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A resident strategy is evolutionarily stable if every distinct mutant earns lower expected fitness when introduced at sufficiently small positive frequency. The upper bound on the frequency may depend on the mutant. In a finite symmetric matrix game, $e(x,y)=x^{T}Ay$ is the expected payoff, with the second argument representing the population strategy. Affineness in this argument yields the two-condition criterion for evolutionary stability. Every evolutionarily stable strategy is a symmetric Nash equilibrium, but a tied best response must also be strictly outperformed against its own mutant population. The Hawk-Dove game supplies an interior stable mixture despite equality of all payoffs against the resident mixture.

Under cooperative initial memories, matches involving only Tit for tat and always cooperate contain only cooperation, so all four payoffs equal $R$. The two-condition criterion for evolutionary stability has both differences zero. Thus Tit for tat is not an evolutionarily stable strategy against a strategy space containing unconditional cooperation, even if it is a Nash equilibrium. This counterexample is valid for finite horizons and discounting as well as infinite averages.

Under the standard two-condition criterion for evolutionary stability, the strict claim fails. Tit for tat and always cooperate both cooperate in every round when playing either each other or their own type under the specified initial memories. HenceBoth stability comparisons tie. The neutrality between Tit for tat and unconditional cooperation proves that Tit for tat is not an evolutionarily stable strategy in the stated strategy space, and cannot strictly invade every strategy. The final differently spelled name is interpreted as the same $(1,0)$ rule, since no second strategy is defined.

There is a useful qualified invasion result. For a resident $M=(p,q)$ with $|p-q|<1$, put $z=q/(1-p+q)$. The long-run payoff of reactive strategies givesThus, if the fraction of Tit for tat players is $\epsilon$, their payoff advantage isMoreover $R-g(z)=(1-z)[R-P+(R+P-S-T)z]$. Under the additional cooperative-efficiency condition $2R\ge S+T$, this is positive whenever $z<1$. Tit for tat invasion of a reactive resident then occurs from every positive frequency, although its invasion exponent at frequency zero vanishes. The replicator equation is $\dot{\epsilon }=[R-g(z)]{\epsilon }^2(1-\epsilon )$, so the initial increase is slow and frequency-dependent. Residents with $p=1$ remain cooperative on all reached histories and tie instead. If $S+T>2R$, some almost-cooperative residents have $g(z)>R$, and even this qualified invasion conclusion fails.

For the exceptional opposite-response resident $(0,1)$, its self-payoff is $h=(R+P)/2$ while its payoff in either order against Tit for tat is $g=(R+S+T+P)/4$. The advantage is $(1-\epsilon )(g-h)+\epsilon (R-g)$. Under $2R\ge S+T$, invasion at arbitrarily small frequency requires $S+T\ge R+P$; otherwise it needsFinally, taking an infinite undiscounted average before the rare-mutant limit is essential. In an $L$-round match against always defect, the first round contributes $S$ to Tit for tat and $T$ to the defector, followed by $L-1$ rounds of payoff $P$. The finite-horizon invasion threshold of Tit for tat against unconditional defection isprovided the denominator is positive and the threshold is below one. Thus a finite horizon generally prevents invasion from an arbitrarily rare introduction. The source's absent payoff table prevents choosing which extra payoff inequalities were intended, but the symbolic cases and the strict-ESS counterexample do not depend on guessing it.

In a two-player symmetric finite game, represent mixed strategies by probability vectors and let $e(x,y)=x^{T}Ay$ be the payoff to a player using $x$ against one using $y$. A resident strategy $x^{\ast }$ is an evolutionarily stable strategy if every distinct mutant strategy $y$ has an invasion barrier ${\epsilon }_0(y)>0$ such that, for $0<\epsilon <{\epsilon }_0(y)$,Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.

