Simon Fraser University
Spring 2014
Econ 302 D200 Final Exam Solution
This brief solution guide does not have the explanations necessary for full marks.
NE = Nash equilibrium, SPE = subgame perfect equilibrium, PBE = perfect
Bayesian equilibrium
1. (10 points) Consider the following simultaneous-move game (Table 1).
W
X
Y
Z
A
1, 7
4, 3
1, 2
6, 2
B
8, 6 4, 6
2, 5
5, 5
C
4, 1
5, 2
3, 3
4, 2
D
5, 5
4, 3
2, 4
9, 3
Table 1
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 NE (pure and mixed) in this game, and for each NE that you find,
calculate the two players’ expected payoffs in the equilibrium.
Solution:
Part i, ISD:
Round 1: Z is strictly dominated by 1/2X + 1/2Y . Delete Z.
Round 2: both A and D are strictly dominated by 1/2B + 1/2C. Delete A and D.
The strategies that survive: B, C for player 1, and W, X, Y for player 2.
Part ii, NE:
For NE we can look at the smaller game given by ISD.
Clearly, the pure-strategy NE are (B, W ) and (C, Y ). The payoffs in these pure-strategy
NE can be read off the table.
If player 1 plays pB + (1 − p)C, where 0 < p < 1, then W cannot be a best response of
player 2, because W is weakly dominated by X among the ISD strategies; if player 1 does
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not play W with positive probability, then the only best response of player 2 is C, which
means that pB + (1 − p)C is not a best response. So pB + (1 − p)C, where 0 < p < 1, cannot
be a part of a NE.
If player 1 plays C, then the best response of player 2 is Y , so we are back to the
pure-strategy NE.
If player 1 plays B, then the best responses of player 2 are W and X. For mixed-strategy
qW + (1 − q)X, we want B to be a best response, i.e., 8q + 4(1 − q) ≥ 4q + 5(1 − q), i.e.,
q ≥ 1/5.
Thus, the non-trivial mixed-strategy NE are (B, qW + (1 − q)X), where 1/5 ≤ q < 1. In
these mixed-strategy NE player 1 gets 8q + 4(1 − q) and player 2 gets 6.
2. (10 points) There are two players each with a wealth w = $3. An object that the
two players each value at v = $3 is sold at an auction. In the auction, the two players
alternatively have the opportunity to bid; a bid must be a positive integer greater than the
previous bid and less than or equal to w. On his turn, a player may pass rather than bid,
in which case the game ends and the other player receives the object; both players pay their
last bids (if any). Player 1 moves initially. If player 1 passes initially, for example, player
2 receives the object and makes no payment (net payoff v − 0 = 3) while player 1 receives
nothing and pays nothing (net payoff 0 − 0 = 0); if player 1 bids 1, player 2 bids 3, and then
player 1 passes, player 2 obtains the object and pays 3 (net payoff v − 3 = 0) while player 1
receives nothing and pays 1 (net payoff 0 − 1 = −1).
Part i: draw the game tree.
Part ii: find and completely describe a pure-strategy SPE.
Solution: Part i: see Figure 1.
Part ii:
A SPE is:
Player 1: Bid $3 whenever there is an opportunity (i.e., initially, and when the previous bid of
bidder 2 is less than $3), and pass when the previous bid of bidder 2 is $3.
Player 2: Bid $3 whenever there is an opportunity (i.e., when the previous bid of bidder 1 is less
than $3), and pass when the previous bid of bidder 1 is $3.
3. (10 points) Part i: suppose that you have an utility over money u(m) =
2
√
m = m0.5
Figure 1
and start with $10000. Consider a risky project with probability 0.25 of gaining $1000,
probability 0.25 of gaining $2000, probability 0.25 of gaining $10000, and probability 0.25
of losing $10000. Find the expected value of this project (on the total wealth). Find your
expected utility if you take this project, and find your certainty equivalent and risk premium
for this project. Would you take the project, and why?
Part ii: repeat part i with u(m) = m0.9 . Are you more or less risk averse with u(m) = m0.9
√
compared to u(m) = m, and why?
In your calculations keep two decimal places.
Solution:
Part i:
Expected wealth of taking the project: 0.25×11000+0.25×12000+0.25×20000+0.25×0 =
$10750.
