ECON3072 Advanced Microeconomics
Problem Set 1
(Static Games of Incomplete Information)
Spring 2019
Problem 1 Consider a Cournot duopoly operating in a market with inverse demand P (Q) = a Q, where
Q = q1 + q2 is the aggregate quantity on the market. Both …rms have total costs ci (qi ) = cqi , but demand
is uncertain: it is high (a = aH ) with probability and low (a = aL ) with probability 1
. Furthermore,
information is asymmetric: …rm 1 knows whether demand is high or low, but …rm 2 does not. All of this
is common knowledge. The two …rms simultaneously choose quantities. (For simplicity, assume that all
equilibrium quantities are positive by making appropriate assumptions concerning the parameters.) What is
the pure-strategy Bayesian Nash equilibrium of this game?
Solution Let q1 (aH ) and q1 (aL ) denote …rm 1’s quantity choice as a function of demand, and q2 denote
…rm 2’s single quantity choice. If demand is high, …rm 1 will choose q1 (aH ) to solve
max [(aH
q1
q1
q2 )
c] q1 .
q1
q2 )
c] q1 .
Similarly, if demand is low, q1 (aL ) will solve
max [(aL
q1
Firm 2 chooses q2 to solve
max [(aH
q2
q1 (aH )
q2 )
c] q2 + (1
) [(aL
q1 (aL )
q2 )
c] q2
so as to maximize expected pro…t. The …rst-order conditions for these three optimization problems are
and
=
q1 (aL )
=
aH
q2
2
q2
2
aL
c] + (1
2
The solutions to the three …rst-order conditions are
q2 =
[aH
q1 (aH )
q1 (aH )
q1 (aH )
=
q1 (aL )
=
and
aH
2
aL
2
c
3
c
3
c
c
;
;
) [aL
q1 (aL )
aH + (1
6
aH + (1
6
) aL
) aL
c]
:
;
;
) aL c
:
3
(q1 (aH ) ; q1 (aL ) ; q2 ) is Bayesian Nash equilibrium of this game.
q2 =
aH + (1
Problem 2 Consider a static Bayesian game. Nature determines whether the payo¤ s are as in the left
matrix or as in the right, each matrix being equally likely. Player 1 learns whether nature has drawn the left
or the right matrix, but player 2 does not. Player 1 chooses either T or B; player 2 simultaneously chooses
either L or R. Find all the pure-strategy Bayesian Nash equilibrium in this game.
T
B
L
1; 1
0; 0
R
0; 0
0; 0
T
B
1
L
0; 0
0; 0
R
0; 0
2; 2
Solution First, let’s check whether there is a pure-strategy Bayesian Nash equilibrium in which player 2
chooses L. Then, player 1’s best response is T T (the …rst T refers to the action player 1 chooses if nature
draws the left matrix, and the second refers to the action player 1 chooses if nature draws the right matrix)
or T B. Is (T T; L) a pure-strategy Bayesian Nash equilibrium? We still need to check whether L is a best
response to T T . Given player 1’s strategy T T , player 2’s expected payo¤ of L is 21 1 + 21 0 = 12 , higher
than player 2’s expected payo¤ of R which is 12 0 + 12 0 = 0. Therefore, the answer is yes. Is (T B; L) a
pure-strategy Bayesian Nash equilibrium? We still need to check whether L is a best response to T B. Given
player 1’s strategy T B, player 2’s expected payo¤ of L is 12 1 + 21 0 = 12 , lower than player 2’s expected
payo¤ of R which is 12 0 + 12 2 = 1. Therefore, the answer is no. Second, let’s check whether there is a
pure-strategy Bayesian Nash equilibrium in which player 2 chooses R. Similarly, you can show that (T B; R)
and (BB; R) are both pure-strategy Bayesian Nash equilibria.
Problem 3 Consider a static Bayesian game. Player 1 chooses the row and player 2 chooses the column.
The type space for player i is Ti = fti1 ; ti2 g (i = 1; 2). For any given type combination, the payo¤ is given
as follows:
t21
t22
L
R
L
R
t11 U 8; 1 4; 0
U 1; 3 1; 0
D 4; 1 3; 0
D 2; 0 2; 2
t12
U
D
L
4; 0
8; 0
R
4; 1
3; 1
U
D
L
2; 1
1; 0
R
1; 0
0; 1
Nature draws the type combination according to the following prior probability distribution, and reveal player
i’s type only to player i but not to the other player:
t11
t12
t21
0
1=6
t22
1=3
1=2
Find all the pure-strategy Bayesian Nash equilibrium in this game.
Solution Note that p1 (t22 jt11 ) = 1. So the player type t11 will choose D in the Bayesian Nash equilibrium
since D is always strictly better o¤ than U . Similarly, the player type t21 will choose R in the Bayesian
Nash equilibrium. Given that the player type t21 chooses R in the Bayesian Nash equilibrium, the player
type t12 will choose U in the Bayesian Nash equilibrium since U is always strictly better o¤ than D. The
last step is to …gure out what action the player type t22 will choose in the Bayesian Nash equilibrium. Note
that p2 (t11 jt22 ) = 52 . Given other player types’ actions in the Bayesian Nash equilibrium, the player type
t22 ’s expected payo¤ of L is 25 0 + 35 1 = 53 , lower than the expected payo¤ of R which is 25 2 + 35 0 = 54 .
Therefore, there is one unique pure-strategy Bayesian Nash equilibrium in which t11 plays D, t12 plays U ,
t21 plays R, and t22 plays R.
