Simon Fraser University
Spring 2013
Econ 302 Final Exam
Instructor: Songzi Du
Section D200
Wednesday April 24, 2013, 8:30 – 11:30 AM
This exam has two parts. Write your name, SFU ID number, and tutorial section number
on each part.
Suggested Solution
• Name:
• SFU ID number:
• Tutorial section number:
1. This is a closed-book exam.
2. You may use a non-graphing calculator.
3. If you use decimals in calculation, keep two decimal places.
4. There is no separate exam booklet. Write your solution in the space following each
question.
5. Show your work! Partial credits are given. Answers without proper explanation/calculation
will be penalized.
6. We will accept a request for regrade only if the solution is written with a pen.
7. The final exam has 135 points (45% of the class grade).
8. 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. (15 points) Consider the following simultaneous-move game with two players. Each
player simultaneously chooses a bid from the set {1 dollar , 50 dollars, 100 dollars }. The
player with the higher bid wins a prize of 100 dollars; the other player (the loser) does not
receive any prize. If the two bids are the same, neither player receives the prize. Each
player must pay his own bid, whether or not he wins the prize. (The loser pays too.) The
payoff/utility of each player is simply his net winnings. For example, if player 1 bids 100
and player 2 bids 50, then player 1 gets 100 − 100 = 0 and player 2 gets 0 − 50 = −50; if
player 1 bids 50 and player 2 bids 50, then each gets 0 − 50 = −50.
(i) Write the normal form (matrix) of this game.
(ii) Calculate a symmetric mixed strategy Nash equilibrium. Keep two decimal places or
use fraction.
Solution:
Part (i):
The normal form is
$1
$50
$100
$1 −1, −1
−1, 50
−1, 0
$50 50, −1 −50, −50
−50, 0
$100 0, −1
0, −50
−100, −100
(5 points)
Part(ii):
Let us solve for the symmetric mixed-strategy Nash equilibrium. Suppose that each
player bids $1 with probability p, $50 with probability q, and $100 with probability 1 − p − q.
To have incentive to use this mixed strategy, each player must be indifferent between the
three actions if he believes that the other player is following this mixed strategy.
Thus, we have two conditions for this symmetric equilibrium:
(−1)p + (−1)q + (−1)(1 − p − q) = 50p + (−50)q + (−50)(1 − p − q)
{z
} |
{z
}
|
payoff from $1
payoff from $50
(0)p + (0)q + (−100)(1 − p − q) = 50p + (−50)q + (−50)(1 − p − q)
|
{z
} |
{z
}
payoff from $100
payoff from $50
2
Solving these equations gives:
1
49
,q = .
100
2
In summary, the symmetric mixed-strategy Nash equilibrium is bidding $1 with probability 49/100, bidding $50 with probability 1/2, and bidding $100 with probability 1/100.
(10 points: 5 points for setting up the mixed-strategy equilibrium condition, and 5 points
for solving the mixing probabilities; a student can get partial points for this part even if
his/her normal form is incorrect.)
p=
3
2. (15 points) In Burnaby Mountain there is nothing to do at night but look at the
stars or go to the local bar. Let us define the utility of looking at stars as 0, and let the cost
of walking to the bar be 1. Suppose that the utility from being at the bar is 2 (so the net
utility is 2 − 1 = 1) if there are fewer than 3 people (including oneself) at the bar, and 1/2
(the net utility is 1/2 − 1 = −1/2) if there are 3 or more people (including oneself) at the
bar. Suppose that there are only 3 people in Burnaby Mountain at night, and they make
decisions (looking at stars or going to the bar) simultaneously.
(i) Calculate the three pure-strategy Nash equilibria.
(ii) Calculate the symmetric mixed-strategy Nash equilibrium. And calculate the utility
of each player in this equilibrium. Keep two decimal places or use fraction/square root.
