Nº 536
ISSN 0104-8910
The set of equilibria of first-price auctions
Paulo Klinger Monteiro
Março de 2004
The set of equilibria of first-price auctions
Paulo Klinger Monteiro
FGV-EPGE
Praia de Botafogo 190 sala 1103
Rio de Janeiro RJ Brazil
March, 2004
Abstract
In this note I specify the class of functions that are equilibria of symmetric
first-price auctions.
1
Introduction
Suppose we somehow obtained a bidding function b (·). It could, for example,
be originated from some laboratory auction or be a linear interpolation of
some auction data. Is this bidding function theoretically possible? If so,
does it come from a model adequate to the situation at hand? In this paper
I study this problem in a sealed bid first-price auction set up. To be more
concrete we are in the independent private values model and suppose we have
3 bidders with signals in the interval [0, 1] and that our estimated bidding
function is b (x) = x2 . If the distribution of signals is uniform with three
bidders the equilibrium is b∗ (x) = 2x
. With two bidders the equilibrium is
3
x
∗
exactly b (x) = 2 . Do we have a dummy bidder? Collusion? Here I focus on
the distribution of signals. The uniform distribution that we supposed in this
example is usually used more for convenience than for theoretical
reasons. In
√
this example if we change the distribution to F (x) = x we have that the
equilibrium bidding function is exactly b (x) . I show in this paper that for
any number of bidders and practically any strictly increasing function b (·) it
is possible to find a strictly increasing continuous distribution function such
that the equilibrium bidding function is exactly b (·) . This result is similar in
spirit to the Sonneschein-Mantel-Debreu theorem on excess demand. I also
analyze a second aspect of this problem. If we insist that the distribution of
signals is given and vary the bidder valuation from Vi = xi to Vi = u (xi ) .
The conditions to find an appropriate u (·) are however harder to be met.
2
The model
We consider first-price sealed bid auctions. There are n bidders with independent private values and each bidders’ signal are in the interval [0, v̄]. We
consider distribution of bidders signals in the set
F = {F : [0, v̄] → [0, 1] ; F is continuous strictly increasing and onto} .
Thus F is the set of strictly increasing distributions with support [0, v̄] . Let
F ∈ F be the distribution of bidder i ≤ n signal xi . If bidder i has signal xi
he valuates the object as Vi = xi . Define bF (0) = 0 and if x ∈ (0, v̄],
R x n−1
F
(v) dv
bF (x) = x − 0 n−1
.
(1)
F
(x)
We first prove the
Lemma 1 bF is continuous and strictly increasing.
1
Proof : The continuity of bF for x′ > 0 is immediate. At x′ = 0 it follows
from bF (x) < x. Let us now prove that it is strictly
Suppose that
R y increasing.
n−1
(v) dv < F n−1 (y) y
0 ≤ x < y ≤ v̄. If x = 0 it is immediate from 0 F
that bF (y) > 0 = bF (x) . If x > 0 then:
R y n−1
R x n−1
F
(u)
du
F
(u) du
bF (y) − bF (x) = y − 0 n−1
− x + 0 n−1
=
F
(y)
F
(x)
R y n−1
¶Z x
µ
F
(u) du
1
1
x
F n−1 (u) du > 0.
+
− n−1
y−x−
n−1
n−1
F
(y)
F
(x) F
(y)
0
R y n−1
(u) du ≤ F n−1 (y) (y − x) .
QED
Note that x F
We now show that bF is an equilibrium bidding function.
Proposition 1 Let F ∈ F. If there are n bidders with independent signals
distributed accordingly to F then bF (·) is a symmetric equilibrium of the
first-price auction.
Proof : Define b = bF and x = xi . Suppose bidder j 6= i with signal xj bids
b (xj ). We have to prove that for any y ∈ [0, v̄] ,
µ
¶
µ
¶
(x − b (x)) Pr b (x) ≥ max b (xj ) ≥ (x − b (y)) Pr b (y) ≥ max b (xj ) .
j6=i
j6=i
Since b is strictly increasing and the signals are independent,
µ
¶ Y
Pr b (x) ≥ max b (xj ) =
Pr (b (x) ≥ b (xj )) =
j6=i
Y
j6=i
j6=i
Pr (x ≥ xj ) = F n−1 (x) .
Therefore
µ
¶ Z
(x − b (x)) Pr b (x) ≥ max b (xj ) =
j6=i
x
F n−1 (v) dv.
0
Thus we have to prove that for every y ∈ [0, v̄] ,
Z
Z x
n−1
n−1
F
(v) dv ≥ (x − y) F
(y) +
y
F n−1 (v) dv.
0
0
This is equivalent to
Z
y
x
F n−1 (u) du ≥ (x − y) F n−1 (y) .
Considering separately the cases x > y and y ≥ x we see that this inequality
is true.
QED
2
Remark 1 Note that I do not suppose the differentiability of the distribution.
This is of essence. If the bidding function is piecewise linear the distribution
cannot be differentiable. See also the example below.
Define
B = {bF (·) ; F ∈ F} .
We may now prove our main theorem.
Theorem 1 Suppose b : [0, v̄] → R . Then b ∈ B if and only if:
1. b (·) is strictly increasing;
2. b (0) = 0;
3. b (x) < x if x > 0;
4. limx→0+ (x − b (x)) e
R v̄
dy
x y−b(y)
= ∞.
Proof : Suppose b ∈ B. It is clear that it satisfies (1), (2) and (3). To see
that it also satisfy (4) let F ∈ F be such that b = bF . Then
µ µZ y
¶¶′
1
F n−1 (y)
n−1
= log
= R y n−1
F
(v) dv
.
y − b (y)
F
(v)
dv
0
0
Therefore
!
