Uniform second moment for Euler–Maruyama on predictable grids

For $R>0$ define ${\rho }_R=\inf \{t\le T:\mathrm{\mid }{Y}_t\mathrm{\mid }\ge R\}\wedge T$, taking ${\rho }_R=T$ if the set is empty. For an integer $k\ge 1$, set ${\theta }_{R,k}={\rho }_R\wedge {\tau }_k$ and $Y_t^{R,k}=Y_{t\wedge {\theta }_{R,k}}$. Continuity, adaptedness and [F8] make ${\rho }_R$ a stopping time, so ${\theta }_{R,k}$ is a stopping time. For the first $k$ intervals, multiply each left-endpoint coefficient by $1_{\{\mathrm{\mid }{Y}_n\mathrm{\mid }\le \mathrm{\mid }{X}_0\mathrm{\mid }\vee R\}}$. These factors are measurable at ${\mathcal{F}}_{{\tau }_n}$, the truncated coefficients are in $L^2$ by [F5], and they agree with the original coefficients before ${\theta }_{R,k}$, except at time endpoints which do not affect the integrals. Thus [F3] gives their finite-step integral representation, to which [F2] applies. Before ${\theta }_{R,k}$, every active left-endpoint value is bounded by $\mathrm{\mid }{X}_0\mathrm{\mid }\vee R$; [F5] and $X_0\in L^2$ therefore give finite expected coefficient energy. Applying [F2] at ${\theta }_{R,k}$ to the finite-step integral representation yields $Y_t^{R,k}=X_0+{\int }_0^t{1}_{[0,{\theta }_{R,k}]}(s){\bar{b}}_{s}ds+{\int }_0^t{1}_{[0,{\theta }_{R,k}]}(s){\bar{\sigma }}_{s}d{W}_s$, where ${\bar{Y}}_s=Y_n$, ${\bar{b}}_s=b({\tau }_n,Y_n)$ and ${\bar{\sigma }}_s=\sigma ({\tau }_n,Y_n)$ on active intervals, with arbitrary values at $s=0$. Also ${\sup }_{u\le t}\mathrm{\mid }{Y}_u^{R,k}\mathrm{\mid }\le \mathrm{\mid }{X}_0\mathrm{\mid }\vee R$, so $U_{R,k}(t):=\mathbb{E}{\sup }_{u\le t}\mathrm{\mid }{Y}_u^{R,k}{\mathrm{\mid }}^2$ is finite and continuous by [F9], since the path suprema converge pointwise with $t$ and are bounded by the integrable variable $(\mathrm{\mid }{X}_0\mathrm{\mid }\vee R{)}^2$.

Write each coordinate of the stopped martingale as $M^i={\sum }_{j=1}^m{M}^{ij}$, with $M_t^{ij}={\int }_0^t{1}_{[0,{\theta }_{R,k}]}(s){\bar{\sigma }}_s^{ij}d{W}_s^j$. For every $i,j$, [F4] applies to the scalar Brownian coordinate $W^j$. The inequalities ${\sup }_t\mathrm{\mid }{\sum }_j{M}_t^{ij}{\mathrm{\mid }}^2\le m{\sum }_j{\sup }_t\mathrm{\mid }{M}_t^{ij}{\mathrm{\mid }}^2$ and $\mathrm{\mid }a+b+c{\mathrm{\mid }}^2\le 3(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2)$ give the martingale bound $4m\mathbb{E}{\int }_0^t{1}_{[0,{\theta }_{R,k}]}\mathrm{\parallel }{\bar{\sigma }}_s{\mathrm{\parallel }}_F^{2}ds$. Cauchy–Schwarz bounds the squared drift supremum by $T{\int }_0^t{1}_{[0,{\theta }_{R,k}]}\mathrm{\mid }{\bar{b}}_s{\mathrm{\mid }}^{2}ds$. On an active interval with $s\le {\theta }_{R,k}$, its left endpoint satisfies ${\tau }_n<s\le {\theta }_{R,k}$, so ${\bar{Y}}_s=Y_n=Y_{{\tau }_n}^{R,k}$ and $\mathrm{\mid }{\bar{Y}}_s{\mathrm{\mid }}^2\le {\sup }_{u\le s}\mathrm{\mid }{Y}_u^{R,k}{\mathrm{\mid }}^2$. Outside this range the stopped-integral indicator is zero; using [F5] in the representation of step 1.1 yields $U_{R,k}(t)\le 3\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+(3T+12m)K^2{\int }_0^t(1+U_{R,k}(s))ds$.

Apply [F6] to $1+U_{R,k}$ with $A=1+3\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2$ and $C=(3T+12m)K^2$. Then $U_{R,k}(T)\le (1+3\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)e^{(3T+12m)K^{2}T}-1$, uniformly in $R,k$.

Since $N<\mathrm{\infty }$ almost surely, for almost every path ${\tau }_k=T$ for all sufficiently large $k$. Thus $Y^{R,k}$ eventually equals $Y^{{\rho }_R}$ uniformly on $[0,T]$; [F7] passes the bound of step 3.1 to $\mathbb{E}{\sup }_{t\le T}\mathrm{\mid }{Y}_{t\wedge {\rho }_R}{\mathrm{\mid }}^2$. Each continuous path $Y$ is bounded on $[0,T]$, so $Y^{{\rho }_R}\to Y$ uniformly almost surely as $R\to \mathrm{\infty }$, and [F7] gives the stated bound. The cases $X_0=0$ or $K=0$ are covered by the same inequality, and no lower mesh bound or deterministic bound on $N$ is used. AC is inherited only through [A1].