# Evolutionary Stable Strategies and Population Games

## 7.1 Evolution

#### Problems

**7.1** In the currency game (Example 7.2), derive the same result we obtained but using the equivalent definition of ESS: *X** is an evolutionary stable strategy if for every strategy *X* = (*x*, 1 − *x*), with *x* ≠ *x**, there is some *p*_{x} ∈ (0, 1), which depends on the particular choice *x*, such that

*u*(*x**, *px* + (1 − *p*)*x**) > *u*(*x*, *px* + (1 − *p*)*x**), for all 0 < *p* < *p*_{x}·

Find the value of *p*_{x} in each case an ESS exists.

**7.1 Answer:** Let’s look at the equilibrium *X*_{1} = (1, 0). We need to show that for *x* ≠ 1, *u*(1, *px* + (1 − *p*)) > *u*(*x*, *px* + (1 − *p*)) for some *p*_{x}, and for all 0 < *p* < *p*_{x}. Now *u*(1, *px* + (1 − *p*)) = 1 − *p* + *px*, and *u*(*x*, *px* + (1 − *p*)) = *p* + *x* − 3*px* + 2*px*^{2}. In order for *X*_{1} to be an ESS, we need 1 > 2*p*(*x* − 1)^{2}, which implies 0 < *p* < 1/(2(*x* − 1)^{2}). So, for 0 ≤ *x* < 1, we can take *p*_{x} = 1/(2(*x* − 1)^{2}) and the ESS requirement will be satisfied. Similarly, the equilibrium *X*_{2} = (0, 1), can be shown to be an ESS. For *X*_{3} = , we have

In order for *X*_{3} to be an ESS, we need

which becomes 0 > , for 0 < *p* < *px*. This is clearly impossible, so ...