Moment-weighted local defect bounds for BAOAB

First suppose $\mathbb{E}\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2<\mathrm{\infty }$. Shift time to $t$ and use $W_{t+u}-W_t$ with filtration ${\mathcal{F}}_{t+u}$. Brownian motion relative to this filtration has future increments independent of ${\mathcal{F}}_t$, so the initial state has the independence required in [F5]. The drift $b$ is globally Lipschitz with linear growth and $G$ is constant; [F5] therefore supplies the exact solution and finite second moments used in the following estimates. We first establish the estimates under this extra integrability condition, then remove it.

Set $s=h\mathrm{/}2$, $a=e^{-\gamma h}$, $\mathrm{\Delta }W=W_{t+h}-W_t$, and $\eta ={\int }_0^h{e}^{-\gamma (h-r)}Sd{W}_{t+r}$. Writing the BAOAB stages from [F2] gives [F1, F2, F3, F4, given, algebra] $q^{\mathrm{\prime }}=q+s(1+a)M^{-1}p+s^2(1+a)M^{-1}F(q)+s{M}^{-1}\eta ,p^{\mathrm{\prime }}=ap+asF(q)+\eta +sF(q^{\mathrm{\prime }})\mathrm{.}$ Subtracting $z+hb(z)+G\mathrm{\Delta }W$ yields $R_q^B=s(a-1)M^{-1}p+s^2(1+a)M^{-1}F(q)+s{M}^{-1}\eta ,$ $R_p^B=(a-1+\gamma h)p+s((a-1)F(q)+F(q^{\mathrm{\prime }})-F(q))+\eta -S\mathrm{\Delta }W\mathrm{.}$ Since $\mathrm{\mid }a-1\mathrm{\mid }\le Ch$ and $\mathrm{\mid }a-1+\gamma h\mathrm{\mid }\le C{h}^2$, the deterministic terms have size at most $C{h}^2(1+\mathrm{\mid }z\mathrm{\mid })$. The displayed formula and linear growth give $\mathbb{E}[\mathrm{\mid }{q}^{\mathrm{\prime }}-q\mathrm{\mid }\mid {\mathcal{F}}_t]\le Ch(1+\mathrm{\mid }z\mathrm{\mid })$ and $\mathbb{E}[\mathrm{\mid }{q}^{\mathrm{\prime }}-q{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le C{h}^2(1+\mathrm{\mid }z{\mathrm{\mid }}^2)$. Lipschitz continuity of $F$ and [F4] therefore give $\mathrm{\mid }\mathbb{E}[R^B\mid {\mathcal{F}}_t]\mathrm{\mid }\le C{h}^2(1+\mathrm{\mid }z\mathrm{\mid })$ and $\mathbb{E}[\mathrm{\mid }{R}^B{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le C{h}^3(1+\mathrm{\mid }z{\mathrm{\mid }}^2)$.

For two starting states driven by the same $\eta$, subtract the expression for $q^{\mathrm{\prime }}$ in step 1.2. The result and the Lipschitz bound on $F$ give $\mathrm{\mid }{q}^{\mathrm{\prime }}(z)-q^{\mathrm{\prime }}(\tilde{z})\mathrm{\mid }\le C\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$. Subtracting the two $R_q^B$ formulas then gives $\mathrm{\mid }{R}_q^B(z)-R_q^B(\tilde{z})\mathrm{\mid }\le C{h}^2\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$. The $R_p^B$ formula gives the same bound for its linear and $(a-1)F$ terms; its last force difference is bounded by $sL(\mathrm{\mid }{q}^{\mathrm{\prime }}(z)-q^{\mathrm{\prime }}(\tilde{z})\mathrm{\mid }+\mathrm{\mid }q-\tilde{q}\mathrm{\mid })\le Ch\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$. Combining components proves $\mathrm{\mid }{R}^B(z)-R^B(\tilde{z})\mathrm{\mid }\le Ch\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$.

