At leading order, $\mathcal L_0^*[p_0(y\mid x)p_0(x,t)]=0$. For fixed $x$, the conditional stationary law therefore satisfies
$$
\boxed{0=\alpha_3V[p_0(y-1\mid x)-p_0(y\mid x)]
+\frac{\alpha_4x}{V}[(y+1)p_0(y+1\mid x)-yp_0(y\mid x)].}
$$
This is an <immigration--death process>, whose stationary distribution is Poisson with mean
$$
\boxed{q(x)=\langle Y\mid X=x\rangle
=\frac{\alpha_3V^2}{\alpha_4x}.}
$$
The mean is finite only for $x>0$. The assumption $X(0)\ne0$ 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.

Solution

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

An immigration--death process has constant birth rate $b$ and linear death rate $dy$ in state $y$. Its stationary distribution is Poisson with mean $b/d$.


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 Incoming links: Immigration--death process