For fixed $y\ne x^{\ast }$ putBy bilinearity the resident-minus-mutant fitness difference isIf this is positive for all sufficiently small positive $\epsilon$, taking $\epsilon \downarrow 0$ gives $a_y\ge 0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<{\epsilon }_0\le a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\epsilon (a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\epsilon b_y>0$ for every $0<\epsilon <1$. This proves the two-condition criterion for evolutionary stability: for every $y\ne x^{\ast }$, eitherorThe weak first comparison is the symmetric Nash equilibrium condition; the second excludes neutrally competitive mutants when that comparison is an equality.

For the Hawk-Dove game, let $x^{\ast }=(1/2,1/2)$ and $y=(p,1-p)$, $0\le p\le 1$. Against the resident, both pure strategies give $1/2$, soDirect multiplication of the payoff matrix givesConsequentlyThe equality case of the two-condition criterion for evolutionary stability therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\epsilon (p-1/2{)}^2$, positive for every $0<\epsilon <1$. Hence $\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$, with the common invasion barrier one.

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 $V\le 0$.

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.

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.

The replicator equation says that a strategy grows in frequency when its expected payoff exceeds the population's current mean payoff. It is an ordinary differential equation for a composition, rather than for total population size. The subtraction of the mean is what makes relative success matter: even a type growing in absolute number can shrink as a fraction if other types grow faster.

Let $x_i(t)$ be the frequency of pure strategy $i$, with $n$ available pure strategies andIn a symmetric bimatrix game, let the payoff matrix $A$ have entry $a_{ij}$ equal to the payoff to strategy $i$ against $j$. The other player's payoff in that encounter is $a_{ji}$: symmetry between player roles does not require $A=A^T$. Uniform random encounters give the expected payoffs and their population meanAfter choosing the time unit so that the growth-rate scale is one, the continuous-time replicator equation isHere a payoff is measured on a scale proportional to net growth per unit time, not merely as an arbitrary preference ranking. More generally one can specify locally Lipschitz frequency-dependent payoffs $f_i(x)$ directly; the same equation with $\bar{f}={\sum }_i{x}_i{f}_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in Taylor and Jonker (1978), sections 2–3; the formulas here also follow by the following direct normalization.

Suppose the number $N_i$ of type $i$ obeys ${\dot{N}}_i=N_i[b(t)+f_i(x)]$, where the baseline $b(t)$ is identical for all types. Set $N={\sum }_i{N}_i>0$ and $x_i=N_i/N$. The quotient rule givesThus the common baseline cancels. Adding the same constant to every entry of $A$ leaves the replicator equation unchanged; multiplying $A$ by a positive constant changes only the speed along trajectories. The same payoff matrix can therefore describe frequency evolution at different growth-rate baselines and time scales.

The probability simplex is invariant in forward time. On ${\sum }_i{x}_i=1$,Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,An absent type remains absent, and a positive frequency remains positive at every finite time. For the fixed payoff matrix model, bounded rates on the compact probability simplex give global forward solutions. Frequencies can still tend to zero as $t\to \infty$. Each face is invariant, so this model has no spontaneous introduction of new pure strategies; mutation or outside introductions would require additional terms. For two present types the logarithm of their frequency ratio satisfies $\frac{d}{dt}\log (x_i/x_j)=f_i-f_j$, directly expressing the accumulated payoff advantage.

For two pure strategies, write the composition as $(x,1-x)$, where $x$ is the frequency of strategy 1, and letTheir payoff difference and the resulting scalar ordinary differential equation areIn the interior $0<x<1$, the sign of the payoff difference determines the direction of motion. The endpoints are always equilibrium points, and an additional interior equilibrium point exists precisely when the affine payoff difference has a zero strictly between them. At a simple interior zero $x_{\ast }$, $F^{\prime }(x_{\ast })=x_{\ast }(1-x_{\ast })\Delta f^{\prime }(x_{\ast })$: a negative value gives asymptotic stability, a positive value gives instability. At the endpoints $F^{\prime }(0)=b-d$ and $F^{\prime }(1)=c-a$, with stability understood relative to $[0,1]$. A zero derivative requires examining the actual flow rather than this strict sign test.

