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?

  • $\begingroup$ You seem to be using \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. $\endgroup$ Feb 29 at 2:46
  • $\begingroup$ @Jean Daviau: This problem is ill posed. Without specifying the form of $f(x_t, y_t)$, the application of Itos Lemma is incorrect. $\endgroup$
    – user522465
    Feb 29 at 5:00
  • $\begingroup$ @Damien It should be possible to relate information about $\phi$ to information about $f$ or vice versa via Itô's lemma. Linear growth and Lipschitz conditions are satisfied for the drift and diffusion terms and we can impose such conditions on $f$ w.r.t both arguments to ensure existence and uniqueness of $y_t$ and $x_t$. $\endgroup$ Mar 2 at 4:15

1 Answer 1


There are some issues with the statements:

  1. $\phi(x)$ is defined as the stationary conditional expectation $\mathbb{E}[y_t | x_t=x]$. This is a function of $x$ only and does not actually depend on time. Therefore, taking the time derivative inside the expectation doesn't make sense.

  2. The conditional expectation $\mathbb{E}[\frac{dy}{dt}| x_t=x]$ is not generally equal to the derivative of the unconditional expectation $\frac{d}{dt}\, \mathbb{E}[y_t]$. The claim seems to be equating these two quantities, which is incorrect.

  3. Even if we look at the derivative of the conditional expectation $\frac{d}{dt} \, \mathbb{E}[y_t | x_t=x]$, there is no reason why this should satisfy the proposed PDE in the stationary limit. The conditional expectation satisfies its own dynamics that need to be derived carefully.

  • $\begingroup$ 1: the evolution equation was referring to $\phi(x_t)=\mathbb{E}\left[y_t\mid x_t\right]$ in the stationary limit of $y_t$. 2: Agreed and I am sceptical of whether this would hold in nontrivial cases (I did read a paper a while back that demonstrated fairly general conditions - strong stationarity, mean-square differentiability etc. - where $d\mathbb{E}\left[y_t\right]/dt=\mathbb{E}\left[\dot{y}\right]$ but forget the name of it for the moment). 3: I'm trying to avoid this by 'neglecting' $y_t$ when defining $\phi$ ... I don't doubt that there are inconsistencies with this formulation. $\endgroup$ Mar 2 at 5:05
  • $\begingroup$ Intuitively, there is some function $\phi$ that relates $x_t$ to where we believe $y_t$ should be, on average, when stationary - mean of the stationary density $p(y_t\mid x_t)$. This function will obviously change over time w.r.t $x_t$ and the change in the expected $y_t$ over time is only dependent on changes in $x_t$ through $\phi$. Can the function $\phi$ somehow be related to $f$? My hope is something like $g(\phi,\phi',\phi'')=f(x,\phi)$. $\endgroup$ Mar 2 at 5:10

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