171 Econ Spring 2010
Problem Set 1
Solutions to the two-star Problems
In some cases Ive provided more explanation than was asked of you.
** Problem 3
Find the rationalizable strategies for this game. For each iteration, list the eliminated
A
B
C
W
5, 6
4, 3
6, 2
X
4, 0
3, 4
2, 0
Y
2, 1
6, 0
0, 5
Z
1, 2
2, 1
5, 7
strategy and which strategy dominates it.
The unique rationalizable strategy is (C, Z ). The logic is as follows:
All actions are rational for Player 1. A is optimal if she believes 2 will play X
with certainty, B is optimal if she believes Y , and C is optimal if she believes W
will be played.
Only W , X and Z are rational for Player 2 because Y is dominated by Z . Y is
eliminated.
If Player 1 knows that 2 is rational, she will never play B . She knows Y wont
be played. A is optimal if she thinks X will be played and C is optimal given W .
Is B ever optimal? Well, if 1 believes that Pr(Z ) > 1/4 then she prefers C to B
and if she believes that Pr(Z ) < 1/2 then she prefers A to B . This means that
B is never optimal, which in this type of game tells us that it is dominated. Any
1
strategy that places probability q on A and 1 q on C , with q ( 2 , 3 ) dominates
4
B.
If Player 2 only knows that 1 is rational, then she cant be certain that 1 will
eliminate any action, but if she knows that 1 knows that she is rational, then
she knows 1 wont play B . This means that X is dominated by W and can be
eliminated.
If Player 1 knows this, then she knows that 2 will only play W or Z , in which
case A is dominated by C and is eliminated.
Knowing this, W is dominated by Z and is eliminated.
** Problem 4
Find all Nash equilibria for these two games.
(a) There is no pure strategy Nash Equilibrium. Any mixed strategy NE has to
involve Player 2 mixing, because otherwise Player 1 would not mix either. Let
pA , pB , and pC be the probabilities that 1 plays A, B , and C , respectively. Let
q and 1 q be the probabilities with which 2 plays L and R, respectively. For
Player 2 to be willing to mix, it is necessary that pB = pC , because 2 strictly
prefers L if C is more likely and prefers R if B is more likely.
1
A
B
C
L
2, 0
3, 0
0, 1
R
2, 0
0, 1
3, 0
A
B
C
a)
L
2, 0
5, 0
0, 2
R
2, 0
0, 3
5, 0
b)
Case 1 Suppose pB = pC > 0. Player 1 will only be willing to mix B and C if L
and R are equally likely (q = 1/2), but this would mean that A is preferred
to both. There is no NE in this case.
Case 2 Suppose pB = pC = 0. A is preferred to both B and C for 1/3 q 2/3.
In this case, pA = 1, to all which q [0, 1] is a best response for 2. Thus
2
(A, q ) is a NE for all q [ 1 , 3 ]. These are all of the NE of the game.
3
(b) Again, there is no pure strategy NE. One way to begin is to note that A is not
rationalizable. It is dominated by an even mix between B and C . Let p be the
probability with which Player 1 plays B and let q be the probability that 2 plays
L. For mixing to be sustained in equilibrium, each players strategy must make
the other indierent between the two pure strategies in that persons mix. Player
2 is indierent when 2(1 p) = 3p, or p = 2/5. Player 1 is indierent (between
B and C ) when q = 1/2. So the unique NE involves 1 playing (0, 2/5, 3/5) and 2
playing (1/2, 1/2).
** Problem 5
Watson 9.7
That (B, X ) is a Nash equilibrium does not put any restrictions on x. (A, Z ) is
1
ecient as long as x 4. For Y to be a best response to 1 = ( 2 , 1 ), we need
2
u2 (1 , Y ) = 3 x/2 + 1 = u2 (1 , Z ), which is satised for x 4. Thus, for all three
statements to be true it must be the case that x = 4.
** Problem 6
Watson 9.9
(a) The Nash equilibria occur where the curves intersect, i.e. (2, 1), (5/2, 2), and
(3, 3).
(b) The set of rationalizable strategy proles is [2, 3] [1, 3]. Explanation (not required): Looking at BR1 we see that no matter Player 1s beliefs, [0, 1) and (4, 5]
are never best responses, and thus not rationalizable. For Player 2, anything in
the interval (4, 5] is never a best response. Player 1, knowing Player 2 is rational,
knows that s2 4 which eliminates s1 > 3.5. Player 2, knowing 1 is rational,
knows that 1 s1 4, which eliminates s2 > 3.5 and s2 less than some value
that looks to be around 0.8. We see from here that each successive iteration will
further restrict the remaining set of proles until it converges to the one stated
above.
** Problem 7
Watson 11.3
(a) (N, L) and (L, N ).
2
(b) Firm Y chooses q so that Firm X is indierent between L and N . This yields
25x
5q +(x 15)(1 q ) = 10 10q . Rearranging yields q = 20x . Firm X chooses p so
that Firm Y is indierent between L and N . This yields 5p +15 15p = 10 10p.
Rearranging yields p = 1/2.
(c) The probability of (L, N ) equals p(1 q ) = (1/2)
20x25+x
20x
= (1/2)
5
x20
.
(d) As x increases, (L, N ) becomes a better outcomes for Firm X, and no worse for
Firm Y, but it becomes less likely to be played. This may seem counterintuitive,
but cases like these force us to rene/relearn our intuition, and draw more insights
in the future. Can you nd the intuition for this result?
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