define Q+ (·, λX ) as follows. If
J λX (H(w|α∗ ), α∗ ) > J
λX
(H(w|α∗ ), α∗ ) for w ∈ (w, w) with w < w and let (w, w) be chosen
22
maximally, then let
Q+ (w, λX ) ≡
1
[H(w|α∗ )n − H(w|α∗ )n ]
n
,
H(w|α∗ ) − H(w|α∗ )
∀w ∈ (w, w).
Otherwise, let Q+ (w, λX ) ≡ H(w|α∗ )n−1 . Define Q− (·, λX ) as follows. If ϕλX (H(·|α∗ ), α∗ ) is
constant on (w, w) with w < w and let (w, w) be chosen maximally, then let
Q− (w, λX ) ≡
1
[H(w|α∗ )n − H(w|α∗ )n ]
n
,
H(w|α∗ ) − H(w|α∗ )
∀w ∈ (w, w).
Otherwise, let Q− (z, λX , α∗ ) ≡ H(z|α∗ )n−1 . Clearly, Q+ and Q− are non-decreasing, implementable and satisfy conditions (i)-(iii). I demonstrate in Appendix A.2.2 that Q+ is
the “steepest” allocation rule and Q− is the “least steep” allocation rule among all nondecreasing implementable Q’s satisfying conditions (i)-(iii) in the following sense: for all
non-decreasing implementable Q’s satisfying conditions (i)-(iii),
∗
∗
∗
Hα (w|α∗ )
Hα (w|α∗ ) −
Hα (w|α∗ ) +
α .
Q
(w,
λ
)
α
Q(w)
α
Q
(w,
λ
)
≥
E
−
≥
E
−
E −
X
X
h(w|α∗ )
h(w|α∗ )
h(w|α∗ )
Hence, there exists a non-decreasing implementable Q satisfying conditions (i)-(iii) and (IA0 )
with equality if and only if λX is such that
∗
∗
∗
Hα (w|α∗ ) +
H
(w|α
)
α
−
α ≥ C 0 (α∗ ) ≥ E −
α .
E −
Q
(w,
λ
)
Q
(w,
λ
)
X
X
h(w|α∗ )
h(w|α∗ )
(9)
Lemma 5 proves that such a λX exists and is unique:
Lemma 5 Suppose the first-order approach is valid and Assumption 1 holds. Suppose, in
addition, that α∗ is such that (7) is violated. There exists a unique λX > 0 such that
inequality (9) holds.
The main result of this section is Theorem 2, which demonstrates that the allocation
rules I have derived informally are indeed optimal solutions of (P 0 -α∗ ):
23
Theorem 2 Suppose the first-order approach is valid and Assumption 1 holds. Suppose, in
addition, that α∗ is such that (7) is violated. Let λX > 0 be such that inequality (9) holds
and Q be a non-decreasing implementable allocation rule. Then Q solves (P 0 -α∗ ) if and only
if Y (w(0, α∗ )) = 0, and Q satisfies (IA0 ) with equality and the following two properties:
1. (pooling property) If J λX (H(w|α∗ ), α∗ ) > J
λX
(H(w|α∗ ), α∗ ) for all w ∈ (w, w) with
w < w and let (w, w) be chosen maximally, then Q is constant on (w, w).
2. If Y (w) > 0 for all w ∈ (w, w) with w < w and let (w, w) be chosen maximally, then
ϕλX (H(·|α∗ ), α∗ ) is constant on (w, w).
My interpretation of the optimal pooling areas is as follows: Optimally, pooling occurs
where ϕλX (H(·|α∗ ), α∗ ) is not strictly increasing, i.e.,
∂
ws,α (s, α∗ )
Hα (w(s, α∗ )|α∗ )
1
≡
−
≥
ws (s, α∗ )
∂w
h(w(s, α∗ )|α∗ )
λX
Recall that if an information structure is more precise, then w(s, α) changes more dramatically as s changes, i.e., ws (s, α) is larger. Hence, one can interpret ws (s, α) as a local measure
of the information structure’s precision around s. Then, ws,α (s, α)/ws (s, α) is the percentage
change of the information structure’s precision around s as α increases. Intuitively, the most
effective way to discourage agents from acquire information is to introduce randomization
to where the information structure’s precision increases most. Assumption 2 implies that
ws,α (s, α)/ws (s, α) is a constant, i.e., when α increases, the information structure becomes
more precision uniformly over [0, 1].
Though intuitive, the proof of Theorem 2 is difficult because of the presence of both
the non-standard constraint (IA0 ) and the monotonicity constraint. In this paper, I use the
following approach first proposed by Reid (1968) and later introduced into the mechanism
design problem by Mierendorff (2009). I first solve (P 0 -α∗ ) under an additional restriction,
24
that Q is Lipschitz continuous with global Lipschitz constant K:
|Q(w) − Q(w0 )| ≤ K|w − w0 |, ∀w, w0 ∈ W.
Denote the above maximization problem by (P K -α∗ ). I show that the solutions of (P K -α∗ )
converge to that of (P 0 -α∗ ) as K → ∞. This allows one to characterize the solutions of
(P 0 -α∗ ).
The proof is not a straightforward modification of Mierendorff (2009). Let Q and QK
denote the solutions to (P K -α∗ ) and (P 0 -α∗ ), respectively. In Mierendorff (2009), Q is
discontinuous at exactly one point, and it can be shown that for K sufficiently large, the slope
of QK is equal to K only in a neighborhood around the discontinuity point. In my model,
however, Q could be discontinuous at most countably many points. If Q is discontinuous
at w, then it is possible that every neighborhood of w includes another discontinuity point.
Hence, it is non-trivial to characterize Q as the limit of QK .
Lastly, given Theorem 2, it is straightforward that there exists an optimal Q which takes
the following relatively simple form:
Corollary 1 There exists ξ ∈ [0, 1] such that Q̂(wi ) ≡ ξQ+ (wi , λX ) + (1 − ξ)Q− (wi , λX )
solves (P 0 -α∗ ).
3.3
Optimal α∗
Given optimal solutions of (P 0 -α∗ ), one can now study the optimal information choices,
i.e., α∗ ∈ argmaxα∈A π s (α) ≡ (1 − γ)V (α) − C(α). I show in Appendix A.2.2 that
s
Z
w(1,α)
π (α) = (1 − γ)
ϕλX (H(w|α), α)H(w|α)n−1 h(w|α)dw + (1 − γ)λX C 0 (α) − C(α),
w(0,α)
where, as demonstrated in Theorem 2, λX depends on α in a complex way. In general, it
is hard to solve the optimal α∗ . In this section, I first give a condition that the optimal α∗
25
must satisfy and then solve it under Assumption 2.
In Appendix A.3, I show that if the second-order condition of the agents’ optimization
problem is satisfied, then
Z
w(1,α)
−Hα (w|α)H(w|α)n−1 dw − C 0 (α) is non-increasing in α.
(10)
w(0,α)
Let
α̂ ≡ inf {α ∈ A|(7) holds for α} .
(11)
Then α̂ is independent of γ and limn→∞ α̂ = α. By (10), inequality (7) holds for all α > α̂.
I claim the socially optimal α∗ ≤ α̂. Note first that for all α > α̂, the solution to (P 0 -α∗ ) is
Q(w) = H(w|α)n−1 . Hence, the average social surplus is
s
Z
w(1,α)
wH(w|α)n−1 h(w|α)dw − C(α).
π (α) = (1 − γ)
w(0,α)
Taking derivative with respect to α gives
s0
Z
w(1,α)
π (α) = (1 − γ)
−Hα (w|α)H n−1 (w|α)dw − C 0 (α).
w(0,α)
Since C 0 (α) is strictly increasing, π s0 (α) is strictly decreasing by (10). By construction,
π s0 (α̂) = −γC 0 (α̂) ≤ 0. Hence, π s0 (α) < 0 for all α > α̂ and at optimum α∗ ≤ α̂. This result
is summarized in the following proposition:
Proposition 2 Suppose the first-order approach is valid and Assumption 1 holds. At optimum, α∗ ≤ α̂, where α̂ is such that (7) holds with equality if α = α̂.
Proposition 2 states the social planner will not encourage agents to acquire more information than they do under the ex post efficient mechanism, which is not surprising given
the results of Bergemann et al. (2009). This also implies that (IA’) always holds with equality when α∗ is chosen optimally. Hence, it is sufficient to consider the one-sided first-order
26
condition.
To obtain further results about the socially optimal information choice α∗ , I assume that
Assumption 2 holds for the rest of this section. In this case, the average social surplus is
π s (α) = (1 − γ)
hµ
n
i
+ b(α)C 0 (α) − C(α), for α ∈ [α, α̂],
Then
π s0 (α) = [(1 − γ)b0 (α) − 1] C 0 (α) + (1 − γ)b(α)C 00 (α).
Hence, π s0 (α) ≤ 0 if and only if 1 − C 0 (α)/ [b0 (α)C 0 (α) + b(α)C 00 (α)] ≤ γ. Assume that
C 0 (α)/ [b0 (α)C 0 (α) + b(α)C 00 (α)] is strictly increasing. Then there exists a unique α◦ ∈ A
such that π s0 (α) ≶ 0 if and only if α ≷ α◦ . Furthermore, α◦ is strictly decreasing in γ
and independent of n, and limn→∞ α◦ = α. Then the socially optimal information choice is
α∗ = min{α◦ , α̂}. This result is summarized by the following proposition:
Proposition 3 Suppose the first-order approach is valid and Assumption 2 holds. Suppose,
in addition, C 0 (α)/ [b0 (α)C 0 (α) + b(α)C 00 (α)] is strictly increasing. At optimum, α∗ (n, γ) =
min{α◦ (γ), α̂(n)}, where α◦ is strictly decreasing in γ, limn→∞ α◦ = α and limn→∞ α̂ = α.
Proposition 3 implies that the ex post efficient mechanism is also ex ante efficient if the
level of interdependency is low. As the level of interdependency increases, the socially optimal
information choice decreases, and an ex ante efficient mechanism introduces more randomization into the allocation rule to discourage agents from acquiring information. Proposition
3 also implies that the ex post efficient mechanism is also ex ante efficient for a sufficiently
large number of agents. Intuitively, when there is a large number of agents, the incentive for
each agent to acquire information is already small and there is no need for the social planner
to further discourage them from acquiring information by distorting the allocation rule.
27
Example 4 (The Linear Experiments) Assume that F (θ) = θ with support [0, 1]. Assume that the cost function (used in Persico (2000)) is of the form
C(α) = K (α − α)2 , ∀α ∈ [α, 1],
where 0 < α < 1 and K > 0. Assume 8Kα ≥ 1. Then, as I demonstrate in Appendix A.3,
the first-order approach is valid. In this case α̂ is such that
2K(α − α) =
n−1
.
2n(n + 1)
Note that the left-hand side of the above equation is strictly increasing in α; and the righthand side is strictly decreasing in n for n ≥ 2 and converges to 0 as n goes to infinity. Hence,
α̂ is strictly decreasing in n and goes to α as n grows to infinity. Finally,
π s0 (α) = 2K [γα − (2γ − 1)α] .
If γ ≤ 12 , then π s0 (α) ≥ 0 and therefore α∗ = α̂. If γ > 12 , then π s0 (α) is strictly decreasing
and therefore
∗
α = min
γα
, α̂ .
2γ − 1
Thus, α∗ = α̂ for γ sufficiently small and/or n sufficiently large, the ex post efficient mechanism is also ex ante efficient. For γ sufficiently large and/or n sufficiently small, the optimal
α∗ is strictly decreasing in γ and goes to α when γ increases to 1.
4
Efficient Asymmetric Mechanisms
In this section, I study ex ante efficient mechanisms without imposing symmetry re-
strictions. First, I show that the result in Theorem 1 is still valid for general asymmetric
mechanisms, i.e., an ex ante efficient mechanism never withholds the object. Second, I derive
28
ex ante efficient mechanisms under Assumption 2 and present conditions under which the
socially optimal information choices are the same for all agents and there exists a symmetric
ex ante efficient mechanism.
As in the symmetric case, I consider the relaxed problem of the social planner by replacing the (IA) constraint by the (IA0 ) constraint and focus on reduced form auctions. Let
Q ≡ (Q1 , . . . , Qn ), where Qi : [θ, θ] → [0, 1] is non-decreasing for all i. By Theorem 3 in
Mierendorff (2011), Q is implementable if and only if it satisfies
n Z
X
w(1,α∗i )
Qi (zi )dH(zi |αi∗ ) ≤ 1 −
wi
i=1
n
Y
H(wi |αi∗ ), ∀w ∈
n
Y
[w(0, αi∗ ), w(1, αi∗ )].
(AF0 )
i=1
i=1
Thus, given the (MON) constraint, we can replace constraint (F) by constraint (AF0 ). Finally, as in the symmetric case, an ex ante efficient mechanism never withholds the object:
Theorem 3 Suppose α∗ and (q, t) solve the relaxed problem of the social planner, then
X
n
Y
qi (w) = 1 for almost all w ∈
[w(0, αi∗ ), w(1, αi∗ )].
i
(12)
i=1
Using Theorem 3 and Law of iterated expectations, we can rewrite the social planner’s
objective function as a function of Q:
"
E
X
i
(wi + γ
X
j6=i
#
X
wj )qi (w) αi = αi∗ ∀i =
E [(1 − γ)wi Qi (wi )| αi = αi∗ ] + nγE[θ].
i
As in the symmetric case, we ignore the second term, nγE[θ], which is a constant, from
hereon. To summarize, the social planner’s relaxed problem, denoted by (P 0 ), becomes:
#
X
X
∗
max
E
(1
−
γ)w
Q
(w
)
α
=
α
∀i
−
C(αi∗ ),
i i
i
i
i
α∗ ,Q
"
i
i
29
subject to
n Z
X
i=1
w(1,α∗i )
Qi (zi )dH(zi |αi∗ ) ≤ 1 −
wi
n
Y
H(wi |αi∗ ), ∀w ∈
i=1
n
Y
[w(0, αi∗ ), w(1, αi∗ )].
(AF0 )
i=1
Qi (wi ) is non-decreasing in wi ,
Hαi (wi |αi∗ )
∗
E −
Qi (wi ) αi = αi ≤ C 0 (αi∗ ).
∗
h(wi |αi )
(MON)
(IA0 )
As in the symmetric case, I solve (P 0 ) in two steps. First, for each α∗ ∈ An , I solve the
following sub-problem, denoted by (P 0 -α∗ ):
"
V (α∗ ) ≡ max E
Q
X
i
#
wi Qi (wi ) α∗ subject to (AF0 ), (MON) and (IA0 ),
Second, I solve α∗ ∈ arg maxα∈An π s (α) ≡ (1 − γ)V (α) −
P
i
C(αi ).
Fix α∗ . If α∗ is such that
"
E
Hαi (wi |αi∗ ) Y
∗ ∗
−
H(wj |αj ) α
h(wi |αi∗ ) j6=i
#
≤ C 0 (αi∗ ), ∀i,
then clearly the ex post efficient mechanism, i.e., Qi (wi ) =
Q
j6=i
(13)
H(wi |αi∗ ) for all i and all
w, solves (P 0 -α∗ ). Assume α∗ is such that (13) is violated.
Suppose there exists 0 < k ≤ n such that the (IA0 ) constraints bind for the first k
agents. Then we can ignore the (IA0 ) constraints for the last n − k agents. Let λi denote the
Lagrangian multiplier associated with the (IA0 ) constraint for agent i (i ≤ k). By a similar
argument to that in Section 3.2, we have λi = b(αi∗ ). The Lagrangian relaxation becomes
L =
XZ
i≤k
w(1,α∗i )
w(0,α∗i )
µQi (wi )h(wi |αi )dwi +
XZ
i>k
w(1,α∗i )
wi Qi (wi )h(wi |αi )dwi +
w(0,α∗i )
Then, for i > k, it is optimal to set Qi (wi ) =
Q
30
j>k,j6=i
X
b(αi∗ )C 0 (αi∗ ).
i≤k
H(wi |αj∗ ) if wi > µ and Qi (wi ) = 0 if
wi < µ. Hence,
∗
V (α ) = µ
Y
H(µ|αi∗ )
i>k
i>k
Z
=θ−
+
XZ
θ
Y
w(1,α∗i )
wi
µ
Y
H(wi |αj∗ )h(wi |αi )dwi +
j>k,j6=i
X
b(αi∗ )C 0 (αi∗ ).
i≤k
H(w|αi∗ )dw,
µ i>k
where the second line follows from integration by parts. Finally, (IA0 ) holds for i > k if and
only if
Z
w(1,α∗i )
Y
−Hαi (wi |αi∗ )
µ
H(wi |αj∗ )dwi ≤ C 0 (αi∗ ).
(14)
j>k,j6=i
Consider an agent i (i > k). I show that if α∗ is chosen optimally, then (14) holds with
equality. Suppose to the contrary that (14) holds with strict inequality, then
∂π s (α∗ )
= −(1 − γ)
∂αi
Z
w(1,α∗i )
µ
Hα∗i (wi |αi∗ )
Y
H(wi |αj∗ )dwi − C 0 (αi∗ ) < −γC 0 (αi∗ ) ≤ 0,
j>k,j6=i
a contradiction to the optimality of αi∗ . Hence, (14), i.e., the (IA0 ) constraints hold with
equality for all i > k. By Assumption 2, we have
"
π s (α) = (1 − γ) µ +
n
X
#
b(αi )C 0 (αi ) −
i=1
X
C(αi ).
i
Let α◦ ∈ arg maxα∈A (1 − γ)b(α)C 0 (α) − C(α). Clearly, if α◦ ≤ α̂, where α̂ is define by (11),
then the socially optimal information choices are αi∗ = α◦ for all i. This result is summarized
by the following proposition:
Proposition 4 Suppose the first-order approach is valid and Assumption 2 holds. Suppose,
in addition, α◦ ≤ α̂ where α◦ ∈ arg maxα∈A (1 − γ)b(α)C 0 (α) − C(α) and α̂ is define by (11).
Then then the socially optimal information choices are the same for all agents: αi∗ = α◦ for
all i.
Note that if the assumptions in Proposition 4 are satisfied, then there exists a symmetric
31
ex ante efficient mechanism.
5
Conclusion
I have studied ex ante efficient mechanisms in an auction setting in which agents have
positively interdependent values and information is endogenous. Specifically, I assume agents
are initially uncertain about the value of the object on sale, and they are able to pay a cost
to acquire information about this value before participating in a mechanism. In an earlier
paper, Bergemann and Välimäki (2002) find that using the ex post efficient mechanism will
lead to ex ante over investment in information by agents. This suggests the potential gain in
ex ante efficiency by adjusting the ex post efficient mechanism in a way to discourage agents
from gathering information.
In the main body of the paper I restrict attention to symmetric mechanisms and symmetric equilibria in which all agents acquire the same amount of information. First, I demonstrate an ex ante efficient mechanism never withholds the object. Intuitively, whenever the
object is withheld, one can instead allocate it randomly among agents. This improves the
allocation efficiency without giving agents additional incentive to acquire information. This
result also allows me to write the social planner’s problem as a function of interim allocation
rule, which is more convenient to work with.
Second, I fully characterize ex ante efficient mechanisms. When the interdependence is
low and/or the number of agents is large, the ex-post efficient mechanism is also ex ante
efficient. When the interdependence is high and/or the number of agents is small, an ex
ante efficient mechanism involves randomization. Specifically, an ex ante efficient interim
allocation rule randomizes in areas in which the accuracy of an agent’s posterior estimate
can be significantly improved if an additional piece of information in acquired.
In this paper, I assume all agents simultaneously acquire information prior to the auction.
One interesting direction for future research is to allow for the possibility of sequential
32
information acquisition. It is likely that the efficiency can be improved if agents are asked
to acquire information in turn, and one’s information acquisition decision can depend on
the signals received by those who take action earlier. Another interesting direction for
future research is to consider the impact of initial private information. In this paper I only
considered static mechanisms in which agents only report their private information once.
In general, one can consider a dynamic mechanism in which agents report their private
information both before and after acquiring information.
A
Appendix
Proof of Lemma 1.
Note that by construction H(w(s, α)|α) = s. Taking derivative of
both sides of the equation with respect to α yields
h(w(s, α)|α)wα (s, α) + Hα (w(s, α)|α) = 0.
Hence,
−
Hα (w(s, α)|α)
= wα (s, α).
h(w(s, α)|α)
(15)
Suppose −Hα (w|α)/h(w|α) is strictly increasing in w. Then wα (s, α) is strictly increasing
in s. Hence, for all s, s0 ∈ (0, 1), s0 > s and α0 > α00 we have
0
0
0
Z
00
α0
wα (s0 , α)dα
w(s , α ) − w(s , α ) =
α00
Z
α0
>
wα (s, α)dα
α00
= w(s0 , α0 ) − w(s0 , α00 ).
Lemma 6 Let (c, d) be an interval of the real line, and J and H be two non-decreasing
33
functions. Assume that for some measure m on R we have
Z
d
H(w)dm(w) = 0.
c
Then
Rd
c
H(w)J(w)dm(w) ≥ 0.
Proof. This lemma is a corollary of Lemma 1 in Persico (2000).
Lemma 7 Suppose J : [θ, θ] → R is non-decreasing on [w(0, αi ), w(1, αi )], then
Hαi (wi |αi )
J(wi ) αi ≥ 0,
−
h(wi |αi )
(16)
where the equality holds if J is constant.
Proof. Since J and −
Hαi (wi |αi )
h(wi |αi )
are non-decreasing on [w(0, αi ), w(1, αi )], it suffices to show
that
Z
w(1,αi )
Hαi (wi |αi )dwi = 0.
(17)
w(0,αi )
On the one hand since
Z
w(1,αi )
Z
w(1,αi )
H(wi |αi )dwi = wi H(wi |αi ) w(0,αi ) −
w(1,αi )
wi dH(wi |αi ),
w(0,αi )
w(0,αi )
= w(1, αi ) − E[θ].
Taking derivative with respect to αi yields
∂
∂αi
Z
w(1,αi )
H(wi |αi )dwi = wαi (1, αi ).
(18)
w(0,αi )
On the other hand, by the chain rule, we have
∂
∂αi
Z
w(1,αi )
Z
w(1,αi )
H(wi |αi )dwi = wαi (1, αi ) +
w(0,αi )
Hαi (wi |αi )dwi .
w(0,αi )
34
(19)
Comparing (18) and (19) proves (17). By Lemma 6, inequality (16) holds. If J is constant,
the equality holds by (17).
Proof of Lemma 2.
Define w[ ≡ sup {wi |Y (wi0 ) > 0 ∀w(0, α∗ ) ≤ wi0 ≤ wi }. By the
continuity of Y , we have Y (w[ ) = 0 and w[ > w(0, α∗ ). There are four cases to consider.
Case I: Suppose there exists wi0 ∈ (w(0, α∗ ), w[ ) such that Q is discontinuous at wi0 .
0
0
Let Q(wi+ ) denote the right-hand limit of Q at wi0 , and Q(wi− ) the corresponding left-hand
n
o
0+
0−
Y (wi )
0
limit. Let 0 < ε ≤ min minw(0,α∗ )≤wi ≤wi H(w0 |α∗ ) , Q(wi ) − Q(wi ) . Let Q̂ be such that
i
∗
Q̂(wi ) = Q(wi ) for wi ≤ w(0, α ) and for wi > w(0, α∗ )
Q̂(wi ) = Q(wi ) + εχ{wi ≤wi0 } ,
where χ{wi ≤wi0 } is an indicator function. By construction, Q̂(w) ≥ Q(w) for all w ∈ W
and the inequality holds strictly on a positive measure set. We now verify that Q̂ satisfies
(MON), (IA0 ) and (F0 ). Clearly, Q̂ is non-decreasing, i.e., satisfies (MON). Since χ{wi ≤wi0 } is
non-increasing on [w(0, α∗ ), w(1, α∗ )], by Lemma 7, we have
Hαi (wi |α∗ )
∗
Q̂(wi ) αi = α ,
E −
h(wi |α∗ )
∗
H
(w
|α
)
Hαi (wi |α∗ )
α
i
i
∗
∗
Q(wi ) αi = α + εE −
χ{wi ≤wi0 } αi = α ,
=E −
h(wi |α∗ )
h(wi |α∗ )
≤C 0 (α∗ ) + 0 = C 0 (α∗ ).
Hence, Q̂ satisfies (IA0 ). Finally, let
Z
θ
Ŷ (wi ) ≡
h
i
H(z|α∗ )n−1 − Q̂(z) h(z|α∗ )dz.
wi
If wi ≤ wi0 , then Ŷ (wi ) = Y (wi ) − ε[H(wi0 |α∗ ) − H(wi |α∗ )] ≥ Y (wi ) − εH(wi0 |α∗ ) ≥ 0. If
wi > wi0 , then Ŷ (wi ) = Y (wi ) ≥ 0. Hence, Q̂ satisfies (F0 ).
Case II: Suppose Q is continuous on [w(0, α∗ ), w[ ]. We first show that there exists
35
wi0 ∈ (w(0, α∗ ), w[ ) such that Q(wi0 ) < Q(w[ ). Suppose to the contrary that Q(wi ) = Q(w[ )
R w[
for all wi ∈ (w(0, α∗ ), w[ ). If Q(w[ ) ≥ H(w[ |α∗ )n−1 , then Y (w(0, α∗ )) = w(0,α∗ ) [H(z|α∗ )n−1 −
Q(z)]h(z|α∗ )dz < 0, a contradiction. If Q(w[ ) < H(w[ |α∗ )n−1 , then, by the continuity of Q
and H, there exists δ > 0 such that Q(wi ) < H(wi |α∗ )n−1 for all wi ∈ [w[ , w[ + δ]. Moreover,
[
Z
w[ +δ
0 = Y (w ) =
[H(z|α∗ )n−1 − Q(z)]h(z|α∗ )dz + Y (w[ + δ) > Y (w[ + δ),
w[
a contradiction. Thus, there exists wi0 ∈ (w(0, α∗ ), w[ ) such that Q(wi0 ) < Q(w[ ).
By the continuity of Q, there exists wi00 ∈ (wi0 , w[ ) such that Q(wi00 ) =
n
o
Y (wi )
00
0
Let 0 < ε ≤ min minw(0,α∗ )≤wi ≤wi00 H(w
,
Q(w
)
−
Q(w
)
. Let
00 |α∗ )
i
i
1
2
Q(wi0 ) + Q(w[ ) .
i