Solution:
Part (i):
There are three pure-strategy Nash equilibria: one of the three people looks at the star,
the other two go to the local bar. (5 points)
Part (ii):
Consider a symmetric mixed-strategy Nash equilibrium in which each person goes to the
bar with probability p. One gets 0 from looking at the stars. Assumes that others play the
mixed strategy, by going to the bar one gets −1/2 with probability p2 (the others are all at
the bar), and gets 1 with probability 1 − p2 . Thus, the condition for this mixed strategy to
be an equilibrium is:
1
0 = − p2 + (1 − p2 ).
2
Solving this equation gives:
r
p=
√
2
6
=
≈ 0.82.
3
3
In summary, the symmetric mixed-strategy Nash equilibrium is that each person goes to
√
the bar with probability 36 . By construction, each person is indifferent between looking at
the stars and going to the bar in this equilibrium, thus he gets an expected utility of 0 in
the equilibrium, since looking at the stars gives 0 utility.
(10 points: 5 points for setting up the mixed-strategy equilibrium condition, 3 points for
solving it, and 2 points for calculating/noting the utility in the equilibrium.)
4
3. (15 points) Find all pure-strategy subgame perfect equilibria of the following game.
Use game matrices to justify your answers.
1
a
2
c
0
1
e
b
2
d
1
2
c
2
d
1
4
0
3
f
2
1
g
0
1
Solution:
Player 2 is indifferent between actions f and g in the node after e. Thus, both f and
g are admissible in a subgame perfect equilibrium. So we consider two possibilities. In the
first possibility, player 2 plays f . Converting the rest of the game into the normal form, we
have:
c, f d, f
a 0, 1 1, 2
b 1, 4 0, 3
e 2, 1 2, 1
Clearly, the pure-strategy equilibria here are (e; c, f ) and (e; d, f ), which are subgame
perfect equilibria for the whole game.
In the second possibility, player 2 plays g. Converting the rest of the game into the
normal form, we have:
c, g d, g
a 0, 1 1, 2
b 1, 4 0, 3
e 0, 1 0, 1
Clearly, the pure-strategy equilibria here are (a; d, g) and (b; c, g), which are also subgame
perfect equilibria for the whole game.
5
4. (15 points) Suppose that crude oil is a homogenous good, and the market (inverse)
demand function for crude oil is P (Q) = 20 − Q, where Q is the total quantity of crude oil on
the market. Country 1 is the traditional producer of crude oil and has a constant marginal
cost of 1 for producing each unit of oil. Country 1 faces competition from new entrants who
are Country 2 and 3 with superior production technology: Country 2 and 3 both have zero
marginal cost for producing crude oil. The game is as follows: Country 1 first sets/commits
to its quantity of production q1 ; Country 2 and 3 observe the commitment of Country 1 (q1 )
and then simultaneously choose their quantities q2 and q3 . Each country then sells its oil at
the market price P (q1 + q2 + q3 ).
Construct a subgame perfect equilibrium in this game. And calculate the equilibrium
price and quantities.
Solution:
In the subgame in which Country 1 commits to q1 , Country 2 and 3 play a symmetric
equilibrium of q2 = q3 = q ∗ (q1 ). The equilibrium condition is (say from the perspective of
Country 2):
q2 = q ∗ solves max q2 (20 − q1 − q ∗ − q2 ),
q2
which is equivalent to the first order condition of
20 − q1 − q − 2q2 ∗
= 0.
q2 =q ∗
Solving the first order condition gives:
q ∗ (q1 ) =
20 − q1
.
3
Now, Country 1 anticipates the strategy of q ∗ (q1 ) =
subgame. Thus, Country 1 solves
20−q1
3
2(20 − q1 )
max q1 20 − q1 −
− q1 .
q1
3
6
by Country 2 and 3 in the
The first condition of Country 1 is thus:
20 − 2q1 −
40 4
+ q1 − 1 = 0.
3
3
Solving the above equation gives q1 = 17
.
