Chapter 12  Nash equilibrium and Evolutionary stable strategies
Recap
In the previous chapter:
 We considered population games;
 We proved a result concerning a necessary condition for a population to be evolutionary stable;
 We defined Evolutionary stable strategies and looked at an example in a game against the field.
In this chapter we’ll take a look at pairwise contest games and look at the connection between Nash equilibrium and ESS.
Pairwise contest games
In a population game when considering a pairwise contest game we assume that individuals are randomly matched. The utilities then depend just on what the individuals do:
As an example we’re going to consider the “HawkDove” game: a model of predator interaction. We have \(S={H,D}\) were:
 \(H\): Hawk represents being “aggressive”;
 \(D\): Dove represents not being “aggressive”.
At various times individuals come in to contact and must choose to act like a Hawk or like Dove over the sharing of some resource of value \(v\). We assume that:
 If a Dove and Hawk meet the Hawk takes the resources;
 If two Doves meet they share the resources;
 If two Hawks meet there is a fight over the resources (with an equal chance of winning) and the winner takes the resources while the loser pays a cost \(c>v\).
If we assume that \(\sigma=(\omega,1\omega)\) and \(\chi=(h,1h)\) the above gives:
It is immediate to note that no pure strategy ESS exists. In a population of Doves (\(h=0\)):
thus the best response is setting \(\omega=1\) i.e. to play Hawk.
In a population of Hawks (\(h=1\)):
thus the best response is setting \(\omega=0\) i.e. to play Dove.
So we will now try and find out if there is a mixedstrategy ESS: \(\sigma^*=(\omega^*,1\omega^*)\). For \(\sigma^*\) to be an ESS it must be a best response to the population it generates \(\chi^*=(\omega^*,1\omega^*)\). In this population the payoff to an arbitrary strategy \(\sigma\) is:
 If \(\omega^*<v/c\) then a best response is \(\omega=1\);
 If \(\omega^*>v/c\) then a best response is \(\omega=0\);
 If \(\omega^*=v/c\) then there is indifference.
So the only candidate for an ESS is \(\sigma^*=\left(v/c,1v/c\right)\). We now need to show that \(u(\sigma^*,\chi_{\epsilon})>u(\sigma,\chi_{\epsilon})\).
We have:
and:
This gives:
which proves that \(\sigma^*\) is an ESS.
We will now take a closer look the connection between ESS and Nash equilibria.
ESS and Nash equilibria
When considering pairwise contest population games there is a natural way to associate a normal form game.
Definition
The associated two player game for a pairwise contest population game is the normal form game with payoffs given by: \(u_1(s,s’)=u(s,s’)=u_2(s’,s)\).
Note that the resulting game is symmetric (other contexts would give non symmetric games but we won’t consider them here).
Using this we have the powerful result:
Theorem relating an evolutionary stable strategy to the Nash equilibrium of the associated game
If \(\sigma^*\) is an ESS in a pairwise contest population game then for all \(\sigma\ne\sigma^*\):
 \(u(\sigma^*,\sigma^*)>u(\sigma,\sigma^*)\) OR
 \(u(\sigma^*,\sigma^*)=u(\sigma,\sigma^*)\) and \(u(\sigma^*,\sigma)>u(\sigma,\sigma)\)
Conversely, if either (1) or (2) holds for all \(\sigma\ne\sigma^*\) in a two player normal form game then \(\sigma^*\) is an ESS.
Proof
If \(\sigma^*\) is an ESS, then by definition:
which corresponds to:
 If condition 1 of the theorem holds then the above inequality can be satisfied for \(\epsilon\) sufficiently small. If condition 2 holds then the inequality is satisfied.

Conversely:

If \(u(\sigma^*,\sigma^*)<u(\sigma,\sigma^*)\) then we can find \(\epsilon\) sufficiently small such that the inequality is violated. Thus the inequality implies \(u(\sigma^*,\sigma^*)\geq u(\sigma,\sigma^*)\).

If \(u(\sigma^*,\sigma^*)= u(\sigma,\sigma^*)\) then \(u(\sigma^*,\sigma)> u(\sigma,\sigma)\) as required.

This result gives us an efficient way of computing ESS. The first condition is in fact almost a condition for Nash Equilibrium (with a strict inequality), the second is thus a stronger condition that removes certain Nash equilibria from consideration. This becomes particularly relevant when considering Nash equilibrium in mixed strategies.
To find ESS in a pairwise context population game we:
 Write down the associated twoplayer game;
 Identify all symmetric Nash equilibria of the game;
 Test the Nash equilibrium against the two conditions of the above Theorem.
Example
Let us consider the HawkDove game. The associated twoplayer game is:
Recalling that we have \(v<c\) so we can use the Equality of payoffs theorem to obtain the Nash equilibrium:
Thus we will test \(\sigma^*=(\frac{v}{c},1\frac{v}{c})\) using the above theorem.
Importantly from the equality of payoffs theorem we immediately see that condition 1 does not hold as \(u(\sigma^*,\sigma^*)=u(\sigma^*,H)=u(\sigma^*,D)\). Thus we need to prove that:
We have:
After some algebra:
Giving the required result.