Generator and martingale problem for reversed Gaussian smoothing

The Gaussian forward process is Markov by independent increments of its deterministic Itô integral. Bayes' formula gives the reversed conditional law of $X_{T-v}$ given $X_{T-u}=y$ as $K_{u,v}(y,dx)$; the density supplier's marginal identity normalizes it, and [F3] gives reverse Chapman–Kolmogorov. Thus $Y$ is Markov with these kernels.

Put $t=T-u$ and $\delta v=v(t)-v(t-h)$ for $h>0$. For $u+h\le T-ϵ$, the reverse transition formula gives ${\displaystyle K_{u,u+h}\phi (y)=\frac{1}{p_t(y)}\int \phi (x)p_{t-h}(x){\varphi }_{\delta v}(y-x)dx\mathrm{.}}$ The heat-kernel expansion, justified on the compact support of $\phi$ by [F1, F4], is ${\displaystyle \int \phi (x)p_{t-h}(x){\varphi }_{\delta v}(y-x)dx=p_t(y)\phi (y)+\frac{\delta v}{2}(\mathrm{\Delta }(p_t\phi )(y)-\phi (y)\mathrm{\Delta }{p}_t(y))+o(\delta v)\mathrm{.}}$ Since $\delta v\mathrm{/}h\to g(t{)}^2$, the product rule gives ${\displaystyle \underset{h\downarrow 0}{\lim }\frac{K_{u,u+h}\phi (y)-\phi (y)}{h}=g(t{)}^2(\mathrm{\nabla }\log p_t(y)\cdot \mathrm{\nabla }\phi (y)+\frac{1}{2}\mathrm{\Delta }\phi (y))={\tilde{L}}_u\phi (y)\mathrm{.}}$ The remainder is locally uniform in $y$ and dominated on the compact test support because $t\ge ϵ$.

Fix $u$, put $b=T-u$, and write $P_{c}f={\varphi }_c\ast f$. For $r>u$, put $a=T-r$. The explicit kernel gives $K_{u,r}\phi (y)=P_{v(b)-v(a)}(p_a\phi )(y)\mathrm{/}{p}_b(y)$. Differentiate in $r$, using the heat equation and integrating the Gaussian Laplacian onto the compactly supported function $p_a\phi$, to obtain ${\displaystyle {\mathrm{\partial }}_r{K}_{u,r}\phi (y)=\frac{g(a{)}^2}{2p_b(y)}{P}_{v(b)-v(a)}\text{\hspace{0.17em}⁣}(\mathrm{\Delta }(p_a\phi )-\phi \mathrm{\Delta }{p}_a)(y)=K_{u,r}({\tilde{L}}_r\phi )(y)\mathrm{.}}$ Away from $r=u$, differentiation is justified by [F4]: the variance is positive, the spatial support is fixed compact, and $p_a,{\mathrm{\partial }}_a{p}_a$ and the needed state derivatives are uniformly bounded there for $a\ge ϵ$. The product rule gives the last equality. By [F2], $\mathrm{\mid }{\tilde{L}}_r\phi (x)\mathrm{\mid }\le C$ uniformly in $x,r$ on the truncated interval, so the probability-kernel property bounds the derivative by $C$. Integrate from $u+\delta$ to $v$ and let $\delta \downarrow 0$: step 1.2 gives $K_{u,u+\delta }\phi (y)\to \phi (y)$, while the bound $C$ controls the omitted integral. Thus $K_{u,v}\phi -\phi ={\int }_u^v{K}_{u,r}({\tilde{L}}_r\phi )dr$. For $s<v$, Markovness and bounded conditional Fubini give ${\displaystyle E\text{\hspace{0.17em}⁣}[\phi (Y_v)-\phi (Y_s)-{\int }_s^v{\tilde{L}}_r\phi (Y_r)dr\mid {\mathcal{G}}_s]=K_{s,v}\phi (Y_s)-\phi (Y_s)-{\int }_s^v{K}_{s,r}({\tilde{L}}_r\phi )(Y_s)dr=0.}$ This proves the martingale claim.