Solutions Manual to Accompany Game Theory: An Introduction, 2nd Edition

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 X₁ = (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 − 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 p_x = 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 ...