State-dependent diffusion needs a divergence correction

Define ${\displaystyle \psi (y)=\{\begin{array}{ll}\exp \text{\hspace{0.17em}⁣}(-\frac{1}{1-y^2}),& \mathrm{\mid }y\mathrm{\mid }<1,\\ 0,& \mathrm{\mid }y\mathrm{\mid }\ge 1.\end{array}}$ Every derivative extends continuously by zero at $y=\pm 1$ because $u^{-m}{e}^{-1\mathrm{/}u}\to 0$ as $u\downarrow 0$ for every integer $m\ge 0$; hence $\psi \in C_c^{\mathrm{\infty }}(\mathbb{R})$. Its first and second derivatives are bounded, and ${\psi }^{\mathrm{\prime }\mathrm{\prime }}$ is not identically zero, since a compactly supported function with zero second derivative would be affine and therefore zero. Choose $\alpha >0$ with $\alpha \mathrm{\parallel }{\psi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}2$ and set $h(y)=y+\alpha \psi (y)$. Then $h^{\mathrm{\prime }}(y)=1+\alpha {\psi }^{\mathrm{\prime }}(y)>1\mathrm{/}2$, $h$ equals the identity outside $[-1,1]$, and it is strictly increasing and onto; the inverse function theorem makes it a smooth diffeomorphism.

Set $X_t=h(W_t)$ and $z=h^{-1}(x)$. By Itô's formula [F1], ${\displaystyle d{X}_t=b(X_t)dt+\sigma (X_t)d{W}_t,b(x)=\frac{1}{2}{h}^{\mathrm{\prime }\mathrm{\prime }}(h^{-1}(x)),\sigma (x)=h^{\mathrm{\prime }}(h^{-1}(x))\mathrm{.}}$ Because $h^{\mathrm{\prime }}$ is bounded below by $1\mathrm{/}2$, and $h^{\mathrm{\prime }\mathrm{\prime }}$ and $h^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ are bounded and vanish outside a compact set, both $b$ and $\sigma$ are globally Lipschitz and bounded. Thus this diffusion is nonexplosive and uniformly elliptic, with $a(x)=\sigma (x{)}^2$.

For $t>0$, $W_t$ has density ${\varphi }_t$ by [F5], and the change-of-variables formula [F2] gives ${\displaystyle p_t(x)=\frac{{\varphi }_t(h^{-1}(x))}{h^{\mathrm{\prime }}(h^{-1}(x))}\mathrm{.}}$ Likewise, for $0\le s<t$ its transition density is ${\displaystyle q(s,x;t,y)=\frac{{\varphi }_{t-s}(h^{-1}(y)-h^{-1}(x))}{h^{\mathrm{\prime }}(h^{-1}(y))}\mathrm{.}}$ These densities are positive and smooth. They satisfy the marginal and Chapman–Kolmogorov identities because they are the images under $h$ of Brownian marginal and transition laws.

Write $z=h^{-1}(x)$. Differentiating $a(x)=h^{\mathrm{\prime }}(z{)}^2$ and using $dz\mathrm{/}dx=1\mathrm{/}{h}^{\mathrm{\prime }}(z)$ gives ${\displaystyle a^{\mathrm{\prime }}(x)=2h^{\mathrm{\prime }}(z)h^{\mathrm{\prime }\mathrm{\prime }}(z)\frac{1}{h^{\mathrm{\prime }}(z)}=2h^{\mathrm{\prime }\mathrm{\prime }}(z)\mathrm{.}}$ Since $h^{\mathrm{\prime }\mathrm{\prime }}=\alpha {\psi }^{\mathrm{\prime }\mathrm{\prime }}$ is nonzero at some $y_{\ast }$, $a^{\mathrm{\prime }}(h(y_{\ast }))\mathrm{\ne }0$.

Fix $0<ϵ<T$ and reindex the actual process on $[ϵ,T]$ as ${\bar{X}}_s=X_{ϵ+s}$ on $[0,T-ϵ]$. Write ${\bar{p}}_s=p_{ϵ+s}$ and $\bar{q}(r,x;s,y)=q(ϵ+r,x;ϵ+s,y)$. The explicit formulas in step 3.1 are positive and smooth on this shifted interval, satisfy the marginal and Chapman–Kolmogorov identities because they are images of Brownian transition laws under the diffeomorphism $h$, and solve the forward density equations for the coefficients in step 2.1 by direct differentiation and the chain rule. Since $h$ equals the identity outside a compact set, these densities and their differentiated integrands have Gaussian tails; differentiation under transition integrals is absolutely integrable and integration by parts has no boundary term. The new forward analytic conditions can also be checked directly: for $f\in C_c^{\mathrm{\infty }}$, put $F=f\circ h$ and $z=h^{-1}(x)$. Then $F\in C_c^{\mathrm{\infty }}$ and $g_{t,f}(r,x)=\int {\varphi }_{t-r}(w-z)F(w)dw$. Differentiation of the Gaussian convolution, followed by two integrations by parts against $F$, gives ${\mathrm{\partial }}_r{g}_{t,f}=-L_r{g}_{t,f}$ and $\mathrm{\mid }{L}_r{g}_{t,f}\mathrm{\mid }\le \frac{1}{2}\mathrm{\parallel }{F}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$; the chain rule gives $L_{r}g=\frac{1}{2}{\mathrm{\partial }}_{zz}(g\circ h)$. The densities and all weighted coefficients $pb$, $pa$, and ${\mathrm{\partial }}_x(pa)$ are jointly bounded on compact subsets of $[ϵ,T]\times \mathbb{R}$, so the displayed local weighted integrals are finite; the spatial adjoint identity follows by compact-support integration by parts. The process is continuous, nonexplosive and Markov. Introduce reverse time $u\in [0,T-ϵ]$ and $Y_u={\bar{X}}_{T-ϵ-u}=X_{T-u}$. The local generator lemma applies to this actual reversed process and gives drift ${\displaystyle {\tilde{b}}_u(x)=-b(x)+a^{\mathrm{\prime }}(x)+a(x){\mathrm{\partial }}_x\log p_{T-u}(x)\mathrm{.}}$ For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $h^{-1}(x)=x$, $h^{\mathrm{\prime }}(x)=1$, $h^{\mathrm{\prime }\mathrm{\prime }}(x)=0$, and $p_t(x)={\varphi }_t(x)$, so this drift is $-x\mathrm{/}(T-u)$. On $\mathrm{\mid }x\mathrm{\mid }\le 1$ it is smooth with bounded state derivative uniformly for $u\in [0,T-ϵ]$, because $p_t$ has a positive minimum there. Thus $\tilde{b}$ is globally Lipschitz with linear growth on the truncated interval; the reverse diffusion coefficient $\sigma$ is bounded, globally Lipschitz, and bounded away from zero.

At $x_{\ast }=h(y_{\ast })$, the actual reverse generator from [F3] has drift differing from the score-only expression $-b+a{\mathrm{\partial }}_x\log p_t$ by the nonzero amount $a^{\mathrm{\prime }}(x_{\ast })$. Choose $\varphi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ equal to $x-x_{\ast }$ near $x_{\ast }$; then ${\varphi }^{\mathrm{\prime }}(x_{\ast })=1$ and ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}(x_{\ast })=0$, so the two proposed generators have different values at $x_{\ast }$. By continuity, the drift difference remains nonzero on an open interval about $x_{\ast }$, and the reversed marginal density is positive there. Thus the score-only coefficient is not the actual reverse generator on a set of positive reversed probability. This refutes the claim.