Simon Fraser University
Spring 2013
Econ 302 Midterm Exam — Solution
Instructor: Songzi Du
Section D200
Tuesday March 12, 2013, 8:30 – 10:20 AM
This exam has two parts. Write your name, SFU ID number, and tutorial section number
on each part.
• Name:
• SFU ID number:
• Tutorial section number:
1. This is a closed-book exam.
2. You may use a non-graphing calculator.
3. There is no separate exam booklet. Write your solution in the space following each
question. There are also some blank pages in the back.
4. Show your work! Partial credits are given. Answers without proper explanation/calculation
will be penalized.
5. We will accept a request for regrade only if the solution is written with a pen.
6. The quiz has 75 points (25% of the class grade).
7. Stay in your seat if less than 20 minutes remain in the exam. You may leave early if
you finish 20 minutes before the exam is over.
1
1.
Part i (7 points): 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 (8 points): 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
5, 0
0, 1
2, 1
3, 4
B
1, 3
3, 4
1, 7
10, 5
C
2, 3
1, 5
3, 1
4, 2
D
1, 10
1, 0
1, 5
4, 5
Answer:
Part i: Iterated deletion of strictly dominated strategies (ISD)
First iteration of deletion: D is strictly dominated by 1/2B + 1/2C. (Note: D is not
strictly dominated by B or C alone.)
Second iteration of deletion: W is strictly dominated by 1/2X + 1/2Z.
Third iteration of deletion: A is strictly dominated by C.
Fourth iteration of deletion: Z is strictly dominated by 1/2X + 1/2Y .
Strategies that survive iterated deletion of strictly dominated strategies: {B, C} and
{X, Y }.
Part ii:
A Nash equilibrium cannot involve a strategy that does not survive iterated deletion
of strictly dominated strategies. Therefore, we can restrict our attentions to {B, C} and
{X, Y }. Clearly, there is no pure-strategy Nash equilibrium. So let us solve for a mixedstrategy Nash equilibrium of the form: (pB + (1 − p)C, qX + (1 − q)Y ).
For player 1 to play pB + (1 − p)C as a best response to qX + (1 − q)Y , player 1 must
be indifferent between B and C, i.e.,
3q + 1(1 − q) = 1q + 3(1 − q),
or equivalently,
q = 1/2.
For player 2 to play qX + (1 − q)Y as a best response to pB + (1 − p)C, player 2 must
be indifferent between X and Y , i.e.,
4p + 5(1 − p) = 7p + 1(1 − p),
2
or equivalently,
p = 4/7.
Therefore, the mixed-strategy Nash equilibrium is ( 74 B + 37 C, 21 X + 12 Y ). In this equilibrium, player 1 gets an expected payoff of 3 × 12 + 1 × 12 = 2, and player 2 gets an expected
payoff of 4 × 47 + 5 × 37 = 31
.
7
3
2.
(10 points) The market (inverse) demand function for a homogeneous good is
P (Q) = 30 − 2Q. There are 2 firms, each with a constant marginal cost of 3 for producing
one unit of the good. The two firms compete by setting their quantities of production, and
the price paid by the consumer is determined by the market demand function given the total
quantity. Assume that the government imposes a sales (percentage) tax of 10% for firm 1
and a sales tax of 20% for firm 2; the taxes do not impact the firms’ costs. (Firm 1 is a
domestic company and therefore receives special favors.) Calculate the Nash equilibrium in
this game, and calculate the prices received by the two firms as well as the consumer price
in this equilibrium. (Hint: there is only one consumer price.)
Answer:
Suppose that firm 1 produces q1 , and firm 2 produces q2 . The consumer’s price is then
1
P (q1 + q2 ), and firm 2 gets the price
P (q1 + q2 ) = 30 − 2(q1 + q2 ), and firm 1 gets the price 1.1
1
P (q1 + q2 ).
1.2
Firm 1’s profit is
1
(30 − 2(q1 + q2 ))q1 − 3q1 .
1.1
Firm 1 maximizes over q1 , thus his first order condition is
10
(30 − 2q2 − 4q1 ) − 3 = 0.