For an original coordination game, imagine two conventions that reward matching, with convention 1 giving the larger matching reward. ChooseThe crossing is $x_{\ast }=2/5$. Below it, convention 1 loses frequency; above it, convention 1 gains frequency. The endpoint equilibrium points $0$ and $1$ have one-sided asymptotic stability, while $2/5$ is unstable: the corresponding derivatives are $-2,-3$ and $6/5$, respectively. Every initial $0<x(0)<2/5$ tends to $0$, every initial $2/5<x(0)<1$ tends to $1$, and exactly $x(0)=2/5$ stays there. Even the convention with the larger matching reward can disappear if it starts below the threshold. This is bistability, not selection of a universally best convention.

The compositions giving a symmetric Nash equilibrium in this coordination game are the two pure conventions and the mixed composition $(2/5,3/5)$. Both pure conventions are evolutionarily stable strategies because a mutant convention has strictly lower payoff against the pure resident. The mixed symmetric Nash equilibrium is not an evolutionarily stable strategy: against that resident both conventions tie, but after a pure convention-1 mutant is introduced its payoff advantage is $5(x-2/5)>0$. Equal payoffs at a symmetric Nash equilibrium need not provide resistance to invasion or asymptotic stability.

For an original coexistence model, choose payoffs making each convention advantageous when sufficiently rare:Here $x_{\ast }=3/4$ attracts every initial composition with both types present: $\dot{x}>0$ below it and $\dot{x}<0$ above it. The interior derivative is $F^{\prime }(3/4)=-3/4$, while $F^{\prime }(0)=3$ and $F^{\prime }(1)=1$ make both endpoints unstable to introduction of the other type. The endpoints nevertheless remain stationary if the other type is exactly absent; the attraction statement excludes those boundary initial states.

The sole symmetric Nash equilibrium in this coexistence model is $p=(3/4,1/4)$. It is also an evolutionarily stable strategy. For any different mixed strategy $q=(y,1-y)$, the first comparison in the two-condition criterion for evolutionary stability ties because $Ap=(3/2,3/2)$. Its second comparison is strictly positive:Thus restoring frequency feedback can protect a mixture even though all present types earn equal payoffs at the mixture. Coexistence here means different individuals using different pure strategies; it does not require every individual to randomize with a mixed strategy.

Frequency evolution need not increase the mean payoff. In the coexistence example $\bar{f}=5x-4x^2$, whose maximum occurs at $x=5/8$, whereas the attracting composition is $3/4$. For $5/8<x<3/4$, the frequency $x$ increases while $\frac{d}{dt}\bar{f}=(5-8x)\dot{x}<0$. Individual payoff advantages drive the equation; maximizing collective mean payoff is a different optimization problem.

More generally, a rest point only equalizes payoffs among present types: $x_i>0$ implies $(Ax{)}_i=x^{T}Ax$. A symmetric Nash equilibrium additionally requires $(Ax{)}_i\le x^{T}Ax$ for every absent type. All vertices are rest points even when an absent type could invade. For finite symmetric bimatrix games, a rest point with Lyapunov stability on the full probability simplex is a symmetric Nash equilibrium; an interior trajectory that converges has a symmetric Nash equilibrium as its limit. These are conditional results, not a convergence theorem for every trajectory. See Cressman and Tao (2014), Theorem 1.

The rock paper scissors payoff matrix gives an explicit obstruction to universal convergence:Since $A^T=-A$, $x^{T}Ax=0$, and in the interior $\frac{d}{dt}\log (x_1{x}_2{x}_3)={\sum }_i(Ax{)}_i=0$. Thus $x_1{x}_2{x}_3$ is conserved. Its maximum on the probability simplex occurs only at $(1/3,1/3,1/3)$, the unique symmetric Nash equilibrium, so an interior trajectory starting elsewhere cannot converge to that composition. Every other interior product level is a closed curve without a rest point, giving a periodic orbit. The existence of these cycles is also discussed in Cressman and Tao (2014), Example 1. The useful lesson is to inspect the specified payoff feedback and dynamical system, rather than infer convergence from the existence of an evolutionarily stable strategy or a Nash equilibrium.