2
In summary, the subgame perfect equilibrium is Country 1 commits to q1 = 17
, and
2
20−q1
∗
Country 2 and 3 play the symmetric strategy of q (q1 ) = 3 . In this equilibrium, Country
= 8.5, and Country 2 and 3 each produces q ∗ (17/2) = 23/6 = 3.83. The
1 produces q1 = 17
2
equilibrium price is 20 − 17
− 2 × 23/6 = 23/6 = 3.83
2
(6 points for solving the subgame of Country 2 and 3; 6 points for solving Country 1’s
problem; 3 points for solving the equilibrium quantities and prices.)
7
√
5. (15 points) You have an utility over money u(m) = m and start with $10000.
Consider a risky project with probability 0.5 of gaining $5000, probability 0.3 of losing
$1000, and probability 0.2 of losing $4000. Find the expected value of this project (on the
total wealth). Find your expected utility if you take this project, and find the certainty
equivalent and risk premium for the this project. Would you take the project, and why? In
your calculations keep two decimal places.
Solution:
The expected value of the project is 0.5 × 15000 + 0.3 × 9000 + 0.2 × 6000 = 11400 dollars.
(3 points)
√
√
√
The expected utility from taking the project is 0.5× 15000+0.3× 9000+0.2× 6000 =
105.19. (3 points)
√
Solving m = 105.19, we have the certainty equivalent m = (105.19)2 = 11064.87 dollars.
(3 points)
The risk premium is 11400 − 11064.87 = 335.13 dollars. (3 points)
You would take this project, because the certainty equivalent of the project is larger than
your wealth from not taking the project (which is $10000). Alternatively, your expected
utility from taking this project is larger than your utility from not taking the project (which
√
is 10000 = 100). (3 points)
8
6. (15 points) The principal is considering hiring a salesman to make to an important
sale. If the sale is successful, it will generate $1,000,000 to the principal. The salesman can
exert effort (e = 1) or no effort (e = 0) in the sales job; if he exert effort (e = 1), the sale will
succeed with probability p; if he does not exert effort (e = 0), the sale will fail for sure. The
salesman is risk averse and has utility (w)1/3 − 10e, where w is the wage he receives from
the principal, and e is either 0 or 1. The principal is risk neutral and has an utility which
is the money from the sale (if any) minus the wage of the salesman. Both the principal and
the salesman maximizes their expected utilities.
Suppose that the salesman has an outside option of $1000 (from another source of employment), which translates to an utility of 10001/3 = 10; the principal has an outside option
of $0.
(i) Suppose that p = 0.2. Solve the principal’s first best problem when he can observe the
effort of the salesman and offers a flat wage, and solve the principal’s second best problem
when he cannot observe the effort of the salesman and offers wages contingent on the success
or failure of the sale. Your solution must specify the optimal wage(s) and whether the
salesman is hired. For the second-best problem you may assume that the (IR) and (IC)
constraints bind for the optimal solution. Exhibit your final answers with two decimal
places.
(ii) Suppose that p = 0.1. Solve the principal’s first best and second best problems.
Solution:
Part (i):
Suppose that the principal offers a flat wage of w. The first best problem is:
max 0.2 × 1000000 − w subject to: (w)1/3 − 10 ≥ 10
w
Clearly, the constraint binds at the optimum: w = (20)3 = $8000. By offering this wage
the principal gets 200000 − 8000 = $192000 > 0. So the principal hires the salesman without
moral hazard.
(2 points: 1 point for setting up the first best problem, and 1 point for solving it and
checking if the principal hires.)