max{Q(wi0 ) + ε, Q(wi )}



Q̂(wi ) ≡
Q(wi ) + ε




 Q(wi )
if wi > wi0 ,
if w(0, α∗ ) < wi ≤ wi0 ,
if wi ≤ w(0, α∗ ).
Note that if wi ≥ wi0 0 then Q(wi ) ≥ Q(wi0 0 ) ≥ Q(wi0 ) + ε. Hence, Q̂(wi ) = Q(wi ) for
wi ≥ wi0 0 . By construction, Q̂(w) ≥ Q(w) for all w ∈ W and the inequality holds strictly
on a positive measure set. We now verify that Q̂ satisfies (MON), (IA0 ) and (F0 ). Clearly,
Q̂ is non-decreasing, i.e., satisfies (MON). It is easy to verify that Q̂ − Q is non-increasing
on [w(0, α∗ ), w(1, α∗ )] and therefore Q̂ satisfies (IA0 ) by Lemma 7. Finally, if wi ≥ wi0 0 , then
Ŷ (wi ) = Y (wi ). If wi < wi0 0 , then
Z
w[
Ŷ (wi ) =
h
i
H(z|α∗ )n−1 − Q̂(z) h(z|α∗ )dz,
wi
Z
wi0 0
=Y (wi ) −
h
i
Q̂(z) − Q(z) h(z|α∗ )dz,
wi
≥Y (wi ) − ε [H(wi0 0 |α∗ ) − H(wi |α∗ )] ,
≥Y (wi ) − εH(wi0 0 |α∗ ) ≥ 0.
36
Hence, Q̂ satisfies (F0 ).
Case III: Suppose Q is continuous on [w(0, α∗ ), w[ ) and Q(w[− ) < H(w[ |α∗ )n−1 . Define
R(wi ) ≡ Y (wi )/(H(w[ |α∗ ) − H(wi |α∗ )) for wi < w[ . Then by L’Hopital’s rule,
lim R(wi ) = H(w[ |α∗ )n−1 − Q(w[− ) > 0.
wi →w[−
Let 0 < ε ≤ min inf w(0,α∗ )≤wi <w[ R(wi ), Q(w[+ ) − Q(w[− ) . Let Q̂ be such that Q̂(wi ) =
Q(wi ) for wi ≤ w(0, α∗ ) and Q̂(wi ) = Q(wi ) + εχ{wi <w[ } for wi > w(0, α∗ ). By construction,
Q̂(w) ≥ Q(w) for all w ∈ W and the inequality holds strictly on a positive measure set. One
can verify that Q̂ satisfies (MON) and (IA0 ) following the arguments in Case I. Finally, if wi <
w[ , then Ŷ (wi ) = Y (wi )−ε[H(w[ |α∗ )−H(wi |α∗ )] ≥ Y (wi )−R(wi )[H(w[ |α∗ )−H(wi |α∗ )] = 0.
If wi ≥ w[ , then Ŷ (wi ) = Y (wi ) ≥ 0. Hence, Q̂ satisfies (F0 ).
Case IV: Suppose Q is continuous on [w(0, α∗ ), w[ ) and Q(w[− ) ≥ H n−1 (w[ |α∗ ). We first
show that Q(w[− ) = H n−1 (w[ |α∗ ). Suppose to the contrary that Q(w[− ) > H n−1 (w[ |α∗ ).
Then by the continuity of Q and H on [w(0, α∗ ), w[ ), there exists δ > 0 such that Q(wi ) >
H n−1 (wi |α∗ ) for all wi ∈ (w[ − δ, w[ ). Then
[
Z
w[
Y (w − δ) =
[H(z|α∗ )n−1 − Q(z)]h(z|α∗ )dz + Y (w[ ) < 0,
w[ −δ
a contradiction. Hence, Q(w[− ) = H n−1 (w[ |α∗ ). Second, we show that there exists wi0 ∈
(w(0, α∗ ), w[ ) such that Q(wi0 ) < Q(w[− ). Suppose to the contrary that Q(wi ) = Q(w[− ) for
R w[
all wi ∈ (w(0, α∗ ), w[ ), then Y (w(0, α∗ )) = w(0,α∗ ) [H n−1 (z|α∗ )−Q(z)]h(z|α∗ )dz +Y (w[ ) < 0,
a contradiction. Hence, there exists wi0 ∈ (w(0, α∗ ), w[ ) such that Q(wi0 ) < Q(w[− ). The rest
of the proof follows that of Case II.
37
A.1
Proof of Lemma 4
The proof follows from a similar argument to that in Border (1991). Before proceeding
to the proof, I first introduce some notations and definitions. Let σi,j : W n → W n denote
the function that interchanges the ith and the jth coordinates, i.e.,
σi,j (w1 , . . . , wn ) = (w1 , . . . , wi−1 , wj , wi+1 , . . . , wj−1 , wi , wj+1 , . . . , wn ), ∀w = (w1 , . . . , wn ),
and σi ≡ σ1,i . We say an allocation rule q is symmetric if q1 is such that q1 (w) = q1 (σi,j (w))
P
for all i, j 6= 1, qi (w) = q1 (σi (w)) and i qi (w) ≤ 1 for all w. We say a mechanism (q, t) is
symmetric if its allocation rule q is symmetric.
Let D1 denote the set of symmetric allocation rules q satisfying
P
qi (w) ≥ ρ(w) for
almost all w ∈ W n . Let D collect all Q : W → [0, 1] such that there exists q ∈ D1 that
R
implements Q. Let hf, gi denote f (wi )g(wi )dm(wi ). Define for each measurable set A ⊂ W
1
B(A) ≡
n
Z
ρ(w)dmn (w),
An
and
B(A) ≡
mn (∪j σj (A × W n−1 ))
.
n
I break the proof down into several lemmas. The first lemma states that condition (6) is
necessary.
Lemma 8 For all measurable A ⊂ W , all Q ∈ D
hχA , Qi ≥ B(A).
Proof. Since Q ∈ D, there exists a symmetric allocation rule q satisfying
38
P
qi (w) ≥ ρ(w)
for almost all w ∈ W n and implementing Q. For each i,
Z
hχA , Qi =
Q(wi )dm(wi ),
A
Z
qi (w)dmn (w),
=
σi
(A×W n−1 )
Z
qi (w)dmn (w).
≥
(20)
An
Hence,
nhχA , Qi ≥
XZ
qi (w)dmn (w),
An
=
Zi X
An
Z
≥
qi (w)dmn (w),
i
ρ(w)dmn (w).
(21)
An
That is, hχA , Qi ≥ B(A). Inequality (21) holds with equality if
P
i qi (w)
= ρ(w) for all
w ∈ An .
Each hχA , ·i defines a function on D, and according to Lemma 8, this function is bounded
below by B(A). We now show that this bound is achieved by a hierarchical reduced form.
Lemma 9 Let A1 , . . . , AL , AL+1 , . . . , AL+K be a partition of W , and define the hierarchical
auctions generated by A1 , . . . , AL , AL+1 , . . . , AL+K , denoted q A1 ,...,AL+K , by
A1 ,...,AL+K
qi
(w) =






