459PS1Sol

459PS1Sol - ECON-459: Applied Game Theory Problem Set 1 -...

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Unformatted text preview: ECON-459: Applied Game Theory Problem Set 1 - Solutions 1. Problems 1.5, 1.7 and 1.8 from Gibbons. Gibbons #1.5 The question asks you to express the Cournot duopoly game as a Prisoners’ Dilemma where the only two available quantities are the monopoly quantity, q m and the Cournot equilibrium quantity q c . To do this, you will need to calculate the payoffs to each player under each combination of these strategies. That is for each player i we know that the profit function is π i ( q i ,q j ) = q i [ a- c- ( q i + q j )] So for player 1, the following values must be calculated. π m = π 1 ( q m 2 , q m 2 ) = ( a- c ) 2 8 π c = π 1 ( q c ,q c ) = ( a- c ) 2 9 π t = π 1 ( q c , q m 2 ) = 5( a- c ) 2 36 π s = π 1 ( q m 2 ,q c ) = 5( a- c ) 2 48 It can be verified that π t > π m > π c > π s which corresponds to the payoffs for the Prisoners’ Dilemma game. These payoffs and strategies can be represented in matrix form as below. q m 2 q c q m 2 π m ,π m π s ,π t q c π t ,π s π c ,π c The second part of the questions asks you to extend the game by adding a third possible quan- tity, q . The game must be such that the Cournot output is still the unique Nash equilibrium, and there are no strictly dominant strategies. There is no single solution to this problem, but one way to proceed is as follows. First set up a hypothetical matrix of this game as below. 1 q m 2 q c q q m 2 π m ,π m π s ,π t A,B q c π t ,π s π c ,π c C,D q B,A D,C E,E where A = π 1 ( q m 2 ,q ) B = π 1 ( q , q m 2 ) C = π 1 ( q c ,q ) D = π 1 ( q ,q c ) E = π 1 ( q ,q ) Next, note that in order for ( q c ,q c ) to be a Nash equilibrium, π c > D . Then, in order for ( q c ,q c ) to be a unique Nash equilibrium q cannot be a best response for either player, when the other is playing q [otherwise ( q ,q ) is also a NE]. So either A > E or C > E or both. For there to be no strictly dominant strategies, q must either be a best response for one of the other strategies, or allow q m 2 to be a best response to q . Let us assume that q m 2 is a best response to q for both players. Then A > C > E . Consequently, for ( q c ,q c ) to be a unique Nash equilibrium it must be the case that B < π t . Using this information, we can now calculate a value for q and then verify that it satisfies all the requirements above. The best response function for firm 1 in the Cournot duopoly game is R 1 ( q 2 ) = 1 2 ( a- c- q 2 ). So for q m 2 to be a best response to q , q must satisfy R 1 ( q ) = q m 2 = 1 2 ( a- c- q ) This yields q = a- c 2 , which implies A = π 1 ( q m 2 ,q ) = ( a- c ) 2 16 B = π 1 ( q , q m 2 ) = ( a- c ) 2 8 C = π 1 ( q c ,q ) = ( a- c ) 2 18 D = π 1 ( q ,q c ) = ( a- c ) 2 12 E = π 1 ( q ,q ) = 0 All of which satisfy the requirements above....
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This note was uploaded on 01/27/2011 for the course ECON 459 taught by Professor Zeca during the Spring '08 term at UConn.

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459PS1Sol - ECON-459: Applied Game Theory Problem Set 1 -...

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