√
√
Expected utility of taking the project: EU = 0.25 × 11000 + 0.25 × 12000 + 0.25 ×
√
√
20000 + 0.25 × 0 ≈ 88.96.
√
Certainty equivalent: CE = EU , so CE ≈ $7914.18.
Risk premium = 10750 − 7914.18 = $2835.82.
You would not take this project, because the certainty equivalent $7914.18 < $10000.
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Part ii:
Expected wealth of taking the project: 0.25×11000+0.25×12000+0.25×20000+0.25×0 =
$10750.
Expected utility of taking the project: EU = 0.25 × (11000)0.9 + 0.25 × (12000)0.9 + 0.25 ×
(20000)0.9 + 0.25 × (0)0.9 ≈ 4114.39.
Certainty equivalent: (CE)0.9 = EU , so CE ≈ $10372.77.
Risk premium = 10750 − 10372.77 = $377.23.
You would take this project, because the certainty equivalent $10372.77 > $10000.
√
You are less risk averse with u(m) = m0.9 compared to u(m) = m, because the certainty
√
equivalent of u(m) = m0.9 is larger than that of u(m) = m.
4. (10 points) Two firms (firm 1 and 2) compete in quantity of production and face
a market demand function of P = 9 − Q. Each firm has a marginal cost of 3 per unit of
production.
Part i: Find the monopolist quantity, price, and payoff when the two firms merge into
one monopoly firm.
Part ii: Find the NE quantity, price, and payoff when the two firms compete by simultaneously choosing quantities.
Part iii: Suppose that the competition is infinitely repeated, and each firm has a discount
P
t
factor of δ. So firm i’s payoff is ∞
t=0 δ (Pt − 3)qi,t , where qi,t is firm i’s production quantity
in period t, and Pt = 9 − (q1,t + q2,t ) is the market price in period t. In each period, the
two firms set their quantities simultaneously, and observe the quantities from the previous
periods. Describe a grim trigger strategy, and find the range of δ such that both firm playing
the grim-trigger strategy is a SPE of the infinitely repeated game. What is the total payoff
of each firm when both firms follow the grim-trigger strategy?
Solution:
Part i:
The monopolist solves:
max(9 − Q − 3)Q,
Q
the first order condition is
6 − 2Q = 0,
so the monopolist’s quantity Qm = 3, price is 9 − 3 = 6, and payoff is (9 − 3 − 3) × 3 = 9.
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Part ii:
Suppose that firm 2 produces q ∗ , firm 1 solves
max(9 − (q1 + q ∗ ) − 3)q1 ,
q1
the first order condition is
6 − 2q1 − q ∗ = 0.
In the symmetric NE, firm 1 and 2 produce the same amount, so q1 = q ∗ . Therefore, the NE
quantity of each firm is q ∗ = 2, price is 9 − (2 + 2) = 5, and payoff is (9 − (2 + 2) − 3) × 2 = 4.
Part iii:
The following is a grim trigger strategy:
In each period, produce half of the monopolist quantity Qm /2 = 1.5 (collusion) if both
firms had produced Qm /2 in all past periods; produce the stage-game NE quantity q ∗ = 2
(price war) if one of the firms had deviated from Qm /2 in any past period.
Clearly, if grim quantity q ∗ is triggered, it is a subgame perfect equilibrium for both firms
to produce q ∗ forever.
When the other firm is producing Qm /2 = 1.5 in the current period, then the best possible
deviation in the current period, qd , solves
max(9 − (q + 1.5) − 3)q,
q
i.e.,
4.5 − 2qd = 0,
So qd = 2.25, which gives a profit of (9 − (2.25 + 1.5) − 3) × 2.25 = 5.0625 in the current
period. But playing qd will trigger the play of q ∗ and per-period payoff of 4 in the subsequent
periods. If the firm does not deviate from grim trigger, then he gets half of the monopolist’s
profit 9/2 = 4.5 in every period.
Therefore, for a firm to not have an incentive to deviate to qd given the grim trigger
strategy, we must have:
5.0625 + 4δ + 4δ 2 + 4δ 3 + · · · ≤ 4.5 + 4.5δ + 4.5δ 2 + 4.5δ 3 + · · · ,
i.e.,
5.0625 + 4
4.5
δ
≤
,
1−δ
1−δ
5
or
5.0625 − 4.5
≈ 0.52.
