Solution. The distribution in the question could be illustrated in the table below
V X1X2Probability
1 1 1 2/9
1 0 0 1/18
1 1 0 1/9
1 0 1 1/9
First of all, given bidder 2’s strategy, bidder 1 can not win by bidding below
bidder 2’s lowest bid, 1
5, and he does not bid above bidder 2’s highest bid, 8
15 :So bids
out of 1
5;8
15 can not be a pro…table deviation.
If bidder 1 with value X1= 0 bids b21
5;8
15 , his expected payo¤ is
 (b; 0) = 1
18 (1 b) + 1
9(1 b)G(b) + 2
9(0 b) + 1
9(0 b)G(b)
=1
Because b21
5;8
15 ;the expected payo¤ is negative unless b=1
5, hence bidder 1
with value 0 will not deviate from the strategy and bid 1
5:
If bidder 1 has value X1= 1, we need to show that any bid in 1
5;8
15 gives him
the same expected payo¤. Suppose the bidder bid any b21
5;8
15 , his expected payo¤
is
 (b; 1) = 2
9(1 b)G(b) + 1
9(1 b) + 1
18 (0 b)G(b) + 1
9(0 b)
=2
15
So bidder with value 1 also does not have any incentive to deviate. So the strategy
constitutes a symmetric equilibrium.
8 Asymmetries and Other Complications
Problem 8.1 (Reserve price) Consider a …rst-price auction with Nbidders who
have private values X1; : : : ; XN:The vector (X1; : : : ; XN)is distributed over [0;1]N
according to a density function fwhich is a¢ liated and symmetric. Suppose that the
seller sets a small reserve price r > 0:Show that the following constitutes a symmetric
34
equilibrium. First, bidders with value x < r do not participate. Second, bidders with
value xrbid
(x; r) = xZx
r
L(yjx)dy (8.4)
where L(yjx)is de…ned in (8.4).
Solution. First, any bidder with value strictly less than rdoes not participate the
auction because he can not win the object with a price lower than his value.
Second, let us consider a bidder with value xr. He does not bid less than
rbecause then he has no chance of winning. Suppose that he bids (z)r, his
expected payo¤ is
Taking the derivative with respect to zyields
@
When z=xwe have
@
@z x=z
=G(zjz)(z(z)) g(zjz)
G(zjz)0(z)
= 0
Similar as in Proposition 6.3, when z < x we have
@
@z > G (zjx)(z(z)) g(zjz)
G(zjz)0(z)
=@
= 0
where the inequality comes from the fact that
g(zjx)
because of a¢ liation.
It can also be veri…ed similarly that @
@z <0when z > x. Thus  (z; x)is
maximized by choosing z=x:
35
Problem 8.2 (Increase in number of bidders) Consider a …rst-price auction with N
bidders who have private values X1; : : : ; XN:The values X1; : : : ; XNare symmet-
rically distributed over [0;1]Nand are a¢ liated. Speci…cally, the joint distribution
of values is determined as follows. First, a random variable Z21
10 ;2is drawn
and each value of Zis equally likely. Next, each bidder’s value Xiis drawn from
the distribution FXjZ(xjz) = exp z1x1over (0;1]:As in the previous prob-
lem, suppose that the seller sets a reserve price r= 0:5:Let (n)(x; r)denote the
symmetric equilibrium bidding strategy (as in (8.4)) when there are nbidders. Now
suppose that the number of bidders increases to n+ 1 and let (n+1) (x; r)denote
the symmetric equilibrium bidding strategy when there are n+ 1 bidders (again as in
(8.4)). Show that for some x>r,(n)(x; r)> (n+1) (x; r) ; that is, an increase in
the number of bidders may cause bids to decrease.
(Note: This problem is computationally intensive. A symbolic computation program
will be of great help.)
