Strong error bound for Euler–Maruyama on a predictable grid

Put $\eta (s)={\tau }_n$ and ${\bar{Y}}_s=Y_n$ on each active interval ${\tau }_n<s\le {\tau }_{n+1}$, and put ${\bar{b}}_s=b({\tau }_n,Y_n)$ and ${\bar{\sigma }}_s=\sigma ({\tau }_n,Y_n)$ there. Values at the countable grid endpoints do not affect time or It^o integrals. Each coefficient step is predictable by [F4]; the countable sum is predictable. Growth [F1], ${\sum }_{n<N}{h}_n=T$, and [F3] give $\mathbb{E}{\int }_0^T(\mathrm{\mid }{\bar{b}}_s{\mathrm{\mid }}^2+\mathrm{\parallel }{\bar{\sigma }}_s{\mathrm{\parallel }}_F^2)ds<\mathrm{\infty }$. Truncating after the first $k$ intervals gives finite-step integral identities by [F4]. Their integrands converge in $L^2(dsdP)$ to the full coefficients: the tail vanishes almost surely because $N<\mathrm{\infty }$, and the increasing partial energy sums converge to the integrable total energy by [F9], so the expected tail energy tends to zero. Isometry [F6] and Cauchy–Schwarz for the drift pass these identities to the limit. Combining with the strong-solution identity gives $e_t:=X_t-Y_t={\int }_0^t(b(s,X_s)-{\bar{b}}_s)ds+{\sum }_{j=1}^m{\int }_0^t({\sigma }_{\cdot j}(s,X_s)-{\bar{\sigma }}_{\cdot j,s})d{W}_s^j$, where ${\sigma }_{\cdot j}$ denotes column $j$.

On interval $n$, put $b_n=b({\tau }_n,Y_n)$, ${\sigma }_n=\sigma ({\tau }_n,Y_n)$, and $h_n={\tau }_{n+1}-{\tau }_n$. The interpolation increment is $Y_s-{\bar{Y}}_s=b_n(s-{\tau }_n)+{\sigma }_n(W_s-W_{{\tau }_n})$. Its drift contribution is ${\int }_{{\tau }_n}^{{\tau }_{n+1}}\mathrm{\mid }{b}_n{\mathrm{\mid }}^2(s-{\tau }_n{)}^{2}ds=\frac{1}{3}\mathrm{\mid }{b}_n{\mathrm{\mid }}^2{h}_n^3\le \frac{1}{3}{h}^2\mathrm{\mid }{b}_n{\mathrm{\mid }}^2{h}_n$. For the diffusion put $M_s^n={\sigma }_n(W_{s\wedge {\tau }_{n+1}}-W_{s\wedge {\tau }_n})$. Each entry is the stopped It^o integral of ${\sigma }_n^{ij}{1}_{({\tau }_n,{\tau }_{n+1}]}$; [F4], [F6], and [F8] imply $\mathbb{E}{\sup }_{s\le T}\mathrm{\mid }{M}_s^n{\mathrm{\mid }}^2\le 4m\mathbb{E}[\mathrm{\parallel }{\sigma }_n{\mathrm{\parallel }}_F^2{h}_n]$. Therefore $\mathbb{E}{\int }_{{\tau }_n}^{{\tau }_{n+1}}\mathrm{\mid }{\sigma }_n(W_s-W_{{\tau }_n}){\mathrm{\mid }}^{2}ds\le 4mh\mathbb{E}[\mathrm{\parallel }{\sigma }_n{\mathrm{\parallel }}_F^2{h}_n]$. Sum over $n$ using [F9], ${\sum }_{n<N}{h}_n=T$, the growth bound [F1], and moment bound [F3]: ${\int }_0^T\mathbb{E}\mathrm{\mid }{Y}_s-{\bar{Y}}_s{\mathrm{\mid }}^{2}ds\le C_{1}h$, where $C_1$ is independent of the controller and mesh; here $h^2\le Th$.

The state-Lipschitz and time-modulus bounds in [F1], together with $\mathrm{\mid }{X}_s-{\bar{Y}}_s\mathrm{\mid }\le \mathrm{\mid }{e}_s\mathrm{\mid }+\mathrm{\mid }{Y}_s-{\bar{Y}}_s\mathrm{\mid }$, imply for a constant $C_2$ fixed by the coefficient bounds, $\mathrm{\mid }b(s,X_s)-b(\eta (s),{\bar{Y}}_s){\mathrm{\mid }}^2+\mathrm{\parallel }\sigma (s,X_s)-\sigma (\eta (s),{\bar{Y}}_s){\mathrm{\parallel }}_F^2\le C_2(\mathrm{\mid }{e}_s{\mathrm{\mid }}^2+\mathrm{\mid }{Y}_s-{\bar{Y}}_s{\mathrm{\mid }}^2+(1+\mathrm{\mid }{\bar{Y}}_s{\mathrm{\mid }}^2)(s-\eta (s)))$. Since $s-\eta (s)\le h$ and [F3] bounds $\mathbb{E}{\sup }_{s\le T}\mathrm{\mid }{Y}_s{\mathrm{\mid }}^2$, the time-modulus term has integral at most $C_{3}h$.

Put $v(t)=\mathbb{E}{\sup }_{r\le t}\mathrm{\mid }{e}_r{\mathrm{\mid }}^2$. The moment bounds [F3] and [F5], with [F10], make $v$ finite and continuous: pathwise running suprema are continuous and dominated by the integrable variable $2{\sup }_{r\le T}(\mathrm{\mid }{X}_r{\mathrm{\mid }}^2+\mathrm{\mid }{Y}_r{\mathrm{\mid }}^2)$. Apply Cauchy–Schwarz to the drift in step 1.1 and [F8] to each scalar diffusion coordinate, using $\mathrm{\mid }{\sum }_{j=1}^m{z}_j{\mathrm{\mid }}^2\le m{\sum }_{j=1}^m\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$. Then $v(t)\le 2T{\int }_0^t\mathbb{E}\mathrm{\mid }b(s,X_s)-{\bar{b}}_s{\mathrm{\mid }}^{2}ds+8m{\int }_0^t\mathbb{E}\mathrm{\parallel }\sigma (s,X_s)-{\bar{\sigma }}_s{\mathrm{\parallel }}_F^{2}ds$. Insert steps 1.2 and 1.3 and use $\mathbb{E}\mathrm{\mid }{e}_s{\mathrm{\mid }}^2\le v(s)$ to obtain $v(t)\le C_4{\int }_0^{t}v(s)ds+C_{5}h$, with constants independent of the controller and mesh. Gronwall [F7] gives $v(T)\le C_5{e}^{C_{4}T}h$, hence also ${\sup }_{t\le T}\mathbb{E}\mathrm{\mid }{e}_t{\mathrm{\mid }}^2\le Ch$.

The pathwise inequality $\mathrm{\mid }{e}_{{\tau }_n}{\mathrm{\mid }}^2\le {\sup }_{t\le T}\mathrm{\mid }{e}_t{\mathrm{\mid }}^2$ and step 2.1 give $\mathbb{E}\mathrm{\mid }{e}_{{\tau }_n}{\mathrm{\mid }}^2\le Ch$ at each random grid endpoint. The constant does not depend on $N$, the lower step lengths, or the controller. The endpoint-refinement mode of [F2] does not meet the predictability hypothesis and is not covered. This proof claims no asymptotic cost law. AC is used only through [A1].