Minimum-intensity coupling of reaction networks

Inputs. Two finite-state reaction systems $X$ and $Y$ with the same finite channel labels $j=1,\dots ,R$. Their state sets may differ, and channel $j$ has respective increments ${\nu }_j^X,{\nu }_j^Y$ and propensities $a_j^X(x),a_j^Y(y)$ as in Finite-channel reaction propensity systems and their generator. Supply an initial pair $(x_0,y_0)$, a horizon $T>0$, and an optional candidate budget $B\in {\mathbb{N}}_0\cup \{\mathrm{\infty }\}$, defaulting to $\mathrm{\infty }$, where ${\mathbb{N}}_0:=\{0,1,2,\dots \}$.

For each channel set $M_j:=\max \{{\max }_{x\in E_X}{a}_j^X(x),{\max }_{y\in E_Y}{a}_j^Y(y)\},M:={\sum }_{j=1}^R{M}_j\mathrm{.}$ These are finite because both state sets are finite. Supply mutually independent $E_n\sim \mathrm{Exp}(1)$ and $U_n\sim \mathrm{Unif}[0,1]$ for $n\ge 1$. The countable input product uses Countable Choice and Dependent Choice; the exponential law is that of Exponential and Weibull survival specifications define probability laws. For the uniform law, Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume makes the set function in Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$ a measure, and A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included gives its restriction to $[0,1]$ mass one and interval probabilities equal to lengths.

Procedure. Set $t=0$, $(x,y)=(x_0,y_0)$, and $n=0$. If $M=0$, return the two constant paths through $T$ with status absorbed. Otherwise define fixed mark blocks $I_j=[b_{j-1},b_j)$, where $b_0=0$ and $b_j=M^{-1}{\sum }_{k=1}^j{M}_k$; omit empty blocks when $M_j=0$, and include the endpoint $1$ in the last nonempty block. While $t<T$:

1. If $n=B$, return the pair of path prefixes with status budget-exhausted.
2. Set $\tau =E_{n+1}\mathrm{/}M$. If $t+\tau \ge T$, return both paths held through $T$ with status horizon.
3. Set $t\leftarrow t+\tau$. Find the unique nonempty block $I_j$ containing $U_{n+1}$ and set $v=(U_{n+1}-b_{j-1})\mathrm{/}(M_j\mathrm{/}M)$. Fire channel $j$ in $X$ if $v<a_j^X(x)\mathrm{/}{M}_j$, and fire it in $Y$ if $v<a_j^Y(y)\mathrm{/}{M}_j$. If both fire, record a shared event and update both states; if only one fires, record its residual event and update only that state; if neither fires, reject the candidate and leave both states fixed. Increment $n$ once for every candidate, including a rejected candidate. Endpoint marks have probability zero, so the last-block convention at $U_{n+1}=1$ does not affect the path law.

The candidate mark block for channel $j$ has length $M_j\mathrm{/}M$. Within it the two acceptance intervals start at the same endpoint and have respective lengths $a_j^X(x)\mathrm{/}M$ and $a_j^Y(y)\mathrm{/}M$. Thus their overlap has candidate intensity $\min (a_j^X(x),a_j^Y(y))$, and the two residual portions have intensities $(a_j^X(x)-a_j^Y(y){)}_{+}$ and $(a_j^Y(y)-a_j^X(x){)}_{+}$. The budget counts candidates, so an exhausted budget returns only an incomplete prefix. No logarithmic event-selection cost is claimed.