# 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*. Now

_{x}*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*= 1/(2(

_{x}*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 ...

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