1
#{j:wj ∈Ak }
if k = 1, . . . , L, wi ∈ Ak and wj ∈
/ A1 ∪ · · · ∪ Ak−1 ∀j 6= i,
ρ(w)
if k = L + 1, . . . , L + K, wi ∈ Ak and wj ∈
/ A1 ∪ · · · ∪ Ak−1 ∀j 6= i,
#{j:wj ∈Ak }




 0
otherwise.
Let Q∗ be the reduced form of the hierarchical auction q A1 ,...,AL+K . For each j = 1, . . . , L+K,
set Fj = A1 ∪ · · · ∪ Aj ; then for each j = 1, . . . , L,
hχFj , Q∗ i = B(Fj );
39
(22)
and for each j = L, . . . , L + K − 1,
hχFjc , Q∗ i = B(Fjc ).
(23)
Proof. Condition (22) follows from Lemma 5.2 in Border (1991). For each i, if w ∈
σi (Fjc × W n−1 )\(Fjc )n , then wi ∈
/ Fj and there exists ι 6= i such that wι ∈ Fj . By definition,
A1 ,...,AL+K
(w) = 0. Hence, inequality (20) holds with equality if A = Fjc . Furthermore,
P
A ,...,AL+K
simple calculations imply that ni=1 qi 1
equals to ρ on (Fjc )n for each j = L, . . . , L +
qi
K − 1. Hence, inequality (21) holds with equality if A = Fjc for each j = L, . . . , L + K − 1,
and so
hχFjc , Q∗ i = B(Fjc ).
This proves (23).
Lemma 8 shows that each Q ∈ D satisfies condition (6). The next lemma shows that if
a simple function separates Q from D, then Q violates (6) on some measurable set of types.
Lemma 10 Let Q : W → [0, 1] be implementable and suppose the simple function
PL+K
j=1 βj χAj separates Q from D. That is, for all Q ∈ D,
*L+K
X
+
βj χEj , Q
<
j=1
*L+K
X
+
βj χEj , Q .
(24)
j=1
Then for some measurable A ⊂ W , hχA , Qi < B(A).
Proof. Without loss of generality, take the Ej ’s to a partition of W and numbered so that
β1 > · · · > βK > 0 ≥ βK+1 > · · · > βK+L . To see that at least one βj > 0 consider a
P
symmetric allocation rule q implementing Q. Let ρ(w) = i q i (w), which is invariant to
permutations of w. Define an allocation rule q by
q1 (w) =



q̄1 (w)
ρ(w)
if ρ(w) > 0,


1
n
if ρ(w) = 0,
40
and qi (w) = q1 (σi (w)). Clearly, q is symmetric. Moreover,
P
i qi (w)
=
P
i (1/n)
= 1 ≥ ρ(w)
for w such that ρ(w) = 0 and
X
qi (w) =
i
X q̄1 (σi (w))
i
ρ̄(σi (w))
,
X q (w)
i
=
,
ρ(w)
i
1 X
=
q̄i (w),
ρ(w) i
= 1 ≥ ρ(w),
for w such that ρ(w) > 0. Let Q be the reduced form of q, then Q ∈ D and Q ≥ Q. This
implies that at least one βj > 0.
Let Aj = EK+L+1−j for j = 1, . . . , K + L and Q∗ be the reduced form of hierarchical
auctions generated by A1 , . . . , AL+K . If for any k = 1, . . . , K, we have hχE1 ∪···∪Ek , Q − Q∗ i <
0, then we are done, since
hχE1 ∪···∪Ek , Q∗ i = hχ(A1 ∪···∪AL+K−k )c , Q∗ i,
= B((A1 ∪ · · · ∪ AL+K−k )c ),
= B(E1 ∪ · · · ∪ Ek ),
where the second equality holds by Lemma 9. Suppose to the contrary that
hχE1 ,∪···∪Ek , Q − Q∗ i ≥ 0
(25)
for all k = 1, . . . , K. To simplify notation, let Q∗j = hχEj , Q∗ i and Qj = hχEj , Qi. Then (24)
implies that
L+K
X
βj (Q∗j − Qj ) > 0.
j=1
41
Taking the j = 1 term to the right and dividing by β1 > 0 yields
L+K
X
j=2
βj ∗
(Qj − Qj ) > Q1 − Q∗1 ≥ 0,
β1
where the second inequality follows from (25) for k = 1. If K > 1, then α2 > 0, so β1 /β2 > 1,
and multiplying the left-hand side of the above inequality by β1 /β2 strengthens it. Then,
taking the j = 2 term to the right and dividing by β2 > 0 yields:
L+K
X
j=3
βj ∗
(Q − Qj ) > (Q1 − Q∗1 ) + (Q2 − Q∗2 ) ≥ 0,
β2 j
where the second inequality follows from (25) for k = 2. Continue in this fashion until
arriving at
L+K
X
βj ∗
(Qj − Qj ) > (Q1 − Q∗1 ) + . . . (QK − Q∗K ) ≥ 0,
β
K
j=K+1
(26)
where the second inequality follows from (25) for k = K. Note that if L = 0, then the
left-hand side of (26) is equal to 0, a contradiction. Suppose that L ≥ 1. For k = 1, . . . , L
L+K
X
Q∗j = hχEL+K+1−k ∪···∪EL+K , Q∗ i,
j=L+K+1−k
= hχA1 ∪···∪Ak , Q∗ i,
= B̄(A1 ∪ · · · ∪ Ak ),
where the second equality holds by Lemma 9. For k = 1, . . . , L,
L+K
X
Qj = hχEL+K+1−k ∪···∪EL+K , Qi
j=L+K+1−k
= hχA1 ∪···∪Ak , Qi
≤ B̄(A1 ∪ · · · ∪ Ak ),
42
where the last equality holds by Proposition 3.1 in Border (1991). Hence,
PL+K
∗
j=L+K+1−k (Qj −
Qj ) ≥ 0 for k = 1, . . . , L. Since βK > 0 ≥ βK+1 > · · · > βK+L , the left-hand side of (26)
satisfies
L+K
X
βj ∗
(Qj − Qj )
β
K
j=K+1
L+K
L+K
βK+2 − βK+1 X
βK+L − βK+L−1 ∗
βK+1 X
∗
(Qj − Qj ) +
(Q∗j − Qj ) + · · · +
(QK+L − QK+L )
=
βK j=K+1
βK
βK
j=K+2
≤0,
a contradiction. Hence, hχE1 ,∪···∪Ek , Qi < hχE1 ,∪···∪Ek , Q∗ i = B(E1 ∪ · · · ∪ Ek ) for some
k = 1, . . . , K, completing the proof of the lemma.
To complete the proof of Theorem 4, we must show that if Q ∈
/ D, then it is separated
from D by a simple function. This will be accomplished using a separating hyperplane
argument, after establishing some topological preliminaries. The set D of implementable
functions is clearly convex. Consider D as a subset of L∞ (m), the set of m-essentially
bounded measurable functions on W and D1 as a subset of L∞ (mn )n . For 1 ≤ p < ∞, let
Lp (m) denote the set of measurable functions whose absolute value raised to the p-th power
has finite integral. For brevity denote Lp (mn )n by Lnp and Lp (m) by Lp .
Since L∞ is the dual of L1 under the duality hf, gi = Ewi [f (wi )g(wi )|αi = α∗ ], topologize
L∞ with its weak∗ , or σ(L∞ , L1 ), topology. Similarly give Ln∞ its σ(Ln∞ , Ln1 ) topology.
Equation (2) with αj = α∗ for all j 6= i defines a function Λ : D1 → D mapping each q
to its reduced form. The next lemma describes the main topological results.
Lemma 11
D1 is σ(Ln∞ , Ln1 ) compact.
(27)
Λ : D1 → D is σ(Ln∞ , Ln1 ), σ(L∞ , L1 ) continuous.
(28)
D is σ(L∞ , L1 ) compact.
(29)
43
Proof. Since D1 is a subset of the unit ball of Ln∞ under the norm given by ||q|| =
maxi ||qi ||∞ , by the Banach-Alaoglu theorem, for (27) we need only prove that D1 is
σ(Ln∞ , Ln1 ) closed. Let q ν be a net in D1 converging in the σ(Ln∞ , Ln1 ) topology to q. First
we verify that qi (w) = q1 (σi (w)). Observe that σi : W n → W n is measurable, σi = (σi )−1 ,
and mn ◦ σi = mn . Since each q ν ∈ D1 , for any f ∈ Ln1 , we have
Z
Wn
f (w)qiν (w)dmn (w)
Z
=
Wn
Z
=
Wn
f (w)q1ν (σi (w))dmn (w),
f (σi (w))q1ν (w)dmn (w).
Taking limits on each side yields
Z
Z
n
f (σi (w))q1 (w)dmn (w).
f (w)qi (w)dm (w) =
Wn
Wn
Transforming variables on the right-hand side, we get
Z
Z
n
f (w)q1 (σi (w))dmn (w).
f (w)qi (w)dm (w) =
Wn
Wn
Since f is arbitrary, conclude that qi = q1 ◦ σi . Similar arguments show that q1 (w) =
P
q1 (σi,j (w)) for all i, j 6= 1 and i qi ≥ ρ, and so q ∈ D1 . This completes the proof of (27).
To prove continuity of Λ, let q ν → q in the σ(Ln∞ , Ln1 ) topology. Let f ∈ L1 and define
f˜ ∈ Ln1 by f˜(w) = f (w1 ). Then
ν
Z
hΛq , f i =
Wn
f˜(w)q1ν (w)dmn (w).
Since q ν → q, the right-hand side converges to
R
f˜q1 dmn = hΛq, f i. Since f is arbitrary,
Λq ν → Λq in the σ(L∞ , L1 ) topology, proving (28).
This shows that D is compact, since continuous images of compact sets are compact.
Lemma 8 shows that the condition (6) is necessary for implementability. To see the con-
44
verse, suppose Q ∈
/ D. Since D1 is σ(Ln∞ , Ln1 ) compact and convex, a separating hyperplane
theorem implies that there exists a nonzero f ∈ L1 satisfying hf, Qi > max{hf, Qi : Q ∈ D}.
Since simple functions are norm dense in L1 , we may take f to be simple. By Lemma 10
then hχA , Qi > B(A) for some measurable A ⊂ W , violating the feasibility inequality (6).
Thus we have shown that if Q satisfies (6) for all measurable A ⊂ W then Q ∈ D ⊂ L∞ .
Strictly speaking, an element of L∞ is not a measurable function, but an equivalent class
of measurable functions, where two functions are equivalent if they differ only on a set of
measure zero. That is, we have shown that if Q satisfies (6) for all measurable A ⊂ W then
P
there exists a symmetric allocation rule q ∗ satisfying i qi∗ (w) ≥ ρ(w) whose reduced form
Q∗ agrees with Q almost everywhere. The following argument shows that we can modify q ∗
on a set of measure zero to implement Q.
Let A∗ ≡ {wi : Q(wi ) 6= Q∗ (wi )} and define q1 : W → [0, 1] by



q ∗ (w)


 1
q1 (w) =
Q(w1 )





