Anderson reverse-time diffusion — Examples

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.

The compact support of $\rho$ makes $X_0$ almost surely bounded, and its independence from $W$ is part of the Example. The restart map is explicit: for $s\le r\le T$, $X_r=X_s+(W_r-W_s)$. The SDE has $b=0$ and $\sigma =I_d$, so it is a globally Lipschitz strong SDE; the convolution and heat equation in [F1] give the stated positive-time marginal density. The separate augmented joint and conditional uniqueness requirements are explicitly assumed, exactly as listed in [A1].

Apply Anderson's theorem with $t_0=0$, $b=0$, $a=I_d$, $\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot a=0$, and $p_t$ from [F1]. Its reverse drift is $-b+\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot a+a\mathrm{\nabla }\log p_t=\mathrm{\nabla }\log p_t$, and the reverse diffusion covariance is $I_d$. Since $p_t$ is only used for $t>0$, this gives the stated equation for $u<T$ and asserts nothing about a score at $u=T$.

Applying [F1] to $e^{\theta t}(X_t-\mu )$ gives, for $t\ge s$, ${\displaystyle X_t-\mu =e^{-\theta (t-s)}(X_s-\mu )+\sigma {\int }_s^t{e}^{-\theta (t-r)}d{W}_r\mathrm{.}}$ The last term is independent of ${\mathcal{F}}_s$ and by [F2] is centered normal with variance ${\displaystyle {\sigma }^2{\int }_s^t{e}^{-2\theta (t-r)}dr=\frac{{\sigma }^2}{2\theta }(1-e^{-2\theta (t-s)})\mathrm{.}}$ With $q={\sigma }^2\mathrm{/}(2\theta )$ and $X_0-\mu \sim N(0,q)$, the mean of $X_t$ is $\mu$ and its variance is $e^{-2\theta t}q+q(1-e^{-2\theta t})=q$. Independence and Gaussian addition [F3, F4] therefore give the same marginal law at every $t$.

Fix $s<t$ and write $r=e^{-\theta (t-s)}$, $U=X_s-\mu$, and $V=X_t-\mu =rU+\eta$, where $U\sim N(0,q)$ and $\eta \sim N(0,q(1-r^2))$ is independent of $U$ by step 1.1. Their joint density is proportional to ${\displaystyle \exp \text{\hspace{0.17em}⁣}[-\frac{1}{2}(\frac{u^2}{q}+\frac{(v-ru{)}^2}{q(1-r^2)})],\frac{u^2}{q}+\frac{(v-ru{)}^2}{q(1-r^2)}=\frac{(u-rv{)}^2}{q(1-r^2)}+\frac{v^2}{q}\mathrm{.}}$ Thus $X_s\mid X_t=y$ has law $N(\mu +r(y-\mu ),q(1-r^2))$, exactly the forward transition law over the same time gap. Bayes' formula shows that the reversed transition kernels equal the forward OU transition kernels.

By [F5] and step 2.1, the reversed process starts from the stationary law and has the same transition kernels as the forward OU process. Indeed, the forward joint density along an ordered time list is the stationary density times the product of the forward transition densities. The Bayes identity in step 2.1 gives detailed balance for each transition factor, so reversing this product gives the Markov factorization with the same kernels in reverse-time order. Thus all finite-dimensional distributions of $Y$ equal those of the forward OU process. Extend each path constantly after $T$; the extended paths are random elements of $C([0,\mathrm{\infty }),\mathbb{R})$ because each coordinate evaluation is measurable. By [F7], their Borel laws are equal. On a forward OU path define ${\displaystyle \mathrm{\Gamma }(X{)}_u=\frac{X_u-X_0+\theta {\int }_0^u(X_r-\mu )dr}{\sigma },0\le u\le T\mathrm{.}}$ The integral equation makes $\mathrm{\Gamma }(X)=W$; because the natural filtration of $X$ is contained in the forward filtration, this driver is Brownian relative to that natural filtration. The map from a continuous path to $\mathrm{\Gamma }(X)$ is continuous in the uniform norm on $[0,T]$, so equality of path laws transfers this Brownian property to $B_u:=\mathrm{\Gamma }(Y{)}_u$. Rearranging its definition gives $d{Y}_u=-\theta (Y_u-\mu )du+\sigma d{B}_u$ in the natural reversed filtration. Finally, the stationary score is ${\mathrm{\partial }}_x\log p(x)=-(x-\mu )\mathrm{/}q=-2\theta (x-\mu )\mathrm{/}{\sigma }^2$. With $b(x)=-\theta (x-\mu )$ and $a(x)={\sigma }^2$, the score-form drift equals $-b(x)+a(x){\mathrm{\partial }}_x\log p(x)=\theta (x-\mu )-2\theta (x-\mu )=-\theta (x-\mu )$, as stated.