PS5_Sol

# PS5_Sol - 14.12 Problem Set 5 Solution Ruitian Lang...

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14.12: Problem Set 5 Solution Ruitian Lang December 3, 2010 1 (a) There are two players, 1 and 2, with A 1 = { U,D } and A 2 = { L,R } . The type spaces are T 1 = {- 1 , 1 } ,T 2 = {- 1 , 1 } , and the beliefs are p 1 ( t 2 | t 1 ) = p 2 ( t 1 | t 2 ) = 1 / 2 for all t 1 T 1 and t 2 T 2 . The payoﬀ function is as follows: u i ( D,L ; t 1 ,t 2 ) = u i ( U,R ; t 1 ,t 2 ) = 0 for i = 1 , 2 and all t 1 ,t 2 ; u 1 ( U,L ; t 1 ,t 2 ) = 2 t 1 ,u 1 ( D,R ; t 1 ,t 2 ) = t 1 ,u 2 ( U,L ; t 1 ,t 2 ) = t 2 and u 2 ( D,R ; t 1 ,t 2 ) = 2 t 2 . In the notation of the problem, t 1 is θ , and t 2 is λ . (b) Suppose that in a BNE Player 2 always plays L . Then Player 1’s best response is U when θ = 1 and D when θ = - 1, which must coincide with Player 1’s strategy in the BNE. Now if Player 2 sees that λ = 1, her payoﬀ by playing L is θ T 1 u 2 ( s 1 ( θ ) ,L ; θ,λ = 1) p 2 ( θ | λ = 1) = 1 / 2 while her payoﬀ by playing R is θ T 1 u 2 ( s 1 ( θ ) ,R ; θ,λ = 1) p 2 ( θ | λ = 1) = 1. Therefore, her best response is R , a contradiction. Similarly, we can see that Player 2 cannot always play R in a BNE, since then Player 1 plays U when θ = - 1 and D when θ = 1 and Player 2’s best response is to play L when λ = - 1. Therefore, in a pure strategy Nash equilibrium, Player 2 must play diﬀerent strategies for diﬀerent types. Since p 1 ( λ | θ ) = 1 / 2 for all λ and θ , λ T 2 u 1 ( a 1 ,s 2 ( λ ); θ,λ ) p 1 ( λ | θ ) = 1 2 u 1 ( a 1 ,L ; θ,λ ) + 1 2 u 1 ( a 1 ,R ; θ,λ ). In other words, Player 1 always believes that Player 2 plays L with probability 1 / 2. Therefore, Player 1’s best response is to play U when θ = 1 and D when θ = - 1. Given this, Player 2’s best response is to play L when λ = - 1 and R when λ = 1. Therefore, the only possible BNE is that s 1 ( θ = 1) = U,s 1 ( θ = - 1) = D,s 2 ( λ = 1) = R and s 2 ( λ = - 1) = L . We have also seen that this is indeed a BNE.

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PS5_Sol - 14.12 Problem Set 5 Solution Ruitian Lang...

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