Evolutionarily Stable Strategy (ESS)
From Theory
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.
- u1(a1,a2) = u2(a2,a1)), where A1 = A2, Δ1 = Δ2.
Contents |
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 σ is an ESS if for all
either:
(1)
(2)
and
This implies that (σ,σ) 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 σis an ESS of a symmetric 2 player game.
Then (σ,σ) is a NE and for all
such that
is the Best Response.
Let β be the mutant fraction in a population and (1 − β) be the fraction of healthy population.
for all
=
Proof
(σ,σ) is a NE by definition.
Regrouping above Theorem,
=
So, though the population may drift slightly from NE it will eventually move back.
Existence
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.
