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Unformatted text preview: Homework #2 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&s information set is redundant if we assume that the student is rational. 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&s info sets)}. 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: 1 n 2 TTTT TTLT TLTT TLLT LTTT LLTT LTLT LLLT M1, 11, 11, 11, 1 1,1 1,1 1,1 1,1 W 0,0 0,0 2,2 2,22,22,2 0,0 0,0 F 1, 1 1,11, 11, 11, 1 1,13,33,3 This can be further reduced to: 1 n 2 TT*T TL*T L*TT L*LT M1,11,1 1,1 1,1 W 0,0 2,22,2 0,0 F 1,11,11,13,3 Here, there are no pure strategy Nash equilibria. Let&s look for mixed strat egy Nash eqa. Assume that the professor mixes with probabilities p 1 ;p 2 ; 1 & p 1 & p 2 , while the student mixes with probabilities q 1 ; q 2 ;q 3 ; 1 & q 1 & q 2 & q 3 . The professor will choose his probabilities so as to make the student indi/er ent between choosing between his strategies. This means that the p&s that the professor chooses must be such that the expected payo/s for the student from each of his possible strategies must be the same. EU S ( TT ¡ T ) = 1 p 1 + 0 p 2 & 1(1 & p 1 & p 2 ) = 2 p 1 + p 2 & 1 EU S ( TL ¡ T ) = 1 p l & 2 p 2 + 1(1 & p 1 & p 2 ) = & 3 p 2 + 1 EU S ( L ¡ TT ) = & 1 p 1 + 2 p 2 + 1(1 & p 1 & p 2 ) = & 2 p 1 + p 2 + 1 EU S ( L ¡ LT ) = & 1 p 1 + 0 p 2 + 3(1 & p 1 & p 2 ) = & 4 p 1 & 3 p 2 + 3 Setting the ¡rst two equal together, EU S ( TT ¡ T ) = EU S ( TL ¡ T ) 2 p 1 + p 2 & 1 = & 3 p 2 + 1 4 p 2 = 2 & 2 p 1 p 2 = 1 2 & 1 2 p 1 1 Setting the second and third equal to one another, & 3 p 2 + 1 = & 2 p 1 + p 2 + 1 2 p 1 = 4 p 2 p 1 = 2 p 2 Then, plugging back into p 2 = 1 2 & 1 2 p 1 , we have that p 2 = 1 2 & 1 2 (2 p 2 ) 2 p 2 = 1 2 p 2 = 1 4 Then, from above we had that p 2 = 1 2 & 1 2 p 1 1 4 = 1 2 & 1 2 p 1 1 2 p 1 = 1 4 p 1 = 1 2 We can check that all three expected payo/s are the same: EU S = 1 4 So the professor will mix with strategy ( 1 2 ; 1 4 ; 1 4 ) over strategies Monday, Wednesday, Friday....
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 Spring '08
 staff
 Econometrics, Game Theory, Nash equilibria, best response, Nash equillibria

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