Euler–Maruyama uniform moments and interpolation increments

The recursion starts from $X_0\in L^p$. If $Y_k$ is ${\mathcal{F}}_{t_k}$-measurable and in $L^p$, the one-step estimate [L2] applies with $U=Y_k$ and $\delta =h$, so $Y_{k+1}$ is in $L^p$. Since $W_{t_{k+1}}-W_{t_k}$ is measurable at $t_{k+1}$ and independent of ${\mathcal{F}}_{t_k}$, the recursion also makes $Y_{k+1}$ ${\mathcal{F}}_{t_{k+1}}$-measurable. Induction gives adapted grid values with finite $p$th moments.

In particular the grid values are square-integrable. On each interval, ${\bar{Y}}_s^{-}=Y_k$ and the coefficient vectors $b(t_k,Y_k)$ and ${\sigma }_{\cdot j}(t_k,Y_k)$ are ${\mathcal{F}}_{t_k}$-measurable. The growth bound gives $E{\int }_0^T(\mathrm{\mid }b({\eta }_h^{-}(s),{\bar{Y}}_s^{-}){\mathrm{\mid }}^2+\mathrm{\parallel }\sigma ({\eta }_h^{-}(s),{\bar{Y}}_s^{-}){\mathrm{\parallel }}_F^2)ds\le K^2{\sum }_{k=0}^{N-1}h(1+E\mathrm{\mid }{Y}_k{\mathrm{\mid }}^2)<\mathrm{\infty }\mathrm{.}$ Each scalar coordinate integrand is therefore predictable and square-integrable, although its coefficients need not be bounded. By [F3], its Itô integral has the finite-sum version; the ordinary drift integral is the corresponding finite sum as well. These are exactly the recursion and interpolation formulas in [L1]. For each $t\in [t_k,t_{k+1}]$, $Y_k$ is ${\mathcal{F}}_t$-measurable and $W_t-W_{t_k}$ is ${\mathcal{F}}_t$-measurable, so the interpolation is adapted. Brownian path continuity and the endpoint recursion make it continuous.

Put $u_k=\mathbb{E}\mathrm{\mid }{Y}_k{\mathrm{\mid }}^p$. The one-step estimate [L2] gives $u_{k+1}\le (1+C_{p}h)u_k+C_{p}h$, hence $1+u_{k+1}\le (1+C_{p}h)(1+u_k)$. Iteration and [F2] yield $1+u_k\le (1+C_{p}h{)}^k(1+u_0)\le \exp (C_{p}kh)(1+\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^p)\le \exp (C_{p}T)(1+\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^p)$. This bound is uniform in $k$ and $h$.

For $t\in [t_k,t_{k+1})$, apply [L2]'s first one-step estimate with $U=Y_k$ and $\delta =t-t_k\le h$ to obtain $\mathbb{E}\mathrm{\mid }{Y}_t{\mathrm{\mid }}^p\le (1+C_p\delta )\mathbb{E}\mathrm{\mid }{Y}_k{\mathrm{\mid }}^p+C_p\delta$. The step process equals $Y_k$ there. The grid estimate from step 2.2 therefore gives the two asserted uniform pointwise moment bounds; at $T$, $Y_T={\bar{Y}}_T=Y_N$ is covered by the grid bound.

With $p=2$, [L2]'s second one-step estimate gives $\mathbb{E}(\mathrm{\mid }{Y}_t-{\bar{Y}}_t{\mathrm{\mid }}^2\mid {\mathcal{F}}_{t_k})\le C_{2,m,K}(t-t_k)(1+\mathrm{\mid }{Y}_k{\mathrm{\mid }}^2)$ on each grid interval. Taking expectations, using step 2.2, and using $t-t_k\le h$ proves $\mathbb{E}\mathrm{\mid }{Y}_t-{\bar{Y}}_t{\mathrm{\mid }}^2\le C_{T,K,d,m}h(1+\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)$. At grid endpoints the difference is zero by [L1].

The induction and estimates use only finite Brownian increments and the explicitly given hypotheses; [A1] is inherited from the interfaces, not spent as a choice principle. A conclusion for a larger $p$ requires $X_0\in L^p$; it does not follow from $X_0\in L^2$ alone.