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

The payoff $e(x,z)$ is the expected fitness of an individual using strategy $x$ against a population using strategy $z$. The resident $x^{\ast }$ is initially common; the mutant $y$ is introduced at small frequency $ϵ$. Convexity of the strategy set makes their population mixture admissible. If residents have higher expected fitness in every sufficiently small invasion by a fixed mutant, selection works against that mutant: this is an evolutionarily stable strategy. The allowed upper bound on $ϵ$ may depend on $y$.

For expected payoffs of mixed strategies, $e(x,z)$ is affine in the population strategy $z$; in a finite matrix game it is $x^{T}Az$. This property is the basis of the two-condition criterion for evolutionary stability. For fixed $y\ne x^{\ast }$, putAffineness gives the exact invasion differenceIf this is positive for all sufficiently small $ϵ>0$, taking $ϵ\downarrow 0$ gives $A_y\ge 0$. If $A_y=0$, the difference is $ϵB_y$, so necessarily $B_y>0$. Conversely, if $A_y>0$, it remains positive for sufficiently small $ϵ$: for example take $ϵ<A_y/(A_y+|B_y|)$. If $A_y=0$ and $B_y>0$, it is positive for every $ϵ>0$. Applying this argument separately to every $y$ provesThus an evolutionarily stable strategy is a symmetric Nash equilibrium with an additional condition against tied mutants. For a general nonlinear interaction function on a convex set, affineness would be an additional necessary hypothesis for this particular equivalence; here it follows from the mixed-strategy expected-payoff interpretation. For example, on $X=[0,1]$ take $e(x,z)=xz(x/2-z)$ and resident $x^{\ast }=0$. For every $y>0$, $A_y=0$ and $B_y=y^3/2>0$, but at population $ϵy$ the mutant earns $ϵy^3(1/2-ϵ)>0$ while the resident earns zero. Thus the two conditions alone would not suffice for an arbitrary nonlinear $e$; the expected-payoff assumption is used explicitly.

For the Hawk-Dove game, let $p$ be an individual's Hawk probability and $q$ the population's Hawk probability. The given matrix yieldsWe use this formula in each case. In the usual interpretation, resource value $V$ and damage cost $D$ are positive. The parts below also specify the boundary or negative-value alternatives because positivity is not written explicitly in the PDF.

Take pure Hawk, $p^{\ast }=1$. For every mutant $p<1$, the payoff difference against an all-Hawk population isThe strict first condition of the two-condition criterion for evolutionary stability holds for every mutant. ThereforeThis conclusion holds even without the usual positivity assumption on the two parameters.

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 givesConsequentlyThe actual invasion advantage is $ϵV(1-p{)}^2/2>0$, despite the tie at invasion frequency zero.

If the equal parameters are negative, pure Dove is instead an evolutionarily stable strategy: for a mutant $p>0$, $e(0,0)-e(p,0)=-pV/2>0$. 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 $V=D=0$.

In the usual positive-parameter case $0<V<D$, set $p^{\ast }=V/D\in (0,1)$. Every mutant ties the resident against the resident population, because $V-D{p}^{\ast }=0$. Against a mutant population, however,The second condition of the two-condition criterion for evolutionary stability holds, soIndeed its advantage at invasion frequency $ϵ$ is exactly $ϵD(p^{\ast }-p{)}^2/2$. The interior frequency balances the two pure-strategy payoffs in the Hawk-Dove game.

For completeness, if $V<D$ but $V<0$, pure Dove is an evolutionarily stable strategy by the strict inequality $e(0,0)-e(p,0)=-pV/2>0$ for every mutant $p>0$. If $V=0<D$, that difference is zero, but $e(0,p)-e(p,p)=D{p}^2/2>0$, so Dove again satisfies the second condition. Thus for the remaining allowed values $\boxed{x^{\ast }=(0,1)\text{ when }V\le 0<D\text{ or }V<D\le 0}$; equivalently, Dove works whenever $V<D$ and $V\le 0$.