0
if w ∈ (W \A∗ )n ,
if w ∈ A∗ × (W \A∗ )n−1 ,
(30)
otherwise.
Then, for all w1 ∈ W ,
Z
q1 (w1 , w2 , . . . , wn )dmn−1 (w2 , . . . , wn ) = Q(w1 ).
(31)
W n−1
To see this, first note that if w1 ∈
/ A∗ , then q1 agrees with q1∗ unless some w2 , . . . , wn belongs
to A∗ , a mn−1 -measure zero event. Since q ∗ implements Q∗ and Q∗ agrees with Q outside
A∗ , (31) holds. If w1 ∈ A∗ , then q1 (w) = Q∗ (w1 ) unless some w2 , . . . , wn belongs to A∗ , a
mn−1 -measure zero event, and therefore (31) holds.
Finally, we show that setting qi (w) = q1 (σi (w)) yields a symmetric allocation rule q :
P
W n → [0, 1]n satisfying i qi (w) ≥ ρ(w) for almost all w and implementing Q. Clearly, q
P
is symmetric, i qi (w) ≥ ρ(w) for almost all w and by (31) it implements Q. It remains
45
to show that q is feasible. If w ∈ (W \A∗ )n , then q(w) = q ∗ (w) by construction and
P
P ∗
/ (W \A∗ )n , there are two cases. If more than one
i qi (w) =
i qi (w) ≤ 1. If w ∈
of w1 , . . . , wn belongs to A∗ , then q1 (w) = 0 by construction and hence each qi (w) = 0. If
exactly one of w1 , . . . , wn belongs to A∗ , say wi ∈ A∗ , then qi (w) = Q(wi ) ≤ 1 and qj (w) = 0
for i 6= j. This shows that the modified q is feasible and completes the proof.
A.2
Solving (P 0 -α∗ )
Recall that the sub-problem (P 0 -α∗ ) is
V (α∗ ) ≡ max E [wQ(w)| α∗ ] ,
Q
subject to
Z
θ
Y (w) ≡
[H(z|α∗ )n−1 − Q(z)]h(z|α∗ )dz ≥ 0, ∀w ∈ [θ, θ].
(F0 )
Q(w) is non-decreasing in w,
∗
Hα (w|α∗ )
α ≤ C 0 (α∗ ).
Q(w)
E −
∗
h(w|α )
(MON)
w
(IA0 )
To simplify notation, denote w(0, α∗ ) by w, w(1, α∗ ) by w, h(z|α∗ ) by h(z), H(z|α∗ )
Rw
by H(z) and Hαi (z|α∗ ) by Hαi (z). Let X(w) ≡ 0 Hα (z)Q(z)dz. Then this is a control
problem with state variables X, Y and Q, and a control variable a ≥ 0. The evolution of
the state variables is governed by
X 0 (w) = Hα (w)Q(w),
(32)
Y 0 (w) = −[H(w)n−1 − Q(w)]h(w),
(33)
Q0 (w) = a(w),
(34)
where the last equality holds if Q(w) is differentiable at w. The non-negativity constraint
46
for a guarantees that Q is non-decreasing. This implies some regularity on Q, but still leaves
some problems to apply control theory directly. First, we have to allow for (upward) jumps in
the state variable Q. Second, Q is not guaranteed to be piecewise continuous and piecewise
continuously differentiable.
These problems can be circumvented by solving the maximization problem under the
restriction that Q is Lipschitz continuous with global Lipschitz constant K:
Q ∈ LK = {Q : W → [0, 1]||Q(z) − Q(z 0 )| ≤ K|z − z 0 | ∀z, z 0 ∈ [0, 1]} .
We define the maximization problem (P K -α∗ ) as (P 0 -α∗ ) subject to an additional constraint
Q ∈ LK . It will be shown that optimal solutions of (P K -α∗ ) converge to an optimal solution
of (P 0 -α∗ ) as K → ∞. Using Lipschitz functions is convenient to show existence because
LK is sequentially compact.
We call a Q that satisfies (MON), (F0 ) and (IA0 ) a solution of (P 0 -α∗ ) and a Q that
maximizes E [wQ(w)| α∗ ] subject to (MON), (F0 ) and (IA0 ) an optimal solution of (P 0 -α∗ ).
Similarly, we call a Q ∈ LK that satisfies (MON), (F0 ) and (IA0 ) a solution of (P K -α∗ ) and a
Q ∈ LK that maximizes E [ wQ(w)| α∗ ] subject to (MON), (F0 ) and (IA0 ) an optimal solution
of (P K -α∗ ). The following proof of existence is based on Mierendorff (2009).
Lemma 12
1. An optimal solution of (P 0 -α∗ ) exists.
2. For every K > 0, an optimal solution of (P K -α∗ ) exists.
Proof.
1. Let {Qν } be a sequence of solutions of (P 0 -α∗ ) such that
Z
w
zQν (z)h(z)dz → V (α∗ ).
w
By Helly’s selection theorem, there exists a subsequence {Qνκ } and a non-decreasing
function Q such that Qνκ converges pointwise to Q. Let D collect all Q : W → [0, 1]
47
that satisfies (F0 ), (MON) and (IA0 ). Consider D as a subset of L2 (m). Recall that m
is the probability measure on corresponding to H(z). Then D is σ(L2 , L2 ) compact by
a proof similar to that of Lemma 11. Therefore, after taking subsequences again, Qνκ
converges to Q in σ(L2 , L2 ) topology and Q ∈ D. Since z ∈ L2 (m) and h ∈ L2 (m),
the weak convergence of {Qνκ } implies that
Z
w
zQ(z)h(z)dz = V (α∗ ).
w
2. Let {Qν } be a sequence of solutions of (P K -α∗ ) such that
Z
w
zQν (z)h(z)dz → V K (α∗ ).
w
After taking subsequences we can assume that Qν converges to Q pointwise and in
σ(L2 , L2 ) topology, and Q ∈ D as in part 1. Since Qν ∈ LK , for all z, z 0 ∈ W ,
|Q(z) − Q(z 0 )| = lim |Qν (z) − Qν (z 0 )| ≤ K|z − z 0 |.
ν→∞
Hence Q ∈ LK .
The next step is to show that the optimal solutions of (P K -α∗ ) converges to an optimal
solution of (P 0 -α∗ ). The proof is based on Reid (1968) and Mierendorff (2009).
Lemma 13 Let {Qν } be a sequence of optimal solutions of (P K -α∗ ) where Kν → ∞ as
ν → ∞. Then there exists a solution Q of (P 0 -α∗ ) and a subsequence Qνκ such that Qνκ
converges to Q for almost every z ∈ W . Furthermore, this solution is optimal, i.e.,
Z
w
zQ(z)h(z)dz = V (α∗ ).
w
Proof. After taking a subsequence, we can assume that Qν converges pointwise to a solution
48
Q̂ of (P 0 -α∗ ) (see the proof of Lemma 12). To show optimality of Q̂, let Q be an optimal
solution of (P 0 -α∗ ). We can extend Q to R by setting Q(z) ≡ 0 for z < w and Q(z) ≡ 1 for
z > w. Define Qd : R → [0, 1] as
1
Qd (z) ≡
d
z
Z
Q(ζ)dζ.
z−d
By the Lebesgue differentiation theorem (see, e.g., Theorem 3.21 in Folland (1999)), Qd (z) →
Q(z) for almost every z ∈ W as d → 0. Since Q is non-decreasing and Q(z) ∈ [0, 1], Qd is
1
non-decreasing and Qd (z) ∈ [0, 1]. Furthermore, Qd ∈ L d : For all z > z 0 ,
1
0 ≤ Qd (z) − Qd (z 0 ) =
d
Z
z
Z
!
z0
Q(ζ)dζ −
Q(ζ)dζ
z 0 −d
z−d
Z z
Z z−d
1
Q(ζ)dζ
Q(ζ)dζ −
=
d
z 0 −d
z0
Z
1 z
≤
Q(ζ)dζ
d z0
1
≤ (z − z 0 ).
d
Finally, Qd is feasible: For all w ∈ W ,
Z
w
H(z)n−1 − Qd (z) h(z)dz
w
Z w
Z
1 z
n−1
Q(ζ)dζ h(z)dz
=
H(z)
−
d z−d
w
Z w
≥
H(z)n−1 − Q(z) h(z)dz ≥ 0,
w
where the first inequality holds since Q is non-decreasing, and the second inequality holds
Rw
since Q satisfies (F0 ). Next, define Q̃d ≡ Qd if − w Hαi (z)Qd (z)dz ≤ C 0 (α∗ ) and otherwise
Q̃d = βd Qd + (1 − βd )/n, where
βd =
C 0 (α∗ )
−
Rw
w
Hαi (z)Qd (z)dz
49
.
Then by Lemma 7, −
Rw
w
1
Hαi (z)Q̃d (z)dz ≤ C 0 (α∗ ). Thus Q̃d is a solution of P d -α∗ . Since
βd → 0, Q̃d → Q almost everywhere as d → 0. By the dominated convergence theorem,
w
Z
w
Z
d
zQ (z)h(z)dz →
zQ(z)h(z)dz,
w
w
and
w
Z
w
Z
ν
zQ (z)h(z)dz →
z Q̂(z)h(z)dz.
w
w
Let dν = 1/Kν . Then for all ν,
Z
w
w
Z
dν
zQν (z)h(z)dz.
zQ (z)h(z)dz ≤
w
w
Hence,
Z
w
Z
w
z Q̂(z)h(z)dz =
w
A.2.1
zQ(z)h(z)dz.
w
Solution in the class LK
The problem (P K -α∗ ) can be summarized as follows:
Z
w
max
X,Y,Q,a
zQ(z)h(z)dz,
w
50
subject to
X 0 (z) = Hαi (z)Q(z),
(32)
Y 0 (z) = −[H(z)n−1 − Q(z)]h(z),
(33)
Q0 (z) = a(z),
(34)
X(w) = 0, X(w) ≥ −C 0 (α∗ ),
(35)
Y (w) = 0, Y (w) = 0,
(36)
Q(w) ≥ 0, Q(w) ≤ 1,
(37)
0 ≤ a(z) ≤ K,
(38)
Y (z) ≥ 0.
(39)
We say that some property holds virtually everywhere if the property is fulfilled at all z except
for a countable number of z’s. We use the following abbreviation for “virtually everywhere”:
v.e. By Theorem 6.7.15 in Seierstad and Sydsæter (1987), we have
Lemma 14 Let (X, Y, Q, a) be an admissible pair that solves (P K -α∗ ). Then there exist a
number λ0 , vector functions (λX , λ̌Y , λQ ) and (η a , η a ) and a non-decreasing function ηY , all
51
having one-sided limits everywhere, such that the following condition holds:
λ0 = 0 or λ0 = 1,
(40)
(λ0 , λX (z), λ̌Y (z), λQ (z), ηY (w) − ηY (w)) 6= 0, ∀z,
(41)
λQ (z)a(z) ≥ λQ (z)a, ∀a ∈ (0, K), v.e.
(42)
λQ (z) − η a (z) + η a (z) = 0.
(43)
ηY (z) is constant on any interval where Y (z) > 0.
(44)
λX (z) and λQ (z) are continuous.
(45)
λ0X (z) = 0, v.e.
(46)
Hαi (z)
0
+ λ̌Y (z) h(z) + ηY (z)h(z), v.e.
λQ (z) = − λ0 z + λX (z)
h(z)
(47)
λ̌Y (z) + ηY (z) is continuous,
λ̌0Y (z) + ηY0 (z) = 0, v.e.
(48)
(49)
λX (w) ≥ 0(= 0 if X(w) > −C 0 (α∗ )),
(50)
λQ (w) ≤ 0(= 0 if Q(w) < 1),
(51)
λQ (w) ≤ 0(= 0 if Q(w) > 0).
(52)
η a (z) ≥ 0(= 0 if a(z) > 0),
(53)
η a (z) ≥ 0(= 0 if a(z) < K).
(54)
In what follows, I assume that (X, Y, Q, a) is an admissible pair that solves (P K -α∗ ) and
(X, Y, Q, a, λ0 , λX , λ̌Y , λQ , η a , η a , ηY ) satisfy the conditions in Lemma 14.
Since λX is continuous and λ0X (z) = 0 virtually everywhere, λX (z) is constant on [w, w],
denoted by λX . Then (50) is equivalent to
λX ≥ 0(= 0 if X(w) > −C 0 (α∗ )).
Similarly, since λ̌Y + ηY is continuous and λ̌0Y (z) + ηY0 (z) = 0 virtually everywhere, λ̌Y (z) +
52
ηY (z) is constant on [w, w]. We can assume without loss of generality that λ̌Y (z) + ηY (z) = 0
for all z ∈ [w, w]. Let λY ≡ 2λ̌Y . Then ηY = −λY /2 and condition (44) is equivalent to
λY (z) is constant on any interval where Y (z) > 0,
and (47) is equivalent to
λ0Q (z)
Hαi (z)
= − λ0 z + λX (z)
+ λY (z) h(z), v.e.
h(z)
Furthermore, ηY is non-decreasing if and only if λY is non-increasing. Since λY has one-sided
limits everywhere, we can assume without loss of generality that λY (w) = limz→w λY (z) and
λY (w) = limz→w λY (z). Finally, (42), (43), (53) and (54) can be simplified to for virtually
all z ∈ (w, w): If 0 < a(z) < K, then λQ (z) = η a (z) = η a (z) = 0. If a(z) = 0, then η a (z) = 0
and −η a (z) = λQ (z) ≤ 0. If a(z) = K, then η a (z) = 0 and η a (z) = λQ (z) ≥ 0.
Lemma 15 λQ (w) = λQ (w) = 0.
Proof. By the transversality condition (52), λQ (w) ≤ 0 and equality holds if Q(w) > 0.
Suppose to the contrary that λQ (w) < 0. Then Q(w) = 0. By continuity there exists δ > 0
such that λQ (z) < 0 for all z ∈ (w, w + δ). Hence a(z) = 0 for all z ∈ (w, w + δ). This
implies that Q(z) = 0 for all z ∈ (w, w + δ). Let z ∈ (w, w + δ), then
Z
z
0 = Y (w) =
H(ζ)n−1 h(z)dz + Y (z) > Y (z),
w
a contradiction. Hence λQ (w) = 0. A similar argument proves that λQ (w) = 0.
Lemma 16 (Non-triviality) λ0 = 1.
Proof. Suppose to the contrary that λ0 = 0. Then
λ0Q (z)
Hαi (z)
= − λX
+ λY (z) h(z), v.e.
h(z)
53
Hence,
Z
w
λQ (w) =λQ (w) −
w
w
Hαi (z)
λX
+ λY (z) h(z)dz,
h(z)
Z
=λQ (w) −
λY (z)h(z)dz.
w
Since λQ (w) = λQ (w) = 0, we have
Z
w
λY (z)h(z)dz = 0.
w
Then it must be that λY (w) ≥ 0 and λY (w) ≤ 0. Suppose λX > 0. Then since Hαi (z)/h(z)
is strictly decreasing and λY (z) is non-increasing, λQ (H −1 (·)) is strictly convex. Hence
λQ (z) < 0 for all z ∈ (w, w) and therefore a(z) = 0 for all z ∈ (w, w). That is, Q is constant.
However, when Q is constant X(w) > −C 0 (α∗ ), a contradiction to that λX > 0. Suppose
λX = 0. Then
Z
z
λQ (z) = −
λY (ζ)dζ.
w
Suppose λY (w) = λY (w) = 0, then λY (z) = 0 for all z ∈ (w, w) and therefore λQ (z) = 0 for
all z ∈ (w, w). Then
(λ0 , λX (z), λY (z), λQ (z), ηY (w) − ηY (w)) = 0, ∀z.
A contradiction. Hence λY (w) > 0 and λY (w) < 0. Thus λQ (z) < 0 for all z ∈ (w, w) and
therefore Q is constant over z ∈ (w, w). This implies that λY is constant over z ∈ (w, w), a
contradiction to that λY (w) > 0 and λY (w) < 0.
To summarize, we have the following corollary:
Corollary 2 Let (X, Y, Q, a) be an admissible pair for (P K -α∗ ). If (X, Y, Q, a) is optimal,
then there exist a constant λX , a continuous and piecewise continuously differentiable func-
54
tion λQ , and a non-increasing function λY such that the following holds:
λX ≥ 0 (= 0 if X(w̄) > −C 0 (α∗ )).
Hαi (z)
0
λQ (z) = − z + λX
+ λY (z) h(z), v.e.
h(z)
(55)
λY is constant on any interval Y (z) > 0.
(57)
λQ (w) ≤ 0(= 0 if Q(w) < 1).
(58)
λQ (w) ≤ 0(= 0 if Q(w) > 0).



