Euler–Maruyama strong rate for square-integrable initial data

By [F4] and [F7], the exact process has its finite-coordinate Itô representation; [F1] supplies the Euler interpolation and its predictable representative ${\bar{Y}}^{-}$ with left-endpoint map ${\eta }_h^{-}$. Use that representative in the Euler integral form: $X_t=X_0+{\int }_0^{t}b(s,X_s)ds+{\int }_0^t\sigma (s,X_s)d{W}_s,$ $Y_t^h=X_0+{\int }_0^{t}b({\eta }_h^{-}(s),{\bar{Y}}_s^{-})ds+{\int }_0^t\sigma ({\eta }_h^{-}(s),{\bar{Y}}_s^{-})d{W}_s\mathrm{.}$ The grid-valued coefficients are predictable on $(t_k,t_{k+1}]$; the grid coefficient energy is a finite sum bounded by [F2], while Tonelli [F12] and the stated $\mathbb{E}\sup \mathrm{\mid }X{\mathrm{\mid }}^2$ bound give finite expected energy for the exact coefficients. Thus these are the required integral representations. In subsequent estimates, $\bar{Y}$ and ${\eta }_h$ may replace their predictable versions because they differ only at finitely many grid times, which do not affect the integrals. For $R>0$ define ${\tau }_R=\inf \{t\le T:\mathrm{\mid }{e}_t\mathrm{\mid }\ge R\}\wedge T$, with $\inf \mathrm{\varnothing }=\mathrm{\infty }$. For $t<T$, $\{{\tau }_R\le t\}=\{{\sup }_{s\le t}\mathrm{\mid }{e}_s\mathrm{\mid }\ge R\}$, which is in ${\mathcal{F}}_t$ by continuity and [F9]; for $t\ge T$ the event is all of $\mathrm{\Omega }$. Thus [F8] makes ${\tau }_R$ a stopping time. Subtracting the displays and using [F5] gives the stopped error as the sum of the two stopped coefficient integrals.

Put $\delta b_s=b(s,X_s)-b({\eta }_h(s),{\bar{Y}}_s),\delta {\sigma }_s=\sigma (s,X_s)-\sigma ({\eta }_h(s),{\bar{Y}}_s)\mathrm{.}$ The state Lipschitz bounds, time modulus, and $\mathrm{\mid }{X}_s-{\bar{Y}}_s\mathrm{\mid }\le \mathrm{\mid }{e}_s\mathrm{\mid }+\mathrm{\mid }{Y}_s^h-{\bar{Y}}_s\mathrm{\mid }$ give $\mathrm{\mid }\delta b_s{\mathrm{\mid }}^2+\mathrm{\parallel }\delta {\sigma }_s{\mathrm{\parallel }}_F^2\le C(\mathrm{\mid }{e}_s{\mathrm{\mid }}^2+\mathrm{\mid }{Y}_s^h-{\bar{Y}}_s{\mathrm{\mid }}^2+h(1+\mathrm{\mid }{\bar{Y}}_s{\mathrm{\mid }}^2))\mathrm{.}$ Consequently, with $v_R(t)=\mathbb{E}{\sup }_{s\le t}\mathrm{\mid }{e}_{s\wedge {\tau }_R}{\mathrm{\mid }}^2$, Tonelli [F12] and [F2] bound the expected drift-integrand error and the expected squared diffusion-integrand error, after stopping, by $C{\int }_0^t{v}_R(s)ds+Ch\mathrm{.}$

Cauchy–Schwarz for time integrals bounds the squared supremum of the stopped drift integral by $T{\int }_0^t\mathbb{E}[1_{\{s\le {\tau }_R\}}\mathrm{\mid }\delta b_s{\mathrm{\mid }}^2]ds$. For the vector stochastic integral, write each state coordinate as a sum of the $m$ scalar Brownian integrals and use $\mathrm{\mid }{\sum }_{j=1}^m{z}_j{\mathrm{\mid }}^2\le m{\sum }_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$. Applying [F3] to each scalar term yields the bound $4m\mathbb{E}{\int }_0^t{1}_{\{s\le {\tau }_R\}}\mathrm{\parallel }\delta {\sigma }_s{\mathrm{\parallel }}_F^{2}ds$. Using Tonelli [F12] and step 1.2 in both bounds gives $v_R(t)\le C{\int }_0^t{v}_R(s)ds+Ch,$ where $C$ is independent of $R$ and $h$.

The stopped error is bounded by $R$, so $v_R$ is finite and continuous; applying [F6] to step 2.1 gives $v_R(T)\le C_{T}h$ uniformly in $R$. Since $e$ is continuous on the compact interval, ${\tau }_R=T$ for all sufficiently large $R$ on each sample path. Taking $R=n\in \mathbb{N}$, the stopped suprema increase to ${\sup }_{s\le T}\mathrm{\mid }{e}_s{\mathrm{\mid }}^2$; monotone convergence [F11] yields $\mathbb{E}{\sup }_{t\le T}\mathrm{\mid }{e}_t{\mathrm{\mid }}^2\le C_{T}h$. Taking square roots gives the grid-point root-mean-square rate. AC enters only through [A1], and no arbitrary-index choice is used in the estimates.