Euler–Maruyama strong rate under Higham–Mao–Stuart hypotheses

Induction in the recursion shows that $Y_k^h$ is ${\mathcal{F}}_{t_k}$-measurable: this holds for $Y_0=X_0$, and each update uses ${\mathcal{F}}_{t_k}$-measurable coefficients and a Brownian increment measurable at $t_{k+1}$. The interpolation is therefore adapted and continuous. For integration use the predictable step representative ${\widehat{Y}}_0^h=Y_0$ and ${\widehat{Y}}_s^h=Y_k^h$ on $(t_k,t_{k+1}]$; its coefficient processes $f({\widehat{Y}}^h)$ and $g({\widehat{Y}}^h)$ are predictable because each $Y_k^h$ is ${\mathcal{F}}_{t_k}$-measurable and $f,g$ are continuous. It agrees with ${\bar{Y}}^h$ outside the finite set of grid times. The recursion and interpolation give the same integral equation with these predictable coefficients. Setting $y=0$ in the coefficient inequalities yields $\mathrm{\mid }f(x){\mathrm{\mid }}^2\le 2\mathrm{\mid }f(0){\mathrm{\mid }}^2+2D(1+\mathrm{\mid }x{\mathrm{\mid }}^q)\mathrm{\mid }x{\mathrm{\mid }}^2$ and $\mathrm{\parallel }g(x){\mathrm{\parallel }}_F^2\le 2\mathrm{\parallel }g(0){\mathrm{\parallel }}_F^2+2c\mathrm{\mid }x{\mathrm{\mid }}^2$. The all-moment premise thus gives finite expected time-integral energy for these Euler coefficients; it gives the same for $f(X)$ and $g(X)$. The definition [F7] then makes the Euler interpolation and exact solution continuous Brownian Itô processes, and its linear-combination clause makes their difference $e$ one as well. In Itô's formula for $\mathrm{\mid }{e}_t{\mathrm{\mid }}^2$, the drift is $2⟨e_t,f(X_t)-f({\widehat{Y}}_t^h)⟩+\mathrm{\parallel }{G}_t{\mathrm{\parallel }}_F^2$ and the martingale term is $M_t=2{\sum }_{j=1}^m{\int }_0^t⟨e_s,G_{s,\cdot j}⟩d{W}_s^j$, where $G_s=g(X_s)-g({\widehat{Y}}_s^h)$. Since $e$ is continuous adapted and $G$ is predictable, each martingale integrand is predictable. Its finite expected energy follows from ${\sum }_j\mathrm{\mid }2⟨e_s,G_{s,\cdot j}⟩{\mathrm{\mid }}^2\le 4\mathrm{\mid }{e}_s{\mathrm{\mid }}^2\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^2$ and $\mathbb{E}{\int }_0^T\mathrm{\mid }{e}_s{\mathrm{\mid }}^2\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^{2}ds\le T(\mathbb{E}{\sup }_{s\le T}\mathrm{\mid }{e}_s{\mathrm{\mid }}^4{)}^{1\mathrm{/}2}(\mathbb{E}{\sup }_{s\le T}\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^4{)}^{1\mathrm{/}2}<\mathrm{\infty }$ by the all-moment and coefficient-growth premises.

Split the drift difference at $Y_t^h$. The one-sided Lipschitz bound and $2ab\le a^2+b^2$ give $2⟨e,f(X)-f(\widehat{Y})⟩\le (2\mu +1)\mathrm{\mid }e{\mathrm{\mid }}^2+\mathrm{\mid }f(Y)-f(\widehat{Y}){\mathrm{\mid }}^2$ almost everywhere in time. The polynomial local-Lipschitz condition bounds the final term by $D(1+\mathrm{\mid }Y{\mathrm{\mid }}^q+\mathrm{\mid }\widehat{Y}{\mathrm{\mid }}^q)\mathrm{\mid }Y-\widehat{Y}{\mathrm{\mid }}^2$. Also, splitting $g(X)-g(\widehat{Y})$ at $g(Y)$ and using its global Lipschitz bound gives $\mathrm{\parallel }G{\mathrm{\parallel }}_F^2\le 2c\mathrm{\mid }e{\mathrm{\mid }}^2+2c\mathrm{\mid }Y-\widehat{Y}{\mathrm{\mid }}^2$. By [F2] with $r=2,4$, the all-moment premise, and [F5], $\mathbb{E}\mathrm{\mid }{Y}_s-{\widehat{Y}}_s{\mathrm{\mid }}^2\le Ch$ and $\mathbb{E}[(1+\mathrm{\mid }{Y}_s{\mathrm{\mid }}^q+\mathrm{\mid }{\widehat{Y}}_s{\mathrm{\mid }}^q)\mathrm{\mid }{Y}_s-{\widehat{Y}}_s{\mathrm{\mid }}^2]\le Ch$ for almost every $s$, uniformly in $h$. Tonelli [F8] then bounds the expected integrated drift by $C{\int }_0^t\mathbb{E}\mathrm{\mid }{e}_s{\mathrm{\mid }}^{2}ds+Ch$.

Apply [F4] to the vector integrand $K_s^j=2⟨e_s,G_{s,\cdot j}⟩$. Since ${\sum }_j\mathrm{\mid }{K}_s^j{\mathrm{\mid }}^2\le 4\mathrm{\mid }{e}_s{\mathrm{\mid }}^2\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^2$, $\mathbb{E}{\sup }_{s\le t}\mathrm{\mid }{M}_s\mathrm{\mid }\le 32\sqrt{m}\mathbb{E}\left[{\sup }_{s\le t}\mathrm{\mid }{e}_s\mathrm{\mid }{\left({\int }_0^t\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^{2}ds\right)}^{1\mathrm{/}2}\right]\le 32\sqrt{m}{(v(t)\mathbb{E}{\int }_0^t\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^{2}ds)}^{1\mathrm{/}2},$ where $v(t)=\mathbb{E}{\sup }_{s\le t}\mathrm{\mid }{e}_s{\mathrm{\mid }}^2$. From [F2], Tonelli [F8], and the Lipschitz estimate for $g$, $\mathbb{E}{\int }_0^t\mathrm{\parallel }{G}_s{\mathrm{\parallel }}_F^{2}ds\le C{\int }_0^t\mathbb{E}\mathrm{\mid }{e}_s{\mathrm{\mid }}^{2}ds+Ch\mathrm{.}$ Young's inequality therefore bounds the martingale term by $\frac{1}{2}v(t)+C{\int }_0^{t}v(s)ds+Ch$.

Take the supremum over times up to $t$ in Itô's identity from step 1.1, take expectations, and insert the drift estimate from step 1.2 and martingale estimate from step 1.3. Moving $\frac{1}{2}v(t)$ to the left gives $v(t)\le C{\int }_0^{t}v(s)ds+Ch\mathrm{.}$ The all-moment assumption makes $v$ finite; pathwise continuity of $e$ and dominated convergence make $v$ continuous. Applying [F6] with the nondecreasing constant term $Ch$ yields $v(T)\le C^{\mathrm{\prime }}h$. The constants depend on $T,d,m,\mu ,c,D,q$ and the all-moment bounds, but not on $h$.

Since $Y_T^h=Y_N^h$, the grid endpoint is included in the supremum bound. Taking square roots gives the stated root-mean-square order. AC is inherited exactly through [A1]; no choice principle is used in the numerical estimates.