5.0625 − 4
So when the above inequality on δ holds, both player playing the grim-trigger strategy
is a SPE. In this SPE, each player gets 4.5/(1 − δ) in total payoff.
δ≥
Write Question 5 – 7 on booklet #2
5. (10 points) There are two players. Player 1 is either strong or weak (those are his
types), and chooses a meal: beer or quiche. Player 2 observes this food choice, and chooses
to fight or retreat.
Player 2 cannot tell if player 1 is strong or weak, and believes that player 1 is strong with
probability 0.1 and weak with probability 0.9. (So Nature chooses player 1 to be strong with
probability 0.1 and weak with probability 0.9.) Player 2 gets 1 if he fights the weak player,
−1 if he fights the strong player, and 0 if he retreats.
The strong player 1 prefers beer; the weak player 1 prefers quiche. Player 1, strong or
weak, gets 2 if Player 2 doesn’t fight him, 0 otherwise; player 1, strong or weak, gets an
additional 1 if he eats his preferred meal.
(i) Draw the game tree.
(ii) Find all separating PBE (if any), and explain.
(iii) Find all pooling PBE (if any), and explain.
For part ii and iii, focus on pure-strategy PBE, and include conditional beliefs (such as
P(strong | beer) = 1 or P(strong | beer) ≤ 1/2.) in your description of PBE.
Solution: Part i: see Figure 2.
Part ii:
Player 2’s best response is to fight if his conditional probability (given either beer or
quiche) on the strong type is less than 1/2, and to retreat if his conditional probability on
the strong type is greater than 1/2.
There is no separating PBE in pure strategies.
If player 1 uses S-Beer and W-Quiche, then player 2 will use B-Retreat and Q-Fight in
best response (since P(S | Beer) = 1 and P(S | Quiche) = 0), which means that player 1 is
better off with W-Beer instead of W-Quiche. Hence S-Beer and W-Quiche cannot be a part
of a PBE.
If player 1 uses S-Quiche and W-Beer, then player 2 will use B-Fight and Q-Retreat in
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Figure 2
best response (since P(S | Beer) = 0 and P(S | Quiche) = 1), which means that player 1 is
better off with W-Quiche instead of W-Beer. Hence S-Quiche and W-Beer cannot be a part
of a PBE.
Part ii:
There is no pooling PBE in pure strategies.
If player 1 uses S-Beer and W-Beer, then player 2 will fight if he sees beer (B-Fight),
because P(S | Beer) = 0.1. Then whatever player 2 does after seeing quiche, player 1 is
strictly better off with W-Quiche. Hence S-Beer and W-Beer cannot be a part of a PBE.
If player 1 uses S-Quiche and W-Quiche, then player 2 will fight if he sees quiche (QFight), because P(S | Quiche) = 0.1. Then whatever player 2 does after seeing beer, player 1
is strictly better off with S-Beer. Hence S-Quiche and W-Quiche cannot be a part of a PBE.
7. (10 points) Consider a market for used cars with risk-neutral participants, like we
did in Lecture 10. Suppose there are four qualities for cars: A, B, C and D. Let the sellers’
valuations for the four qualities be $10000 (quality A), $9000 (quality B), $5000 (quality C)
and $3000 (quality D), and let the buyers’ valuations be $20000 (quality A), $15000 (quality
B), $7000 (quality C) and $4000 (quality D). Assume that there is an equal number of cars
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in each of the four categories.
Suppose only sellers know the quality of each car. Is the sales of quality A cars feasible?
If so what prices for quality A cars could you observe? What about quality B, C, and D
(feasibility of sales and if so the price range)? Explain.
Solution: Since a buyer does not know the quality, he is willing to pay at most (20000 +
15000 + 7000 + 4000)/4 = $11500 for a car, since there is an equal number of cars in each of
the four quality categories. Since 11500 > 10000, the sales of quality A cars is feasible, and
the possible prices range from $10000 to $11500. If the sales of quality A cars are feasible,
so are the sales of quality B, C and D, because the buyers cannot distinguish these qualities.
The prices for quality B cars range from $9000 to $11500. The prices for quality B cars
range from $5000 to $11500. The prices for quality B cars range from $3000 to $11500.
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