Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 3

Use stochastic mass-action propensitiesWriting $E_x^{r}f(x,y)=f(x+r,y)$ and similarly for $E_y$, the fast and slow forward operators areEach shift operator acts on everything to its right, including the propensity. Birth of $Y$ and consumption of $Y$ by $X+Y\to X$ are fast; $2Y\to X$ and $2X\to X$ are slow. The slow species is $X$.

At leading order, ${\mathcal{L}}_0^{\ast }[p_0(y\mid x)p_0(x,t)]=0$. For fixed $x$, the conditional stationary law therefore satisfiesThis is an immigration--death process, whose stationary distribution is Poisson with meanThe mean is finite only for $x>0$. The assumption $X(0)\ne 0$ and the pair-coalescence propensity $x(x-1)$ ensure that the slow process cannot remove its last $X$ molecule, so the reduced model remains in that domain.

Averaging the slow propensities over the conditional Poisson distribution uses $\mathbb{E}[Y(Y-1)\mid X=x]=q(x{)}^2$. Thus the effective birth and death rates of $X$ areConsequently

Applying the reduced Markov jump-process generator to $f(x)=x$ gives the exact moment equationUnder the stated moment closure this becomes the same expression evaluated at $X=m$. Set $m=V\bar{x}$ and let $V\to \infty$ to obtainIts positive stable equilibrium is

The stochastic quasi-steady-state simulation evolves only $X$:

1. At the current integer $x\ge 1$, compute $q(x)={\alpha }_3{V}^2/({\alpha }_{4}x)$ and the averaged rates ${\lambda }_1(x)={\alpha }_{1}q(x{)}^2/V$ and ${\lambda }_2(x)={\alpha }_{2}x(x-1)/V$.
2. Draw a waiting time $\Delta t\sim \mathrm{Exp}({\lambda }_1+{\lambda }_2)$.
3. Set $x\leftarrow x+1$ with probability ${\lambda }_1/({\lambda }_1+{\lambda }_2)$; otherwise set $x\leftarrow x-1$.
4. Advance time by $\Delta t$ and repeat.

This is a Gillespie algorithm for the averaged slow master equation. It samples the fast conditional equilibrium analytically through its factorial moment and never simulates individual fast $Y$ births or deaths.