Weak order one for finite-state Poisson tau-leaping with an exit bound

At most $NR$ scalar Poisson draws are needed for $N$ steps, so the stated budget never exhausts; absorption or an invalid endpoint instead holds the state through $T$. The terminal leap law is therefore $K_h^N$. Both $K_h$ and $Q_h$ are Markov kernels: they preserve nonnegative functions and constants, so each is a contraction in the supremum norm. Outside $E$, both hold the state fixed by the stated extension.

The finite algebraic identity $K_h^N-Q_h^N={\sum }_{r=0}^{N-1}{K}_h^{N-1-r}(K_h-Q_h)Q_h^r$ follows by cancellation of consecutive terms. Apply it to $g$, use the contraction from step 1.1 and the one-step bound [F3] to obtain $\mathrm{\parallel }(K_h^N-Q_h^N)g{\mathrm{\parallel }}_{\mathrm{\infty }}\le N4A^2{h}^2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}=4A^{2}Th\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Evaluating at $x_0$ proves the weak-order bound.

While the leap state is in $E$, any single channel count produces a state in $E$ by [F1]. Therefore an invalid endpoint requires at least two total channel firings in that step. If $M_n={\sum }_j{K}_{j,n}$, independence [F2] and the factorial moment [F4] give $\mathbb{E}[M_n(M_n-1)\mid Y_n]={\sum }_j(a_j(Y_n)h{)}^2+2{\sum }_{i<j}{a}_i(Y_n)a_j(Y_n)h^2=(\lambda (Y_n)h{)}^2\le (Ah{)}^2\mathrm{.}$ Since ${\mathbf{1}}_{\{M_n\ge 2\}}\le M_n(M_n-1)\mathrm{/}2$, the conditional invalid-step probability is at most $A^2{h}^2\mathrm{/}2$. A union bound over $N=T\mathrm{/}h$ steps gives $\mathbb{P}(\text{invalid by }T)\le A^{2}Th\mathrm{/}2$.