Second and fourth moments of bounded-coefficient SDE increments

Put $Y_s=X_s-X_t$, and let $B$ bound $\mathrm{\mid }b\mathrm{\mid }$ and $S^2$ bound ${\sum }_j\mathrm{\mid }{\sigma }_j{\mathrm{\mid }}^2$. The integral equation, $\mathrm{\mid }{\int }_t^{t+r}b(X_s)ds{\mathrm{\mid }}^2\le B^2{r}^2$, the finite-sum inequality $\mathrm{\mid }{\sum }_j{v}_j{\mathrm{\mid }}^2\le m{\sum }_j\mathrm{\mid }{v}_j{\mathrm{\mid }}^2$, and [F2] give $\mathbb{E}[{\mathbf{1}}_A\mathrm{\mid }{Y}_{t+r}{\mathrm{\mid }}^2]\le 2B^2{r}^{2}P(A)+2S^{2}rP(A)\le CrP(A),$ since $r\le 1$. The event-weighted identity applies to each integrand ${\sigma }_j(X_s){\mathbf{1}}_{(t,t+r]}(s)$, whose squared norm is bounded by $\mathrm{\parallel }{\sigma }_j{\mathrm{\parallel }}_{\mathrm{\infty }}^2$.

For $R>0$, use the continuous adapted version from [F1], modifying it on its exceptional null set if necessary; usual completeness makes this an adapted null modification and leaves all moment and SDE assertions unchanged. Let ${\theta }_R=\inf \{u\ge 0:\mathrm{\mid }{Y}_{t+u}\mathrm{\mid }\ge R\}$. This is a stopping time: for fixed $u$, $\{{\theta }_R\le u\}=\{{\sup }_{0\le v\le u}\mathrm{\mid }{Y}_{t+v}\mathrm{\mid }\ge R\}$, and continuity makes the supremum equal to the supremum over rational $v\in [0,u]$ together with $u$, hence ${\mathcal{F}}_{t+u}$-measurable. Put $Y_u^R=X_{t+(u\wedge {\theta }_R)}-X_t=Y_{t+(u\wedge {\theta }_R)}$, $q_R(u)=\mathbb{E}[{\mathbf{1}}_A\mathrm{\mid }{Y}_u^R{\mathrm{\mid }}^4]$, and $p_R(u)=\mathbb{E}[{\mathbf{1}}_A\mathrm{\mid }{Y}_u^R{\mathrm{\mid }}^2]$. By the stopped representation in [F1], the drift and diffusion integrands are multiplied by ${\mathbf{1}}_{s-t\le {\theta }_R}$. Its drift square is at most $B^2{u}^2$, and applying [F2] coordinatewise with the finite-sum bound gives $p_R(u)\le 2B^2{u}^{2}P(A)+2S^{2}uP(A)\le CuP(A)$, by the same bound as step 1.1. Apply [F3] to $f(y)=\mathrm{\mid }y{\mathrm{\mid }}^4$ and the stopped process. The Hessian formula gives drift $4\mathrm{\mid }{Y}^R{\mathrm{\mid }}^2⟨Y^R,b(X)⟩+2\mathrm{\mid }{Y}^R{\mathrm{\mid }}^2{\sum }_j\mathrm{\mid }{\sigma }_j(X){\mathrm{\mid }}^2+4{\sum }_j⟨Y^R,{\sigma }_j(X){⟩}^2$ before the stopping time. Bounded coefficients and $4B\mathrm{\mid }{Y}^R{\mathrm{\mid }}^3\le C(\mathrm{\mid }{Y}^R{\mathrm{\mid }}^4+\mathrm{\mid }{Y}^R{\mathrm{\mid }}^2)$ bound its absolute value by $C(\mathrm{\mid }{Y}^R{\mathrm{\mid }}^4+\mathrm{\mid }{Y}^R{\mathrm{\mid }}^2)$. The stopped stochastic integrand is bounded, so [F4] makes its expectation after multiplication by ${\mathbf{1}}_A$ zero. Consequently $q_R(u)\le C{\int }_0^u{q}_R(v)dv+C{u}^{2}P(A),0\le u\le r\mathrm{.}$ The second term follows from $p_R(v)\le CvP(A)$; all constants are independent of $R$.

Define $Q_R(u)={\int }_0^u{q}_R(v)dv$. The inequality in step 2.1 implies $Q_R^{\mathrm{\prime }}(u)\le C{Q}_R(u)+C{u}^{2}P(A)$ for almost every $u$, with $Q_R(0)=0$. Multiplying by $e^{-Cu}$ and integrating gives $Q_R(u)\le C^{\mathrm{\prime }}{u}^{3}P(A)$ for $u\le 1$. Returning to step 2.1 gives $q_R(u)\le C^{\mathrm{\prime }\mathrm{\prime }}{u}^{2}P(A)$. As $R\to \mathrm{\infty }$, path continuity implies $Y_u^R\to Y_{t+u}$ almost surely; [F5] therefore yields the asserted fourth-moment bound.

Steps 1.1 and 3.1 are exactly the asserted event-weighted second- and fourth-moment estimates, with constants independent of $t,r,A$. The stopped argument and finite-dimensional It^o calculation add no axiom beyond [A1].