Reverse Gaussian smoothing on a truncated time interval

By [F1], the reversed Gaussian kernels are normalized and compose, and [F2] gives the reverse martingale problem for $Y_u=X_{T-u}$ on $0\le u\le T-ϵ$. In particular the reverse generator has drift ${\beta }_u(x)=g(T-u{)}^2\mathrm{\nabla }\log p_{T-u}(x)$ and diffusion matrix $a_u(x)=g(T-u{)}^2{I}_d$.

First pass to the usual augmentation of the reversed filtration and normalize the continuous processes on their common null exceptional set in its initial sigma-algebra. The continuous compact-test martingales in [F2] remain martingales there: on the finite horizon each is a conditional expectation of its integrable terminal value, hence uniformly integrable; passing its event identities along times decreasing to a fixed time gives the right-continuous extension, and completion adds only null-event identities. All stopping and product-formula uses below are on these normalized representatives. Fix $R_1$ and stop when $\mathrm{\mid }Y\mathrm{\mid }$ reaches $R_1$. Compactly supported cutoffs equal to $x_i$ and $x_i{x}_j$ on that ball turn the identities in [F2] into coordinate and coordinate-product martingales. Subtracting the finite-variation drift gives continuous local martingales $M^i$. The product formula in [F4] makes $Y_i{Y}_j-Y_i(0)Y_j(0)-\int ({\beta }_i{Y}_j+{\beta }_j{Y}_i)dr-[M_i,M_j]$ a local martingale. The coordinate-product identity makes the same expression with $\int g(T-r{)}^2{\delta }_{ij}dr$ in place of $[M_i,M_j]$ a local martingale. Their difference is a continuous finite-variation local martingale, hence constant by [F4]; it starts at zero. Thus $[M{]}_u={\int }_0^{u}g(T-r{)}^2{I}_{d}dr$. Letting $R_1\uparrow \mathrm{\infty }$ is valid because $Y$ has continuous paths on the compact time interval.

The process ${\sigma }_r=g(T-r)I_d$ is invertible with bounded inverse. Apply the inverse-diffusion construction in [F4] to define $B_u={\int }_0^{u}g(T-r{)}^{-1}d{M}_r$. Its bracket is $u{I}_d$, so vector Lévy characterization makes $B$ Brownian in the usual augmented reversed filtration in which step 2.1 was carried out. The decomposition of $M$ gives the displayed reverse SDE.

By [F3], ${\beta }_u$ is globally Lipschitz in $x$ uniformly on the truncated interval, with constant at most $g_{\max }^{}(1\mathrm{/}v(ϵ)+R^2\mathrm{/}v(ϵ{)}^2)$; its linear-growth bound follows by multiplying the supplied score bound by $g_{\max }^{}$. The diffusion coefficient $g(T-u)I_d$ is deterministic and bounded. The initial value $Y_0=X_T$ has finite second moment since $X_0$ is bounded and the Gaussian increment has finite second moment. Thus [F6] gives a unique strong solution for the same initial value and Brownian motion $B$ on the usual augmented reversed filtration. Pathwise uniqueness identifies this solution with the weak solution $Y$ from step 3.1.

Since $Y_{T-ϵ}=X_{ϵ}$, the endpoint coupling statement [F5] gives the exact mean-square error and $W_2$ bound. Because $v(ϵ)\le g_{\max }^{}ϵ\to 0$, the coupling proves $L^2$ convergence to $X_0$; the same supplier gives weak convergence of the laws to $\mu$. No part of this step evaluates a score at zero.