With $b,G$ from step 1.2, the exact residual is $R^E={\int }_t^{t+h}(b(Z_r)-b(Z_t))dr$. For $0\le u\le h$, the integral equation, linear growth of $b$, Cauchy–Schwarz in time, and [F4] give, with $m(u)=\mathbb{E}[1+\mathrm{\mid }{Z}_{t+u}{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]$, [A1, F1, F4, F5, F6, step 1.2] $m(u)\le C(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)+C{\int }_0^{u}m(v)dv\mathrm{.}$ Set $M(u)=C(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)+C{\int }_0^{u}m(v)dv$ with the constant enlarged so $m\le M$. Then $M^{\mathrm{\prime }}(u)=Cm(u)\le CM(u)$ almost everywhere, so $(e^{-Cu}M(u){)}^{\mathrm{\prime }}\le 0$ and $m(u)\le C(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)$ for $u\le 1$. Substitution in the integral equation gives $\mathbb{E}[\mathrm{\mid }{Z}_{t+u}-Z_t{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le Cu(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)$, and Cauchy–Schwarz in time gives $\mathbb{E}[\mathrm{\mid }{R}^E{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le C{h}^3(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)$. For its mean, the existence of the second derivative makes $DF$ continuous, so $F\in C^1$. The ordinary chain rule along $Q$ and the linear momentum equation [F6] give $db(Z_r)=a(Z_r)dr+\widehat{G}d{W}_r$, where $a(q,p)=(M^{-1}(F(q)-\gamma p),DF(q)M^{-1}p-\gamma (F(q)-\gamma p))$ and $\widehat{G}=(M^{-1}S,-\gamma S{)}^T$. Bounded $DF$ and linear growth of $F$ give $\mathrm{\mid }a(z)\mathrm{\mid }\le C(1+\mathrm{\mid }z\mathrm{\mid })$. The constant-diffusion integral has conditional mean zero, so the preceding moment bound yields $\mathrm{\mid }\mathbb{E}[b(Z_{t+u})-b(Z_t)\mid {\mathcal{F}}_t]\mathrm{\mid }\le Cu(1+\mathrm{\mid }{Z}_t\mathrm{\mid })$; integration yields $\mathrm{\mid }\mathbb{E}[R^E\mid {\mathcal{F}}_t]\mathrm{\mid }\le C{h}^2(1+\mathrm{\mid }{Z}_t\mathrm{\mid })$.

To remove the temporary $L^2$ assumption, put $D_n=\{\mathrm{\mid }{Z}_t\mathrm{\mid }\le n\}\in {\mathcal{F}}_t$ and $Z_t^{(n)}=1_{D_n}{Z}_t$. Apply [F5] to each bounded start with the same Brownian driver, obtaining $Z^{(n)}$. On $D_n$, for $m\ge n$ the two initial values agree and the constant diffusion cancels from their integral equations. Thus $\mathrm{\mid }{Z}_{t+u}^{(n)}-Z_{t+u}^{(m)}\mathrm{\mid }\le L{\int }_0^u\mathrm{\mid }{Z}_{t+v}^{(n)}-Z_{t+v}^{(m)}\mathrm{\mid }dv$; the integrating-factor argument of step 2.2 with zero initial bound gives equality throughout the interval. Intersect the countably many full events and paste the paths on $D_n\setminus D_{n-1}$, with $D_0=\mathrm{\varnothing }$. These starting-time measurable events cover almost surely, so the pasted process is adapted and continuous and satisfies the additive-noise integral equation. The same difference argument proves uniqueness. The BAOAB outputs and both residuals likewise agree on $D_n$ with their truncated-start versions. Multiplying their conditional estimates from steps 1.2 and 2.2 by $1_{D_n}$ therefore gives the stated bounds on $D_n$ with the original $Z_t$ and the same constants; let $n\to \mathrm{\infty }$. For a merely finite random start, a signed conditional mean here denotes the consistent pasting of $\mathbb{E}[1_{D_n}R\mid {\mathcal{F}}_t]$ on $D_n$; each truncated residual is integrable by the bounds already proved. Conditional second moments are the corresponding nonnegative conditional expectations and are finite almost surely by the displayed bounds. No unconditional moment is asserted for a start outside $L^2$. Step 2.1 gives the pathwise Lipschitz estimate for all finite starting states. Time homogeneity makes all constants uniform in $t$.