First-Price Auctions Without Affiliation ⋆
Paulo Klinger Monteiro ∗
FGV-EPGE, Praia de Botafogo 190, sala 1103, 22250-900, RJ, Brazil
Humberto Moreira
FGV-EPGE, Praia de Botafogo 190, sala 1100, 22250-900, RJ, Brazil
Abstract
We give conditions for equilibrium existence in private value, symmetric, first-price
auctions without affiliation assumptions.
JEL: D44
Key words: first-price auction, affiliation
1
Introduction
The affiliation assumption was cleverly exploited in Milgrom and Weber (1982)
to study general properties of symmetric auction models when types are not independently distributed. Affiliation is a natural generalization of the monotone
likelihood ratio property 1 so useful in statistical models. However it is sometimes too strong. See, for example, Luciano (2004) for a recent analysis of the
shortcomings of the affiliation assumption. In this note we give conditions for
equilibrium existence in private values, symmetric, first-price auctions that are
valid even if the likelihood ratio is decreasing. We focus on strictly increasing
equilibria. This is for a couple of reasons. One practical reason is that a strictly
increasing equilibrium has an explicit formula. Another reason is that most
of the existence result that obtain monotonic equilibria requires affiliation.
It is therefore useful to see that monotonic equilibria exist for non-affiliated
densities. Moreover there are uniqueness results for monotonic equilibria 2 as
⋆ Monteiro and Moreira acknowledges the financial support of CNPq.
∗ Corresponding author: tel: (55)(21) 25595832, Fax: 25538821, email: [email protected]
1
Monotone meaning increasing.
2
Albeit for affiliated densities.
Preprint submitted to Economics Letters
18 May 2005
in Maskin and Riley (2000, 2003) and Rodriguez (2000). Thus it is possible,
although we do not try to prove it, that our equilibrium is unique. We apply
the results to f(a, b) = δ(aθ + bθ ), which has a decreasing likelihood ratio.
2
The Model
We consider a sealed bid first-price auction with n bidders. Bidders are exante symmetric, have private valuations and types are correlated. Thus if
bidder’s i type is given by the realization of the random variable Xi , the joint
distribution of types (X1 , . . . , Xn ) is given by F̄ : [0, 1]n → [0, 1] which has a
continuous, symmetric density f¯ : [0, 1]n → (0, ∞) . If Y = max {X2 , . . . , Xn }
then (X1 , Y ) has a joint distribution F : [0, 1]2 → [0, 1] with a density f =
f(X1 ,Y ) , f : [0, 1]2 → R++ . The conditional density is denoted by f (y|x) and
the conditional distribution by F (y|x) . Thus
y
y
f (x, y)
f (x, u) du
f (y|x) = 1
and F (y|x) =
f (u|x) du = 01
.
0
0 f (x, u) du
0 f (x, u) du
Finally we define the conditional ratio and its reciprocal:
f (y|x)
f (x, y)
1
γ (y|x) =
= y
; g (y|x) =
=
F (y|x)
γ (y|x)
0 f (x, u) du
y
0
f (x, u) du
.
f (x, y)
(1)
Let also γ(x) = γ(x|x) and g(x) = g(x|x). Define b : [0, 1] → R+ by
b (x) = x −
0
x
exp −
x
u
γ (ω) dω du.
(2)
It is immediate to prove the
Lemma 1 The function b defined above is the only solution of the differential
equation


 b′ (x) = (x − b (x)) γ(x), x > 0, with initial condition

 b (0) = 0.
3
(3)
Necessary Conditions.
We begin with the following
Theorem 1 Suppose b : [0, 1] → R+ is a strictly increasing differentiable
function. If b is a symmetric equilibrium of the first-price auction then b is
2
given by (2) and
1 ≥ b′ (x) g2′ (x|x).
(4)
Here, g2′ (x|x) := g2′ (s|x)|s=x .
Proof. The bidder’s payoff when his type is x and bids b(s) is
h(s) = h(s, x) = (x − b(s))F (s|x).
