## 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 “Hawk-Dove” 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 $$\chi=(h,1-h)$$ 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 mixed-strategy 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 $$\sigma^*=\left(v/c,1-v/c\right)$$. 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 $$\sigma\ne\sigma^*$$:

1. $$u(\sigma^*,\sigma^*)>u(\sigma,\sigma^*)$$ OR
2. $$u(\sigma^*,\sigma^*)=u(\sigma,\sigma^*)$$ 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:

1. Write down the associated two-player game;
2. Identify all symmetric Nash equilibria of the game;
3. Test the Nash equilibrium against the two conditions of the above Theorem.

### Example

Let us consider the Hawk-Dove game. The associated two-player 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.