Simon Fraser University
Fall 2015
Econ 302 D200 Final Exam Solution
Instructor: Songzi Du
Wednesday December 16, 2015, 8:30 – 11:30 AM
NE = Nash equilibrium, SPE = subgame perfect equilibrium, PBE = perfect
Bayesian equilibrium
1. There is a shared internet connection which has a maximum capacity of 100. There
are n = 10 users of this connection, and user i uses xi units of data, where 0 ≤ xi ≤ 100.
P
(xi can be a real number, not necessarily an integer.) If ni=1 xi ≥ 100, then the connection
P
is jammed and each user gets zero payoff. If ni=1 xi < 100, then each user i gets a payoff of
P
xi (100 − ni=1 xi ). Suppose that each user chooses his/her xi simultaneously. (Recall that
Pn
i=1 xi = x1 + x2 + · · · + xn .)
P
P
P
i. (5 points) The total payoff of all users is ( ni=1 xi )(100 − ni=1 xi ). Let y = ni=1 xi .
What value of y maximizes the total payoff?
P
ii. (10 points) Calculate the symmetric NE (in pure strategy) such that ni=1 xi < 100.
Would everyone be better off (relative to the NE outcome) if xi = y/n, where y is
determined in part (i), and why or why not?
Solution:
Part i: we maximize the total payoff y(100 − y). The first order condition is 100 − 2y = 0,
i.e., y = 50 maximizes the total payoff.
Part ii: suppose xi = x is the symmetric NE. Let us pick user 1 and look at his best
response: player 1 maximizes x1 (100 − x1 − 9x) since all other users uses x. Player 1’s first
order condition is then:
100 − 2x1 − 9x
= 0,
x1 =x
i.e., x = 100/11 is the symmetric NE.
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In this NE, each user gets payoff 100/11(100 − 10 · 100/11) = 10000/112 ≈ 82.64. If
xi = y/10 = 5 for everyone, then each user gets payoff 5(100 − 50) = 250. Clearly, everyone
is much better off with xi = 5. Thus, this is a prisoner’s dilemma.
2.
(10 points) Consider a new version of rock-paper-scissors game which has the
following payoff matrix (player 1 has a big advantage if he uses Rock and player 2 uses
Scissors):
Rock
Scissors
Paper
Rock
0, 0
30, -30
-10, 10
Scissors
-10, 10
0, 0
10, -10
Paper
10, -10
-10, 10
0, 0
As before each player chooses his action simultaneously. Find the mixed-strategy NE
(this equilibrium is asymmetric: the two players use different mixed strategies). What are
players’ expected payoff in this equilibrium?
Solution:
Suppose the mixed-strategy equilibrium is player 1 using pR + qS + (1 − p − q)P and
player 2 using xR + yS + (1 − x − y)P . Player 1 must be indifferent between R, S and P :
i.e.,
30y − 10(1 − x − y) = −10x + 10(1 − x − y),
30y − 10(1 − x − y) = 10x − 10y,
i.e., x = 1/3, y = 1/5.
Player 2 must be indifferent between R, S and P :
−10q + 10(1 − p − q) = 30p − 10(1 − p − q),
−10q + 10(1 − p − q) = −10p + 10q,
i.e., p = 1/5, q = 1/3.
Notice that in equilibrium player 1 places less probability on Rock, relative to the symmetric version of the game.
3. There are 4 houses in a neighborhood. For a resident in a house, the cost of installing
an alarm system is 13, and the cost of his/her house being burglarized is 50. If n (0 ≤ n ≤ 4)
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houses have the alarm, then these n houses will not be burglarized (each gets a payoff of
−13), while for the remaining 4−n houses each will be burglarized with probability 1/(4−n)
(each gets a payoff of −50/(4 − n)).
Suppose the residents sequentially decide whether or not to install the alarm system:
resident 1 moves first and either installs the alarm or does not; after observing what resident
1 did, resident 2 decides; after observing what resident 1 and 2 did, resident 3 decides; after
observing what resident 1, 2 and 3 did, resident 4 decides.
i. (7 points) Draw the game tree (write the first payoff for resident 1’s payoff, the second
payoff for resident 2’s payoff, and so on).
ii. (8 points) Find and describe the SPE in pure strategy.
Solution:
Part i: (I omit the game tree because it is straightforward.)
Part ii: notice that 13 > 50/(4 − n) if and only if n = 0. Here is the SPE which is solved
by backward induction:
• Let 0 ≤ n4 ≤ 3 be the number of houses that have installed the alarm before resident
4. Resident 4 installs the alarm if n4 > 0 and does not install the alarm if n4 = 0.
• Let 0 ≤ n3 ≤ 2 be the number of houses that have installed the alarm before resident
3. Resident 3 installs the alarm if n3 > 0 and does not install the alarm if n3 = 0.
• Let 0 ≤ n2 ≤ 1 be the number of houses that have installed the alarm before resident
2. Resident 2 installs the alarm if n2 > 0 and does not install the alarm if n2 = 0.
• Resident 1 does not install the alarm in the beginning of the game.
4. A pharmaceutical company (player 1) introduces a new cold medicine. The medicine
may either be highly effective (H type) or have little effect (L type). The company knows
the type of the drug, but the consumer (player 2) does not know the type (he/she only
knows the medicine is H type with probability 0.3 and is L type with probability 0.7). The
company can choose either to advertise the drug (action A), at a cost of c, or not to advertise
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(action N ), which costs nothing. The consumer decides whether or not to buy the drug after
observing whether the company advertised the drug.