11
(1)
Firm 2’s profit is
1
(30 − 2(q1 + q2 ))q2 − 3q2 .
1.2
Firm 2 maximizes over q2 , thus his first order condition is
5
(30 − 2q1 − 4q2 ) − 3 = 0.
6
(2)
Solving equations (1) and (2), we get the Nash equilibrium of q1 = 9/2 and q2 = 87/20.
Consumer price is 30 − 2(9/2 + 87/20) = 123/10. Firm 1’s price is 123/11, and firm 2’s price
is 123/12 = 41/4.
4
3. (10 points) Find all pure-strategy subgame perfect equilibria of the following game.
1
a
2
c
2
1
e
b
2
d
−1
0
2
c
0
0
g
f
d
0
1
1
1
h
i j
k
1
4
0
3
2
3
0
4
Answer:
Clearly, in a subgame perfect equilibrium player 1 must play (h, k), and player 2 must
play f , when those nodes are reached. To solve for the subgame perfect equilibria in the
part of the tree with information sets, we set up the following matrix:
(c, f )
(a, h, k) 2, 1
(b, h, k) 0, 0
(e, h, k) 1, 4
(d, f )
-1, 0
0, 1
1, 4
Clearly, the pure-strategy subgame perfect equilibria are {(a, h, k), (c, f )} and {(e, h, k), (d, f )}.
5
4. (10 points) Tim Hortons is the only coffee place in WMC and makes a yearly profit of
$100,000. Random-Coffeehouse contemplates entering the coffee market in WMC, incurring
setup-costs of $25,000, which will force Tim Hortons to share its profits 50-50 with them.
Tim Hortons vows to sell coffee at cost, if necessary, to preserve its monopoly position.
The sequence of moves is that Random-Coffeehouse first decides to enter or not enter
the market; if not entering, the payoffs of Random-Coffeehouse (Player 1) and Tim Hortons
(Player 2) are (0, 100000). If Random-Coffeehouse enters, Tim Hortons then decides to fight
(leading to payoffs of (−25000, 10000)) or accommodate (leading to payoffs of (25000, 50000))
Random-Coffeehouse. Draw the game tree, and find the subgame perfect equilibrium. Is
Tim Hortons’s threat to fight Random-Coffeehouse credible?
Answer:
(See Figure 1 of the last page for the game tree.)
If Random-Coffeehouse enters the market, then Tim Hortons should accommodate instead of fight (50000 > 10000). Thus, Random-Coffeehouse should enter the market (25000 >
0). Therefore, the subgame perfect equilibrium is (Enter, Accommodate). In the equilibrium, Tim Hortons accommodates after Random-Coffeehouse enters the market, so Tim’s
threat to fight is not credible.
6
5.
(10 points) Player 1 and 2 are bargaining over a dollar. Player 1 moves first,
proposing a division x (where 0 ≤ x ≤ 1) in which he gets x and Player 2 gets 1 − x, and
Player 2 responds with Accept or Reject. If Player 2 accepts, they get the proposed division
of (x, 1 − x), and the game ends; if Player 2 rejects, the game moves to the next stage. In
the next stage, the dollar is shrunk 10% to 0.9 dollar, and Player 2 gets to propose a division
y (where 0 ≤ y ≤ 0.9) in which he gets y and Player 1 gets 0.9 − y; Player 1 then responds
with Accept or Reject, and the game ends. If Player 1 accepts, then they get the proposed
division (0.9 − y, y); if Player 1 rejects, they get (0, 0).
Find a subgame perfect equilibrium of this game. (Hint: draw the game tree to understand what is going on.)
Answer:
In the second stage, Player 1 should accept any y < 0.9, since he gets zero otherwise.
If y = 0.9, Player 1 is indifferent between accepting and rejecting; assume that he accepts.
Therefore, in the second stage, Player 2 will offer y = 0.9, and this offer is accepted by Player
1.
In the first stage, Player 2 should accept any x < 0.1 and reject any x > 0.1, since he
gets 0.9 from the second stage. When x = 0.1, Player 2 is indifferent between accepting and
rejecting; assume that he accepts. Then Player 1 should offer x = 0.1 in the first stage.
