= Replicator equation
{title2=$\dot x_i=x_i[(Ax)_i-x^TAx]$}
= Replicator dynamics
{synonym}
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> and
$$
x\in\Delta_n=\left\{x\in\mathbb R^n:x_i\ge0,\ \sum_{i=1}^n x_i=1\right\}.
$$
In 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 <mean>
$$
f_i(x)=(Ax)_i=\sum_j a_{ij}x_j,\qquad \bar f(x)=\sum_i x_if_i(x)=x^TAx.
$$
After choosing the time unit so that the growth-rate scale is one, the continuous-time <replicator equation> is
$$
\dot x_i=x_i\bigl(f_i(x)-\bar f(x)\bigr)=x_i\bigl((Ax)_i-x^TAx\bigr),\qquad i=1,\ldots,n.
$$
Here 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 $\bar f=\sum_i x_if_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in https://static1.squarespace.com/static/5dc9917c4cdb1467e1e54c4c/t/6526a5647942e33b89ab7af1/1697031525653/1978%2Bevolutionarily%2Bstable%2Bstrategies%2Band%2Bgame%2Bdynamics.pdf[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 $N=\sum_iN_i>0$ and $x_i=N_i/N$. The <quotient rule> gives
$$
\dot N=N[b(t)+\bar f(x)],\qquad \dot x_i=\frac{\dot N_iN-N_i\dot N}{N^2}=x_i(f_i-\bar f).
$$
Thus 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 $\sum_i x_i=1$,
$$
\sum_i\dot x_i=\sum_i x_if_i-\bar f\sum_i x_i=0.
$$
Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,
$$
x_i(t)=x_i(0)\exp\left(\int_0^t [f_i(x(s))-\bar f(x(s))]\,ds\right).
$$
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 let
$$
A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\qquad f_1=ax+b(1-x),\qquad f_2=cx+d(1-x).
$$
Their <payoff> difference and the resulting scalar <ordinary differential equation> are
$$
\Delta f(x)=f_1-f_2=(b-d)+(a-b-c+d)x,\qquad \dot x=F(x)=x(1-x)\Delta f(x).
$$
In 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 $x_*$, $F'(x_*)=x_*(1-x_*)\Delta f'(x_*)$: a negative value gives <asymptotic stability>, a positive value gives <instability>. At the endpoints $F'(0)=b-d$ and $F'(1)=c-a$, 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. Choose
$$
A_{\mathrm{coord}}=\begin{pmatrix}3&0\\0&2\end{pmatrix},\qquad f_1=3x,\quad f_2=2(1-x),\quad \dot x=x(1-x)(5x-2).
$$
The crossing is $x_*=2/5$. 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.
\Image[/replicator-coordination.png]
{title=Toy coordination model: an unstable frequency threshold separates attraction to the two pure conventions}
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:
$$
A_{\mathrm{coex}}=\begin{pmatrix}1&3\\2&0\end{pmatrix},\qquad f_1=3-2x,\quad f_2=2x,\quad \dot x=x(1-x)(3-4x).
$$
Here $x_*=3/4$ attracts every initial composition with both types present: $\dot x>0$ below it and $\dot x<0$ above it. The interior <derivative> is $F'(3/4)=-3/4$, while $F'(0)=3$ and $F'(1)=1$ 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.
\Image[/replicator-coexistence.png]
{title=Toy coexistence model: both rare types can invade, and every interior composition approaches frequency three quarters}
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:
$$
p^TAq-q^TAq=(p-q)^TAq=(3/4-y)(3-4y)=4(y-3/4)^2>0.
$$
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 $(Ax)_i=x^TAx$. A <symmetric Nash equilibrium> additionally requires $(Ax)_i\le x^TAx$ 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 https://pmc.ncbi.nlm.nih.gov/articles/PMC4113915/[Cressman and Tao (2014), Theorem 1].
The <rock paper scissors> <payoff matrix> gives an explicit obstruction to universal convergence:
$$
A=\begin{pmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{pmatrix}.
$$
Since $A^T=-A$, $x^TAx=0$, and in the interior $\frac{d}{dt}\log(x_1x_2x_3)=\sum_i(Ax)_i=0$. Thus $x_1x_2x_3$ 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 https://pmc.ncbi.nlm.nih.gov/articles/PMC4113915/[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>.
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