Solution. The density of joint distribution of X1; : : : ; XNconditional on Z=zis
h(x1; : : : ; xNjz) = fXjZ(x1jz): : : fXjZ(xNjz)
and the joint density of X1; : : : ; XNis
h(x1; : : : ; xN) = 1
2h(x1; : : : ; xNjz= 0:1) + 1
2h(x1; : : : ; xNjz= 2)
=1
Conditional on bidder 1’s value X1=x ; the cumulative distribution function of the
highest bid of the other bidders is
G(yjx) = Ry
0: : : Ry
0h(x; t2; : : : ; tN)dt2: : : dtN
R1
0: : : R1
0h(x; t2; : : : ; tN)dt2: : : dtN
=Ry
0: : : Ry
0
1
2fXjZ(xj0:1) fXjZ(t2j0:1) : : : fXjZ(tNj0:1) dt2: : : dtN
+Ry
0: : : Ry
0
1
2fXjZ(xj2) fXjZ(t2j2) : : : fXjZ(tNj2) dt2: : : dtN
R1
1
36
and the corresponding density function is
g(yjx) = 1
2fXjZ(xj0:1) (N1) FXjZ(yj0:1)N2fXjZ(yj0:1)
+1
1
2fXjZ(xj0:1) + 1
2fXjZ(xj2)
The expression of G(yjx)and g(yjx)give us the ratio
g(tjt)
G(tjt)= fXjZ(xj0:1) (N1) FXjZ(yj0:1)N2fXjZ(yj0:1)
+fXjZ(xj2) (N1) FXjZ(yj2)N2fXjZ(yj2) !
where the second equality comes from the substitution of the function form of FXjZ.
According to the de…nition in (6:7), we have
L(yjx) = exp Zx
y
g(tjt)
G(tjt)dt
= exp 0
B
B
B
B
@Zx
y
0:01t2(N1) exp 0:11t1N
+4t2(N1) exp 21t1N!
0:1 (exp (0:1 (1 t1)))N+ 2 (exp (2 (1 t1)))Ndt1
C
C
C
C
A
y2 (N1) t2dt
where the fourth equality comes from changing variable s= exp 0:1N1t1.
37
Now applying formula (8.4) yields the bidding function with Nbidders
N(x) = xZx
0:5
L(yjx)dy
N
The following …gure plots the bidding functions with number of bidders N= 2 and
N= 6, where the solid line represents the bidding function with 2 bidders, while the
dash line is the bidding function with 6 bidders. Note that the dash line is below the
solid line for small values, which means for those values the bids are lower with more
bidders.
0.5 0.6 0.7 0.8 0.9 1.0
0.5
0.6
0.7
0.8
0.9
1.0
x
bid
Figure S8.1
Problem 8.3 (A second-price auction without a monotone equilibrium) Consider a
second-price auction with three bidders. Each bidder ireceives a private signal Xi.
Bidder 1’s value V1=X1+X2is the sum of his own signal and bidder 2’s signal, while
bidder 2’s and 3’s values are private and equal their own signals— that is, V2=X2
and V3=X3. Each bidder receives either a “high” or a “low” signal. Speci…cally,
X12 f1;2g,X22 f0;4g, and X32 f0;3g. Bidders 1and 3have perfectly correlated
signals, both being low with probability 1=2or both being high with probability 1=2.
Bidder 2’s signal is independent of the others’ signals, being low with probability p
and high with probability 1p, where p2(2=3;1).
a. Verify that the bidders’signals are a¢ liated.
38
b. Show that with the speci…ed information structure, there does not exist a pure-
strategy equilibrium of the second-price auction such that the equilibrium strategies
are increasing and undominated.
Solution. Part a.
By the de…nition of a¢ liation in Appendix D, we need to show
Pr x0_x00Pr x0^x00Pr x0Pr x00for all x0and x00 2 X (13)
If bidder 2’s values are both 0in x0and x00;(13) becomes
1
If bidder 2’s values are both 1in x0and x00;(13) becomes
1
If bidder 2’s values are di¤erent in x0and x00;(13) becomes
1
So the signals are a¢ liated.
Part b. It is easy to see that bidding their values is a undominated strategy
for bidders 2 and 3. Therefore bidder 2 and 3 each employ a pure-strategy which is
increasing. It remains only to compute an undominated strategy for bidder 1.
When X1= 1, bidder 1 knows that 3(X3) = X3= 0 . Since 1’s value is
v1= 1 + X2, his value is 1 if 2(X2) = X2= 0 ;while it is 5 is 2(X2) = X2= 4.
When X1= 2;bidder 1 knows that 3(X3) = X3= 3 . Consequently, a bid above
4 still wins with probability one, but bidder 1 now pays 3 when 2(X2) = X2= 0 and
Hence, an undominated pure-strategy equilibrium exists, but in no such equilib-
rium is bidder 1’s strategy nondecreasing.
39