=0
if λQ (z) ≤ 0,



a(z) =
∈ [0, K] if λQ (z) = 0, v.e.




 =K
if λQ (z) ≥ 0.
(59)
For each λX ≥ 0 and t ∈ [0, 1], define
ϕλX (t) ≡ H −1 (t) + λX
Hαi (H −1 (t))
,
h(H −1 (t))
and
J
λX
t
Z
ϕλX (τ )dτ.
(t) ≡
0
For each w ∈ W , define
Z
w
mY (w) ≡ −
λY (z)h(z)dz.
w
It follows from (56) that for z, z ∈ W
z
Hαi (z)
λQ (z) =λQ (z) −
z + λX
+ λY (z) h(z)dz,
h(z)
z
Z z
λ
=λQ (z) −
ϕ X (H(z)) + λY (z) h(z)dz.
Z
z
When z = w, we have
λQ (z) = λQ (w) − J λX (H(z)) + mY (z).
55
(56)
(60)
Hence, for virtually all z ∈ (w, w),



=0



a(z) =
∈ [0, K]




 =K
if λQ (w) + mY (z) ≤ J λX (H(z)),
if λQ (w) + mY (z) = J λX (H(z)),
if λQ (w) + mY (z) ≥ J λX (H(z)).
Lemma 17 (Reid) Suppose λQ (w) + mY (H −1 (t)) = J λX (t) for t ∈ {t, t}. Let a, b ∈ R and
l(t) = a + bt. If J λX (t) ≥ l(t) for all t ∈ [t, t], then
λQ (w) + mY (H −1 (t)) ≥ l(t), ∀t ∈ [t, t̄].
Proof. Suppose that λQ (w) + mY (H −1 (t)) < l(t) for some t ∈ (t, t). Then there exists ε > 0
and t < t1 < t2 < t such that λQ (w) + mY (H −1 (τ )) < l(τ ) − ε for τ ∈ (t1 , t2 ), and
λQ (w) + mY (H −1 (t1 )) = l(t1 ) − ε,
λQ (w) + mY (H −1 (t2 )) = l(t2 ) − ε.
This implies that λY ((H −1 (τ )) = −m0Y (H −1 (τ )) cannot be constant on (t1 , t2 ). On the other
hand, λQ (w) + mY (H −1 (τ )) < l(τ ) − ε < J λX (t) for τ ∈ (t1 , t2 ). Hence a(H −1 (τ )) = 0 for
τ ∈ (t1 , t2 ), which implies that Y (H −1 (τ )) > 0 over the interval (t1 , t2 ). To see this, note that
Y 0 (z) = Q(z) − H n−1 (z) is strictly decreasing if Q is constant. Thus, Y is strictly concave on
(H −1 (t1 ), H −1 (t2 )). For any τ ∈ (t1 , t2 ) there exists λ ∈ (0, 1) such that H −1 (τ ) = λH −1 (t1 )+
(1 − λ)H −1 (t2 ). Then Y (H −1 (τ )) > λY (H −1 (t1 )) + (1 − λ)Y (H −1 (t2 )) ≥ 0. By (57),
Y (H −1 (τ )) > 0 on (t1 , t2 ) implies that λY (H −1 (τ )) is constant on (t1 , t2 ), a contradiction.
λX
(t), where
An immediate implication of the Lemma is that λQ (w) + mY (H −1 (t)) ≥ J[t,t]
λX
J[t,t]
(t) denotes the convex hull of J λX restricted to [t, t]:
λX
J[t,t]
(t) ≡ min βJ(t1 ) + (1 − β)J(t2 )|t1 , t2 ∈ [t, t], βt1 + (1 − β)t2 = t .
56
Furthermore, λQ (w) + mY (H −1 (t)) is convex because λY is non-increasing. This yields the
following corollary:
Corollary 3 If λQ (w)+mY (H −1 (t)) ≤ J λX (t) for all t ∈ [t, t], with equality at the endpoints
λX
of the interval, then λQ (w) + mY (H −1 (t)) = J[t,t]
(t) for all t ∈ [t, t].
For each t ∈ [0, 1], let J
λX
λX
(t) ≡ J[0,1]
(t). Since J
λX
is convex, it is continuously differen-
tiable virtually everywhere. Define ϕλX as follows. For all t ∈ (0, 1) such that dJ
exists, let ϕλX (t) ≡ dJ
λX
λX
(t)/dt
(t)/dt, and extend ϕλX to all [0, 1] by right continuity. Note that
ϕλX (t) is non-decreasing. Combining Lemma 15 and Lemma 17 yields:
Corollary 4 For all t ∈ [0, 1],
λQ (w) + mY (H −1 (t)) ≥ J
λX
(t).
Lemma 18 (interior solution) Suppose a(z) ∈ (0, K) for z ∈ (z, z), then λY (z) =
−ϕλX (H(z)) for virtually every z ∈ (z, z).
Proof. If a(z) ∈ (0, K) for z ∈ (z, z), then λQ (w) + mY (z) = J λX (H(z)) for virtually every
z ∈ (z, z). Differentiating this with respect to z yields for virtually every z ∈ (z, z):
−λY (z)h(z) = ϕλX (H(z))h(z),
i.e. −λY (z) = ϕλX (H(z)) since h(z) > 0.
Lemma 19 (constant Q) Suppose a(z) = 0 on (z, z) with z < z and let (z, z) be chosen
maximally. Then
λQ (z) = 0,
λQ (w) + mY (z) = J λX (H(z)).
57
for z = z if z > w and z = z if z < w. Furthermore,
ϕλX (H(z)) + λY (z − ) ≥ 0, if z > w,
ϕλX (H(z)) + λY (z + ) ≤ 0, if z < w.
Proof. Since a(z) = 0 on (z, z), then
λQ (w) + mY (z) ≥ J λX (H(z)), v.e. z ∈ (z, z).
Suppose z > w and let S− ≡ {z < z|a(z) > 0}. Since (z, z) is chosen maximally, Q(z) < Q(z)
for all z < z. Furthermore, since Q is absolutely continuous, S− ∩ [z − δ, z] has positive
measure for every δ > 0. Hence, there exists a sequence {zk } ∈ S− converging to z with
λQ (w) + mY (zk ) ≥ J λX (H(zk )) for all k. By continuity, λQ (w) + mY (z) = J λX (H(z)) if
z > w. A similar argument proves that λQ (w) + mY (z) = J λX (H(z)) if z < w.
Furthermore, then for virtually all z ∈ S−
0 = λQ (z)
z
= λQ (z) −
ϕλX (H(ζ)) + λY (ζ) h(ζ)dζ
z
Z z
λ
≥−
ϕ X (H(ζ)) + λY (ζ) h(ζ)dζ.
Z
z
Thus, there exists a sequence {zk } ∈ S− converging to z such that
Z
z
ϕλX (H(ζ)) + λY (ζ) h(ζ)dζ ≥ 0, ∀k.
zk
Hence,
ϕ(H(z)) + λY (z − ) ≥ 0, if z > w.
58
A similar argument proves that
ϕ(H(z)) + λY (z + ) ≤ 0, if z < w.
Lemma 19 implies that there cannot be an interval where Q is constant if ϕλX (H(·)) is
strictly increasing.
Lemma 20 (a(z)=K) Suppose a(z) = K on (z, z) with z < z and let (z, z) be chosen
maximally. Then
λQ (z) = 0,
λQ (w) + mY (z) = J λX (H(z)).
for z = z if z > w and z = z if z < w. Furthermore,
ϕ(H(z)) + λY (z − ) ≤ 0, if z > w,
ϕ(H(z)) + λY (z + ) ≥ 0, if z < w,
and
ϕ(H(z)) + λY (z + ) ≤ 0, if z = w,
ϕ(H(z)) + λY (z − ) ≥ 0, if z = w.
Proof. The proofs for the case z > w and the case z < w are very similar to that of Lemma
19 and neglected here. We now show that the third inequality holds for z = w. Note that
59
by the transervasality condition, λQ (w) ≤ 0. For virtually all z ∈ (w, z),
0 ≤ λQ (z)
z
= λQ (w) −
ϕλX (H(ζ)) + λY (ζ) h(ζ)dζ
w
Z z
λ
≤−
ϕ X (H(ζ)) + λY (ζ) h(ζ)dζ,
Z
w
i.e.,
Z
z
[ϕ(H(ζ)) + λY (ζ)] h(ζ)dζ ≤ 0, v.e. z ∈ (w, z).
w
Let z goes to w and this proves the third inequality for z = w. A similar argument proves
the fourth inequality for z = w.
Lemma 21 If K > K ≡ maxz∈W (n − 1)H(z)n−2 h(z), then
−1
Hαi (z)
∂
−
.
λX ≤ λX = min
z∈W ∂z
h(z)
Proof. Suppose to the contrary that λX > λX . Then ϕλX (H(z)) is strictly decreasing.
Suppose there exists an interval (z, z) such that a(z) ∈ (0, K) for z ∈ (z, z). Then by Lemma
18 λY (z) = −ϕλX (H(z)) for virtually every z ∈ (z, z). Thus, λY is strictly increasing on
(z, z), a contradiction. Since a is piecewise continuous by assumption, a(z) ∈ {0, K} for
almost every z ∈ W .
Suppose there exists an interval (z, z) such that a(z) = K on (z, z) and let (z, z) be
chosen maximally. Then Y 0 (z) = Q(z) − H(z)n−1 which is strictly increasing since K >
maxz∈W (n − 1)H(z)n−2 h(z), and therefore Y (z) is strictly convex on [z, z]. This implies
that Y (z) > 0 on [z, z] except at most one point. Suppose Y (z) > 0 for all z ∈ (z, z).
Rt Then λY is constant on (z, z), and λQ (H −1 (t)) = λQ (z) − H(z) ϕλX (τ ) + λY (H −1 (τ )) dτ is
strictly convex on (H(z), H(z)). By Lemma 19 and the transversality condition, λQ (z) ≤ 0
and λQ (z) ≤ 0. Then the strict convexity of λQ (H −1 (t)) implies that λQ (z) < 0 for all
z ∈ (z, z). However, a(z) = K on (z, z) implies that λQ (z) ≥ 0 for virtually every z ∈ (z, z),
60
a contradiction. Hence, there exists a unique z0 ∈ (z, z) such that Y (z0 ) = 0, and therefore
Y (z) > 0 and Y (z) > 0. Since Y (w) = Y (w) = 0 we have w < z < z < w. Note that this
also implies that λY is constant on a neighborhood of and therefore continuous at z ∈ {z, z}.
By Lemma 20, we have
ϕλX (H(z)) + λY (z) ≤ 0,
ϕλX (H(z)) + λY (z) ≥ 0.
Hence,
λY (z) ≤ −ϕλX (H(z)) < −ϕλX (H(z)) ≤ λY (z),
where the second inequality holds since ϕλX is strictly decreasing and H is strictly increasing.
However, this is a contradiction to that λY is non-increasing. Hence, a(z) = 0 for almost all
z ∈ W.
Since Q is absolutely continuous, this implies that Q is constant on W . However, by
Lemma 7, X(w) = 0 > −C 0 (α∗ ) when Q is constant on W , which implies that λX = 0, a
contradiction to the supposition that λX > λ̄X > 0. Hence, λX ≤ λ̄X .
A.2.2
Proof of Theorem 2
Let Qν be a sequence of optimal solutions of P Kν where K < Kν → ∞ and let Q∞ denote
the almost everywhere limit of the sequence. Denote the joint variables in these solutions
ν
∞
by λν , and let λ∞
is an optimal solution. I show that any
X = limν→∞ λX . By Lemma 13, Q
Q satisfying condition in Theorem 2 yields the same expected social surplus as Q∞ .
Lemma 22 Let
ν
ϕ (t) ≡ H
−1
(t) +
Hα (H −1 (t))
λνX i −1
h(H
(t))
61
ν
Z
and J (t) ≡
0
t
ϕν (τ )dτ,
and
∞
ϕ (t) ≡ H
−1
(t) +
Hαi (H −1 (t))
λ∞
X
−1
h(H
(t))
∞
Z
and J (t) ≡
t
ϕ∞ (τ )dτ.
0
Then the sequence ϕν is uniformly convergent with limit ϕ∞ , the sequence J ν is uniformly
convergent with limit J ∞ , and the sequence ϕν0 is uniformly convergent with limit ϕ∞0 .
Proof. Let γ1 ≡ maxz∈W |Hαi (z)/h(z)| > 0,
∂
Hαi (z) γ2 ≡ max −
> 0,
z∈W ∂z
h(z) and γ3 ≡ maxz∈W 1/h(z) > 0. Here γ1 , γ2 and γ3 are well define since H and h are twice
continuously differentiable and W is compact.
ν
∞
|ϕ (t) − ϕ (t)| =
|λνX
Hαi (H −1 (t)) λ∞
|
X −1
−
h(H
(t))
≤ γ1 |λνX − λ∞
X | → 0,
as ν → ∞. Hence, the sequence ϕν is uniformly convergent with limit ϕ∞ .
Z
∞
ν
t
|J (t) − J (t)| =
|ϕν (τ ) − ϕ∞ (τ )| dτ
0
≤ tγ1 |λνX − λ∞
X|
≤ γ1 |λνX − λ∞
X | → 0,
as ν → 0. Hence, the sequence J ν is uniformly convergent with limit J ∞ .
ν0
∞0
|ϕ (t) − ϕ (t)| =
|λνX
−
∂ Hαi (H −1 (t))
1
≤ γ2 γ3 |λνX − λ∞
X | → 0,
−1
−1
∂z h(H (t)) h(H (t)) λ∞
X|
as ν → ∞. Hence, the sequence ϕν0 is uniformly convergent with limit ϕ∞0 .
∞
Lemma 23 Suppose J ∞ (H(z)) > J (H(z)) for z ∈ (z, z) with z < z and let (z, z) be
chosen maximally. Then Q∞ is constant on (z, z).
∞
Proof. For each 0 < δ < (z − z)/2, let ε(δ) ≡ minz∈[z+δ,z−δ] {J ∞ (H(z)) − J (H(z))}. Then
62
ε(δ) is non-increasing in δ and converges to zero as δ converges to zero. Fix δ0 > 0. Let
ε0 ≡ 14 ε(δ0 ) > 0. There exist 0 < δ1 < δ2 < δ0 such that ε(δ1 ) = ε0 and ε(δ2 ) = 2ε0 . Since
the sequence J ν is uniformly convergent with limit J ∞ , there exists ν such that for all ν > ν,
7ε0
if z ∈ [z + δ0 , z − δ0 ],
2
ε0
if z ∈ [z + δ1 , z − δ1 ],
J ν (H(z)) − J¯ν (H(z)) ≥
2
5ε0
∞
if J ∞ (H(z)) − J (H(z)) ≤ 2ε0 .
J ν (H(z)) − J¯ν (H(z)) ≤
2
J ν (H(z)) − J¯ν (H(z)) ≥
We begin by prove the second inequality. Since the sequence J ν is uniformly convergent
with limit J ∞ , there exists ν such that for all ν > ν, |J ν (t) − J ∞ (t)| < ε0 /8 for all t ∈ [0, 1].
∞
Let t ∈ [H(z +δ1 ), H(z −δ1 )]. Then, by construction, J ∞ (t)−J (t) > ε0 . Hence, there exists
β, t1 , t2 ∈ [0, 1] such that βt1 + (1 − β)t2 = t and βJ ∞ (t1 ) + (1 − β)J ∞ (t2 ) < J ∞ (t) − 3ε0 /4.
Then
J¯ν (t) ≤βJ ν (t1 ) + (1 − β)J ν (t2 )
≤βJ ∞ (t1 ) + β|J ν (t1 ) − J ∞ (t1 )| + (1 − β)J ∞ (t2 ) + (1 − β)|J ν (t2 ) − J ∞ (t2 )|
≤βJ ∞ (t1 ) + (1 − β)J ∞ (t2 ) +
≤J ∞ (t) −
ε0
8
3ε0 ε0
+
4
8
≤J ν (t) + |J ν (t) − J ∞ (t)| −
5ε0
8
ε 5ε0
−
8
8
ε
0
=J ν (t) − .
2
≤J ν (t) +
Hence, J ν (H(z)) − J¯ν (H(z)) ≥ ε/2 for all z ∈ [z + δ1 , z − δ1 ]. A similar argument proves the
∞
first inequality. To show the third inequality, let z be such that J ∞ (H(z))−J (H(z)) ≤ 2ε0 .
63
For any β, t1 , t2 ∈ [0, 1] such that βt1 + (1 − β)t2 = t, we have
βJ ν (t1 ) + (1 − β)J ν (t2 )
≥βJ ∞ (t1 ) − β|J ν (t1 ) − J ∞ (t1 )| + (1 − β)J ∞ (t2 ) − (1 − β)|J ν (t2 ) − J ∞ (t2 )|
