国外博弈论课件lecture14.ppt

June 6, 2003 73-347 Game Theory--Lecture 14 June 6, 2003 Lecture 14 Exercise 112.1,138.1 of Osborne Solution Mixed strategy equilibrium: 2-player each with two strategies Theorem 2 Let ((r*, 1-r*), (q*, 1-q*)) be a pair of mixed strategies, where 0 r*1, 0q*1. Then ((r*, 1-r*), (q*, 1-q*)) is a mixed strategy Nash equilibrium if and only if  EU1(s11, (q*, 1-q*)) = EU1(s12, (q*, 1-q*)) EU2(s21, (r*, 1-r*)) = EU2(s22, (r*, 1-r*)) Remark: when you use Theorem 2 to solve a game, if the solution does not satisfy 0 r*1, 0q*1 then you can conclude that there is no Nash equilibrium in which each player assigns positive probability to each of her two pure strategies Exercise 112.1 of Osborne (right game) Player 1’s expected payoff of playing T EU1(T, (q, 1–q)) = q×0 + (1–q)×0 = 0 Player 1’s expected payoff of playing B EU1(B, (q, 1–q)) = q×2 + (1–q)×0 = 2q Player 1 is indifferent between playing T and B EU1(T, (q, 1–q)) = EU1(B, (q, 1–q)) 0 = 2q So q = 0 Hence, we can conclude that there is no Nash equilibrium in which each player assigns positive probability to each of her two pure strategies Exercise 112.1 of Osborne (right game) We can solve this game by appealing to Theorem 4. Since we have found two pure strategy Nash equilibria and checked one case in the previous slide, so we need to consider four more cases Exercise 112.1 of Osborne (right game) Case 1: check whether there is a mixed strategy in which r=1, 0q1 By theorem 4, we should have  EU2(L, (r, 1–r)) = r ×1+(1–r)×2 = 1 =EU2(R, (r, 1–r)) = r ×2+(1–r)×1 = 2 This is a contradiction. Hence, there is no Nash equilibrium in this case. Exercise 112.1 of Osborne (right game) Case 2: check whether there is a mixed strategy in which r=0, 0q1 By theorem 4, we should have  EU2(L, (r, 1–r)) = r ×1+(1–r)×2 = 2 =EU2(R, (r, 1–r)) = r ×2+(1–r)×1 = 1 This is a contradiction. Hence, there is no Nash equilibrium in this case. Exercise 112.1 of Osborne (right game) Case 3: check whether there is a mix