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Unformatted text preview: 6.254 Game Theory with Engr App Midterm Solutions Thursday, April 8, 2010 Problem 1 (35 points) For each one of the statements below, state whether it is true or false. If the answer is true, explain why. If the answer is false, give a counterexample. Explanations and counterexamples are required for full credit. (7 points each) (1) In the following game, a strategy is strictly dominated if and only if it is a neverbest response. A B C 4, 2 0, 3 D 3, 1 1, 0 E 0, 0 2, 2 True. In a two player finite game, a strategy is strictly dominated if and only if it is a neverbest response. We can easily verify that in the given game there is no strictly dominated strategy, and also no strategy is a neverbest response. Note that D is a best response when player 2 mixes with probability ( 1 3 , 2 3 ) over strategies A and B . (2) In the following game, the set of correlated equilibria is the same as the set of Nash equilibria. H T H 1, 1 1, 1 T 1, 1 1, 1 True. In any zero sum game the set of correlated equilibria is the same as the set of Nash equilibria. In the matching pennies game, the unique NE is when both players mix with probability ( 1 2 , 1 2 ) over strategies H and T . To find the set of correlated equilibria, we set up the system of linear constraints as follows, 1 P ( H , H ) 1 P ( H , T ) ≥  1 P ( H , H ) + 1 P ( H , T ) 1 P ( T , H ) + 1 P ( T , T ) ≥ 1 P ( T , H ) 1 P ( T , T ) 1 P ( H , H ) + 1 P ( T , H ) ≥ 1 P ( H , H ) 1 P ( T , H ) 1 P ( H , T ) 1 P ( T , T ) ≥  1 P ( H , T ) + 1 P ( T , T ) where the first 2 inequalities are needed to guarantee that player 1 does not have incentive to deviate given the public signal, and similarly the latter 2 are needed for player 2 in a correlated equilib rium. Note also that the probabilities are nonnegative, and satisfies P ( H , H ) + P ( H , T ) + P ( T , H ) + P ( T , T ) = 1. Solving the system above, yields, P ( H , H ) = P ( H , T ) = P ( T , H ) = P ( T , T ) = 1 4 , which is the same equilibria as the unique NE. (3) Suppose that function f satisfies strictly increasing differences. That is, suppose that for all x > x and θ > θ , we have f ( x , θ ) f ( x , θ ) > f ( x , θ ) f ( x , θ ) . Let X * ( θ ) = argmax x ∈ R { f ( x , θ ) + g ( x ) } be nonempty for each θ . For θ > θ if z ∈ X * ( θ ) and y ∈ X * ( θ ) then y ≥ z . True. We will prove by contradiction: suppose y < z . Since y ∈ X * ( θ ) and z ∈ X * ( θ ) , we have f ( y , θ ) + g ( y ) ≥ f ( z , θ ) + g ( z ) f ( z , θ ) + g ( z ) ≥ f ( y , θ ) + g ( y ) , which implies f ( z , θ ) f ( y , θ ) ≤ g ( y ) g ( z ) ≤ f ( z , θ ) f ( y , θ ) , which contradicts strictly increasing difference property....
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This note was uploaded on 05/08/2010 for the course CS 6.254 taught by Professor Asuozdaglar during the Spring '10 term at MIT.
 Spring '10
 AsuOzdaglar

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