#2 Homework Solutions 1. The set of strategies for the professor is {Monday, Wednesday, Friday} We can see from the past HW1 solutions that there are 7 information sets for the student: one on Monday, two on Wednesday, and four on Friday. But Friday information set is redundant if we assume that the student is rational. s The student will always pick "Today" on Friday if he should reach one of those Friday information sets. Therefore, we can write out a generalized strategy for the student as follows: {decision on Monday, decision on Wednesday if at the upper info set, decision on Wednesday if at the lower info set, Today (for Friday info sets)}. s So the set of strategies for the student can be reduced to: {TTTT, TTLT, TLTT, LTTT, TLLT, LTLT, LLTT, LLLT} The normal form game can then be written as: 1n2 TTTT TTLT TLTT TLLT LTTT LLTT LTLT LLLT M -1, 1 -1, 1 -1, 1 -1, 1 1,-1 1,-1 1,-1 1,-1 W 0,0 0,0 2,-2 2,-2 -2,2 -2,2 0,0 0,0 F 1, -1 1,-1 -1, 1 -1, 1 -1, 1 1,-1 -3,3 -3,3 This can be further reduced to: 1n2 TT*T TL*T L*TT L*LT M -1,1 -1,1 1,-1 1,-1 W 0,0 2,-2 -2,2 0,0 F 1,-1 -1,1 -1,1 -3,3 Here, there are no pure strategy Nash equilibria. Let look for mixed strats egy Nash eqa. Assume that the professor mixes with probabilities p1 ; p2 ; 1 p1 p2 , while the student mixes with probabilities q1; q2 ; q3; 1 q1 q2 q3 . The professor will choose his probabilities so as to make the student indierent between choosing between his strategies. This means that the p that the s professor chooses must be such that the expected payos for the student from each of his possible strategies must be the same. EUS (T T T ) = 1p1 + 0p2 1(1 p1 p2 ) = 2p1 + p2 1 EUS (T L T ) = 1pl 2p2 + 1(1 p1 p2 ) = 3p2 + 1 EUS (L T T ) = 1p1 + 2p2 + 1(1 p1 p2 ) = 2p1 + p2 + 1 EUS (L LT ) = 1p1 + 0p2 + 3(1 p1 p2 ) = 4p1 3p2 + 3 Setting the rst two equal together, EUS (T T T ) = EUS (T L T ) 2p1 + p2 1 = 3p2 + 1 4p2 = 2 2p1 11 p1 p2 = 22 1
Setting the second and third equal to one another, 3p2 + 1 = 2p1 + p2 + 1 2p1 = 4p2 p1 = 2p2 Then, plugging back into p2 =
1 2 1 2 p1 ,
we have that 1 (2p2 ) 2
p2 2p2 p2
= = =
1 2 1 2 1 4
Then, from above we had that 1 2 1 2 1 4 1 2 1 p1 2 1 p1 2
p2 1 4 1 p1 2 p1
= = = =
We can check that all three expected payos are the same: EUS = 1 4
So the professor will mix with strategy ( 1 ; 1 ; 1 ) over strategies Monday, 244 Wednesday, Friday. Likewise, the student will choose his probabilities over strategies such that the professor will be indierent between each of his strategies: M, W, F. EUP (M ) = 1q1 1q2 + 1q3 + 1(1 q1 q2 q3 ) = 2q1 2q2 + 1 EUP (W ) = 0q1 + 2q2 2q3 + 0(1 q1 q2 q3 ) = 2q2 2q3 EUP (F ) = 1q1 1q2 1q3 3(1 q1 q2 q3 ) = 4q1 + 2q2 + 2q3 3 If you try to nd a mixed strategy of q1 , q2 , q3 ; q4 ;you will miserably waste time (as your TA did!). (Also, you can check to see that there isn a Nash equit libria where the student mixes over only two strategies.) The Gambit software informs us that there are a couple of mixed strategy Nash equilibria where the student mixes over only three strategies. Let look for those. s
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First, consider the case where the student mixes over the rst three of his strategies: TT*T TL*T L*TT Then, EUp (M ) = 1q1 1q2 + 1(1 q1 q2 ) = 1 2q1 2q2 EUP (W ) = 0q1 + 2q2 2(1 q1 q2 ) = 2q1 + 4q2 2 EUP (F ) = 1q1 1q2 1(1 q1 q2 ) = 2q1 1 Setting the rst two equal to one another, we get: 1 2q1 2q2 4q1 q1 = = = 2q1 + 4q2 3 6q2 33 q2 42 2
