Density-dependent Markov jump process

With scale $N$, such a process has finitely many possible jump vectors $\ell /N$ occurring at rates $N{\beta }_{\ell }(x)$ depending on the current scaled state. Its drift is ${\sum }_{\ell }\ell {\beta }_{\ell }(x)$, while the predictable covariance rate of its martingale fluctuations is $N^{-1}{\sum }_{\ell }\ell {\ell }^T{\beta }_{\ell }(x)$.

// Target: probability-and-statistics.bigb

In a population of $N$ playing rock paper scissors, let losers copy winners. With pair-selection events of rate $(N-1)/2$, the proportion process has jumps $(e_1-e_3)/N$, $(e_2-e_1)/N$, $(e_3-e_2)/N$ at rates $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. Its fluid limit obeys ${\dot{x}}_i=x_i(x_{i-1}-x_{i+1})$, with indices interpreted cyclically. The product of numbers $x_1{x}_2{x}_3$ is conserved in this ordinary differential equation, whereas the finite chain eventually reaches a monochromatic absorbing state.

For the cyclic imitation chain, its infinitesimal generator satisfies ${\mathcal{L}}_N(x_1{x}_2{x}_3)=-3x_1{x}_2{x}_3/N$. Therefore $\mathbb{E}[X_1^N(t)X_2^N(t)X_3^N(t)]=e^{-3t/N}\mathbb{E}[X_1^N(0)X_2^N(0)X_3^N(0)]$. While all three types remain, the product of numbers is at least $(N-2)/N^3$ for $N\ge 3$, so this identity proves that one type is lost almost surely. The remaining pair has a fixed winner, which eventually occupies the whole population. The conserved product in the deterministic fluid limit therefore describes fixed finite horizons, not the chain's infinite-time behavior.

With jumps $\ell /N$ at rates $N{\beta }_{\ell }(z)$, compensate each jump count by its integrated rate. The drift is $F(z)={\sum }_{\ell }\ell {\beta }_{\ell }(z)$ and the predictable quadratic variation matrix of the martingale term is $N^{-1}{\int }_0^t{\sum }_{\ell }\ell {\ell }^{\mathsf{T}}{\beta }_{\ell }(Z_s^N)ds$. On a compact region with bounded rates this bracket is of order $1/N$. A Lipschitz drift and the Gronwall inequality transfer martingale concentration to the fluid limit.