Euler–Maruyama weak order one under backward regularity

By [F6], the exact solution has the continuous Itô decomposition; apply [F5] to $u(t,X_t)$. The backward equation cancels its drift. Moreover, $\mathrm{\mid }\mathrm{\nabla }u(t,x)\mathrm{\mid }\le C(1+\mathrm{\mid }x{\mathrm{\mid }}^q)$ and linear growth of $\sigma$ give $\mathbb{E}{\int }_0^T\mathrm{\parallel }\mathrm{\nabla }u(s,X_s)\sigma (s,X_s){\mathrm{\parallel }}_F^{2}ds<\mathrm{\infty }$ by Tonelli [F10], because $p\ge 2q+8$ and [F8] supplies the exact process moments. Thus the stochastic integral is a square-integrable mean-zero martingale by [F6]. With $u(T,\cdot )=\phi$, this proves $\mathbb{E}\phi (X_T)=\mathbb{E}u(0,X_0)\mathrm{.}$

Fix $t<T$, $x\in {\mathbb{R}}^d$, and $0<h\le \min (1,T-t)$; set $\mathrm{\Delta }W\sim N_m(0,h{I}_m)$ and $\delta =b(t,x)h+\sigma (t,x)\mathrm{\Delta }W$. Taylor's formula along the spatial segment gives $u(t,x+\delta )={\sum }_{k=0}^3\frac{1}{k!}{D}_x^{k}u(t,x)[{\delta }^{\otimes k}]+R_4,$ $u_t(t,x+\delta )={\sum }_{k=0}^2\frac{1}{k!}{D}_x^k{u}_t(t,x)[{\delta }^{\otimes k}]+R_3^t,$ where $D_x^{k}u[z^{\otimes k}]$ means the $k$th spatial derivative applied to $z$ in all $k$ slots. The integral remainders are $R_4=\frac{1}{3!}{\int }_0^1(1-r{)}^3{D}_x^{4}u(t,x+r\delta )[{\delta }^{\otimes 4}]dr,$ $R_3^t=\frac{1}{2!}{\int }_0^1(1-r{)}^2{D}_x^3{u}_t(t,x+r\delta )[{\delta }^{\otimes 3}]dr\mathrm{.}$ Time Taylor expansion gives $u(t+h,x+\delta )=u(t,x+\delta )+h{u}_t(t,x+\delta )+R_{tt},R_{tt}=h^2{\int }_0^1(1-r)u_{tt}(t+rh,x+\delta )dr\mathrm{.}$ The derivative bounds in the Statement and $\mathrm{\mid }x+r\delta {\mathrm{\mid }}^q\le C_q(\mathrm{\mid }x{\mathrm{\mid }}^q+\mathrm{\mid }\delta {\mathrm{\mid }}^q)$ imply $\mathrm{\mid }{R}_4\mathrm{\mid }\le C(1+\mathrm{\mid }x{\mathrm{\mid }}^q+\mathrm{\mid }\delta {\mathrm{\mid }}^q)\mathrm{\mid }\delta {\mathrm{\mid }}^4,\mathrm{\mid }{R}_3^t\mathrm{\mid }\le C(1+\mathrm{\mid }x{\mathrm{\mid }}^q+\mathrm{\mid }\delta {\mathrm{\mid }}^q)\mathrm{\mid }\delta {\mathrm{\mid }}^3,\mathrm{\mid }{R}_{tt}\mathrm{\mid }\le C{h}^2(1+\mathrm{\mid }x+\delta {\mathrm{\mid }}^q)\mathrm{.}$ If $q=0$, interpret $1+\mathrm{\mid }x{\mathrm{\mid }}^q$ as the constant $2$.

For each finite real $s>0$, choose an integer $n\ge 1$ with $2n\ge s$. If $Z$ is scalar standard normal, $\mathrm{\mid }Z{\mathrm{\mid }}^s\le 1+\mathrm{\mid }Z{\mathrm{\mid }}^{2n}$, so [F4] gives $E\mathrm{\mid }Z{\mathrm{\mid }}^s\le 1+(2n-1)!!<\mathrm{\infty }$. For a standard $m$-dimensional Gaussian vector $G$, $\mathrm{\mid }G\mathrm{\mid }\le \sqrt{m}{\max }_j\mathrm{\mid }{G}_j\mathrm{\mid }$ gives $\mathrm{\mid }G{\mathrm{\mid }}^s\le m^{s\mathrm{/}2}{\sum }_{j=1}^m\mathrm{\mid }{G}_j{\mathrm{\mid }}^s$, and hence $E\mathrm{\mid }G{\mathrm{\mid }}^s\le m^{1+s\mathrm{/}2}\{1+(2n-1)!!\}$. By the Gaussian scaling convention in [F3] and [F9], $\mathrm{\Delta }W$ has law $\sqrt{h}G$, so $E\mathrm{\mid }\mathrm{\Delta }W{\mathrm{\mid }}^s\le C_{s,m}{h}^{s\mathrm{/}2}$. This also covers $h=0$ when $s>0$; for $s=0$ adopt $\mathrm{\mid }x{\mathrm{\mid }}^0=1$, and its expectation is one. No initial-state moment stronger than the Statement is used in this Gaussian calculation.