à R v̄
¶¶′
Z v̄ µ µZ y
Z v̄
n−1
F
(v)
dv
dy
=
log
F n−1 (v) dv
dy = log R0x n−1
F
(v) dv
x
0
x y − b (y)
0
and
(x − b (x)) e
R v̄
dy
x y−b(y)
=
Rx
0
R v̄ n−1
R v̄
F
(v) dv
F n−1 (v) dv 0 F n−1 (v) dv
0
R
=
.
x
F n−1 (x)
F n−1 (x)
F n−1 (v) dv
0
Thus b satisfy (4). Now suppose b (·) satisfy (1), (2), (3) and (4). I show
that b = bG where G (0) = 0 and if x ∈ (0, v̄],
1
¶ n−1
µ
R v̄
1
1
v̄
−
b
(v̄)
dy
− n−1
x y−b(y)
G (x) = e
.
x − b (x)
First note that (4) imply that G is continuous at 0. And obviously G (v̄) = 1.
It is also clear that G is continuous if x > 0. We now show that G is strictly
increasing. It is equivalent to prove that
Z v̄
1
n−1
φ (x) := log G
(x) = −
dy − log (x − b (x))
x y − b (y)
3
is strictly increasing. If h > 0 then
µ
¶
Z
1 x+h
1
1
x − b (x)
φ (x + h) − φ (x)
=
dy + log
.
h
h x
y − b (y)
h
x + h − b (x + h)
If
x−b(x)
x+h−b(x+h)
≥ 1 then
φ (x + h) − φ (x)
1
≥
h
h
Z
x
x+h
1
dy.
y − b (y)
x+h−b(x+h)
x−b(x)
> 1, then
µ
¶
µ
¶
x + h − b (x + h)
h − b (x + h) + b (x)
h − b (x + h) + b (x)
log
= log 1 +
<
x − b (x)
x − b (x)
x − b (x)
If
and therefore
Z
1 x+h
1 h − b (x + h) + b (x)
1
φ (x + h) − φ (x)
>
dy −
≥
h
h x
y − b (y)
h
x − b (x)
Z
1 x+h
1
1
dy −
.
h x
y − b (y)
x − b (x)
Thus
φ (x + h) − φ (x)
≥0
h→0
h
which implies that φ is increasing (see Saks, theorem 7.2 page 204). To show
that it is strictly increasing suppose not. Then φ is constant in an interval
(c, d) and therefore it is differentiable and hence b (·) is differentiable in (c, d)
as well. Thus from
d
1
1 − b′ (x)
b′ (x)
φ (x) =
−
=
, x ∈ (c, d)
dx
x − b (x)
x − b (x)
x − b (x)
lim inf+
we have that b′ (x) = 0 if x ∈ (c, d) and this contradicts that b (·) is strictly
increasing. It remains only to check that
R x n−1
G
(v) dv
bG (x) = x − 0 n−1
G
(x)
is equal to b (x) . Now note that
¶
µ
Z x R
Z x
1
v̄ − b (v̄)
− uv̄ y−b(y)
dy
n−1
e
G
(u) du =
du =
u − b (u)
0
0
Z x
´
³ R v̄ 1
´¯x
1
d ³ − Ruv̄ y−b(y)
¯
dy
du = (v̄ − b (v̄)) e− u y−b(y) dy ¯ =
(v̄ − b (v̄))
e
0
0 du
(v̄ − b (v̄)) e−
4
R v̄
1
x y−b(y) dy
.
Since
G
n−1
(x) = e
−
R v̄
1
x y−b(y) dy
we conclude that
bG (x) = x −
Rx
0
µ
v̄ − b (v̄)
x − b (x)
¶
=
Rx
0
Gn−1 (u) du
x − b (x)
Gn−1 (u) du
= x − (x − b (x)) = b (x) .
Gn−1 (x)
QED
Example 1 Let us consider b (x) = x/2, x ∈ [0, 1] . Then if there are n
bidders,
R1 2
1
1
1
G (x) = e− n−1 x y dy 1 = x n−1 .
x n−1
√
If n = 3 then G (x) = x.
3
The model with a more general valuation
Suppose now that the set of signals of bidder i is an abstract probability space
(X, T , P ) and if bidder i has a signal xi ∈ X his valuation is Vi = u (xi ) where
u : X → [0, v̄] . If the distribution
Fu (l) = Pr (u (x) ≤ l) , l ∈ [0, v̄]
belongs to F then we can easily see that
bu (x) = bFu (u (x))
(2)
is a symmetric equilibrium bidding function.
Is it possible to fix a distribution F and vary the valuation u (·) to obtain
a pre-specified bidding function b (·)? Suppose the set of signals is X = [0, v̄]
with distribution F (x) with a continuous density f (x) > 0. Then we have
the
Theorem 2 Suppose b : [0, v̄] → R is continuously differentiable, strictly
increasing such that b (0) = 0 and
′
u (x) = b (x) +
b′ (x) F n−1 (x)
(b (x) F n−1 (x))
=
is increasing.
(F n−1 )′ (x)
(F n−1 )′ (x)
Then b (·) is the symmetric equilibrium of the first-price auction if bidders
have valuation Vi (x) = u (x) and the distribution of signals is F .
5
Proof: First note that Fu (u (x)) = F (x) and therefore using (2) that
R u(x)
Fun−1 (l) dl
=
Fun−1 (u (x))
Rx
′
n−1
′
F
(l)
u
(l)
dl
u (l) (F n−1 ) (l) dl
0
0
u (x) −
=
=
F n−1 (x)
F n−1 (x)
Rx
′
(b (l) F n−1 (l)) dl
0
= b (x) .
F n−1 (x)
bu (x) = u (x) −
Rx
0
References
1. Theory of the Integral, S. Saks, Dover Edition, 1964
6
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