Simon Fraser University
Spring 2015
Econ 302 (D200) Quiz
Instructor: Songzi Du
Thursday February 5 2015, 8:30 – 9:20 AM
This brief solution guide may not have the explanations necessary for full marks.
1.
(10 points) Part i: Find the strategies that survive iterative deletion of strictly
dominated strategies (ISD). For each strategy that you delete, write down the strategy that
strictly dominates it.
Part ii: Find all Nash equilibria (pure and mixed) in this game, and for each Nash
equilibrium that you find, find the two players’ expected payoffs in the equilibrium.
W
X
Y
Z
A
10, 1
3, 9
3, 5
5, 2
B
4, 4
10, 4
1, 6
1, 5
C
4, 1
2, 0
5, 11
4, 2
D
5, 1
5, 1
2, 3
8, 4
Solution:
Part (i) ISD:
1. First iteration: W is strictly dominated by Y .
2. Second iteration: A is strictly dominated by 1/2C + 1/2D.
3. Third iteration: X is strictly dominated by Z.
4. Fourth iteration: B is strictly dominated by C.
The strategies that survive ISD: {C, D} for player 1 and {Y, Z} for player 2.
Part (ii) Nash equilibrium:
1
Clearly, the pure-strategy Nash equilibria are (C, Y ) and (D, Z). Players’ payoffs in these
equilibria can be read off the matrix.
For the mixed-strategy Nash equilibrium, suppose they play pC + (1 − p)D and qY +
(1 − q)Z.
For both C and D to be best responses to qY + (1 − q)Z, we must have:
5q + 4(1 − q) = 2q + 8(1 − q),
i.e.,
q = 4/7.
For both Y and Z to be best responses to pC + (1 − p)D, we must have:
11p + 3(1 − p) = 2p + 4(1 − p),
i.e.,
p = 1/10.
Thus, the mixed-strategy Nash equilibrium is (1/10C + 9/10D, 4/7Y + 3/7Z). In this
equilibrium, player 1 gets 5q + 4(1 − q)|q=4/7 = 32/7, and player 2 gets 11p + 3(1 − p)|p=1/10 =
19/5.
2. (10 points) The market demand function for a homogeneous good is Q(P ) = 10−2P .
There are two firms: firm 1 has a constant marginal cost of 2 for producing/selling each unit
of the good, and firm 2 has a constant marginal cost of 4.
Suppose that the two firms compete by simultaneously setting their quantities (q1 and
q2 ), and the price of the good is determined by the market demand function given the total
quantity. Calculate the Nash equilibrium (q1 , q2 ) in this game, and calculate the price in this
equilibrium.
Solution:
First, we solve the inverse demand function:
P =5−
2
Q
.
2
For the best response to q2 , firm 1 solves:
q1 + q2
q2
max 5 −
q1 − 2q1 =⇒ FOC: 5 − q1 − − 2 = 0.
q1
2
2
For the best response to q1 , firm 2 solves:
q1 + q2
q1
max 5 −
q2 − 4q2 =⇒ FOC: 5 − q2 − − 4 = 0.
q2
2
2
The solution to the above two equations are q1 = 10/3 and q2 = −2/3. Since firm 2’s
quantity cannot be negative, the Nash equilibrium is firm 2 producing q2 = 0, and firm 1
producing q1 satisfying 5 − q1 − q22 − 2|q2 =0 = 0, i.e., q1 = 3 (the monopolist’s quantity). In
this equilibrium, firm 2 is forced out of the market because of his high marginal cost. The
price in this equilibrium is the monopolist’s price: P = 5 − 3/2 = 7/2.
3.
(10 points) Suppose that there are n airlines (n ≥ 2). Each airline i chooses a
number si from the set {1, 2, 3, 4, 5, 6, 7}, where each number represents a level of security
expenditure by an airline. Airline i’s payoff is 50 + 20 × min(s1 , s2 , . . . , sn ) − 10si (10 is the
cost for each level of security, and 20 is the benefit; security is as strong as the weakest link).
(For example, if n = 3, and s1 = 1, s2 = 5, and s3 = 2, then min(s1 , s2 , s3 ) = 1, so airline 1
gets 50 + 20 − 10 = 60, airline 2 gets 50 + 20 − 50 = 20 and airline 3 gets 50 + 20 − 20 = 50.)
Consider only pure strategy in this game. Can a non-symmetric strategy profile be a Nash
equilibrium? Find all pure-strategy Nash equilibria. Explain your answers.
Solution:
A non-symmetric strategy profile cannot be a Nash equilibrium, because if one airline
(say airline i) sets its security level si below that of all other airlines, then every other airline
j 6= i will want to lower its security level sj , since lowering the security level will not decrease
the benefit 20 × min(s1 , s2 , . . . , sn ) = 20si but will decrease the cost 10sj .
There are seven pure-strategy Nash equilibrium: s1 = s2 = · · · = sn = 1, s1 = s2 = · · · =
sn = 2, . . . , s1 = s2 = · · · = sn = 7. (Add explanations.)
3