Mixed Strategies Krzysztof R. Apt CWI, Amsterdam, the Netherlands , - PowerPoint PPT Presentation

Mixed Strategies Krzysztof R. Apt CWI, Amsterdam, the Netherlands , University of Amsterdam Mixed Strategies – p. 1/13

Overview Mixed strategies. Mixed extension of a finite game. Nash Theorem. Minimax Theorem. Mixed Strategies – p. 2/13

Example: Battle of Sexes F B F 2 , 1 0 , 0 B 0 , 0 1 , 2 Suppose player 1 (man) chooses F with the probability 1 2 and B with the probability 1 2 . We write it as 1 2 F + 1 2 B . Similarly, suppose player 2 (woman) chooses F with the probability 1 4 and B with the probability 3 4 . We write it as 1 4 F + 3 4 B . These are examples of mixed strategies. We call the previous strategies pure strategies. Mixed Strategies – p. 3/13

Example: Battle of Sexes (2) Suppose: player 1 (man) chooses 1 2 F + 1 2 B , player 2 (woman) chooses 1 4 F + 3 4 B . F B F B 1 3 F F 2 , 1 0 , 0 8 8 1 3 B B 0 , 0 1 , 2 8 8 The payoffs: p 1 ( m ) = 1 82 + 3 80 + 1 80 + 3 81 = 5 8 . p 2 ( m ) = 1 81 + 3 80 + 1 80 + 3 82 = 7 8 . Mixed Strategies – p. 4/13

Mixed Extension of a Finite Game Probability distribution over a finite non-empty set A : π : A → [0 , 1] such that � a ∈ A π ( a ) = 1 . Notation: ∆ A . Fix a finite strategic game G := ( S 1 , . . ., S n , p 1 , . . ., p n ) . Mixed strategy of player i in G : m i ∈ ∆ S i . Joint mixed strategy: m = ( m 1 , . . ., m n ) . Mixed Strategies – p. 5/13

Mixed Extension of a Finite Game (2) Mixed extension of G : (∆ S 1 , . . ., ∆ S n , p 1 , . . ., p n ) , where m ( s ) := m 1 ( s 1 ) · . . . · m n ( s n ) and � p i ( m ) := m ( s ) · p i ( s ) . s ∈ S Theorem (Nash ’50) Every mixed extension of a finite strategic game has a Nash equilibrium. Mixed Strategies – p. 6/13

2 Examples Matching Pennies H T H 1 , − 1 − 1 , 1 T − 1 , 1 1 , − 1 ( 1 2 · H + 1 2 · T, 1 2 · H + 1 2 · T ) is a Nash equilibrium. The payoff to each player in the Nash equilibrium: 0. The Battle of the Sexes F B F 2 , 1 0 , 0 B 0 , 0 1 , 2 (2 / 3 · F + 1 / 3 · B, 1 / 3 · F + 2 / 3 · B ) is a Nash equilibrium. The payoff to each player in the Nash equilibrium: 2 / 3 . Mixed Strategies – p. 7/13

Lemma Consider a finite ( S 1 , . . . , S n , p 1 , . . . , p n ) . The following are equivalent: m is a Nash equilibrium in mixed strategies, i.e., p i ( m ) ≥ p i ( m ′ i , m − i ) for all i ∈ { 1 , . . . , n } and all m ′ i ∈ ∆ S i , for all i ∈ { 1 , . . . , n } and all s i ∈ S i p i ( m ) ≥ p i ( s i , m − i ) . This equivalence implies that each Nash equilibrium of the initial game is a pure Nash equilibrium of the mixed extension. Mixed Strategies – p. 8/13

Kakutani’s Fixed Point Theorem Theorem (Kakutani ’41) Suppose A is a compact and convex subset of R n and Φ : A → P ( A ) is such that Φ( x ) is non-empty and convex for all x ∈ A , for all sequences ( x i , y i ) converging to ( x, y ) y i ∈ Φ( x i ) for all i ≥ 0 , implies that y ∈ Φ( x ) . Then x ∗ ∈ A exists such that x ∗ ∈ Φ( x ∗ ) . Mixed Strategies – p. 9/13

Proof of Nash Theorem Fix ( S 1 , . . ., S n , p 1 , . . ., p n ) . Define best i : Π j � = i ∆ S j → P (∆ S i ) by best i ( m − i ) := { m ′ i ∈ ∆ S i | p i ( m ′ i , m − i ) attains the maximum } . Then define best : ∆ S 1 × · · · × ∆ S n → P (∆ S 1 × · · · × ∆ S n ) by best ( m ) := best 1 ( m − 1 ) × · · · × best n ( m − n ) . Note m is a Nash equilibrium iff m ∈ best ( m ) . Mixed Strategies – p. 10/13

Proof of Nash Theorem, ctd best i ( m − i ) := { m ′ i ∈ ∆ S i | p i ( m ′ i , m − i ) attains the maximum } . and best ( m ) := best 1 ( m − 1 ) × · · · × best n ( m − n ) . best ( · ) satisfies the conditions of Kakutani’s Theorem. best ( m ) is non-empty. Extreme Value Theorem Suppose that A is a non-empty compact subset of R n and f : A → R is a continuous function. Then f attains a minimum and a maximum. Mixed Strategies – p. 11/13

Minimax Theorem Note The mixed extension of a zero-sum game is a zero-sum game. Minimax Theorem (Von Neumann, 1928) Consider a finite zero-sum game G := ( S 1 , S 2 , p 1 , p 2 ) . Then for i = 1 , 2 max m − i ∈ M − i p i ( m i , m − i ) = min m − i ∈ M − i max min m i ∈ M i p i ( m i , m − i ) . m i ∈ M i Proof By Zero-sum Theorem and Note. Mixed Strategies – p. 12/13

Historical Remarks First special case of Nash theorem: Cournot (1838). Nash theorem (1950) generalizes von Neumann’s Minimax Theorem (1928). An alternative proof (also by Nash) uses Brouwer’s Fixed Point Theorem. Search for conditions ensuring existence of Nash equilibrium. Mixed Strategies – p. 13/13