Homework 15
1. Consider the following strategic situation. Two opposed armies are poised to
seize an island. Each army’s general can choose either “attack” or “not attack.” In
addition, each army is either strong or weak with equal probability (the draws for
each army are independent), and an army’s type is only known to its general. Payoffs
are as follows: The island is worth M if captured. An army can capture an island
either by attacking when its opponent does not or by attacking when its rival does if
it is strong and the rival is weak. If two armies of equal strength both attack, neither
captures the island. An army also has a “cost” of fighting, which is s if it is strong
and w if it is weak, with s < w. There is no cost of attacking if its rival does not.
Find all pure strategy Bayesian Nash equilibria of this game.
Answer:
Each player’s strategy space contains four strategies:
• Attack if strong, not attack if weak (AN)
• Attach is weak, not attack if strong (NA)
• Always attack (AA)
• Never attack (NN)
The payoffs from each of the strategies against any other strategy can be calculated. For example, the payoff from AN playing against AA is:
1/4*(M-s)+1/4*(-s)+1/4*0+1/4*0=1/4M-1/2*s.
Similarly, we can calculate all other payoffs and put them in a table:
AA
AN
NA
NN
AA
M/4 − (s + w)/2, M/4 − (s + w)/2
M/4 − s/2, M/2 − (s + w)/4
−w/2, 3M/4 − (s + w)/4
0, M
AN
M/2 − (s + w)/4, M/4 − s/2
(M − s)/4, (M − s)/4
(M − w)/4, M/2 − s/4
0, M/2
NA
3M/4 − (s + w)/4, −w/2
M/2 − s/4, M − w/4
(M − w)/4, (M − w)/4
0, M/2
NN
M, 0
M/2, 0
M/2, 0
0, 0
You can then carefully figure out which of these cells can form Nash equilibria
and when. For example, (AN,AA) is a Nash equilibrium if the following inequalities
are satisfied:
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• u(AN, AA) ≥ u(AA, AA) ⇔ M/4 − s/2 ≥ M/4 − (s + w)/2
• u(AN, AA) ≥ u(N A, AA) ⇔ M/4 − s/2 ≥ −w/2
• u(AN, AA) ≥ u(N N, AA) ⇔ M/4 − s/2 ≥ 0
The first inequality is always true, and the second is implied by the third. Thus,
(AN,AA) is a Nash equilibrium if M/2 ≥ s. In the end, we have the following cases:
Case 1: M > w > s and w > M/2 > s. Then (AA,AN) and (AN,AA) are both pure
strategy Bayesian Nash equilibria.
Case 2: M > w > s and M/2 < s. Then (AA,NN) and (NN,AA) are both pure strategy
Bayesian Nash equilibria.
Case 3: w > M > s and M/2 < s. Then (AN,AN), (AA,NN) and (NN,AA) are all pure
strategy Bayesian Nash equilibria.
Case 4: w > M > s and M/2 > s. Then (AA,AN), (AN,AA) and (AN,AN) are all pure
strategy Bayesian Nash equilibria.
2. Consider the single good auction environment discussed in class. There is a
single good to be allocated among N agents. Monetary transfers can be made. An
outcome is (y1 , ..., yN , t1 , ..., tN ), where each yi ∈ {0, 1} (1 if the agents gets the good
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and 0 otherwise), ti ∈ R, and i ti ≤ 0. Agents’ utility functions are given by:
ui (x, θi ) = yi θi + mi + ti ,
where mi is agent i’s endowment. Show that a social choice function is efficient if,
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for every θ, it gives the good to the highest agent and if i ti (θ) = 0.
P
Answer: Assume that i ti (θ) < 0 for some one θ. Then you can increase any
ti by and increase the associated agent’s utility without hurting anyone else. This
is not efficient.
P
Not assume i ti (θ) = 0 but the good doesn’t go to the agent who values it most
for some θ. Let’s say the good goes to θi and the guy who values it most is θj . Take
the good away from Mr. i and increase his transfer by θi . Give the good to Mr. j
and decrease his transfer by θi . Then Mr. j is better off and no one else is worse off.
Hence the arrangement wasn’t efficient.
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3. Show that the equilibria of first and second price auctions that we studied in
class generate the same revenue for the seller. Note: in first-price auctions, consider
the equilibrium where each buyer bids one half of his valuation.
Answer: Consider a first price auction. The revenue the seller collects from Mr.
1 is
Z 1 Z θ1
θ1 /2 · P r(θ1 ≥ θ2 )dθ2 dθ1 ,
0
0
with the second integration limit going up to θ1 because if θ2 > θ1 , then Mr. 1 doesn’t
get the object. Within the integration limits, P r(θ1 ≥ θ2 ) = 1. Hence, the expected
revenue is
Z 1 2
Z θ1
Z 1
Z 1 Z θ1
θ1
dθ1 = 1/6.
dθ2 dθ1 =
θ1 /2
θ1 /2 dθ2 dθ1 =
0 2
0
0
0
0
Now consider a second price auction. The revenue the seller collects from Mr. 1
is
1
Z
Z
θ1
θ2 · P r(θ1 ≥ θ2 )dθ2 dθ1 .
0
0
Again, P r(θ1 ≥ θ2 ) within the integration limits and hence the revenue is:
Z
1
Z
θ1
Z
[θ2 ]dθ2 dθ1 =
0
0
0
3
1
θ12
dθ1 = 1/6.
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