HW7Sol.pdf - Stat 155 Spring 2020 Homework 7 Solution Solution to Problem 1 1 The payoff matrix for this game is Male coy f aithf ul(2 2 f ast(5 5

# HW7Sol.pdf - Stat 155 Spring 2020 Homework 7 Solution...

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Stat 155 Spring 2020 Homework 7: Solution Solution to Problem 1 1. The payoff matrix for this game is Male faithful philandering ! coy (2 , 2) (0 , 0) Female fast (5 , 5) ( - 5 , 15) . When the female is coy and the male is philandering neither courtship nor mating happen, which is why the payoff is 0 for both. When the male is faithful and the female is fast, they mate without courtship and the male sticks around to care for the chicks, so the payoff is 15 - 10 = 5 for both. When the Male is faithful and the female is coy, they go through a period of courtship, then mate and then raise the babies together, so the payoff is - 3 + 15 - 10 = 2 for both. When the female is fast and the male is philandering, they mate, but then the female raises the chicks alone, so the female has a payoff of 15 - 20 = - 5 and the male has a payoff of 15. 2. To find a strategy for the males that equalizes the payoffs for both possible female actions, we equalize the payoff vector for the female Ay = 2 0 5 - 5 q 1 - q = 2 q 5 q - 5(1 - q ) = 2 q 10 q - 5 . And 2 q = 10 q - 5 q = 5 8 . So the strategy for the male that equalizes the payoffs for the female, is y * = ( 5 8 , 3 8 ) T . 3. Similarly to find the strategy x * for the female, that equalizes the payoff for the two actions of the male, we equalize the entries of the following vector: x T B = p 1 - p 2 0 5 15 = 2 p + 5(1 - p ) 15(1 - p ) = 5 - 3 p 15 - 15 p . 5 - 3 p = 15 - 15 p gives p = 5 6 . So x * = ( 5 6 , 1 6 ) T . 4. If both females and males follow these strategies, the average payoffs for the females is x * T Ay * = ( 5 6 1 6 ) 2 0 5 - 5 5 8 3 8 = 5 4 . The average payoff for the male is x * T By * = ( 5 6 1 6 ) 2 0 5 15 5 8 3 8 = 5 2 . This outcome is not Pareto-optimal since the outcome is (5 , 5) if the female uses the action fast and the male uses the action faithful, which is better for both players. 1
Solution to Problem 2 We first look for the NE supported on { A, B } . Any such strategy will be of the form ( p 1 - p 0 ) for 0 p 1. Since there are no pure NE, as can be checked using the arrow method, we actually know that 0 < p < 1. Because of this, and because we are looking for symmetric NE, such a strategy must equalize the payoffs for the other player, when that other player uses strategy A or B , and give rise to a lower payoff when the other player uses C . ( p 1 - p 0 ) B = ( p 1 - p 0 ) 0 2 - 1 6 0 9 - 1 3 0 = ( 6(1 - p ) 2 p ( - 1) p + 9(1 - p ) ) = ( 6 - 6 p 2 p 9 - 10 p ) So we need to solve the system 6 - 6 p = 2 p 9 - 10 p 2 p 0 < p < 1 . The equation 6 - 6 p = 2 p leads to p = 3 4 and that also satisfies the inequalities since 9 - 10 3 4 = 3 2 = 2 3 4 . Since the matrix is symmetric the same will hold true for B ( p 1 - p 0 ) . So x = ( 3 4 1 4 0 ) T is the only symmetric NE supported on { A, B } . Similarly any strategy supported on { B, C } will be of the form ( 0 p 1 - p ) with 0 < p < 1 and ( 0 p 1 - p ) B = ( 0 p 1 - p ) 0 2 - 1 6 0 9 - 1 3 0 = ( 6 p - 1(1 - p ) 3(1 - p ) 9 p ) = ( 7 p - 1 3 - 3 p 9 p ) . Here we need the payoffs for using B or C to be equal. Solving 3 - 3 p = 9 p gives p = 1 4 , and 7 1 4 - 1 = 3 4 < 9 1 4 , so the payoff is less when the second player uses A . Thus y = ( 0 1 4 3 4 ) T is a NE supported on { B, C } .