One-step weak defect of finite-state Poisson tau-leaping

For the Poisson leap, the probability of zero total counts is $e^{-\lambda (x)h}$. The probability that exactly channel $j$ fires once is $a_j(x)h{e}^{-\lambda (x)h}$, so its difference from $a_j(x)h$ is at most $a_j(x)\lambda (x)h^2$. Summing these errors over $j$ gives at most $\lambda (x{)}^2{h}^2\le A^2{h}^2$. If $N={\sum }_j{K}_j$, independence and [F5] give $\mathbb{E}[N(N-1)\mid x]={\sum }_j(a_j(x)h{)}^2+2{\sum }_{i<j}{a}_i(x)a_j(x)h^2=(\lambda (x)h{)}^2,$ hence $\mathbb{P}(N\ge 2\mid x)\le (\lambda (x)h{)}^2\mathrm{/}2\le A^2{h}^2\mathrm{/}2$. The zero-count probability differs from $1-\lambda (x)h$ by at most $A^2{h}^2\mathrm{/}2$. A single allowed channel event remains in $E$ by [F1], so only the $N\ge 2$ part can trigger the invalid-state branch. The zero-count, total single-count, and multiple-count contributions are bounded respectively by $G{A}^2{h}^2\mathrm{/}2$, $G{A}^2{h}^2$, and $G{A}^2{h}^2\mathrm{/}2$. Thus $\mathrm{\mid }(K_{h}g)(x)-g(x)-h(Qg)(x)\mathrm{\mid }\le 2G{A}^2{h}^2$.

For the exact chain, the no-event probability is $e^{-\lambda (x)h}$, which differs from $1-\lambda (x)h$ by at most $A^2{h}^2\mathrm{/}2$. The probability of one channel-$j$ event and no later event is ${\int }_0^h{a}_j(x)e^{-\lambda (x)s}{e}^{-\lambda (x+{\nu }_j)(h-s)}ds\mathrm{.}$ Its difference from $a_j(x)h$ is at most $a_j(x)A{h}^2$, since all total rates on $E$ are at most $A$; summing over $j$ gives at most $\lambda (x)A{h}^2\le A^2{h}^2$. The probability of two or more events is at most $A^2{h}^2\mathrm{/}2$: conditional on a first event at time $s$, a later event has probability at most $A(h-s)$, and integration against the first-event rate bounded by $A$ gives the bound. The no-event, total single-event, and multiple-event contributions are bounded respectively by $G{A}^2{h}^2\mathrm{/}2$, $G{A}^2{h}^2$, and $G{A}^2{h}^2\mathrm{/}2$. Therefore $\mathrm{\mid }(Q_{h}g)(x)-g(x)-h(Qg)(x)\mathrm{\mid }\le 2G{A}^2{h}^2$.

Subtract the two first-order expansions from steps 1.1 and 1.2 and use the triangle inequality. This gives $\mathrm{\mid }(K_{h}g)(x)-(Q_{h}g)(x)\mathrm{\mid }\le 4G{A}^2{h}^2$ for every $x\in E$. Outside $E$ both kernels hold the state fixed, so the same bound holds there. Taking the supremum proves the final inequality.