FE450: GAME THEORY
Fall 2011 – Professor Schmitt
Homework 2 – due 22 Sept
Be sure to list the names of all those you work with, show all work and feel free to ask
questions if you are confused.
1. (10 points) Find the Nash Equilibrium of Bertrand’s game when there are two firms, and
the inverse demand is given by: P = A – BQ, and the costs function for each firms is
given by C(qi) = ciqi (that is, each firm has different costs). Assume c1 < c2, which firm
produces more?
(15 points) Illustrate best response functions graphically (be sure to label graph, axis and
intercepts).
Because Bertrand competition is price competition, firms will undercut each other typically
until P = MC. Because firm 1 has a lower MC, they can set P1 = c2 – ε = P2 – ε.
p1
BR2
p1monopoly
BR1
c1
p2monopoly
c2
p2
2. (20 points) Using the following utility function, U i ( xi )  c( w  xi ) 
xi
F ( X ) , where
X
n
F ( X )  AX  BX 2 , X   xi , c = marginal cost, and w = endowment, prove that the
i 1
Nash Equilibrium for the commons game is the same as the Nash Equilibrium for the
Cournot game with n firms, that is, xi 
U i ( xi )  c( w  xi )  xi ( A  B  xi )

n
dU 1 ( x1 )
 c  A  B  xi  2 Bx1  0
dx1
i 2
n
x1 
A  c  B  xi
i2
2B
because it is symmetric:
Ac
.
B(n  1)
x1 
A  c  B(n  1) x1
2B
solving for x1
x1 
Ac
B(n  1)
3. The resource that we analyzed in class was an exhaustible resource. To see that the
phenomenon of the tragedy of the commons can also arise if the resource is renewable,
consider the following variant of the model presented in class. Each player extracts an
amount ci in the first period , i = 1,2. Whatever is not extracted, that is, the amount y-c1c2, regenerates and becomes an amount equal to y  c1  c2 in period 2. The rest of the
model is identical to what was given in class. This means the utility functions of both
players for consumption in a period is ln(c), and the allocation rule in the case that the
total desired is greater than the amount available is to split the resource evenly between
the players. Recall the derivative of ln(x) = 1/x and the derivative of ln(x2) = (1/x)*(2x).
a) (10 points) Write down the best response problem for player 1.
Maximizec1 ln(c1) +ln (
y  c1  c 2
2
)
Show that the best response function is given by c1 
2( y  c 2 )
3
Taking the derivative by c1:
1

c1
1 1
1 1
1
;
* * ( y  c1  c 2) 1 / 2   *
c1 2 y  c1  c 2
y  c1  c 2 2 2
2
Setting this equal to 0 and solving for c1 gets you:
c1 
2
2
( y  c 2) ; which by symmetry implies c 2  ( y  c1)
3
3
b) (10 points) Compute the Nash Equilibrium.
Plug in for c2, and solve—you should get c1=c2= (2/5)*Y
c) (15 points) Set up and solve the social optimality problem in the renewable
resource setting. Show that the socially optimal extraction is Y/3 for each player.
Since they both have the same utility function, the social optimum will have them
consuming the same amount. Let aggregate consumption be called C. Then the SW
problem is:
Maxc ln(c) +ln(
yc
2
)
Taking the derivative by c:
1
1
1 1
1 1
1

* * ( y  c) 1 / 2   *
c 1
2 2
c1 2 y  c
yc
2
Setting this equal to 0 and solving for c gets you:
c
2
( y ) ; since c is aggregate consumption, this means each player gets Y/3
3
4. Use the following preference ordering to answer the following question:
Paul: L > M > H
Sue: H > L > M
Mary: M > H > L
(20 points) Suppose Paul credibly pre-commits himself to voting for M in the event of a tie.
Show that the unique iterated weakly dominant strategy equilibrium is for Paul and Sue to vote
for option L and for Mary to vote for option M.
The question stipulates that Paul can credibly commit to M in the event of a tie.
For Paul, if Sue and Mary vote for the same thing, his vote does not matter so L is a best
response. If they vote for different things but one of their votes is for L, Paul gets L by voting for
it, so L is a best response. If they vote for different things but neither votes for L, If Paul votes
for L he gets M (in the tie break), if he votes for M he gets M (breaking the tie in the initial voting
round), and if he votes for H he gets H (breaking the tie in the initial voting round). He likes M
better than H, and since a vote for L or M both results in M, voting for L is a best response. In
other words, Paul has a weakly dominant strategy to vote for L.
Next, consider Mary. She knows Paul will vote for L. If Sue votes for L, Mary’s vote doesn’t
matter, so voting for M is a best response (weak). If Sue votes for M, Mary’s best response
would be to vote for M (her top choice) and get it. If Sue votes for H, Mary’s best response is to
vote for M so that she gets M (in the tie break). So given that Paul adopts his weakly dominant
strategy, Mary has an iterated weakly dominant strategy to vote for M.
Finally, consider Sue. She knows Paul will vote for L and Mary for M. If she votes for H she
gets M, if she votes for M she gets M, if she votes for L she gets L. So her best response is L
because she likes this better than M. She has an iterated dominant strategy to vote for L.
Thus the strategy profile {L,L,M} is an iterated weakly dominant strategy equilibrium.