Evolutionarily Stable Strategy (ESS)

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Notes for CS 8803 - Game Theory and Computer Science. Spring 2008

Evolutionarily stable strategy, ESS; (or also evolutionary stable strategy) was formalized by Maynard Smith. This is a strategy which, if adopted by a population of players, cannot be invaded by any mutant strategy that is initially rare.

A good motivating example is the Hawk - Dove Game , also sometimes called Chicken.

Hawk Dove
Hawk -10, -10 30, 0
Dove 0, 30 10, 10
Hawk-Dove game

This is a symmetric 2 Player Game.

<math>u_1(a_1,a_2)= u_2(a_2,a_1)), \mbox{ where } {A}_1 = {A}_2</math>, <math> \Delta_1 = \Delta_2</math>.

Hawk-Dove Game

Review: From previous lectures we found that in nature there are 2/3 Hawks and 1/3 Doves.

This can be found by solving this game from Player 2's perspective given that Player 1 is a Hawk with probability p and a Dove with probability(1-p).

The following sets of equations are developed:

Player 2 is a Hawk, expectation is: -10p + 30(1-p)= 30-40p

Player 2 is a Dove, expectation is: 0p + 10(1-p)= 10-10p

Solving simultaneously, Player 2 is a Hawk p = 2/3 and Player 1 is a Dove 1/3 through the symmetric nature of the game.

Theorem 1

Every symmetric finite game has a symmetric Nash Equilibrium. The proof will be a later homework exercise.

Definition. A mixed strategy <math>\sigma</math> is an ESS if for all <math>\bar \sigma \in \Delta, \bar\sigma \not = \sigma </math> either:

(1) <math>u_1(\bar\sigma,\sigma)< u_1(\sigma,\sigma) </math>

(2)<math>u_1(\bar\sigma,\sigma)= u_1(\sigma,\sigma)</math> and <math>u_1(\bar\sigma,\bar\sigma)< u_1(\sigma,\bar\sigma)</math>

This implies that <math>(\sigma,\sigma)</math> is a NE...no other choice is better.

Stability Condition

If you move away from the equilibrium you'll want to move back.

Restated as a Theorem 2:

Theorem 2

Suppose <math>\sigma </math>is an ESS of a symmetric 2 player game.

Then <math>(\sigma,\sigma)</math> is a NE and for all <math>\bar\sigma \not = \sigma </math> such that <math>u_1(\bar\sigma,\sigma)= u_1(\sigma,\sigma)</math> is the Best Response.

Let <math>\beta </math> be the mutant fraction in a population and <math> (1-\beta)</math> be the fraction of healthy population.

for all <math>\beta > 0, u_1(\sigma,\beta \bar\sigma +(1-\beta)\sigma )> u_1(\bar\sigma,\beta\bar\sigma +(1-\beta)\sigma)</math>

= <math>[(\beta) u_1(\sigma, \bar\sigma)+(1-\beta)u_1(\sigma,\sigma)]-[(\beta)u_1(\bar\sigma,\bar\sigma)+(1-\beta)u_1(\bar\sigma, \sigma)] </math>


<math>(\sigma,\sigma)</math> is a NE by definition.

Regrouping above Theorem,

=<math>[\beta(u_1(\sigma,\bar\sigma) - u_1(\bar\sigma,\bar\sigma))] + [(1-\beta)(u_1(\sigma,\sigma)-u_1(\bar\sigma,\sigma))]>0 </math>

So, though the population may drift slightly from NE it will eventually move back.


Although, ESS doesn't always exist.

See the following two games for examples:

Dog vs Dog'

Who gets $10?

Dog Dog'
Dog 0, 0 0, 0
Dog' 0, 0 0, 0
Dog vs Dog'

The production of this material was supported in part by NSF award SES-0734780.