Put $a(t,x)=\sigma (t,x)\sigma (t,x{)}^{\mathsf{T}}$. The coefficient growth bounds and [F3]–[F4], [F9] give $\mathbb{E}\delta =b(t,x)h,\mathbb{E}(\delta {\delta }^{\mathsf{T}})=a(t,x)h+h^{2}b(t,x)b(t,x{)}^{\mathsf{T}},$ $\mathrm{\parallel }\mathbb{E}({\delta }^{\otimes 3})\mathrm{\parallel }\le C{h}^2(1+\mathrm{\mid }x{\mathrm{\mid }}^3),\mathbb{E}\mathrm{\mid }\delta {\mathrm{\mid }}^s\le C_s{h}^{s\mathrm{/}2}(1+\mathrm{\mid }x{\mathrm{\mid }}^s)$ for each finite exponent $s>0$ used in the Taylor remainders. Indeed, $\mathrm{\mid }b(t,x)\mathrm{\mid }+\mathrm{\parallel }\sigma (t,x){\mathrm{\parallel }}_F\le C(1+\mathrm{\mid }x\mathrm{\mid })$ and $\mathrm{\mid }\delta \mathrm{\mid }\le h\mathrm{\mid }b(t,x)\mathrm{\mid }+\mathrm{\parallel }\sigma (t,x){\mathrm{\parallel }}_F\mathrm{\mid }\mathrm{\Delta }W\mathrm{\mid }$; [F9] and $h\le 1$ give the stated bound. When $q=0$, its $\mathrm{\mid }\delta {\mathrm{\mid }}^q$ factor is the constant $1$. For the third tensor moment, write $\delta =hb+Z$ with $Z=\sigma (t,x)\mathrm{\Delta }W$. The centered Gaussian $Z$ has covariance $ha(t,x)$ and zero third moment, so expanding the tensor cube leaves three terms with one $hb$ and two $Z$ factors, plus $(hb{)}^{\otimes 3}$; each has expectation $O(h^2(1+\mathrm{\mid }x{\mathrm{\mid }}^3))$. Substituting these moments in step 1.2, the first-order terms combine to $h({\mathrm{\partial }}_{t}u+b\cdot \mathrm{\nabla }u+\frac{1}{2}a:D_x^{2}u)(t,x)=0$. Every other Taylor term is bounded by $C{h}^2(1+\mathrm{\mid }x{\mathrm{\mid }}^{q+4})$: the fourth spatial remainder uses $\mathbb{E}\mathrm{\mid }\delta {\mathrm{\mid }}^4(1+\mathrm{\mid }x{\mathrm{\mid }}^q+\mathrm{\mid }\delta {\mathrm{\mid }}^q)$, the time-space terms use $\mathbb{E}\delta$ and $\mathbb{E}\delta {\delta }^{\mathsf{T}}$, and the $u_t$ third-order remainder is $O(h^{5\mathrm{/}2})(1+\mathrm{\mid }x{\mathrm{\mid }}^{q+3})\le C{h}^2(1+\mathrm{\mid }x{\mathrm{\mid }}^{q+4})$. Therefore $\mid \mathbb{E}u(t+h,x+\delta )-u(t,x)\mid \le C{h}^2(1+\mathrm{\mid }x{\mathrm{\mid }}^{q+4})\mathrm{.}$

Condition on ${\mathcal{F}}_{t_k}$. By [F1], $Y_{k+1}^h$ is the one-step update in step 2.1 with $t=t_k$ and $x=Y_k^h$; [F3] makes the Brownian increment independent of ${\mathcal{F}}_{t_k}$. Thus $\mid \mathbb{E}[u(t_{k+1},Y_{k+1}^h)-u(t_k,Y_k^h)\mid {\mathcal{F}}_{t_k}]\mid \le C{h}^2(1+\mathrm{\mid }{Y}_k^h{\mathrm{\mid }}^{q+4})\mathrm{.}$ Since $p\ge 2q+8\ge q+4$, [F2] and the exact pointwise moment premise make the expectation of the right side at most $C^{\mathrm{\prime }}{h}^2$, uniformly in $k,h$.

Sum the conditional one-step errors over $k=0,\dots ,N-1$. The sum telescopes to $\mid \mathbb{E}u(T,Y_N^h)-\mathbb{E}u(0,Y_0^h)\mid \le N{C}^{\mathrm{\prime }}{h}^2=C^{\mathrm{\prime }}Th\mathrm{.}$ Here $u(T,Y_N^h)=\phi (Y_N^h)$ and $Y_0^h=X_0$; step 1.1 gives $\mathbb{E}u(0,X_0)=\mathbb{E}\phi (X_T)$. Combining the equalities proves the desired weak error bound. AC enters only through [A1].