≥βJ ∞ (t1 ) + (1 − β)J ∞ (t2 ) −
∞
≥J (t) −
ε0
8
ε0
8
∞
=J ∞ (t) − [J ∞ (t) − J (t)] −
≥J ∞ (t) − 2ε0 −
ε0
8
ε0
,
8
≥J ν (t) − |J ν (t) − J ∞ (t)| − 2ε0 −
ε0
,
8
ε0
ε0
− 2ε0 − ,
8
8
5ε
0
≥J ν (t) −
.
2
≥J ν (t) −
Hence,
ν
J (t) ≡ min{βJ ν (t1 ) + (1 − β)J ν (t2 )|β, t1 , t2 ∈ [0, 1] and βt1 + (1 − β)t2 = t} ≥ J ν (t) −
5ε0
.
2
Since ε(δ1 ) = ε0 and ε(δ2 ) = 2ε0 , by continuity
∞
mδ ≡ min m({z ∈ [z + δ1 , z + δ2 ]|J ∞ (H(z)) − J (H(z)) ≤ 2ε0 }),
∞
m({z ∈ [z − δ2 , z − δ1 ]|J ∞ (H(z)) − J (H(z)) ≤ 2ε0 }) > 0.
Fix ν > ν such that Kν > 1/mδ . Suppose there exists (b1 , b2 ) ⊂ (z + δ0 , z − δ0 ) such that
aν (z) > 0. Then λQ (b1 ), λQ (b2 ) ≥ 0. Since λνY is non-increasing, we have λνY (b2 ) ≤ λνY (b1 ).
ν
Note that J is linear and therefore ϕν is constant on (z, z). Hence, we have either −λνY (b2 ) ≥
ϕν (H(z)) for all z ∈ (z, z) or −λνY (b1 ) ≤ ϕν (H(z)) for all z ∈ (z, z). Assume without loss
of generality that −λνY (b2 ) ≥ ϕν (H(z)) for all z ∈ (z, z). For any z ∈ [z − δ2 , z − δ1 ] with
64
∞
J ∞ (H(z)) − J (H(z)) ≤ 2ε0 , we have
λνQ (z)
Z
z
[λνY (ζ) + ϕν (H(ζ))] h(ζ)dζ
b2
Z z
Z z
ν
≥
ϕν (H(ζ))h(ζ)dζ
ϕ (H(ζ))h(ζ)dζ −
=
λνQ (b2 )
−
b2
b2
ν
ν
= J (H(z)) − J (H(b2 )) − J ν (H(z)) + J ν (H(b2 ))
ν
ν
= J ν (H(b2 )) − J (H(b2 )) − J ν (H(z)) − J (H(z))
≥
7ε0 5ε0
−
= ε0 > 0.
2
2
∞
That is, aν (z) = Kν for almost every z ∈ [z − δ2 , z − δ1 ] with J ∞ (H(z)) − J (H(z)) ≤ 2ε0 .
However, this is a contradiction to that Kν > 1/mδ since 0 ≤ Q ≤ 1.
Hence, aν (z) = 0 for almost every z ∈ [z − δ0 , z + δ0 ] for ν sufficiently large. Let ν goes
to infinity and we have Q∞ is constant on [z − δ0 , z + δ0 ]. Since this is true for any δ > 0,
we have that Q∞ is constant on (z, z).
∞
ν
Lemma 24 The sequence J is uniformly convergent with limit J .
Proof. Since the sequence J ν is uniformly convergent with limit J ∞ , for any ε > 0 there
exists ν > 0 such that for all ν > ν, |J ∞ (t) − J ν (t)| ≤ ε for all t ∈ [0, 1]. Fix t ∈ [0, 1]. Let
t1 , t2 , β ∈ [0, 1] be such that βt1 + (1 − β)t2 = t. Then for any ν > ν
∞
J (t) ≤βJ ∞ (t1 ) + (1 − β)J ∞ (t2 )
≤βJ ν (t1 ) + (1 − β)J ν (t2 ) + β|J ∞ (t1 ) − J ν (t1 )| + (1 − β)|J ∞ (t2 ) − J ν (t2 )|
≤βJ ν (t1 ) + (1 − β)J ν (t2 ) + ε
Hence,
∞
J (t) ≤ min {βJ ν (t1 ) + (1 − β)J ν (t2 )|βt1 + (1 − β)t2 = t} + ε = J ν (t) + ε.
65
∞
ν
∞
ν
Similarly, we can show that J (t) ≤ J (t) + ε. Hence, |J (t) − J (t)| ≤ ε. Since this holds
∞
ν
for any t ∈ [0, 1], we have that the sequence J is uniformly convergent with limit J .
Lemma 25 Suppose Y ∞ (z) > 0 for all z ∈ (z, z) with z < z and let (z, z) be chosen
maximally. Then ϕ∞ is constant on (H(z), H(z)).
Proof. Suppose to the contrary that ϕ∞ (H(z)− ) > ϕ∞ (H(z)). Since ϕ∞ is non-decreasing
and right-continuous, there exists δ > 0 such that ϕ∞ (H(z − δ)) > ϕ∞ (H(z + δ)) for all
δ ∈ (0, δ). Fix δ ∈ (0, min{δ/2, (z − z)/4}). Since J
∞
is convex and ϕ∞ is non-constant on
(z + δ, z − δ), we have
∞
∞
J (H(z − δ)) − J (z + δ)
J (H(z)) < J (H(z + δ)) + [z − (z + δ)]
, ∀z ∈ (z + δ, z − δ).
z − z − 2δ
∞
∞
Let
(
ε1 ≡
min
∞
∞
J (H(z + δ)) + [z − (z + δ)]
z∈[z+2δ,z−2δ]
ε2 ≡
∞
)
J (H(z − δ)) − J (z + δ)
∞
− J (H(z))
z − z − 2δ
> 0,
Y ∞ (z) > 0,
min
z∈[z+δ,z−δ]
and
M ≡ 2 max |ϕ∞ (H(z))| > 0.
z∈[z,z]
∞
ν
Since the sequence J is uniformly convergent with limit J , the sequence Y ν is uniformly
convergent with limit Y ∞ and the sequence ϕν is uniformly convergent with limit ϕ∞ , there
∞
ν
exists ν such that for ν > ν, |Y ν (z)−Y ∞ (z)| < ε2 /2 for all z ∈ W , |J (t)−J (t)| ≤ ε1 /8 for
all t ∈ [0, 1] and |ϕν (t)−ϕ∞ | ≤ M/2 for all t ∈ [0, 1]. Then for all ν > ν and z ∈ [z+2δ, z−2δ]
66
we have
ν
ν
J (H(z − δ)) − J (H(z + δ))
ν
J (H(z + δ)) + [z − (z + δ)]
− J (H(z))
z − z − 2δ
∞
ν
∞
ν
∞
∞
≥J (H(z + δ)) − J (H(z + δ)) − J (H(z + δ)) − J (H(z)) − J (H(z)) − J (H(z))
i
z − (z + δ) h ∞
ν
∞
ν
∞
∞
+
J (H(z − δ)) − J (H(z − δ)) − J (H(z − δ)) − J (H(z + δ)) − J (H(z + δ)) − J (H(z + δ))
z − z − 2δ
ν
∞
∞
J (H(z − δ)) − J (H(z + δ))
ε1
∞
≥J (H(z + δ)) + [z − (z + δ)]
− J (H(z)) −
z − z − 2δ
2
ε1
≥ .
2
∞
For all ν > ν and z ∈ [z + δ, z − δ] we have
Y ν (z) ≥ Y ∞ (z) − |Y ν (z) − Y ∞ (z)| ≥
ε2
> 0.
2
For all ν > ν and z ∈ [z + δ, z − δ], we have
ϕν (H(z)) ≤ ϕ∞ (H(z)) + |ϕν (H(z)) − ϕ∞ (H(z))| ≤
M1 M1
+
= M1 .
2
2
Finally, let ε3 ≡ minz∈[z,z] h(z) > 0 and
ν
J (H(z − δ)) − J ν (H(z + δ)) M2 ≡ > 0.
z − z − 2δ
Fix ν > ν such that Kν > 1/ min{ε1 /8ε3 M, ε1 /8M2 , δ}. Since Y ν (z) > 0 on [z + δ, z − δ],
λY is constant and therefore mY is affine on [z + δ, z − δ]. By Corollary 4, we have mνY (z) ≥
ν
ν
J (H(z)) for all z ∈ W . In particular, mνY (z + δ) ≥ J (H(z + δ)) and mνY (z − δ) ≥
ν
J (H(z − δ)). Furthermore, mY is affine on [z + δ, z − δ]. Hence, for all z ∈ [z + δ, z − δ],
ν
ν
mνY (z) ≥ J (H(z + δ)) + [z − (z + δ)]
ν
ν
J (H(z − δ)) − J (z + δ)
.
z − z − 2δ
Suppose J ν (H(z0 )) = J (H(z0 )) for some z0 ∈ [z + 2δ, z − 2δ]. Then for all z ∈
67
(z0 , min {z0 + ε1 /8M2 , z − δ}) such that H(z) − H(z0 ) ≤ ε1 /8M1 we have
ν
Z
ν
H(z)
ϕν (τ )dτ
J (H(z)) =J (H(z0 )) +
H(z0 )
ν
≤J (H(z0 )) + M1 (H(z) − H(z0 ))
ν
ν
J (H(z − δ)) − J (H(z + δ)) ε1 ε1
≤J (H(z + δ)) + [z − (z + δ) − (z − z0 )]
− +
z − z − 2δ
2
8
ε1 ε1
≤mνY (z) + (z − z0 )M2 − +
2
8
ε
ε
ε
ε1
1
1
1
≤mνY (z) + − +
= mνY (z) −
8
2
8
4
ν
That is, J ν (H(z)) < mνY (z) and therefore aν (z) = Kν . Since H(z) − H(z0 ) ≥ ε3 (z − z0 ), we
have aν (z) = Kν for all z ∈ (z0 , min{z0 + ε1 /8ε3 M, z0 + ε1 /8M2 , z − δ}), a contradiction to
that Kν > 1/ min{ε1 /8ε3 M, ε1 /8M2 , δ} since 0 ≤ Q ≤ 1.
ν
Hence, J ν (H(z)) > J (H(z)) for all z ∈ [z + 2δ, z − 2δ]. This implies that ϕν is constant
on [H(z − 2δ), H(z + 2δ)]. Clearly, ϕν converges uniformly on [H(z − 2δ), H(z + 2δ)] and
ν
limν→∞ ϕν is constant on [H(z−2δ), H(z+2δ)]. Since, on [H(z−2δ), H(z+2δ)], J converges
∞
ν
uniformly to J , each J is differentiable with derivative ϕν , and ϕν converges uniformly, we
have J
∞
is differentiable on [H(z − 2δ), H(z + 2δ)] and its derivative ϕ∞ (t) = limν→∞ ϕν (t)
for all t ∈ [H(z − 2δ), H(z + 2δ)]. Thus, ϕ∞ is constant on [H(z − 2δ), H(z + 2δ)] and 2δ < δ,
a contradiction to the supposition. Hence, ϕ∞ is constant on (H(z), H(z)).
Corollary 5 Suppose ϕ∞ (H(z)) is constant on (z, z) with z < z and let (z, z) be chosen
maximally. Then Y ∞ (z) = Y ∞ (z) = 0, i.e.,
Z
z
H(ζ)n−1 − Q∞ (ζ) h(ζ)dζ = 0.
z
Proof. This is an immediate corollary of Lemma 25. Suppose to the contrary Y ∞ (z) > 0.
Then by Lemma 25, ϕ∞ (H(·)) is constant on a neighborhood of z, a contradiction. Hence
Y ∞ (z) = 0. Similarly, Y ∞ (z) = 0.
68
Let Q be a non-decreasing implementable allocation rule that satisfies the following two
pooling properties:
∞
1. If J ∞ (H(z)) > J (H(z)) for z ∈ (z, z) with z < z and let (z, z) be chosen maximally,
then Q is constant on (z, z).
2. If Y (z) > 0 for all z ∈ (z, z) with z < z and let (z, z) be chosen maximally. Then ϕ∞
is constant on (H(z), H(z)).
Then
Z
w
z+
w
Z w
=
Hαi (z)
λ∞
X
h(z)
0
∗
Q(z)h(z)dz + λ∞
X C (α )
0
∗
ϕ∞ (H(z))Q(z)dH(z) + λ∞
X C (α )
w
w
Z
∞
Z
∞
w
[ϕ (H(z)) − ϕ (H(z))] Q(z)dH(z) +
=
w
w
Z
=
w
ϕ∞ (H(z))Q(z)dH(z) −
w
w
Z
=
Z
w
0
∗
ϕ∞ (H(z))Q(z)dH(z) + λ∞
X C (α )
∞
n−1
ϕ (H(z))H(z)
=
=
=
0
∗
ϕ∞ (H(z))Y 0 (z)dz + λ∞
X C (α )
dH(z) +
w
w
ϕ∞ (H(z))H(z)n−1 dH(z) −
w
w
Z
w
Z
w
w
Z
∞
∞
0
∗
J (H(z)) − J (H(z)) dQ(z) + λ∞
X C (α )
w
w
w
Z
0
∗
ϕ∞ (H(z))Q(z)dH(z) + λ∞
X C (α )
Z
0
∗
Y (z)dϕ∞ (H(z)) + λ∞
X C (α )
w
0
∗
ϕ∞ (H(z))H(z)n−1 dH(z) + λ∞
X C (α ).
w
The second line follows from the definition of ϕ∞ . The third line follows from integration by
parts. The fourth line holds since Q satisfies the first pooling property. The fifth line follows
from the definition of Y . The sixth line follows from integration by parts. The seventh line
holds since Q satisfies the second pooling property. Suppose in addition that
Z
w
w
Hαi (z)
Q(z)h(z)dz = −C 0 (α∗ ).
h(z)
69
(61)
Then
Z
w
zQ(z)h(z)dz
w
Z w
z+
=
Hαi (z)
λ∞
X
h(z)
w
w
Z
=
0
∗
Q(z)h(z)dz + λ∞
X C (α )
0
∗
ϕ∞ (H(z))H(z)n−1 dH(z) + λ∞
X C (α ).
w
That is, all non-decreasing implementable allocation rules Q satisfying the two pooling properties and equality (61) generate the same expected surplus. By Lemmas 23 and 25, Q∞
satisfies the two pooling properties. Furthermore, Q∞ satisfies equality (61). Hence any
non-decreasing implementable allocation rule Q satisfying the two pooling properties and
equality (61) solves problem (P-α∗ ).
∞
Define Q+ in the following way. If J ∞ (H(z)) > J (H(z)), let
+
Q (z) ≡
1
[H(z)n
n
− H(z)n ]
,
H(z) − H(z)
∞
where (z, z) with z < z < z is the maximal interval on which J ∞ (H(·)) > J (H(·)).
Otherwise, let Q+ (z) ≡ H(z)n−1 . Define Q− in the following way. If there exists (z, z) with
z < z < z on which ϕ(H(·))) is constant and let (z, z) be chosen maximally, then let
Q− (z) ≡
1
[H(z)n
n
− H(z)n ]
.
H(z) − H(z)
Otherwise, let Q− (z) ≡ H(z)n−1 . Note that all Q+ , Q− and Q∞ are implementable, nondecreasing and satisfy the two pooling properties. Hence for Q ∈ {Q+ , Q− , Q∞ }
Z
w
z+
w
Hαi (z)
λ∞
X
h(z)
Q(z)h(z)dz +
0
∗
λ∞
X C (α )
Z
w
=
0
∗
ϕ∞ (H(z))H(z)n−1 dH(z) + λ∞
X C (α ).
w
(62)
Next we show that Y + ≤ Y ∞ ≤ Y − . Let S + = {z ∈ W |Y + (z) > 0}, S − = {z ∈ W |Y − (z) >
70
0} and S = {z ∈ W |Y ∞ (z) > 0}. By construction,
∞
S + ≡ ∪{(z, z)|J ∞ (H(z)) > J (H(z)) ∀z ∈ (z, z)},
and
S − ≡ ∪{(z, z)|ϕ∞ (H(·)) is constant on (z, z)}.
It follows from Lemma 23 and Lemma 25 that S + ⊂ S ⊂ S − . If z ∈
/ S − , then Y + (z) =
Y ∞ (z) = Y − (z) = 0. If z ∈ S − \S, then Y + (z) = Y ∞ (z) = 0 < Y − (z). Consider z ∈ S ⊂
S − , then there exists (z, z) with z < z < z such that ϕ∞ (H(·)) is constant on (z, z). Let
(z, z) be chosen maximally, then by Corollary 5, Y ∞ (z) = Y ∞ (z) = 0. For any z ∈ (z, z),
−
Z
∞
Y (z) − Y (z) =
z
Q∞ (ζ) − Q− (ζ) dH(ζ).
z
Then for any t ∈ (H(z), H(z)),
− −1
0
Y (H (t)) − Y ∞ (H −1 (t)) = Q− (H −1 (t)) − Q∞ (H −1 (t)),
which is non-increasing on (H(z), H(z)) since Q∞ is non-decreasing and Q− is constant on
(z, z) by construction. Hence Y − (H −1 (t)) − Y ∞ (H −1 (t)) is concave on (H(z), H(z)). Since
Y − (z) − Y ∞ (z) = 0 and Y − (z) − Y ∞ (z) = 0, we have Y − (z) − Y ∞ (z) ≥ 0 for all z ∈ (z, z).
Thus Y − (z) − Y ∞ (z) ≥ 0 for all z ∈ S. If z ∈ S\S + , then Y − (z) ≥ Y ∞ (z) ≥ 0 = Y + (z).
Finally, consider z ∈ S + ⊂ S, it suffices to show that Y + (z) ≤ Y ∞ (z). By construction
∞
there exists (z, z) with z < z < z such that J (H(z)) < J ∞ (H(z)) for all z ∈ (z, z). Let
(z, z) be chosen maximally, then by construction Y + (z) = Y + (z) = 0. For any z ∈ (z, z)
+
∞
Z
Y (z) − Y (z) =
z
∞
Q (ζ) − Q− (ζ) dH(ζ) − Y ∞ (z).
z
71
Then for any t ∈ (H(z), H(z)),
+ −1
0
Y (H (t)) − Y ∞ (H −1 (t)) = Q+ (H −1 (t)) − Q∞ (H −1 (t)),
