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# 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 xx*, there is some px ∈ (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 < px·

Find the value of px in each case an ESS exists.

7.1 Answer: Let’s look at the equilibrium X1 = (1, 0). We need to show that for x ≠ 1, u(1, px + (1 − p)) > u(x, px + (1 − p)) for some px, and for all 0 < p < px. Now u(1, px + (1 − p)) = 1 − p + px, and u(x, px + (1 − p)) = p + x − 3px + 2px2. In order for X1 to be an ESS, we need 1 > 2p(x − 1)2, which implies 0 < p < 1/(2(x − 1)2). So, for 0 ≤ x < 1, we can take px = 1/(2(x − 1)2) and the ESS requirement will be satisfied. Similarly, the equilibrium X2 = (0, 1), can be shown to be an ESS. For X3 = , we have

In order for X3 to be an ESS, we need

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

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