Suppose that the principal offers a wage of w1 if the sale is successful, and a wage of w0
if the sale is not successful. The second best problem is:
9
max 0.2 × (1000000 − w1 ) + 0.8(−w0 ) subject to: 0.2(w1 )1/3 + 0.8(w0 )1/3 − 10 ≥ (w0 )1/3
w1 ,w0
0.2(w1 )1/3 + 0.8(w0 )1/3 − 10 ≥ 10
The two constraints bind at the optimum, which gives two equations:
0.2(w1 )1/3 + 0.8(w0 )1/3 − 10 = (w0 )1/3
0.2(w1 )1/3 + 0.8(w0 )1/3 − 10 = 10
Solving these two equations gives
(w1 )1/3 = 60, (w0 )1/3 = 10,
or equivalently,
w1 = $216000, w0 = $1000.
By offering these wages, the principal gets 0.2 × (1000000 − 216000) + 0.8 × (−1000) =
$156000 > 0. So the principal hires the salesman given moral hazard.
(5.5 points: 2 points for setting up the second best problem, 2 points for solving it, 1.5
points for checking if the principal hires.)
Part (ii): Suppose that the principal offers a flat wage of w. The first best problem is:
max 0.1 × 1000000 − w subject to: (w)1/3 − 10 ≥ 10
w
Clearly, the constraint binds at the optimum: w = (20)3 = $8000. By offering this wage
the principal gets 100000 − 8000 = $92000 > 0. So the principal hires the salesman without
moral hazard.
(2 points: 1 point for setting up the first best problem, and 1 point for solving it.)
Suppose that the principal offers a wage of w1 if the sale is successful, and a wage of w0
if the sale is not successful. The second best problem is:
10
max 0.1 × (1000000 − w1 ) + 0.9(−w0 ) subject to: 0.1(w1 )1/3 + 0.9(w0 )1/3 − 10 ≥ (w0 )1/3
w1 ,w0
0.1(w1 )1/3 + 0.9(w0 )1/3 − 10 ≥ 10
The two constraints bind at the optimum, which gives two equations:
0.1(w1 )1/3 + 0.9(w0 )1/3 − 10 = (w0 )1/3
0.1(w1 )1/3 + 0.9(w0 )1/3 − 10 = 10
Solving these two equations gives
(w1 )1/3 = 110, (w0 )1/3 = 10,
or equivalently,
w1 = $1331000, w0 = $1000.
By offering these wages, the principal gets 0.1 × (1000000 − 1331000) + 0.9 × (−1000) =
−$34000 < 0. So the principal does not hire the salesman given moral hazard.
(5.5 points: 2 points for setting up the second best problem, 2 points for solving it, and
1.5 points for noting that the principal does not hire.)
11
7. (15 points) (i) The worker has two types: skilled (s = 1) or unskilled (s = 0); the
probability that that the worker is skilled is 30%. A worker of either type can choose to get
educated (e = 1) or not (e = 0); suppose that a skilled worker gets educated with probability
10%, and a unskilled worker gets educated with probability 80%. Calculate the conditional
probability that a worker is skilled given that he has an education.
(ii) The Census Bureau has estimated the following survival probabilities for men:
• probability that a man lives at least 50 years: 95%;
• probability that a man lives at least 60 years: 85%;
• probability that a man lives at least 70 years: 65%.
What is the conditional probability that a man lives at least 70 years given that he has
just celebrated his 60th birthday?
In your calculations keep two decimal places or use fraction.
Solution:
Part (i)
Using Bayes’s rule:
P(s = 1 | e = 1) =
0.3 × 0.1
≈ 0.05.
0.3 × 0.1 + 0.7 × 0.8
(9 points)
Part (ii)
P (lives 70 | lives 60) =
P (lives 70)
0.65
P (lives 70 and lives 60)
=
=
≈ 0.76.
P (lives 60)
P (lives 60)
0.85
(6 points)
12
8. (15 points) Consider a third -price auction of a single, indivisible object: the highest
bidder gets the object and pays the third highest bid; the other bidders do not get the object
and do not pay.1 Suppose that there are three bidders.
(i) Give an example such that bidder 1 with value v1 does not have an incentive to bid
truthfully given bids b2 and b3 (i.e., specify the values of v1 , b2 and b3 ). Explain.
