Explicit Poisson tau-leaping with an invalid-state status

Inputs. A finite-channel reaction propensity system Finite-channel reaction propensity systems and their generator, an initial state $x_0\in E$, a deterministic grid $0=t_0<t_1<\cdots <t_N=T$ with finite $N$, and a draw budget $B\in \mathbb{N}\cup \{\mathrm{\infty }\}$. Write $h_n=t_{n+1}-t_n$. At each step the count vector has the product law whose $j$th marginal is $\mathrm{Pois}(a_j(x_n)h_n)$, as defined in The scalar Poisson count law. The budget counts scalar Poisson draws.

Output. The recorded grid endpoint pairs, with piecewise-constant interpolation, and a status in {horizon, absorbed, invalid-state, budget-exhausted}. The budget-exhausted output is an incomplete prefix and is not a terminal sample at $T$.

Procedure. Set $x=x_0$, grid index $n=0$, Poisson-draw counter $d=0$, recorded path $P=[(0,x_0)]$, and status horizon.

1. If $n=N$, return $P$ with status horizon. If $x\mathrm{\notin }E$, set status invalid-state and hold $x$ fixed to $T$. If $P$ does not end at $(t_n,x)$, append that pair; if $t_n<T$, append $(T,x)$; then return. If $\lambda (x)=0$, hold $x$ fixed to $T$, append $(T,x)$ to $P$ if needed, and return $P$ with status absorbed.
2. If $B<\mathrm{\infty }$ and $B-d<R$, return the path prefix with status budget-exhausted; it does not define a terminal sample at $T$.
3. Draw conditionally independent $K_{j,n}\sim \mathrm{Pois}(a_j(x)h_n)$, $1\le j\le R$, and set $y=x+{\sum }_{j=1}^R{K}_{j,n}{\nu }_j$. Increase $d\leftarrow d+R$.
4. Advance time to $t_{n+1}$ and append $(t_{n+1},y)$ to $P$. If $y\mathrm{\notin }E$, set $x=y$, set status invalid-state, and keep $y$ fixed through all remaining grid times; append $(T,y)$ to $P$ if $t_{n+1}<T$, then return. If $y\in E$, set $x\leftarrow y$, increment $n\leftarrow n+1$, and continue.