The consumer gets a payoff of 1 if he buys a highly effective drug, -1 if he buys the
little-effect drug, 0 if he does not buy.
If the drug is highly effective, then the company gets 100 if the consumer buys the drug
(i.e., making money from repeated purchases); if the drug has little effect, then the company
gets 10 if the consumer buys the drug (i.e., making money from one time purchase). The
company gets 0 if the consumer does not buy. The net payoff of the company is the payoff
from sales minus the advertisement cost. For example, if the drug has little effect but the
company advertises it, and the consumer buys, then the net payoff of the company is 10 − c.
i. (4 points) Draw the game tree.
ii. (6 points) Suppose c = 20, find all pure-strategy PBE (if any exists).
iii. (10 points) Suppose c = 5, find all pure-strategy PBE (if any exists). Find a mixedstrategy PBE in which the L-type company randomizes between advertising and not
advertising.
In part (ii) and (iii), include conditional beliefs P(H | A) and P(H | N ) in your description
of PBE.
Solution:
Part ii:
Clearly, the consumer’s best response is to buy after advertisement (A-buy) if P(H |
A) ≥ 1/2, and not buy after advertisement (A-not) if P(H | A) ≤ 1/2; to buy after no
advertisement (N-buy) if P(H | N ) ≥ 1/2, and not buy after no advertisement (N-not) if
P(H | N ) ≤ 1/2.
The low-type company clearly always prefers no advertisement (L-N) to advertisement
(L-A). Thus, we can safely ignore L-A. There are only two strategies of the company that
are potentially PBE:
• Suppose the company uses the strategy (H-A, L-N). Then P(H | A) = 1 > 1/2 and
P(H | N ) = 0 < 1/2. So consumer’s best response is (A-buy, N-not). And given
(A-buy, N-not), (H-A, L-N) is clearly also a best response for the company. Thus, this
is a PBE.
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Figure 1 : part i when c = 20.
• Suppose the company uses the strategy (H-N, L-N). Then P(H | A) can be anything,
and P(H | N ) = 0.3. Consumer’s best response must use N-not since 0.3 < 1/2. If
the consumer uses A-buy, then H-N is not a best response for the high-type company,
which is inconsistent with the initial conjecture. If the consumer uses A-not, then H-N
is a best response for the high-type company.
Thus, the PBE here is: the company uses (H-N, L-N); the consumer believes that
P(H | N ) = 0.3 and P(H | A) ≤ 1/2 and uses (A-not, N-not).
Part iii: consumer’s best response as a function of his beliefs is the same as part ii,
because his payoff is not changed. Now both the high-type and the low-type company have
no dominant strategy. So we go through all four strategies of the company:
• Suppose the company uses the strategy (H-A, L-A). Then P(H | A) = 0.3 < 1/2 and
P(H | N ) can be anything. So consumer’s best response must include A-not. But
given A-not by the consumer, H-N is always better for the high-type company than
H-A, so (H-A, L-A) cannot be a PBE.
• Suppose the company uses the strategy (H-A, L-N). Then P(H | A) = 1 > 1/2 and
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P(H | N ) = 0 < 1/2. So consumer’s best response is (A-buy, N-not). And given (Abuy, N-not), L-A is better for the low-type company than L-N, so (H-A, L-N) cannot
be a PBE.
• Suppose the company uses the strategy (H-N, L-A). Then P(H | N ) = 1 > 1/2 and
P(H | A) = 0 < 1/2. So consumer’s best response is (A-not, N-buy). And given (Anot, N-buy), L-N is better for the low-type company than L-A, so (H-N, L-A) cannot
be a PBE.
• Suppose the company uses the strategy (H-N, L-N). Then P(H | A) can be anything,
and P(H | N ) = 0.3. Consumer’s best response must use N-not since 0.3 < 1/2. If
the consumer uses A-buy, then H-N is not a best response for the high-type company,
which is inconsistent with the initial conjecture. If the consumer uses A-not, then H-N
is a best response for the high-type company. And given N-not, L-N is also a best
response for the low-type company.
Thus, the PBE here is: the company uses (H-N, L-N); the consumer believes that
P(H | N ) = 0.3 and P(H | A) ≤ 1/2 and uses (A-not, N-not).
Now for the mixed-strategy equilibrium in which the low-type company is indifferent
between L-A and L-N and randomizes between the two actions. Then, the high-type company
must strictly prefers H-A over H-N, because they have the same advertisement cost but the
high type makes more money when the consumer buys.
Suppose that the low type plays L-A with probability p and the high type plays H-A with
probability 1. Then P(H | A) = 0.3/(0.3 + 0.7p) and P(H | N ) = 0. So the best response of
the consumer must use N-not.
For the low type to be indifferent between L-A and L-N, the consumer must randomize
after advertisement; the only way for that to happen is P(H | A) = 0.3/(0.3 + 0.7p) = 1/2,
i.e., p = 3/7. Suppose the consumer uses A-buy with probability q. For the low type to be
indifferent we must have 5q − 5(1 − q) = 0, i.e., q = 1/2.
Thus, the mixed-strategy PBE is: the company uses (H-A, 3/7 L-A + 4/7 L-N), the
consumer believes that P(H | A) = 1/2 and P(H | N ) = 0, and uses (1/2 A-buy + 1/2
A-not, N-not). This mixed-strategy equilibrium is more efficient than the pure-strategy
equilibrium in which nobody buys.
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