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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-35/6/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 35 / 6 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

For

V = D > 0

, choose pure Hawk. Every mutant

p < 1

ties it against a resident Hawk population, so the second condition must be checked. The Hawk-Dove game formula gives

e (1, p) - e (p, p) = \frac{( 1 - p ) ( V - D p )}{2} = \frac{V}{2} (1 - p)^{2} > 0.

Consequently

x^{*} = (1, 0) is an ESS when V = D > 0.

The actual invasion advantage is

ϵ V (1 - p)^{2} /2 > 0

, despite the tie at invasion frequency zero.

If the equal parameters are negative, pure Dove is instead an evolutionarily stable strategy: for a mutant

p > 0

,

e (0, 0) - e (p, 0) = - p V /2 > 0

. If

V = D = 0

, every payoff is zero and no resident has a strict advantage over any mutant. Thus the full equal-parameter answer is Hawk for $V = D > 0$ , Dove for $V = D < 0$ , and no ESS for $V = D = 0$ .

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