(ii) Give an example such that bidder 1 with value v1 does have an incentive to bid
truthfully given bids b2 and b3 (i.e., specify the values of v1 , b2 and b3 ). Explain.
Solution:
Part (i):
Suppose that v1 = 5, b2 = 6 and b3 = 4. Then bidder 1 has an incentive to bid b1 > 6
because by doing so he wins the object and pays only 4 (the third highest bid), which leads
to an utility of 5 − 4 = 0. By contrast, by bidding truthfully (b1 = 5), bidder 1 does not
win the object and gets an utility of 0. Thus, in this example bidder 1 does not have an
incentive to bid truthfully.
(7 points)
Part (ii):
Suppose that v1 = 5, b2 = 3 and b3 = 4. By bidding more than 4 (b1 > 4), bidder 1 wins
the object and pays only 3, which leads to an utility of 5 − 3 = 2. By bidding less than 4
(b1 < 4), bidder 1 does not win the object, which leads to an utility of 0. And by bidding
exactly at 4 (b1 = 4), bidder 1 wins the object and pays 3 with probability 0.5 (and does
not win with probability 0.5), which leads to an utility of 0.5 × 2 + 0.5 × 0 = 1. Clearly,
bidding truthfully (b1 = v1 = 5) induces the first outcome, which is the best outcome. So in
this example bidder 1 has an incentive to bid truthfully.
(8 points)
1
If there are ties for the highest bid, each of the highest bidders gets the object and pays the third highest
bid with equal probabilities. That is, ties are resolved by a fair randomization.
13
9.
(15 points) Consider an average-price auction of a single, indivisible object: the
highest bidder gets the object and pays the average of all bids; the other bidders do not
get the object and do not pay.2 For example, if b1 = 0.9 and b2 = 0.8, then bidder 1 wins
the object and pays (0.9 + 0.8)/2 = 0.85. (The auctioneer might (naively) use this kind of
auction because he wants the winning bidder to pay the “fair” price.)
Suppose that there are two bidders. The private value vi of each bidder is uniformly
distributed on the interval [0, 1] and is independent of one another.
(i) Construct the symmetric equilibrium bidding strategy b(vi ). Hint #1: b(vi ) = Avi so
your job is just to calculate the constant A. Hint #2: if bidder 1 wins by bidding b(x), then
his expected payment conditional on winning is
1
1
(b(x) + E[b(v2 ) | b(v2 ) ≤ b(x)]) = (Ax + E[Av2 | v2 ≤ x]),
2
2
and
E[v2 | v2 ≤ x] =
x
2
by our assumption of uniform distribution.
(ii) Calculate the expected payment of bidder i of value vi (the overall expected payment,
weighted by the probabilities of winning and losing).
Solution:
Part (i):
Let us solve for the symmetric equilibrium b(vi ). Following the hint, b(vi ) = Avi . Suppose
that bidder 1 of type v1 imitates type x and bids b(x), he gets:
1
x
x v1 −
Ax + A
.
2
2
Therefore, for b(vi ) = Avi to be a symmetric equilibrium, x = v1 must solve
1
x
max x v1 −
Ax + A
.
0≤x≤1
2
2
2
If there are ties for the highest bid, each of the highest bidders gets the object and pays the average bid
with equal probabilities. Hint: don’t worry about ties.
14
The first order condition of this maximization problem is:
3
3 (v1 − Ax) − Ax
= 0,
4
4
x=vi
or after simplification:
3
v1 − Avi = 0.
2
Clearly, this means that A = 2/3. Thus, the symmetric equilibrium bidding strategy is
b(vi ) = 32 vi .
(10 points: 5 points for setting up the maximization problem, and 5 points for solving for
A.)
Part (ii):
The expected payment of bidder 1 of type v1 is v1 × 21 (Av1 + E[Av2 | v2 ≤ v1 ]) =
v1 × 21 A × 23 v1 = 12 (v1 )2 . (5 points)
15