which is constant on (H(z), H(z)) since Q∞ is constant on (z, z) by Lemma 23 and Q− is constant on (z, z) by construction. Hence Y + (H −1 (t)) − Y ∞ (H −1 (t)) is affine on (H(z), H(z)).
Since Y + (z) = 0 ≤ Y ∞ (z) and Y + (z) = 0 ≤ Y ∞ (z), we have Y + (z) − Y ∞ (z) ≤ 0 for all
z ∈ (z, z). Thus Y + (z) − Y ∞ (z) ≤ 0 for all z ∈ S + .
Furthermore, for any implementable allocation rule Q, we have
w
Z
zQ(z)dH(z)
w
Z w
=
Z
0
w
zY (z)dz +
w
w
w
Z
=
zH(z)n−1 dH(z)
zH(z)
n−1
Z
w
dH(z) −
w
Y (z)dz.
(63)
w
Hence
Z
w
Z
+
w
zQ (z)dH(z) ≥
w
∞
Z
w
zQ (z)dH(z) ≥
w
zQ− (z)dH(z).
w
Since λ∞
X > 0, combining this and (62) yields
X + (w) ≤ −C 0 (α∗ ) ≤ X − (w).
Since X is linear in Q, there exists ρ ∈ [0, 1] such that Q̂ = ρQ+ (1 − ρ)Q− is implementable,
non-decreasing and
Z
w
w
Hαi (z)
Q̂(z)h(z)dz = −C 0 (α∗ ).
h(z)
Then both Q∞ and Q̂ yield the same expected surplus.
Lemma 26 Let t ∈ (0, 1). If J(t) = J(t), then J is continuously differentiable at t with
72
derivative ϕ(t) = ϕ(t) and ϕ0 (t) ≥ 0. Furthermore, J(t) = J(t) if and only if
J(τ ) ≥ (τ − t)ϕ(t) + J(t), ∀τ ∈ [0, 1].
(64)
Proof. Suppose J(t) = J(t). Suppose to the contrary that J is not continuously differentiable at t, then ϕ(t− ) < ϕ(t+ ). Then either ϕ(t) > ϕ(t− ) or ϕ(t) < ϕ(t+ ). Assume without
loss of generality that ϕ(t) < ϕ(t+ ). Since ϕ is continuous and ϕ is non-decreasing, there
exists δ > 0 such that ϕ(τ ) < ϕ(t+ ) ≤ ϕ(τ ) for all τ ∈ (t, t + δ). Then
Z
t+δ
Z
t+δ
ϕ(τ )dτ < J(t) +
J(t + δ) = J(t) +
ϕ(τ )dτ = J(t + δ),
t
t
a contradiction. Hence J is continuously differentiable at t. It follows from a similar argument
that ϕ(t) = ϕ(t) with ϕ0 (t) ≥ 0. Furthermore, for all τ ∈ [0, 1],
J(τ ) ≥ J(τ ) ≥ (τ − t)ϕ(t) + J(t) = (τ − t)ϕ(t) + J(t),
where the second inequality holds since J is convex.
Suppose (64) holds. Then (τ − t)ϕ(t) + J(t) is a convex function below J. Since J is the
greatest convex function below J, we have
J(τ ) ≥ (τ − t)ϕ(t) + J(t) ∀τ ∈ [0, 1].
When τ = t, we have J(t) ≥ J(t). Hence J(t) = J(t).
Lemma 27 Let λ0X > λX . Suppose ϕλX is constant on (t, t) with t < t and let (t, t) be chosen
0
maximally. Then there exists δ > 0 such that J λX (t) > J
λ0X
(t) for all t ∈ (t − δ, t + δ).
Proof. Fix t ∈ (t, t). Let β ∈ (0, 1) be such that βt + (1 − β)t = t. It suffices to show that
0
0
0
J λX (t) > βJ λX (t) + (1 − β)J λX (t).
73
Since ϕλX is constant on (t, t), we have
J λX (t) ≥ J
λX
= βJ
(t)
λX
(t) + (1 − β)J
λX
(t)
= βJ λX (t) + (1 − β)J λX (t),
where the last line holds since J
J
λX
λX
(t) = J λX (t) and J
(t) = J λX (t), suppose to the contrary that J
λX
λX
(t) = J λX (t).
To see that
(t) > J λX (t). Then ϕλX (t) is constant in
a neighborhood of t. A contradiction to that (t, t) is chosen maximally. A similar argument
proves that J
λX
(t) = J λX (t). Furthermore,
0 ≤J λX (t) − βJ λX (t) − (1 − β)J λX (t)
Z t
Z t
Z t
−1
−1
H (ζ)dζ − (1 − β)
=
H (ζ)dζ − β
H −1 (ζ)dζ
0
0
0
"Z
#
Z t
Z t
t
−1
−1
Hαi (H (ζ))
Hαi (H (ζ))
Hαi (H −1 (ζ))
+ λX
dζ − β
dζ − (1 − β)
dζ .
h(H −1 (ζ))
h(H −1 (ζ))
h(H −1 (ζ))
0
0
0
Since H −1 (·) is strictly increasing,
Z
t
H
−1
Rt
0
Z
(ζ)dζ − β
0
H −1 (ζ)dτ is strictly convex in t and therefore
t
H
−1
Z
t
(ζ)dζ − (1 − β)
H −1 (ζ)dζ < 0.
0
0
Hence
Z
0
t
Hαi (H −1 (ζ))
dζ − β
h(H −1 (ζ))
Z
0
t
Hαi (H −1 (ζ))
dζ − (1 − β)
h(H −1 (ζ))
74
Z
0
t
Hαi (H −1 (ζ))
dζ > 0.
h(H −1 (ζ))
Then
0
0
0
J λX (t) − βJ λX (t) − (1 − β)J λX (t)
Z t
Z t
Z t
−1
−1
H (ζ)dζ − (1 − β)
H −1 (ζ)dζ
H (ζ)dζ − β
=
0
0
0
"Z
Z t
Z t
t
−1
−1
Hαi (H (ζ))
Hαi (H (ζ))
0
+ λX
dζ − β
dζ − (1 − β)
h(H −1 (ζ))
h(H −1 (ζ))
0
0
0
Z t
Z t
Z t
>
H −1 (ζ)dζ − β
H −1 (ζ)dζ − (1 − β)
H −1 (ζ)dζ
0
0
0
"Z
Z t
Z t
t
Hαi (H −1 (ζ))
Hαi (H −1 (ζ))
+ λX
dζ − β
dζ − (1 − β)
h(H −1 (ζ))
h(H −1 (ζ))
0
0
0
Hαi (H −1 (ζ))
dζ
h(H −1 (ζ))
#
Hαi (H −1 (ζ))
dζ
h(H −1 (ζ))
#
≥J λX (t) − βJ λX (t) − (1 − β)J λX (t)
≥0.
Consider t. Since ϕλX is constant on (t, t) with J λX (t) = J
λX
(t) and J λX (t) = J
λX
(t),
by Lemma 26, we have
J
λX
(t) =J λX (t)
≥(t − t)ϕλX (t) + J λX (t)
=(t − t)ϕλX (t) + J
λX
(t) = J
λX
(t).
Hence, J λX (t) = (t − t)ϕλX (t) + J λX (t). Thus,
Z
"
t
H
−1
(τ )dτ − (t − t)H
−1
(t) = λX
t
Hα (H −1 (t))
(t − t) i −1
−
h(H (t))
Z
t
t
#
Hαi (H −1 (τ ))
.
h(H −1 (τ ))
Since H −1 (t) is strictly increasing, the left-hand side of the above equality is strictly positive.
Hence for λ0X > λX > 0 we have
Z
t
"
t
H −1 (τ )dτ − (t − t)H −1 (t) < λ0X
Hα (H −1 (t))
(t − t) i −1
−
h(H (t))
75
Z
t
t
#
Hαi (H −1 (τ ))
,
h(H −1 (τ ))
i.e.,
0
0
0
J λX (t) < (t − t)ϕλX (t) + J λX (t).
0
By Lemma 26, J λX (t) > J
λ0X
(t).
0
A similar argument proves that J λX (t) > J
0
that J λX (t) > J
λ0X
λ0X
(t). By continuity, there exists δ > 0 such
(t) for all t ∈ (t − δ, t + δ).
Lemma 28 Let [z, z] ⊂ [w, w] with z < z, and z 0 ∈ (z, z). Suppose Q : [θ, θ] → [0, 1] and
Q̌ : [θ, θ] → [0, 1] satisfying that
z
Z
z
Z
Q̌(z)h(z)dz,
Q(z)h(z)dz =
z
(65)
z
and
Q(z) ≥ Q̌(z) if z > z 0 , and Q(z) ≤ Q̌(z) if z < z 0 .
(66)
Then
Z
z
−
z
Hαi (z)
[Q(z) − Q̌(z)]h(z)dz ≥ 0,
h(z)
(67)
where the inequality holds strictly if the set {z ∈ [z, z]|Q(z) 6= Q̌(z)} has a positive measure.
Proof. Since −
Hαi (wi )
h(wi )
Z
z
is strictly increasing in wi , and Q and Q̌ satisfy (66), we have
z
Hαi (z) Hαi (z 0 )
−
+
[Q(z) − Q̂(z)]h(z)dz ≥ 0,
h(z)
h(z 0 )
where the inequality holds strictly if the set {z ∈ [z, z]|Q(z) 6= Q̌(z)} has a positive measure.
This implies inequality (67) by (65).
By Lemmas 27 and 28 we have
Z
w
w
Hα (z)
− i Q+ (z, λ0X )h(z)dz <
h(z)
Hence, λX is unique.
76
Z
w
−
w
Hαi (z) −
Q (z, λX )h(z)dz.
h(z)
A.3
Sufficient Conditions for the First-order Approach
In this section I provide sufficient conditions for the first-order approach to be valid. Let
π(αi ) denote an agent i’s payoff from choosing αi given mechanism (q, t) and αj = α∗ for all
j 6= i. Then
Z
w(1,αi )
[1 − H(wi |αi )] Q(wi )dwi − C(αi ),
π(αi ) ≡ U (w(0, αi )) +
w(0,αi )
where Q is defined by (2) for αj = α∗ for all j 6= i. Then
0
Z
w(1,αi )
π (αi ) =
−Hαi (wi |α)Q(wi )dwi − C 0 (αi ).
w(0,αi )
A sufficient condition for the first-order approach to be valid is that π 0 (αi ) is strictly decreasing. Note that this also implies (10) holds for Q(wi ) = H(wi |α)n−1 . If the support of
the conditional expectation [w(0, αi ), w(1, αi )] is invariant, then π 0 (αi ) is strictly decreasing
if −Hαi (wi |αi ) has the single-crossing property in (αi ; wi ) and C 0 (αi ) is strictly decreasing.
Suppose C is twice continuously differentiable, then
00
Z
w(1,αi )
−
π (αi ) =
w(0,αi )
∂ 2 H(wi |α)
Q(wi )dwi − Hαi (w(1, αi )|αi )wαi (1, αi )Q(w(1, αi ))
∂αi2
+ Hαi (w(0, αi )|αi )wαi (0, αi )Q(w(0, αi )) − C 00 (αi ),
Z w(1,αi )
∂ 2 H(wi |α)
−
≤
Q(wi )dwi − Hαi (w(1, αi )|αi )wαi (1, αi )Q(w(1, αi )) − C 00 (αi ).
2
∂α
w(0,αi )
i
The inequality holds since Hαi (w(0, αi )|αi ) ≥ 0 and wαi (0, αi ) ≤ 0 by Assumption 1. The
following proposition from Shi (2012) gives sufficient condition for π 00 (α) < 0 for the two
leading examples.
Proposition 5 (Shi (2012)) Sufficient conditions for first order approach:
• In the linear experiments, if αi C 00 (αi ) ≥ f θ (θ − µ)2 for all αi , then π 00 (αi ) < 0 when
either F (θ) is convex of F (θ) = θb (b > 0) with support [0, 1].
77
• In the normal experiments, π 00 (αi ) < 0 if
A.4
p
√
β 3 / [αi3 (αi + β)5 ] < 2 2πC 00 (αi ) for all αi .
Efficient Asymmetric Mechanisms
A.4.1
Proof of Theorem 3
As in the symmetric case, I prove Theorem 3 by proving the following two lemmas. Define
Y (w) ≡ 1 −
n
Y
i=1
H(wi |αi∗ )
−
n Z
X
i=1
w(1,α∗i )
Qi (zi )dH(zi |αi∗ ), ∀w
n
Y
∈
[w(0, αi∗ ), w(1, αi∗ )].
wi
i=1
Observe that if αi = αi∗ for all i, then (12) is violated if and only if inf w−i Y (w(0, α∗ ), w−i ) > 0
for some i. Then
Lemma 29 Suppose αi = αi∗ for all i. Let Q be any interim allocation rule satisfying (AF0 ),
(MON), (IA0 ) and inf w−i Y (w(0, α∗ ), w−i ) > 0 for some i, then there exists Q̂ satisfying (F0 ),
(MON) and (IA0 ) such that Q̂j = Qj for j 6= i and
Q̂i (wi ) ≥ Qi (wi ), ∀wi ∈ [w(0, αi∗ ), w(1, αi∗ )],
(68)
and ” > ” holds for a set of wi with positive measure.
Proof. Define Yi (wi ) ≡ inf w−i Y (w) for all wi ∈ [w(0, αi∗ ), w(1, αi∗ )]. By Theorem 3 in
Q
Milgrom and Segal (2002), Yi is differentiable and Yi0 (wi ) = −h(wi |αi∗ ) j6=i H(wj∗ |αj∗ ) +
∗
∗
Qi (wi )h(wi |αi∗ ) where w−i
is such that Y (wi , w−i
) = Yi (wi ) for all wi ∈ (w(0, αi∗ ), w(1, αi∗ )).
By assumption,
there exists i such that Yi (w(0, αi∗ ))
>
0.
Define w[
≡
sup {wi |Yi (wi0 ) > 0 ∀w(0, αi∗ ) ≤ wi0 ≤ wi }. By the continuity of Yi , we have Yi (w[ ) = 0 and
w[ > w(0, αi∗ ). There are four cases to consider.
Case I: Suppose there exists wi0 ∈ (w(0, αi∗ ), w[ ) such that Qi is discontinuous at wi0 . Let
0
0
Qi (wi+ ) denote the right-hand limit of Qi at wi0 , and Qi (wi− ) the corresponding left-hand
78
n
limit. Let 0 < ε ≤ min minw(0,α∗i )≤wi ≤wi0
0
Yi (wi )
, Qi (wi+ )
H(wi0 |α∗i )
o
0
− Qi (wi− ) , and Q̂i be such that
Q̂i (wi ) ≡ Qi (wi ) for wi ≤ w(0, α∗ ) and for wi > w(0, α∗ )
Q̂i (wi ) ≡ Qi (wi ) + εχ{wi ≤wi0 } ,
where χ{wi ≤wi0 } is the indicator function. Let Q̂j ≡ Qj for all j 6= i. By construction, Q̂i (w) ≥
Qi (w) for all wi ∈ Wi and the inequality holds strictly on a positive measure set. By a similar
argument to that in the proof of Lemma 2, Q̂i satisfies (MON) and (IA0 ). We now verify
that Q̂ satisfies (AF0 ). If wi ≤ wi0 , then Ŷ (wi , w−i ) = Y (wi , w−i ) − ε[H(wi0 |α∗ ) − H(wi |α∗ )] ≥
∗
Y (wi , w−i ) − εH(wi0 |α−i
) ≥ 0 for all w−i . If wi > wi0 , then Ŷ (wi , w−i ) = Y (wi , w−i ) ≥ 0 for
all w−i . That is, Q̂ satisfies (AF0 ).
Case II: Suppose Qi is continuous on [w(0, αi∗ ), w[ ].
exists wi0
∈
(w(0, αi∗ ), w[ ) such that Qi (wi0 )
<
We first show that there
Qi (w[ ).
Suppose to the con-
∗
be such that
Let w−i
trary that Qi (wi ) = Qi (w[ ) for all wi ∈ (w(0, αi∗ ), w[ ).
Q
∗
∗
∗ ∗
∗
) = Yi (w[ ) = 0. If Qi (w[ ) ≥
Y (w[ , w−i
j6=i H(wj |αj ), then Y (w(0, αi ), w−i ) =
i
h
R w[
Q
∗ ∗
∗
[
∗
Y (w[ , w−i
) + w(0,α∗ )
j6=i H(wj |αj ) − Qi (z) h(z|αi )dz < 0, a contradiction. If Qi (w ) <
i
Q
∗ ∗
j6=i H(wj |αj ), then, by the continuity of Qi and H, there exists δ > 0 such that
Q
Qi (wi ) < j6=i H(wj∗ |αj∗ ) for all wi ∈ [w[ , w[ + δ]. Moreover,
∗
)=
0 = Y (w[ , w−i
Z
w[ +δ
w[
"
#
Y
∗
∗
H(wj∗ |αj∗ ) − Q(z) h(z|αi∗ )dz +Y (w[ +δ, w−i
) > Y (w[ +δ, w−i
),
j6=i
a contradiction. Thus there exists wi0 ∈ (w(0, αi∗ ), w[ ) such that Qi (wi0 ) < Qi (w[ ).
1
2
By the continuity of Qi , there exists wi00 ∈ (wi0 , w[ ) such that Qi (wi00 ) =
n
o
Yi (wi )
00
0
Qi (wi0 ) + Qi (w[ ) . Let 0 < ε ≤ min minw(0,α∗i )≤wi ≤wi00 H(w
,
Q
(w
)
−
Q
(w
)
. Let
00 |α∗ )
i
i
i
i
i
79
i
Q̂j ≡ Qj for j 6= i and



