Prof. Muhamet Yildiz
Fall 2010
Homework 2 Solutions
Due on 10/12/2010 (in class)
You need to show your work in all questions.
1. Consider the following game:
L
M
N
R
A
(3
;
2)
(0
;
0)
(6
;
0)
(0
;
0)
B
(0
;
0)
(2
;
4)
(1
;
2)
(0
;
0)
C
(1
;
4)
(1
;
3)
(0
;
5)
(
1
;
0)
D
(0
;
0)
(0
;
0)
(0
;
1)
(0
;
0)
(a) Compute the set of rationalizable strategies.
Solution:
C is strictly dominated by 2/5A+3/5C, strike out. Once C has been
eliminated, N is strictly dominated by any mixture of L and M. The remaining
strategies are rationalizable.
(b) Find all Nash equilibria (including those in mixed strategies).
Solution:
L
M
N
R
A
(3
;
2
)
(0
;
0)
(6
;
0)
(0
;
0)
B
(0
;
0)
(2
;
4
)
(1
;
2)
(0
;
0)
C
(1
;
4)
(1
;
3)
(0
;
5
)
(
1
;
0)
D
(0
;
0
)
(0
;
0
)
(0
;
1)
(0
;
0
)
numbers underlined in the table are the best responses of the players for a given
(B,M), and (D,R)).
(excluding non rationalizable ones). That is, (A, B, D) for player 1 and (L, M,
R) for player 2. Next, note that, in fact, D and R won±t be played in a mixed
strategy NE. The argument is as follows. Consider the decision of player 1. For
any arbitrary large probability that player 2 plays R (bounded away from 1),
player 1 strictly prefers to play any mixture of A and B instead of D. Therefore,
in any ²strictly³mixed NE, player 1 does not play D. By the similar argment, in
any ²strictly³mixed NE, player 2 does not play R.
Let us look for mixed strategies with support A and B, and L and M. Indi´erence
condition for player 2:
2
(
A
) + 0
(
B
) = 0
(
A
) + 4
(
B
)
yields
(
A
) = 2
=
3
and
(
B
) = 1
=
3
. Doing the same for player 1 yields the mixed
NE (2/3A+1/3B, 2/5L+3/5M).
To conclude, the NE are (A,L), (B,M), (D,R), and (2/3A+1/3B, 2/5L+3/5M).