Here u j x 1 x 2 x k s 1 s 1 s k s k x 1 s 1 x k s k

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Here, ¯ u j ( x 1 , x 2 , . . . , x k ) := s 1 S 1 ,...,s k S k x 1 ( s 1 ) . . . x k ( s k ) u j ( s 1 , . . . , s k ) , where x i ( s ) is the probability with which player i plays pure strategy s in the mixed strategy x i . Definition 3.3.4. A game is symmetric if the players strategies and pay- o ff s are identical, up to relabelling, i.e., for every i 0 , j 0 { 1 , . . . , k } , there is a permutation π of the set { 1 , . . . , k } such that π ( i 0 ) = j 0 and u π ( i ) ( π (1) , . . . , π ( k ) ) = u i ( 1 , . . . , k ) . (For this definition to make sense, we require that the strategy sets of the players coincide.) We will prove the following result in § Section 3.8. Theorem 3.3.5 ( Nash’s theorem ) . Every finite general sum game has a Nash equilibrium. Moreover, in a symmetric game, there is a symmetric Nash equilibrium. For determining Nash equilibria in (small) games, the following lemma (which we have already applied several times for 2-player games) is useful. Lemma 3.3.6. Consider a k -player game with x i the mixed strategy of
3.3 General-sum games with more than two players 59 player i . For each i , let T i = { s S i | x i ( s ) > 0 } . Then ( x 1 , . . . , x k ) is a Nash equilibrium if and only if for each i , there is a constant c i such that s i T i u i ( s i , x i ) = c i and s i ̸∈ T i u i ( s i , x i ) c i . Exercise: Prove Lemma 3.3.6. Use Lemma 3.3.6 to derive an exponential time algorithm for finding a Nash equilibrium in two-player general sum games using linear programming (Section 2.7). Returning to the ecology game, it is easy to check that the pure equilibria consist of all three firms polluting, or one of the three firms polluting, and the remaining two purifying. Next we consider mixed strategies. Suppose that player i ’s strategy is x i = ( p i , 1 p i ) (i.e. i purifies with probability p i ). It follows from lemma 3.3.6 these strategies are a Nash equilibrium with 0 < p i < 1 if and only if: u i (purify , x i ) = u i (pollute , x i ) . Thus, if player 1 plays a mixed strategy, then p 2 p 3 + p 2 (1 p 3 )+ p 3 (1 p 2 ) + 4(1 p 2 )(1 p 3 ) = 3 p 2 (1 p 3 ) + 3 p 3 (1 p 2 ) + 3(1 p 2 )(1 p 3 ) , or, equivalently, 1 = 3( p 2 + p 3 2 p 2 p 3 ) . (3.1) Similarly, if player 2 plays a mixed strategy, then 1 = 3( p 1 + p 3 2 p 1 p 3 ) , (3.2) and if player 3 plays a mixed strategy, then 1 = 3( p 1 + p 2 2 p 1 p 2 ) . (3.3) Subtracting (3.2) from (3.3), we get 0 = 3( p 2 p 3 )(1 2 p 1 ). This means that if all three firms use mixed strategies, then either p 2 = p 3 or p 1 = 1 / 2. In the first case ( p 2 = p 3 ), equation (3.1) becomes quadratic in p 2 , with two The notation ( s i , x i ) is an abbreviation where we identify the pure strategy s i with the probability vector 1 s i that assigns s i probability 1.
60 General-sum games solutions p 2 = p 3 = (3 ± 3) / 6, both in (0 , 1). Substituting these solutions into the first equation, yields p 1 = p 2 = p 3 , resulting in two symmetric mixed equilibria. If, instead of p 2 = p 3 , we let p 1 = 1 / 2, then the first equation becomes 1 = 3 / 2, which is nonsense. This means that there is no asymmetric equilibrium with at least two mixed strategies. It is easy to check that there is no equilibrium with two pure and one mixed strategy. Thus we