Problem 4 The game of Matching Pennies (a static game of complete information) has no pure-strategy
Nash equilibrium but has one mixed-strategy Nash equilibrium.
H
T
H
1; 1
1; 1
T
1; 1
1; 1
Provide a pure-strategy Bayesian Nash equilibrium of a corresponding game of incomplete information such
that as the incomplete information disappears, the players’ behavior in the Bayesian Nash equilibrium approaches their behavior in the mixed-strategy Nash equilibrium in the original game of complete information.
Solution Consider the following game of incomplete information
H
T
H
1 + tr ; 1
1; 1
2
T
1; 1 + tc
1; 1
in which tr and tc are independent draws from a uniform distribution on [0; x] (x is a very small number),
and privately known by the row player and the column player respectively. Let’s construct a pure-strategy
Bayesian Nash equilibrium in which the row player plays
H
T
if tr r,
otherwise,
T
H
if tc c,
otherwise,
and the column player plays
where r and c are two critical values. Given the column player’s strategy, the row player’s expected payo¤
from playing H and from playing T are
c
(1 + tr ) + 1
x
c
c
( 1) = (2 + tr )
x
x
and
c
c
( 1) + 1
x
x
respectively. Thus, playing H is optimal if and only if
tr
2x
c
1=1
2
1
c
x
4 = r.
Similarly, given the row player’s strategy, the column player’s expected payo¤ from playing H and from
playing T are
r
r
r
1
( 1) +
1=2
1
x
x
x
and
r
r
r
(1 + tc ) + ( 1) = 1 + tc
(2 + tc )
1
x
x
x
respectively. Thus, playing T is optimal if and only if
tc
Solving the system of these two equations yields
p p
2 8 + 4x + x2
r=
2
4r
x
2x
= c.
r
2 and c =
2x
p p
2 8+4x+x2
2
.
+2
The probability that the row player plays T , namely xr , equals to
p p
2 8 + 4x + x2 4
2x
and
p 1
p p
2 2 8 + 4x + x2
2 8 + 4x + x2 4 L’Hospital
lim
=
lim
x!0
x!0
2x
2
c
And the probability that the column player plays H, namely x , equals to
1
2
(4 + 2x)
=
1
.
2
2
p p
2 8+4x+x2
2
and
lim
x!0
+2
2
p p
2 8+4x+x2
2
=
+2
1
.
2
Thus, as the incomplete information disappears, the players’ behavior in the Bayesian Nash equilibrium
approaches their behavior in the mixed-strategy Nash equilibrium in the original game of Matching Pennies.
3
Problem 5 Consider the double auction we’ve discussed in the class. Suppose that the seller’s strategy is
x
1
ps (vs ) =
if 0 vs x,
otherwise,
for some x 2 (0; 1).
1. If the buyer’s strategy is to o¤ er 1 no matter what vb is, what is the expected payo¤ for the buyer?
What are the buyer types who always have negative expected payo¤ (irrespective of the value of x)?
2. Now, instead, consider the one-price equilibrium in which the buyer’s strategy is
pb (vb ) =
x
0
if x vb 1,
otherwise.
Prove that (pb (vb ) ; ps (vs )) is a Bayesian Nash equilibrium.
Solution
1. Note that, given the seller’s strategy, the seller demands x with probability x and 1 with probability
1 x. Hence, the buyer’s expected payo¤ is
1
(1 + x) x + vb
2
vb
1
(1 + 1) (1
2
1 2
x
2
x) = vb
x+2 :
The buyer type vb has negative expected payo¤ if and only if
vb <
Since x 2 (0; 1),
expected payo¤.
1
2
x2
x+2 2
7
8; 1
1 2
x
2
x+2 .
. Therefore, the buyer types vb <
7
8
always have negative
2. First, we show that pb (vb ) is a best response to ps (vs ); that is, for any vb 2 [0; 1], pb (vb ) is optimal
given that the seller’s strategy is ps (vs ).
(a) If vb < x, then according to pb (vb ) the buyer’s price is 0 and there is no trade. Hence, the buyer’s
expected payo¤ by using pb (vb ) is 0. Suppose that the buyer increases the price to a number in
(0; x). Then there is still no trade and the buyer’s expected payo¤ is 0. Suppose that the buyer
increases the price to p 2 [x; 1). Then the trade occurs with probability x and the buyer’s expected
payo¤ is vb 21 (p + x) x < 0 since vb < x < 21 (p + x). Suppose that the buyer increases the
price equal to 1. Then the trade occurs and the buyer’s expected payo¤ is
vb
since vb < x <
1
2
1
(1 + x) x + vb
2
1
(1 + 1) (1
2
x) < 0
(1 + x) < 1.
(b) If vb x, then according to pb (vb ) the buyer’s price is x and the trade occurs with probability x.
Hence, the buyer’s expected payo¤ by using pb (vb ) is vb 12 (x + x) x = (vb x) x 0. Suppose
that the buyer decreases the price below x. Then there is no trade and the buyer’s expected payo¤
is 0. Suppose that the buyer increases the price to p 2 (x; 1]. Then the trade still occurs with
probability x and the buyer’s expected payo¤ is vb 21 (p + x) x < (vb x) x. Suppose that the
buyer increases the price equal to 1. Then the trade occurs and the buyer’s expected payo¤ is
vb
since vb
1
2
(1 + x) < vb
1
(1 + x) x + vb
2
x and vb
1
1
(1 + 1) (1
2
x) < (vb
0.
Second, similarly, we can show that ps (vs ) is a best response to pb (vb ).
4
x) x