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

2F + 1 2B.

Similarly, suppose player 2 (woman) chooses F with the probability 1

4 and B with the probability 3 4.

We write it as 1

4F + 3 4B.

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

2F + 1 2B,

player 2 (woman) chooses 1

4F + 3 4B.

F B F

1 8 3 8

B

1 8 3 8

F B F 2, 1 0, 0 B 0, 0 1, 2

The payoffs:

p1(m) = 1 82 + 3 80 + 1 80 + 3 81 = 5 8. p2(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 := (S1, . . ., Sn, p1, . . ., pn). Mixed strategy of player i in G: mi ∈ ∆Si. Joint mixed strategy: m = (m1, . . ., mn).

Mixed Strategies – p. 5/13

Mixed Extension of a Finite Game (2)

Mixed extension of G:

(∆S1, . . ., ∆Sn, p1, . . ., pn),

where

m(s) := m1(s1) · . . . · mn(sn)

and

pi(m) :=

• s∈S

m(s) · pi(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 (S1, . . . , Sn, p1, . . . , pn). The following are equivalent:

m is a Nash equilibrium in mixed strategies,

i.e.,

pi(m) ≥ pi(m′

i, m−i)

for all i ∈ {1, . . . , n} and all m′

i ∈ ∆Si,

for all i ∈ {1, . . . , n} and all si ∈ Si

pi(m) ≥ pi(si, 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 Rn and

Φ : A → P(A)

is such that

Φ(x) is non-empty and convex for all x ∈ A,

for all sequences (xi, yi) converging to (x, y)

yi ∈ Φ(xi) 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 (S1, . . ., Sn, p1, . . ., pn). Define

besti : Πj=i∆Sj → P(∆Si)

by

besti(m−i) := {m′

i ∈ ∆Si | pi(m′ i, m−i) attains the maximum}.

Then define

best : ∆S1 × · · · × ∆Sn → P(∆S1 × · · · × ∆Sn)

by

best(m) := best1(m−1) × · · · × bestn(m−n).

Note m is a Nash equilibrium iff m ∈ best(m).

Mixed Strategies – p. 10/13

Proof of Nash Theorem, ctd

besti(m−i) := {m′

i ∈ ∆Si | pi(m′ i, m−i) attains the maximum}.

and

best(m) := best1(m−1) × · · · × bestn(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 Rn 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 := (S1, S2, p1, p2). Then for i = 1, 2

max

mi∈Mi

min

m−i∈M−i pi(mi, m−i) =

min

m−i∈M−i max mi∈Mi pi(mi, 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