Minimum-intensity coupling of two birth-death chains

For both systems, a positive birth rate occurs only at $x<2$ and moves to $x+1\in E$; a positive death rate occurs only at $x>0$ and moves to $x-1\in E$. All rates are nonnegative and finite, so [F1] gives valid finite-state systems.

At $(1,1)$ the birth rates are $1,1$, hence [F2] gives shared birth rate $1$. The death rates are $1,2$, hence shared death rate $1$ and $Y$-only death rate $1$; the $X$-only death rate is $0$. There are no other event types, so the total paired-event rate is $3$.

The algorithm's birth envelope is $M_{+}=\max \{{\max }_x{a}_{+}^X(x),{\max }_y{a}_{+}^Y(y)\}=2$, and its death envelope is $M_{-}=\max \{{\max }_x{a}_{-}^X(x),{\max }_y{a}_{-}^Y(y)\}=4$, so the total candidate rate is $M=M_{+}+M_{-}=6$. At the fixed pair $(1,1)$ the three disjoint event-mark portions have total length $3\mathrm{/}6$; [F4] gives first-event survival $e^{-6(3\mathrm{/}6)t}=e^{-3t}$ and assigns each equal-length portion probability $1\mathrm{/}3$ conditional on an event. Their state updates are respectively $(2,2)$, $(0,0)$, and $(1,0)$.

The marginal theorem [F3] applies because the budget is unlimited and the finite rates supply the bounds $M_{+}=2$ and $M_{-}=4$. It identifies each coordinate path with its own Gillespie reaction law, completing the example.