Therefore the following is a subgame perfect equilibrium:
1. Player 1 proposes x = 0.1. In the second stage, then he accepts any y ≤ 0.9.
2. Player 2 accepts x if and only if x ≤ 0.1. In the second stage, Player 2 proposes
y = 0.9.
(If the proposals are in cents (0, 0.01, 0.02, . . .), then there is another pure-strategy subgame perfect equilibrium in which Player 1 accept any y < 0.9 and rejects y = 0.9 in the
second stage, ...)
7
6. (10 points) Nature chooses the ability of the student: with probability 0.5 he is of
high ability, and with probability 0.5 he is of low ability. The student knows his type, and
decides whether to go to the university. A firm then observes the educational background
of the student (but not his ability), and decides whether to hire the student. The firm gets
a payoff of 1 if it hires the high ability student, -1 if it hires the low ability student, and 0
if it does not hire. The high-ability student enjoys university and gets 10 from going to the
university, and 0 from not going to the university; the low-ability student hates university
and gets 0 from going to the university, and 10 from not going to the university. Moreover,
the student, high or low ability, gets an additional payoff of 1 if he is hired by the firm, and
0 if not hired. Thus, if a low ability student does not go to the university and gets hired, he
gets a total payoff of 11, and likewise for the other cases.
Draw the extensive form of this game, and find a subgame perfect equilibrium.
Answer:
(See Figure 2 of the last page for the game tree.)
Clearly, for the low-ability student, not going to university is the strictly dominant action;
likewise, for the high-ability student, going to the university is the strictly dominant action.
And given that the low-ability student does not go to university and the high-ability student
goes to university, the best response of the firm is to hire when there is university education,
and not hire when there is no university education, leading to an expected payoff of 0.5 ×
0 + 0.5 × 1 = 0.5, which is better than hiring whether or not there is university education
(expected payoff of 0.5 × (−1) + 0.5 × 1 = 0), not hiring whether or not there is university
education (expected payoff of 0.5 × 0 + 0.5 × 0 = 0), and hiring if there is no university
education and not hiring if there is university education (expected payoff of 0.5 × (0) + 0.5 ×
(−1) = −0.5).
Therefore, the following is a subgame perfect equilibrium:
1. The low-ability student does not go to university.
2. The high-ability student goes to university.
3. Firm hires the student if he has gone to university, does not hires the student if he has
not gone to university.
8
7. (10 points) The following stage game is repeated twice:
C
D
A
0, 20
-1, 0
B
1, 0
4, 0
Assume that each player’s total payoff is the sum of his payoffs from the two periods.
Construct a subgame-perfect equilibrium in which (A, C) is played in the first period.
Answer:
There are two Nash equilibria in the stage game: the good equilibrium (for Player 1) of
(B, D), and the bad equilibrium (for Player 1) of (B, C). Thus, if Player 1 has played A in
the first period, then the players should play the good equilibrium of (B, D) in the second
period, and if Player 1 has not played A in the first period, then the players should play the
bad equilibrium of (B, C) in the second period. This way, Player 1 has an incentive to follow
the action profile (A, C): playing A leads to 0 + 4 = 4, and playing B leads to 1 + 1 = 2.
Player 2 has incentive to play C if Player 1 plays A.
Therefore, the following is a subgame perfect equilibrium:
1. Player 1 plays A in period 1, and plays B in period 2 regardless of the outcome in
period 1.
2. Player 2 plays C in period 1; in period 2, Player 2 plays D if A was played in period
1, and plays C if B was played in period 1.
9
Random-Coffeehouse
stay out
0
100000
enter
Tim Hortons
fight
−25000
10000
accommodate
25000
50000
Figure 1: Game tree for Q4
(11, 1)
H
(10, 0)
NH
U
(0, 0)
Nature
U
H
(1, 1)
NH
(0, 0)
firm
low [0.5]
H
NH
NU
high [0.5]
firm
(1, −1)
student
student
NU
Figure 2: Game tree for Q6
H
NH
(11, −1)
(10, 0)