2. War of Attrition In our class. we typically stick to very simple games. and we believe a great deal of insight can be gained from these games. One natural concern is that our results from analyzing such simple games might not carry over if the games were made more complicated or realistic, and thus our insights would not apply to the real world. Throughout the class. we will address this concern. One simplication we often make is to model players as having only 2 or 3 choices. when in practice, they may have many choices, sometimes innitely many so. We might be concerned that that our results were an artifact of our decision to model players as having only 2 choices (or any nite number). So, it's often good to double check that our results are robust to making the strategy space continuous. In this question we will show that the results in Question 1 are robust to making the strategy space continuous. In the context of the Hawk-Dove game, assuming players choose from only 2 choices boils down to assuming that players can only choose to be aggressive forever or passive forever. We'll set up a new game, in which players can instead choose to ght for some time and then give up. This game is called the War of Attrition. As in question 1(a)iii of PSI, allowing for continuous strategies prevents us from representing a game as a matrix form game. Instead we must represent our payo's u.- as a function of the player's strategies. With this understm1ding, we can move on and work with these continuous values. Twu players are ghting over some resource worth in Now. instead of simply choosing whether to play dove or hawk, the players must choose for how long to ght before giving up. Let tl be the time player 1 devotes to the ght and t2 be the time player 2 devotes to the ght. Fighting is still costly: fighting costs both players c > 0 per unit of time. The payoffs of the 3 game are thus: Ct1 if t[ t2 Ct2 If {2 t1 For this problem we will consider that each player can choose a pure strategy, but not a mixed strategy, i.e. players are forbidden to randomize. (a) Often, the easiest way to solve problems with a continuous strategy space. is to solve for players" best response (BR) function. The BR function tells you, for any L,- that the other player chooses. what choice of t; would maximize player 12's payoffs? I.e. 3&(t_) = t:, where 'u (t:,t_.-) 2 n(t.t_,-) 1 Vti. Let's solve for player 1's BR function. i. Suppose 2 t2}. ii. Next. suppose t2 = i. What is 331(5)? iii. Finally. if t2 > 3. What is BR1(t2)? (b) \\Vhat is player 2's BR function? (c) Now. recall that in a Nash equilibrium, neither player can benet by deviating. That is, both players must be playing according to their best response. which can only happen if 1 E BR1(t2) and t2 6 BR2(t1). Find all Nash equilibria in this game. That is, nd all pains ( '1', t5) such that if is a best response to t. and t; is a best response to t}