Simon Fraser University
Fall 2014
Econ 302 (D200) Quiz — Solution
Instructor: Songzi Du
Wednesday February 5 2014, 8:30 – 9:20 AM
1.
(10 points) The market (inverse) demand function for a homogeneous good is
P (Q) = 10 − Q. There are three firms: firm 1 and 2 each have a constant marginal cost of
2 for producing each unit of the good, and firm 3 has a constant marginal cost of 1. The
three firms compete by setting their quantities of production, and the price of the good is
determined by the market demand function given the total quantity. Calculate the Nash
equilibrium in this game and the corresponding market price. (Hint: firm 1 and 2 face the
same incentive problem and use the same strategy in equilibrium.)
Solution:
Suppose that in a Nash equilibrium firm 1 produces quantity q1 , firm 2 produces quantity
q1 as well, and firm 3 produces quantity q3 .
Firm 1 solves
max qP (q + q1 + q3 ) − 2q = q(10 − q − q1 − q3 ) − 2q.
q
Hence his first order condition (for q = q1 to be a best response) is:
10 − 2q − q1 − q3 − 2|q=q1 = 0,
i.e.,
3q1 + q3 = 8.
Firm 2 faces exactly the same incentive problem as firm 1, so firm 1’s best reponse is
firm 2’s best response as well.
Firm 3 solves
max qP (q1 + q1 + q) − q = q(10 − q1 − q1 − q) − q.
q
1
Hence his first order condition (for q = q3 to be a best response) is:
10 − 2q1 − 2q − 1|q=q3 = 0,
i.e.,
2q1 + 2q3 = 9.
Solving these two equations gives:
q1 = 7/4, q3 = 11/4.
In summary, the Nash equilibrium is firm 1 and 2 each producing 7/4, and firm 3 producing 11/4. The equilibrium price is P (7/4 + 7/4 + 11/4) = 10 − 25/4 = 15/4.
2.
(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, 10
3, 4
10, 5
8, 6
B
4, 4
10, 10
8, 6
10, 5
C
1, 10
2, 12
3, 11
4, 10
D
5, 11
5, 11
8, 12
8, 2
Solution:
Part (i) ISD:
1. First iteration: C is strictly dominated by A, D is strictly dominated by 1/2A + 1/2B.
2. Second iteration: Y is strictly dominated by 1/2W + 1/2X, Z is strictly dominated by
1/2W + 1/2X.
The strategies that survive ISD: {A, B} and {W, X}.
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Part (ii) Nash equilibrium:
Clearly, the pure-strategy Nash equilibria are (A, W ) and (B, X). In both equilibria, the
two players get (10, 10).
For the mixed-strategy Nash equilibrium, suppose they play pA + (1 − p)B and qW +
(1 − q)X.
For both A and B to be best responses to qW + (1 − q)X, we must have:
10q + 3(1 − q) = 4q + 10(1 − q),
i.e.,
q = 7/13.
For both W and X to be best responses to pA + (1 − p)B, we must have:
10p + 4(1 − p) = 4p + 10(1 − p),
i.e.,
p = 1/2.
Thus, the mixed-strategy Nash equilibrium is (1/2A + 1/2B, 7/13W + 6/13X). In this
equilibrium, player 1 gets 10q+3(1−q)|q=7/13 = 88/13, and player 2 gets 10p+4(1−p)|p=1/2 =
7.
3. (10 points) Recall the bystander game from lecture: n bystanders witness a crime.
Each bystander has two actions: call the police or do not call the police. If the police is
called by someone, each bystander gets a value of v; the bystanders who call the police each
incur a cost of c. The payoff of each bystander is value (if any) minus the cost (if any).
Suppose that v = 10 and c = 1. Find the symmetric, mixed-strategy Nash equilibrium as a
function of n. And calculate the probability that no one calls the police in the equilibrium
when (i) n = 2, (i) n = 10, and (iii) n = 25.
Solution: In the symmetric mixed-strategy Nash equilibrium, calling the police and not
calling the police are both best responses. Let p be the probability of calling the police by
each bystander. Hence we must have
v − c = v(1 − (1 − p)n−1 ),
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since 1 − (1 − p)n−1 is the probability that at least one other bystander calls the police.
Thus, the symmetric mixed-strategy Nash equilibrium is each bystander calling the police
with probability
c 1/(n−1)
.
p=1−
v
The probability that in this equilibrium no one calls the police is
n
(1 − p) =
c n/(n−1)
v
.
Since c = 1 and v = 10, The probability that no one calls the police is (1/10)2 = 0.01
with 2 bystanders, (1/10)10/9 ≈ 0.077 with 10 bystanders, and (1/10)25/24 ≈ 0.091 with 25
bystanders. Note this striking number: with 25 bystanders there is an approximately 10%
chance that no one calls the police, even though the cost of calling the police is much lower
than the value for each bystander (they are all good people).
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