For an equilibrium we must have h (x) ≥ h(s) for all s. Thus we have the first
order condition h′1 (x) = 0 and the second order condition h′′1 (x) ≤ 0. Since
h′1 (s) = −b′ (s) F (s|x) + (x − b (s)) f (s|x)
we see that h′ (x) = 0 implies (3). Therefore from Lemma 1, (2) is true. To
finish the proof note that
φ (s, x) :=
h′1 (s)
= −b′ (s) g (s|x) + x − b (s)
f (s|x)
(5)
is such that φ (x, x) = 0 for every x and therefore φ′1 (x, x) + φ′2 (x, x) = 0.
From h′′1 (x) ≤ 0 we have that φ′1 (x, x) ≤ 0 and therefore
φ′2 (x, x) = −b′ (x) g2′ (x|x) + 1 ≥ 0.
QED
4
Sufficient Conditions.
We now consider sufficient conditions for an equilibrium. We begin with
Lemma 2 Suppose b defined by (2) is such that for every s, x, x = s :
1
g (s|x) − g (s|s)
≥
.
b′ (s)
x−s
(6)
Then b is an equilibrium.
Proof. Suppose x = s. Let φ be defined by (5). Note that φ and h′1 have the
same sign. Thus if we prove that s > x implies that φ(s, x) ≤ 0 and s < x
implies that φ(s, x) ≥ 0 then s = x will be optimum and b an equilibrium.
Now note that
φ(s, x) = −b′ (s)g(s|x) + x − s + s − b(s) = −b′ (s)g(s|x) + x − s + b′ (s)g(s|s) =
b′ (s)(g(s|s) − g(s|x)) + x − s.
3
Therefore
2
φ(s, x)(x − s) = (x − s)
g(s|x) − g(s|s)
1 − b (s)
.
x−s
′
Thus from (6), φ(s, x)(x − s) ≥ 0.
QED
We now use the sufficient condition of Lemma 2 to obtain our equilibrium
existence results. The following is well known and the proof is omitted. See
for example, Milgrom and Weber (1982) page 1107 or Menezes and Monteiro
(2005) section 4.3.2.
Theorem 2 Suppose the density f has a monotone likelihood ratio. Then b
is an equilibrium.
The following theorem is our first result that covers cases with decreasing
likelihood ratio.
Theorem 3 Suppose that (4) is true. Then b is an equilibrium if either (i) or
(ii) are true for every s:
i) g(s|·) is concave;
ii) g (s|·) is convex, and
g(s|1)−g(s|s)
1−s
≤
1
.
b′ (s)
Proof. (i) Consider first s > x. Then
g(s|x) − g(s|s)
g(s|s) − g(s|0)
g(s|s)
b′ (s)g(s|s)
s − b(s)
1
≤
≤
=
=
≤ ′ .
′
′
x−s
s
s
sb (s)
sb (s)
b (s)
Now consider s < x. Then the concavity of g (s|·) and (4) implies that
g(s|x) − g(s|s)
1
≤ g2′ (s|s) ≤ ′ .
x−s
b (s)
(ii) If x < s the convexity of g (s|·) and (4) implies that
g(s|x) − g(s|s)
1
≤ g2′ (s|s) ≤ ′
.
x−s
b (s)
Now suppose x > s. Then
g(s|x) − g(s|s)
g (s|1) − g (s|s)
1
≤
≤ ′
.
x−s
1−s
b (s)
Thus in any case by Lemma 2, b is an equilibrium.
4
QED
Remark 1 From the proof we see that the concavity assumption is used in a
quite weak form. It would be enough if we suppose that
g(s|x) − g(s|s)
g(s|s) ′
≤ max
, g2(s|s) .
x−s
s
Theorem 4 Suppose b is convex and g1′ (s|x)+1 ≥ 0. Then b is an equilibrium.
Proof. Let us differentiate φ defined by (5) with respect to s:
φ′1 (s, x) = −b′′ (s)g(s|x)−b′ (s)g1 (s|x)−b′ (s) = −b′′ (s)g(s|x)−b′ (s) (1 + g1′ (s|x)) .
Therefore since b′′ (s) ≥ 0, φ′1 (s, x) ≤ 0 for every s and therefore φ(s, x) ≤ 0 if
s > x and φ(s, x) ≥ 0 if s < x. Thus s = x is optimum.
QED
In many interesting cases the bidding function is in fact linear:
Theorem 5 Suppose f is homogeneous of degree r ∈ R. Then b is linear.
