The set of equilibria of first-price auctions

The set of equilibria of first-price auctions

Paulo Klinger Monteiro

FGV-EPGE

Praia de Botafogo 190 sala 1103

Rio de Janeiro RJ Brazil

Abstract

1

Introduction

Suppose we somehow obtained a bidding functionb(_·). 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) = x

2. If the distribution of signals is uniform with three

bidders the equilibrium is b∗_{(x) =} 2x

3 . With two bidders the equilibrium is

exactlyb∗_{(x) =} 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 functionb(_·) it is possible to find a strictly increasing continuous distribution function such that the equilibrium bidding function is exactlyb(_·).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 inde-pendent private values and each bidders’ signal are in the interval [0,v]. We¯ consider distribution of bidders signals in the set

F =_{F : [0,v¯]_→[0,1] ;F is continuous strictly increasing and onto_}.

Thus _F is the set of strictly increasing distributions with support [0,v]¯ . Let F _{∈ F} be the distribution of bidder i_≤n signal xi. If bidder ihas signal xi

he valuates the object as Vi =xi. Define bF (0) = 0 and if x∈(0,v],¯

bF(x) =x−

Rx

0 F

n−1_(v)dv

Fn−1_(x) . (1)

We first prove the

Proof: The continuity of bF for x′ > 0 is immediate. At x′ = 0 it follows

frombF (x)< x.Let us now prove that it is strictly increasing. Suppose that

0 _≤ x < y _≤ v.¯ If x = 0 it is immediate from Ry

0 F

n−1_{(v)dv < F}n−1_(y)y

that bF(y)>0 =bF(x). Ifx >0 then:

bF(y)−bF (x) =y−

Ry

0 F

n−1_(u)du

Fn−1_(y) −x+

Rx

0 F

n−1_(u)du

Fn−1_(x) =

y₋x₋

Ry

x F

n−1_(u)du

Fn−1_(y) +

µ

1 Fn−1_(x) −

1 Fn−1_(y)

¶ Z x

0

Fn−1_{(u)du >0.}

Note that Ry

x F

n−1_(u)du≤Fn−1_{(y) (y−x). QED}

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: Defineb =bF andx=xi. Suppose bidder j 6=i with signalxj bids

b(xj). We have to prove that for any y∈[0,v]¯ ,

(x₋b(x)) Pr

µ

b(x)_≥max

j6=i b(xj)

¶

≥(x₋b(y)) Pr

µ

b(y)_≥max

j6=i b(xj)

¶

.

Since b is strictly increasing and the signals are independent,

Pr

µ

b(x)_≥max

j6=i b(xj)

¶

=Y

j6=i

Pr (b(x)_≥b(xj)) =

Y

j6=i

Pr (x_≥xj) =Fn−1(x).

Therefore

(x₋b(x)) Pr

µ

b(x)_≥max

j6=i b(xj)

¶

=

Z x

0

Fn−1(v)dv.

Thus we have to prove that for every y_∈[0,¯v],

Z x

0

Fn−1(v)dv _≥(x₋y)Fn−1(y) +

Z y

0

Fn−1(v)dv.

This is equivalent to

Z x

y

Fn−1_(u)du

≥(x₋y)Fn−1_(y).

Considering separately the cases x > y and y_≥xwe see that this inequality

is true. QED

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,v]¯ _→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¯v x

dy

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 thatb =bF. Then

1

y₋b(y) =

Fn−1_(y) Ry

0 Fn−1(v)dv

=

µ

log

µZ y

0

Fn−1(v)dv

¶¶′

.

Therefore

Z ¯v

x

dy

y₋b(y) =

Z ¯v

x

µ

log

µZ y

0

Fn−1_(v)dv ¶¶′

dy= log

à Rv¯

0 F

n−1_(v)dv

Rx

0 Fn−1(v)dv !

and

(x₋b(x))e

Rv¯ x

dy y−b(y) =

Rx

0 F

n−1_(v)dv

Fn−1_(x) R¯v

0 F

n−1_(v)dv

Rx

0 Fn−1(v)dv

=

Rv¯

0 F

n−1_(v)dv

Fn−1_(x) .

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,v],¯

G(x) = e−n1−1 Rv¯

x 1 y−b(y)dy

µ

¯

v ₋b(¯v) x₋b(x)

¶_n1 −1

.

