ps2sol - 14.12 - Fall 05 Solutions for Problem Set #2...

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14.12 - Fall 05 Solutions for Problem Set #2 Question 1 (30) By inspection, it can easily be seen that there is no pure strategy Nash Equilibrium of the game. Whenever a cell is considered, it is always possible to f nd a deviation by at least one of the players. Thus we conclude that there is no pure strategy Nash Equilibrium. In order to f nd all the possible mixed strategy Nash equilibria, we have to clear the game from the strictly dominated strategies since we know from the lectures and recitations that in equilibrium, none of the strictly dominated strategies are given a positive probability. Let’s try to see if there is any strictly dominated strategies: For player 1 B is a best-response (BR) to L . Similarly C is a BR to M and A is a BR to R . Therefore there is no strictly dominated strategy for player 1 at the f rst round of reasoning. For player 2 playing L is a BR to A . Similarly M is a BR to B and L is a BR to C . Therefore the only possible candidate for strictly dominated strategy is R . Let’s see if there is any possible combination of L and M which dominates R . Calling the probability assigned to L as p, the combination of L and M has to satisfy all of the following inequalities. p 1+(1 p ) 0 > 0 p 1+(1 p ) 2 > 1 p 2+(1 p ) 1 > 1 These inequalities can be reduced to: p> 0 , 1 >p, p> 0 respectively. As a result, any combination of L and M with 0 <p< 1 is doing better than playing R . Therefore R is a strictly dominated strategy for player 2 at the f rst round of reasoning (Do not think that they are playing sequentially. They are just doing the reasoning before playing the simultaneous game). Since player 1 knows that player 2 will not play R , he eliminates his own strategy A since in the reduced game (which excludes the strategy R of player 2) A is strictly dominated by both B or C .Noww ea r el e f t with the following game LM B 2,1 1,2 C 1,2 2,1 Let’s say player 1 plays B with probability µ and player 2 plays L with probability λ. Then the mixed equilibrium should satisfy the following equations µ 1+(1 µ ) 2= µ 2+(1 µ ) 1 λ 2+(1 λ ) 1= λ 1+(1 λ ) 2 Therefore the solution is µ = λ =1 /
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This implies that the only Nash equilibrium is NE = 1 2 B + 1 2 C, 1 2 L + 1 2 M ¸ Remember that a Nash equilibrium is the equilibrium strategies - NOT the payo f s. About grading : It was enough to give just one example for p. The details here were only for instructive purposes. Question 2 (30) a d c b R L D C a 1 a 3 a 2 z y x n m h g f 3 1 2 3 1 2 2 5 1 0 0 0 1 1 1 1 6 0 1 3 3 3 0 5 0 3 0 0 2 2 2 0 0 0 0 2 1 2 1 0 1 1 1 SG 1 SG 2 SG 3 SG 4 SG 5 SG 6 SG 7 Starting from the smallest subgame (SG 1), we are solving the game backwards. All the proper subgames are taken into a circle in the above picture. Once we solve a SG, that particular subgame is replaced with the sequentially rational decision’s payo f . Just to illustrate the substitution process, after solving the f rst 2
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ps2sol - 14.12 - Fall 05 Solutions for Problem Set #2...

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