Proof. Define δ = 1f (1,1) . Recall that
0
f(1,z)dz
γ (s) = γ(s|s) = s
0
f (s, s)
sr f (1, 1)
= s
.
f (s, u)du
0 f (s, u)du
Changing variables z = u/s in the integral we get,
1
r+1 1
0 f(s, zs)sdz = s
0 f(1, z)dz. Therefore
Then
x
u
δ
f(1, 1)
γ(s) = 1
= .
s
s 0 f(1, z)dz
δ
γ(ω)dω = δ log( ux ) = log( xuδ ). Therefore exp (−
b(x) = x −
0
x
x
u
s
0
f(s, u)du =
γ(ω)dω) =
uδ
x
δx
=
.
du = x −
δ
x
δ+1
δ+1
uδ
xδ
and
(7)
We now apply the previous theorems in a few cases.
Example
1 Let f(a, b) = δ (φ(a) + φ(b)) be a density function. Thus
1
2δ 0 φ(x)dx = 1. If Φ(s) = 0s φ(u)du we easily see that
g(s|x) =
Φ(s) + sφ(x)
.
φ(x) + φ(s)
The first and second derivatives of g(s|·) are:
g2′ (s|x) =
φ′ (x)(sφ(s) − Φ(s))
(φ(x) + φ(s))2
5
(8)
and
φ′′ (x)(φ(x) + φ(s)) − 2(φ′ (x))2
= (sφ(s) − Φ(s))
.
(φ(x) + φ(s))3
If φ is increasing, sφ(s) − Φ(s) ≥ 0. Therefore g(s|·) is increasing. Now if φ is
increasing and concave then g(s|·) is concave. If φ is decreasing and concave,
g(s|·) is convex.
′′
g22
(s|x)
Example 2 We now consider f (a, b) = θ+1
(aθ + bθ ), θ > −1. We may apply
2
2(θ+1)x
theorem 5 and using (7) we see that b(x) = 3θ+4 . From (8):
θ+1
sxθ + sθ+1
g(s|x) =
.
xθ + sθ
Also:
g2′ (s|x) =
′′
g22
(s|x) =
s1+θ xθ−1 θ2
> 0 and
(xθ + sθ )2 (1 + θ)
s1+θ xθ−2θ2 sθ (θ − 1) − xθ (1 + θ)
(xθ + sθ )3 (1 + θ)
.
Thus if θ > 1, g (s|·) is neither concave nor convex. We may however apply
theorem 4. Differentiating in s:
θ+1
1+
g1′ (s|x)
2θ
2
y
z
sxθ + sθ+1 θ−1
z θ + θ+1
y + θ+1
=2− θ
θs
=2−θ
= 2−θ
.
(x + sθ )2
(1 + z θ )2
(1 + y)2
Above we made the substitutions, s = zx and y = z θ . The inequality 1 +
g1′ (s|x) ≥ 0 is true for every y if and only if
θ
2−
y2 + (4 − θ) y + 2 ≥ 0, ∀y.
θ+1
Thus the inequality is true if and only if
θ
(4 − θ) − 8 2 −
θ+1
2
= (θ − 7)
θ2
≤ 0.
θ+1
Hence if and only if θ ≤ 7. Therefore b (x) = 2(θ+1)x
is an equilibrium if θ ≤ 7.
3θ+4
If we examine
the
condition
(4)
for
this
example
we
√
√ see that it is true for
θ ≤ 3 + 17 ≃ 7.12. In particular if θ > 3 + 17 there is no increasing
equilibrium.
6
References
Castro, L., 2004, Pure Strategy Equilibria in Auctions Under More General
Assumptions, Phd thesis, Pre-print IMPA, série C 28.
Maskin, E. and J. Riley, 2000, Equilibrium in Sealed High Bid Auctions,
Review of Economic Studies 67, 439-454.
Maskin, E. and J. Riley, 2003, Uniqueness of equilibrium in sealed high-bid
auctions, Games and Economic Behavior 45(2), 395-409.
Menezes, F. and P. K. Monteiro, 2005, An Introduction to Auction Theory
(Oxford University Press, London).
Milgrom, P. and R. J. Weber, 1982, A theory of auctions and competitive
bidding, Econometrica 50(5), 1089-1122.
Rodriguez, G., 2000, First price auctions: Monotonicity and uniqueness, International Journal of Game Theory 29, 413-432.
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