Order-one correction variance for coupled Poisson tau-leap levels

For lattice states $x,z$, write $e=x-z$ and ${\delta }_j=a_j(x)-a_j(z)$. Bounded rates and the integer lattice's minimum nonzero distance one imply $\mathrm{\mid }{\delta }_j\mathrm{\mid }\le L\mathrm{\parallel }e\mathrm{\parallel }$ for a constant $L$; this also holds when one state is outside $E$, where the extended rate is zero. In one half interval, shared counts cancel from the increment difference $\eta$. The independent residual Poisson counts and [F5] give $m:=\mathbb{E}[\eta \mid x,z]=h{\sum }_j{\nu }_j{\delta }_j,\mathbb{E}[\mathrm{\parallel }\eta -m{\mathrm{\parallel }}^2\mid x,z]=h{\sum }_j\mathrm{\mid }{\delta }_j\mathrm{\mid }\mathrm{\parallel }{\nu }_j{\mathrm{\parallel }}^2\mathrm{.}$ Thus $\mathrm{\parallel }m\mathrm{\parallel }\le Ch\mathrm{\parallel }e\mathrm{\parallel }$ and the centered second moment is at most $Ch\mathrm{\parallel }e\mathrm{\parallel }$. If $e=0$, then $\eta =0$. Otherwise $\mathrm{\parallel }e\mathrm{\parallel }\ge 1$, so $h\mathrm{\parallel }e\mathrm{\parallel }\le h\mathrm{\parallel }e{\mathrm{\parallel }}^2$. For $h\le 1$ this proves $\mathbb{E}[\mathrm{\parallel }\eta {\mathrm{\parallel }}^2\mid x,z]\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2\mathrm{.}$

If $x\mathrm{\notin }E$, the invalid-state convention freezes the fine path, so its increment $R$ is zero and the estimates below are immediate. Suppose $x\in E$, and let $R$ be its fine-path increment on the first half interval. By [F3], [F4], and the triangle inequality, $\mathbb{E}\mathrm{\parallel }R\mathrm{\parallel }\le {\sum }_j\mathrm{\parallel }{\nu }_j\mathrm{\parallel }h{a}_j(x)\le Ch\mathrm{.}$ By [F2] and bounded propensities, $\mathbb{E}\mathrm{\parallel }R{\mathrm{\parallel }}^2\le Ch$ for $h\le 1$. If $x^{\mathrm{\prime }}=x+R$, then $\mathbb{E}\mathrm{\parallel }{x}^{\mathrm{\prime }}-z\mathrm{\parallel }\le \mathrm{\parallel }e\mathrm{\parallel }+Ch,\mathbb{E}\mathrm{\parallel }{x}^{\mathrm{\prime }}-z{\mathrm{\parallel }}^2\le C(\mathrm{\parallel }e{\mathrm{\parallel }}^2+h)\mathrm{.}$ For the second inequality, expand $\mathrm{\parallel }e+R{\mathrm{\parallel }}^2$ and use $\mathrm{\parallel }\mathbb{E}R\mathrm{\parallel }\le Ch$ and $\mathbb{E}\mathrm{\parallel }R{\mathrm{\parallel }}^2\le Ch$; the cross term is bounded by $Ch\mathrm{\parallel }e\mathrm{\parallel }\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2$ when $e\mathrm{\ne }0$, and is zero when $e=0$.

Write the two half-interval differences as ${\eta }_i=m_i+{\xi }_i$, where $m_i$ is the conditional mean and ${\xi }_i$ is centered given the starting states of that half interval. Step 1.1 gives $\mathrm{\parallel }{m}_1\mathrm{\parallel }\le Ch\mathrm{\parallel }e\mathrm{\parallel },\mathbb{E}\mathrm{\parallel }{\xi }_1{\mathrm{\parallel }}^2\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2\mathrm{.}$ The second half starts at $x^{\mathrm{\prime }}$ and $z$, so $\mathrm{\parallel }{m}_2\mathrm{\parallel }\le Ch\mathrm{\parallel }{x}^{\mathrm{\prime }}-z\mathrm{\parallel }$. Steps 1.1 and 1.2 therefore give $\mathbb{E}\mathrm{\parallel }{m}_2{\mathrm{\parallel }}^2\le C{h}^2(\mathrm{\parallel }e{\mathrm{\parallel }}^2+h),\mathbb{E}\mathrm{\parallel }{\xi }_2{\mathrm{\parallel }}^2\le Ch(\mathrm{\parallel }e\mathrm{\parallel }+h)\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2+C{h}^2\mathrm{.}$ For the latter inequality, use $\mathbb{E}\mathrm{\parallel }{x}^{\mathrm{\prime }}-z\mathrm{\parallel }\le \mathrm{\parallel }e\mathrm{\parallel }+Ch$ and $\mathrm{\parallel }e\mathrm{\parallel }\le \mathrm{\parallel }e{\mathrm{\parallel }}^2$ when $e\mathrm{\ne }0$. Let ${\mathcal{H}}_1$ be the complete first-half history. The fresh second-half copies are independent of that history, so $\mathbb{E}[{\xi }_2\mid {\mathcal{H}}_1]=0$ and $\mathbb{E}[{\xi }_1\cdot {\xi }_2]=0$.