First note that (4) imply thatGis continuous at 0.And obviouslyG(¯v) = 1. It is also clear that G is continuous if x >0.We now show that Gis strictly increasing. It is equivalent to prove that

φ(x) := logGn−1(x) =₋

Z v¯

x

1

is strictly increasing. If h >0 then

φ(x+h)₋φ(x)

h =

1 h

Z x+h

x

1

y₋b(y)dy+ 1 hlog

µ

x₋b(x)

x+h₋b(x+h)

¶

.

If _x+x_h−₋b_b(₍x_x)_{+h) ≥}1 then

φ(x+h)₋φ(x)

h ≥

1 h

Z x+h

x

1

y₋b(y)dy.

If x+h_x−₋b_b(_(xx+₎h) >1, then

log

µ

x+h₋b(x+h)

x₋b(x)

¶

= log

µ

1 + h−b(x+h) +b(x) x₋b(x)

¶

< h−b(x+h) +b(x) x₋b(x)

and therefore

φ(x+h)₋φ(x)

h >

1 h

Z x+h

x

1

y₋b(y)dy− 1 h

h₋b(x+h) +b(x)

x₋b(x) ≥

1 h

Z x+h

x

1

y₋b(y)dy− 1 x₋b(x). Thus

lim inf

h→0+

φ(x+h)₋φ(x)

h ≥0

which implies thatφ 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 henceb(_·) is differentiable in (c, d) as well. Thus from

d

dxφ(x) = 1

x₋b(x)−

1₋b′_(x)

x₋b(x) =

b′_(x)

x₋b(x), x∈(c, d)

we have that b′_{(x) = 0 if x}

∈(c, d) and this contradicts that b(_·) is strictly increasing. It remains only to check that

bG(x) =x−

Rx

0 G

n−1_(v)dv

Gn−1_(x)

is equal to b(x). Now note that

Z x

0

Gn−1_(u)du=

Z x

0

e−Ruv¯ 1 y−b(y)dy

µ

¯

v₋b(¯v) u₋b(u)

¶

du=

(¯v₋b(¯v))

Z x 0 d du ³ e−

R¯v u

1 y−b(y)dy

´

du= (¯v₋b(¯v))³e−

R¯v u

1 y−b(y)dy

´¯ ¯ ¯

x

0 =

(¯v₋b(¯v))e−Rx¯v 1 y−b(y)dy.

Since

Gn−1_{(x) =e}−Rv¯ x

1 y−b(y)dy

µ

¯

v ₋b(¯v) x₋b(x)

¶

=

Rx

0 G

n−1_(u)du

x₋b(x) we conclude that

bG(x) = x−

Rx

0 G

n−1_(u)du

Gn−1_(x) =x−(x−b(x)) =b(x).

QED

Example 1 Let us consider b(x) = x/2, x _∈ [0,1]. Then if there are n

bidders,

G(x) = e−n1−1 R1

x 2 ydy 1

xn−11

=xn−11.

If n= 3 then G(x) =√x.

3

The model with a more general valuation

Suppose now that the set of signals of bidderiis an abstract probability space (X,_T, P) and if bidderihas a signalxi ∈Xhis valuation isVi =u(xi) where

u:X _→[0,¯v].If the distribution

Fu(l) = Pr (u(x)≤l), l ∈[0,v]¯

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 distributionF and vary the valuationu(_·) to obtain a pre-specified bidding function b(_·)? Suppose the set of signals isX = [0,v]¯ with distribution F (x) with a continuous density f(x) >0. Then we have the

Theorem 2 Suppose b : [0,v]¯ _→ R is continuously differentiable, strictly

increasing such that b(0) = 0 and

u(x) = b(x) + b

′_(x)Fn−1_(x)

(Fn−1₎′_(x) =

(b(x)Fn−1_(x))′

(Fn−1₎′_(x) is increasing.

Then b(_·) is the symmetric equilibrium of the first-price auction if bidders

Proof: First note thatFu(u(x)) =F (x) and therefore using (2) that

bu(x) = u(x)−

Ru(x)

0 F

n−1

u (l)dl

Fn−1

u (u(x))

=

u(x)₋

Rx

0 F

n−1_(l)u′_(l)dl

Fn−1_(x) =

Rx

0 u(l) (F

n−1₎′ (l)dl

Fn−1_(x) =

Rx

0 (b(l)F

n−1_(l))′_dl

Fn−1_(x) =b(x).

References

1. Theory of the Integral, S. Saks, Dover Edition, 1964

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