Consider the modified Ornstein–Uhlenbeck process $$\mathop{dx_t}=\theta(y_t-x_t)\, dt+{}\sigma\,dW_t$$ for a standard Brownian motion $W_t$ and $\theta,\sigma\in\mathbb{R}_{>0}$. Let's define the sufficiently smooth function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ such that $\phi(x):=\lim_{t\rightarrow\infty}\mathbb{E}\left[y_t\mid x_t=x\right]$ and $y_t$ is deterministically dependent on $x_t$ somehow (i.e. $\mathop{dy_t}=f(x_t,y_t)\mathop{dt}$). The limit ensures that we are referring to the stationary conditional expectation only depending on the value of $x_t$, rather than one which also varies over time. Implicitly it is assumed that both $x_t$ and $y_t$ are stationary, mean-square differentiable random processes.

By Itô's lemma, $$\mathop{d\left[\phi(x_t)\right]}=\left(\theta(y_t-x_t)\phi'(x_t)+\frac{\sigma^2}{2}\phi''(x_t)\right)\,dt +\sigma\phi'(x_t)\mathop{dW_t}.$$

Is the claim that $$\lim_{t\rightarrow\infty}\mathbb{E}\left[\left.\frac{dy}{dt}\right|x_t=x\right]=\theta(\phi(x)-x) \phi'(x) + \frac{\sigma^2}{2}\phi''(x)$$ correct and, if not, is it possible to express the above limit in terms of $\phi$ and its derivatives?

**Edit:**
Unless I've made an error, the claim in my question implies that
\begin{align}
\mathbb{E}\left[\left.\frac{dy}{dt}\right|x_t=x\right]&=\mathbb{E}\left[\left.\frac{d}{dt}\mathbb{E}\left[\phi(x_t)\right]\right|x_t=x\right]\\
&=\mathbb{E}\left[\left.\frac{d}{dt}\mathbb{E}\left[\mathbb{E}\left[y_t\mid x_t\right]\right]\right|x_t=x\right]\\
&=\mathbb{E}\left[\left.\frac{d}{dt}\mathbb{E}\left[y_t\right]\right|x_t=x\right]
\end{align}
in the stationary limit. Is this true?

`\mathop{dx}`

to separate $dx$, as in`f(x)\mathop{dx}`

, so that you see $f(x)\,dx$ rather than $f(x)dx.$ However, that messes with the space surrounding the plus sign: In $\mathop{dt}+\sigma,$ coded as`\mathop{dt}+\sigma`

, there is a conspicuous lack of the horizontal space that you see in $dt+\sigma.$ I edited so as to code $f(x)\,dx$ as`f(x)\,dx`

, and then it's back to normal. $\endgroup$