Let $$dX_t=(a+bX_t)dt+dW_t,$$ where $a,b\in\mathbb R$ and $X_0=0$ a.s. I read on wikipedia that contrary to the Brownian motion, $(X_t)$ has a stationnary solution. So, I guess that $\partial _tp(x,t)=0$ where $p(x,t)$ is the density of $X$ at time $t$. I know that such a distribution has to satisfy fokker planck equation, i.e. $$\partial _tp(x,t)=-\partial _x((a+bx)p(x,t))+\partial _{xx}p(x,t),\tag{E}$$ where $p(0,t)=\delta _0$.

Q1) How can I prove that (E) has a stationary solution ?

Q2) I know that the solution of such an equation is unique. So I don't really understand how it can have either a stationary solution and a non stationary solution. Maybe Ornstein Uhlenbeck process has always stationary distribution ?

  • $\begingroup$ We need $b<0$. In the standard Ornstein–Uhlenbeck process, $a=0$. The analysis of the standard process is easily generalized to incorporate a drift ($a\neq0$). $\endgroup$ May 3 '20 at 11:55

By stationary you define, I guess you mean to show $p(x, t) = p(x)$?

Ornstein–Uhlenbeck process is a Gaussian process, which has a Gaussian probability density. Thus you can show its mean and covariance function do not depend on $t$.

You can verify that the mean and covariance are Wiki

$$ \mathbb{E}[X_t] = X_0 \, e^{bt},\\ \mathrm{cov}[X_t, X_t] = -\frac{1}{2b}, $$

provided that your $b<0$, $a=0$, and you start from initial stationary condition $X_0\sim N(0, -\frac{1}{2b})$.

  • $\begingroup$ I'm not so sure why starting from $X_0\sim N(0,-1/2b)$ gives that you have a stationary distribution (yes, by stationnary I mean $p(x,t)=p(x)$). $\endgroup$
    – Walace
    May 7 '20 at 17:22
  • $\begingroup$ If starts from $X_0$, then your $p(x, t) = N(0, -1/2b)$, which is time-homogeneous. If not, the process will experience a non-stationary stage, and converge to N(0, -1/2b) as $t\to\infty$. $\endgroup$ May 7 '20 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.