Fix the starting pair $(x,z)$ during this expansion; the estimates are uniform in the pair and depend on it only through $e=x-z$. The endpoint error is $e^{\mathrm{\prime }}=e+m_1+{\xi }_1+m_2+{\xi }_2$. First, $\mathbb{E}\mathrm{\parallel }e+{\eta }_1{\mathrm{\parallel }}^2=\mathrm{\parallel }e+m_1{\mathrm{\parallel }}^2+\mathbb{E}\mathrm{\parallel }{\xi }_1{\mathrm{\parallel }}^2\le (1+Ch)\mathrm{\parallel }e{\mathrm{\parallel }}^2\mathrm{.}$ The vector $e+m_1+{\xi }_1+m_2$ is ${\mathcal{H}}_1$-measurable, so $\mathbb{E}[{\xi }_2\mid {\mathcal{H}}_1]=0$ makes its expected inner product with ${\xi }_2$ zero. The remaining cross terms satisfy $2\mathrm{\mid }\mathbb{E}({\xi }_1\cdot m_2)\mathrm{\mid }\le 2(\mathbb{E}\mathrm{\parallel }{\xi }_1{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}(\mathbb{E}\mathrm{\parallel }{m}_2{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\le C{h}^{3\mathrm{/}2}\mathrm{\parallel }e\mathrm{\parallel }\sqrt{\mathrm{\parallel }e{\mathrm{\parallel }}^2+h}\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2,$ with value zero if $e=0$, and $2\mathrm{\mid }\mathbb{E}((e+m_1)\cdot m_2)\mathrm{\mid }\le h\mathrm{\parallel }e+m_1{\mathrm{\parallel }}^2+h^{-1}\mathbb{E}\mathrm{\parallel }{m}_2{\mathrm{\parallel }}^2\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2+C{h}^2\mathrm{.}$ The additional terms satisfy $\mathbb{E}\mathrm{\parallel }{m}_2{\mathrm{\parallel }}^2\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2+C{h}^2$ and $\mathbb{E}\mathrm{\parallel }{\xi }_2{\mathrm{\parallel }}^2\le Ch\mathrm{\parallel }e{\mathrm{\parallel }}^2+C{h}^2$ by step 2.1. Combining these expansion terms proves the recurrence conditional on $(x,z)$; averaging over starting pairs with difference $e$ gives $\mathbb{E}[\mathrm{\parallel }{e}^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mid e]\le (1+Ch)\mathrm{\parallel }e{\mathrm{\parallel }}^2+C{h}^2\mathrm{.}$

Let $u_n=\mathbb{E}\mathrm{\parallel }{e}_n{\mathrm{\parallel }}^2$ at coarse endpoints. Since $u_0=0$, step 3.1 gives $u_{n+1}\le (1+Ch)u_n+C{h}^2$. There are $N=T\mathrm{/}(2h)$ coarse intervals, so iteration yields $u_N\le C{h}^2{\sum }_{r=0}^{N-1}(1+Ch{)}^r\le C{h}^{2}N{e}^{ChN}\le C_{T}h\mathrm{.}$

If $e_N\mathrm{\ne }0$, its integer coordinates imply $\mathrm{\parallel }{e}_N{\mathrm{\parallel }}^2\ge 1$, hence $\mathbb{P}(e_N\mathrm{\ne }0)\le \mathbb{E}\mathrm{\parallel }{e}_N{\mathrm{\parallel }}^2\le C_{T}h$. The payoff difference is zero when $e_N=0$ and is at most $2G$ in absolute value otherwise. Its variance is at most its second moment, hence at most $4G^2\mathbb{P}(e_N\mathrm{\ne }0)\le Ch$. For $h>1$ the same conclusion follows directly from $\mathrm{Var}\le 4G^2\le 4G^{2}h$.