# How can I prove that Ornstein Uhlenbeck process has a stationnary distribution?

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 ?

• 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 ($a\neq0$). 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 $$t$$.
$$\mathbb{E}[X_t] = X_0 \, e^{bt},\\ \mathrm{cov}[X_t, X_t] = -\frac{1}{2b},$$
provided that your $$b<0$$, $$a=0$$, and you start from initial stationary condition $$X_0\sim N(0, -\frac{1}{2b})$$.
• 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 $p(x,t)=p(x)$). May 7 '20 at 17:22
• 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 $t\to\infty$. May 7 '20 at 17:42