max{Qi (wi0 ) + ε, Qi (wi )}



Q̂i (wi ) ≡
Qi (wi ) + ε




 Qi (wi )
if wi > wi0 ,
if w(0, αi∗ ) < wi ≤ wi0 ,
if wi ≤ w(0, αi∗ ).
Note that if wi ≥ wi0 0 then Qi (wi ) ≥ Qi (wi0 0 ) ≥ Qi (wi0 ) + ε. Thus, Q̂i (wi ) = Qi (wi ) for
wi ≥ wi0 0 . By construction, Q̂i (wi ) ≥ Q(wi ) for all wi ∈ Wi and the inequality holds
strictly on a positive measure set. By a similar argument to that in the proof of Lemma
2, Q̂i satisfies (MON) and (IA0 ). We now verify that Q̂ satisfies (AF0 ). If wi ≥ wi0 0 , then
Ŷ (wi , w−i ) = Y (wi , w−i ) ≥ 0 for all w−i . If wi < wi0 0 , then for all w−i ,
Z
wi0 0
Ŷ (wi , w−i ) =Y (wi , w−i ) −
h
i
Q̂i (z) − Qi (z) hi (z|αi∗ )dz,
wi
≥Y (wi , w−i ) − ε [H(wi0 0 |αi∗ ) − H(wi |αi∗ )] ,
≥Y (wi , w−i ) − εH(wi0 0 |α∗ ) ≥ 0.
Hence, Q̂ satisfies (AF0 ).
∗
∗
Case III: Let w−i
be such that Y (w[ , w−i
) = Yi (w[ ) = 0. Suppose Qi is continuous on
Q
[w(0, αi∗ ), w[ ) and Qi (w[− ) < j6=i H(wj∗ |αj∗ ). Define R(wi ) ≡ Yi (wi )/(H(w[ |αi∗ ) − H(wi |αi∗ ))
for wi < w[ . Then by Theorem 3 in Milgrom and Segal (2002) and L’Hopital’s rule,
lim R(wi ) =
wi →w[−
Y
Hj (wj∗ |αj∗ ) − Qi (w[− ) > 0.
j6=i
o
n
Let 0 < ε ≤ min inf w(0,α∗i )≤wi <w[ R(wi ), Qi (w[+ ) − Qi (w[− ) . Let Q̂j ≡ Qj for all j 6= i,
Q̂i (wi ) ≡ Qi (wi ) for wi ≤ w(0, αi∗ ) and Q̂i (wi ) ≡ Qi (wi ) + εχ{wi <w[ } for wi > w(0, αi∗ ).
By construction, Q̂i (wi ) ≥ Qi (wi ) for all wi ∈ Wi and the inequality holds strictly on
a positive measure set. One can verify that Q̂i satisfies (MON) and (IA0 ) by an argument
80
similar to that in the proof of Lemma 2. Finally, if wi < w[ , then Ŷ (wi , w−i ) = Y (wi , w−i ) −
ε[H(w[ |αi∗ ) − H(wi |αi∗ )] ≥ Y (wi , w−i ) − R(wi )[H(w[ |α∗ ) − H(wi |α∗ )] = 0 for all w−i . If
wi ≥ w[ , then Ŷ (wi , w−i ) = Y (wi , w−i ) ≥ 0 for all w−i . Hence, Q̂ satisfies (AF0 ).
∗
∗
) = Yi (w[ ) = 0. Suppose Qi is continuous on
be such that Y (w[ , w−i
Case IV: Let w−i
Q
Q
[w(0, αi∗ ), w[ ) and Qi (w[− ) ≥ j6=i H(wj∗ |αj∗ ). We first show that Qi (w[− ) = j6=i H(wj∗ |αj∗ ).
Q
Suppose to the contrary that Qi (w[− ) > j6=i H(wj∗ |αj∗ ). Then by the continuity of Qi
Q
and H on [w(0, α∗ ), w[ ), there exists δ > 0 such that Qi (wi ) > j6=i H(wj∗ |αj∗ ) for all wi ∈
(w[ − δ, w[ ). Then
Z
[
w[
Y (w − δ, w−i ) =
w[ −δ
a contradiction.
exists wi0
∈
"
#
Y
H(wj∗ |αj∗ )
− Qi (z) h(z|αi∗ )dz < 0,
j6=i
Hence, Qi (w[− ) =
Q
j6=i
H(wj∗ |αj∗ ).
(w(0, αi∗ ), w[ ) such that Qi (wi0 )
<
Second, we show that there
Qi (w[− ).
Suppose to the con-
trary that Qi (wi ) = Qi (w[− ) for all wi ∈ (w(0, α∗ ), w[ ), then Y (w(0, α∗ ), w−i ) =
i
hQ
R w[
∗ ∗
∗
0
j6=i H(wj |αj ) − Qi (z) h(z|αi )dz < 0, a contradiction. Hence, there exists wi ∈
w(0,α∗ )
(w(0, αi∗ ), w[ ) such that Qi (wi0 ) < Qi (w[− ). The rest of the proof follows that of Case II.
Lemma 30 Suppose two implementable allocation rules Q and Q̂ satisfy (5). Let q be an
ex-post allocation rule that implements Q, then there exists an ex-post allocation rule q̂ that
implements Q̂ and satisfies
"
E
X
i
(wi + γ
X
j6=i
#
"
#
X
X
∗
∗
wj )q̂i (w) αi = α ∀i > E
(wi + γ
wj )qi (w) αi = α ∀i .
i
j6=i
The proof of Lemma 30 relies on the following technical lemma. For each i, let mi denote
the probability measure on [w(0, αi∗ ), w(1, αi∗ )] corresponding to H(wi |αi∗ ), then
Q
∗
∗
i [w(0, αi ), w(1, αi )]
→ [0, 1] be a measurable
P
function. Then there exists an ex post allocation rule q implementing Q such that i qi (w) ≥
Lemma 31 Let Q be implementable and ρ :
81
ρ(w) for almost all w ∈
Q
∗
∗
i [w(0, αi ), w(1, αi )]
if and only if for each measurable set A =
(A1 , . . . , An ) where Ai ⊂ [w(0, α1∗ ), w(0, αi∗ )] for each i, the following inequality is satisfied
XZ
i
Z
Q(wi )dmi (wi ) ≥
Ai
ρ(w)dm1 (w1 ) . . . dmn (wn ).
(69)
A
The proof of Lemma 4 can be extended to prove Lemma 31 in a similar fashion to that
of Mierendorff (2011) and is neglected here. With Lemma 31 in hand, the proof of Lemma
3 can be readily extended to prove Lemma 30 and is neglected here. Theorem 3 follows
immediately given Lemmas 29 and 30.
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