Next, set the second and third equation equal to one another to obtain: 2q1 + 4q2 2 = 2q1 4q2 = 1 1 q2 = 4 1
Substituting this into the equation above for q1 yields: 3 4 3 4 3 8 3 q2 2 3 8
q1 q1
= = =
3 8 Therefore, the student will mix over his rst three strategies with probabil33 ities ( 8 ; 4 ; 3 ; 0). 8 We can also check another mixed strategy, where the student mixes over TT*T L*TT L*LT The professor expected payos from each of his strategies is then given by: s EUP (M ) = 1q1 + 1q2 + 1(1 q1 q2 ) = 1 2q1 EUP (W ) = 0q1 2q2 + 0(1 q1 q2 ) = 2q2 EUP (F ) = 1q1 1q2 3(1 q1 q2 ) = 4q1 + 2q2 3 1 q1 q2 = Setting the rst two equal to one another yields:
and
3
1
2q1 q1
=
2q2 1 2
= q2 +
Setting the last two equations equal to one another gives: 2q2 q1 Then we have that 1 2 3 4 1 4 1 8 = = 4q1 + 2q2 3 q2 4 3
q2 +
= = =
q2
2q2 q2 Then, q1 = = = and
3 4 3 4 5 8
q2 1 8
1 4 Therefore, we have two mixed strategy Nash equillibria: 1 1 {( 2 ; 1 ; 1 ); ( 3 ; 4 ; 3 ; 0g 44 8 8 111 5 {( 2 ; 4 ; 4 ); ( 8 ; 0; 1 ; 1 )g 84 2. Cournot game: Maximize the payos for both A and B: qA qB )qA and B (qA ; qB ) = (30 qA A (qA ; qB ) = (30 @A B = 30 2qA qB = 0 and @ B = 30 2qB qA = 0 @qA @q The best response of A given B choice of qB is: s 1 q1 q2 = BRA (qB ) qA = 15 1 qB 2
qB )qB
The best response of B given A choice of qA is: s 4
1 qA 2 The Nash equilibria are where these two best response functions intersect: B RB (qA ) qB = 15 1 qB 2 3 qB 2 qB qA
15
= = = =
30 15 10 10
2qB
Therefore, there exists a Nash equilibria at qA = 10; qB = 10: Stackelberg game. Using the hint, we rst nd the best response function of the follower: 1 qA 2 Plugging this into the leader payo function, obtain: we s BRB (qA ) qB = 15
A (qA ; qB )
= =
(30 (30
qA qA
qB )qA 1 15 + qA )qA 2
Taking the derivative of
A
with respect to qA gives: 15 qA qA = = 0 15
Then 1 qA 2
qB
= =
15 7:5
The reason we can use this hint is that the rst mover anticipates the second mover strategy using backward induction. Here, A understands that B choice s s of quantity is not an exogeneous variable in his payo function. Instead, he sees B quantity as a function of his own quantity. s 3.
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1n2 A B C D
A 90,90 120,0 0,120 90,20
B 0,120 90,90 120,0 90,20
C 120,0 0,120 90,90 90,20
D 20,90 20,90 20,90 0,0
a. For each player i , the best responses given player j strategy is s i (Aj ) = B i (Bj ) = C i (Cj ) = A b. There will not be a Nash equilibria in pure strategy form. To check, consider the best response functions for each strategy. There is never a case where i best response to player j strategy is actually the same strategy. Player s s i will always choose something dierent. Therefore, there are no pure strategy nash equilibria. Instead look for a mixed strategy nash equilibrium. Suppose each player mixes over strategies A, B, and C. Then the expected payos to each of Player I strategies is given by: s EU1 (A) = 90q1 + 0q2 + 120(1 q1 q2 ) = 120 30q1 120q2 EU1 (B ) = 120q1 + 90q2 + 0(1 q1 q2 ) = 120q1 + 90q2 EU1 (C ) = 0q1 + 120q2 + 90(1 q1 q2 ) = 90 90q1 + 30q2 Setting the rst two equal to one another, we obtain: 120 30q1 120q2 150q1 q1 = = = 120q1 + 90q2 120 210q2 47 q2 55
Then setting the second and third equal to each other, we get: 120q1 + 90q2 210q1 q1 Then, 47 q2 55 28 49 q2 35 35 13 35 q2 = = = = 32 q2 77 15 10 q2 35 35 39 q2 35 1 3 = 90 = 90 3 = 7 90q1 + 30q2 60q2 2 q2 7
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Then, q1 = = = = and 1 q1 q2 = 1 3 3 21 () 7 73 9 2 21 21 7 21 1 3
