A stationary Ornstein–Uhlenbeck diffusion reverses to itself

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.