Econ 805 Problem Set 3 Solution 1. (Gibbo (Gibbons ns 1.2 1.2). ). In the follow following ing game, game, what what strate strategie giess surviv survivee IESDS IESDS (iterate (iterated d elimination of strictly dominated strategies)? What are the pure strategy Nash equilibria? L C R T 2, 0 1, 1 4, 2 M 3, 4 1, 2 2, 3 B 1, 3 0, 2 3, 0
Answer: Answer: for player player 1, B is strictly dominated by T . So we can delet deletee the row of B . Then for player 2, C is strictly dominated by R. No further further strateg strategies ies can be delete deleted. d. Therefore, the strategies that survive IESDS are: (T , M ) and (L, R). The pure strategy NE of this game are (M, L) and (T , R) (see the underscore in the bi-matrix). 2. Mascolell 8.B.5. Answer: Answer: (i) First, any quantity quantity q > q m = a2bc (the monopoly quantity) is strictly dominated by q m. Then Then the remain remaining ing strateg strategy y set for each each firm is [0, q m ]. Now yo you q a c can calculate the best response to q m . You will will get R(q m) = 2b − 2 .. One One can show show that any q < b(q m ) is strictly dominated by R(q m ). No Now w the relev relevan antt strategy strategy set is [R(q m ), q m ]. No Now w you can calculate calculate the best response response to R(q m ). You can easily get get that ( ) R q Again ain you can prove prove that any quantit quantity y q > R2 (q m ) is strictly R2 (q m ) = a2bc − 2 . Ag )]. If you dominated by R2 (q m ). No Now w the relev relevant strategy strategy set shrinks to [R(q m ), R2 (q m )]. you 2n 1 (q m), continue this process, after 2n rounds of IESDS, the remaining strategy set is [R 2n n )]. No R (q m )]. Now w you only need need to show show that that R (q m ) converges as n goes to infinity. nity. You can show that −
−
m
m
−
−
2n − 2n
1
+ 2n
2
+ ... + (−1)n 2n n R (q m ) = q m 2n 1 1 1 = [1 − + − ... + (−1)n n ]q m 2 4 2 1 1 {1 − 4 [1 + 4 + ... + 1− ]}q m if n is even 4 = 1 1 1 [1 + + + ]q m .. if n is odd − 2 4 −
n
−
−
(
If n is even,
n 2
n
4
n
n
R (q m) = {1 −
1 1 − ( 14 ) 4 1 − 14
2
If n is odd,
1
1
2
1
−
}q m
n−1
1 1 − ( 14 ) n R (q m ) = 2 1 − 14 1
2
q m →
→
2 q m 3
2 q m 3
(ii) If there are 3 firms, one can still show that any q > q m is strictly dominated by q m for any firm. Therefore, we can restrict to the strategy set [0, q m ] for each firm. But no further strategies are strictly dominated in this set. (Note that 0 is best response to 2q m , thus cannot be strictly dominated). IESDS yields very imprecise predictions about how the game will be played. 3. (Gibbons 1.4). Suppose there are n firms in the Cournot oligopoly model. Let q i be firm i’s quantity, and Q = ni=1 q i be the market quantity. The market clearing price P (Q) = a − Q (assuming Q < a, else P = 0). The total cost of firm i is C i (q i ) = cq i , where c < a. Firms choose quantities simultaneously. What is Nash equilibrium? What happens for the Nash equilibrium as n goes to infinity?
P
Answer: First we find firm i’s best response to q i . Firm i’s programming problem is: −
max q i (a − q i − q
− c)
i
−
qi
The first order condition gives you the reaction function a − c − q
i
−
q i = ∗
2
A NE is a quantity pro file (q 1 , q 2 ,...,q n) such that ∗
∗
∗
a − c − q
∗
q i = ∗
i
−
2
for each i
Using summation we get n
X
n
q i
∗
=
i=1
X X
a − c − q
∗
(
2
i=1 n
⇒
i
−
q i =
n
∗
i=1
Then
n+1
)=
n
2
(a − c) −
n−1
2
n
X
q i
∗
i=1
(a − c)
n (a − c) − q i ] a − c − [ n+1 ∗
q i = ∗
2
From here you can get q i = ∗
a−c n+1
for each i
As n goes to infinity, the market equilibrium quantity ni=1 q i converges to a − c, thus the market price converges to c. As a result, each firm’s profit converges to 0 as n goes to infinity, which is exactly the market outcome of perfect competition.