1 Therefore, each player mixes equally with probability 3 over each strategy. c) Now consider the full RPSD game. The best response functions for this game are: i (Aj ) = B i (Bj ) = C i (Cj ) = A i (Dj ) = A; B; C d. Suppose that you believe that the overall average "shares" at the bottom of the posted sheet are representative of what you will face next period. Given that belief, the expected payos for each strategy are: EU (A) = 0:25(90) + 0:19(0) + 0:19(120) + 0:36(20) = 52:5 EU (B ) = 0:25(120) + 0:19(90) + 0:19(0) + 0:36(20) = 54:3 EU (C ) = 0:25(0) + 0:19(120) + 0:19(90) + 0:36(20) = 47:1 EU (D) = 0:25(90) + 0:19(90) + 0:19(90) + 0:36(0) = 56:7 e. Then the best response in (d) is D - it gives the highest expected payo. The best response to an opponent playing D is either A, B, or C. f. There are no pure strategy Nash equilibria. There is a mixed strategy Nash. Let assume it is in the form p,p,p, 1-3p. s Then: EUi (A) = p(90) + p(0) + p(120) + (1 3p)(20) = 20 150p EUi (B ) = p(120) + p(90) + p(0) + (1 3p)(20) = 20 150p
EUi (C ) = p(0) + p(120) + p(90) + (1 EUi (D) = p(90) + p(90) + p(90) + (1
3p)(20) = 20 150p 3p)(0) = 270p
20 + 150p = 270p 20 = 120p 1 p= 6
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Given the symmetry of the game, each player mixes with probabilities ( 1 ; 1 ; 1 ; 1 ): 6662 1 Therefore, we have a mixed strategy Nash equilibrium of {( 1 ; 1 ; 1 ; 2 );( 1 ; 1 ; 1 ; 1 )g 666 6662 g. The RPSD game does not have any other Nash equilibria. If we mix over all 4 strategies with probabilities a,b,c,(1-a-b-c), we nd that EU (A) = a(90) + b(0) + c(120) + (1 a b c)(20) EU (B ) = a(120) + b(90) + c(0) + (1 a b c)(20) EU (C ) = a(0) + b(120) + c(90) + (1 a b c)(20) EU (D) = a(90) + b(90) + c(90) + (1 a b c)(0) If we set the rst two equation equal to one another: a(90) + b(0) + c(120) + (1 a b c)(20) = a(120) + b(90) + c(0) + (1 90a + 120c = 90b + 120a a b c)(20)
and if we set the rst equation equal to the third equation, we have: a(90) + b(0) + c(120) + (1 a b c)(20) = a(0) + b(120) + c(90) + (1 90a + 120c = 120b + 90c a b c)(20)
This system is equivalent to: 3a = 12c 3c = 12b 9b 9a
Substituting the second into the rst, we get: 3a = 12(4b 39a = 39b ie:a = b 3a) 9b
Then, if a=b, we must have that 3c=3b=3a. That is, a Nash equilibrium must have that a=b=c, just as we had solved earlier. Textbook Problems Chapter 6, Problem 1.
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Chapter 6, Problem 3
Chapter 6, Problem 5
Chapter 7, Problem 3 R={(x,c)}. The order does not matter because if a strategy is dominated (not a best response) relative to some set of strategies of the other player, then this strategy will also be dominated relative to a smaller set of strategies for the other player. Chapter 7, Problem 4
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Chapter 8, Problem 3
Chapter 8, Problem 6
Chapter 9, Problem 2
Problem 9, Chapter 5
Problem 9, Chapter 9
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Problem 7, Chapter 6 Strategy {M} is not dominated by any other pure strategies. Let us try a mixed strategy of putting 1/3 probability on {K}; 2/3 probability on {K} and 0 on {M}. When player 2 plays {X} the expected payoff from this mix is: 1 (1 , ) = 1 (1 , ) = 1 2 11 9+ = 3 3 3
When player 2 plays {Y} the expected payoff from this mix is:
Thus no matter what strategy player 2 plays, the mixture of (1/3,2/3,0) yields higher expected payoff than strategy {M}. Thus {M} is dominated by this mixed strategy.
12 13 + 6 = 33 3