P
2
∗
4. (Gibbons 1.12) Find the mixed strategy Nash equilibrium of the following game. L R T 2, 1 0, 2 B 1, 2 3, 0
Answer: Let p be the probability that player 1 plays T , and q be the probability that 2 plays L. In a mixed strategy equilibrium, the following two conditions have to be satis fied: (1) Given q , player 1 should be indi ff erent between playing T and B . That is: 2q = q + 3(1 − q ) ⇒ q =
3 4
(2) Given p, player 2 should be indiff erent between playing L and R: p + 2(1 − p) = 2 p ⇒ p =
2 3
Therefore, the mixed strategy equilibrium of this game is: (( 23 , 13 ), ( 34 , 14 )). 5. Mascollel 8.D.4. Answer: (a) There is no strictly dominated strategies for each player. (b) For each player, submitting any mi > 100 is weakly dominated by mi = 100. (c) Pure strategy Nash equilibria: any (m1 , m2 ) such that m1 + m2 = 100 and both m1 ≥ 0 and m2 ≥ 0. 6. Mascollel 8.D.5. Answer: (a) First we show that in any NE, d1 = d2 (di denotes i’s location). Without loss of generality, suppose d1 > d2. Then given d1, player 2 has incentive to reduce d2 to = d2 cannot be a Nash equilibrium. d1 + ε. Hence any (d1 , d2 ) such that d1 6 = 12 cannot be a NE. Suppose d1 = d2 < 12 . Then Second, we prove that any d1 = d2 6 given d1 , player 2 has incentive to deviate to d2 = d1 + ε. By doing that, his market share can increase from 1/2 to more than 1/2. Therefore, d1 = d2 < 12 cannot be a NE. Similarly, you can prove that any d1 = d2 > 12 is not an NE either. Third, we prove that d1 = d2 = 12 is a NE. Given d1 = 12 , player 2 cannot increase his = 12 . Similarly, given d2 = 12 , d1 = 12 is player 1’s best market share by deviating to d2 6 response. Therefore ( 12 , 12 ) is the unique pure strategy Nash equilibrium of this game. (b) Without loss of generality, we assume d1 ≤ d2 ≤ d3 . We prove the result in several steps. (i) There is no NE if d1 < d2 < d3 . This is because in this case, vendor 1 always has incentive to deviate to d2 − ε. 3
(ii) There is no NE if d1 = d2 < d3 . This is due to the fact that vendor 3 has incentive to deviates to d1 − ε. By similar logic, there is no NE if d1 < d2 = d3. (iii) There is no NE if d1 = d2 = d3 = d. If d < 12 , then player 1 wants to deviate to d + ε. If d > 12 , then player 1 wants to deviate to d − ε. If d = 12 . Then player wants to deviate to d − ε. 7. (Gibbons 1.14) Prove the following proposition: the strategies played with positive probability in a mixed strategy Nash equilibrium survive the process of IESDS. Proof: Let (σ 1 , σ2 ,...,σI ) be a mixed strategy NE. And let S i+ ∈ S i be the set of strategy of player i that is played with positive probability in the NE σ . Now suppose the proposition is false. Let si ∈ S i+ be the first strategy be eliminated by IESDS among all the strategies that are played with positive probability in σ . The it must be the case that there is a σi ∈ ∆(S i ) such that ∗
∗
∗
∗
∗
ui (si , σ i ) < ui (σ i , σ i ) −
−
for any σ i that still can be constructed from the remaining strategies. Since si is the first strategy be eliminated, then σ i still can be constructed from the remaining strategies. Hence we have: ui (si , σ i ) < ui (σ i , σ i ) (1) −
∗
−
But given that si
+
∈ S i
∗
∗
−
−
and σ is a NE, we have ∗
0
ui (si , σ i ) ≥ ui (si , σ i ) ∗
−
for any si 0
∈ S i .
∗
−
This implies that 0
ui (si , σ i ) ≥ ui (σi , σ i ) ∗
−
for any σi 0
∗
−
∈ ∆(S i ). This directly contradicts the inequality (1).
4
