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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 ?

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    $\begingroup$ 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, 2020 at 11:55
  • $\begingroup$ This is correct, the $a\ne0$ case is easily generalized from the $a=0$ case. My comment is very minor: per usual terminology even the $a=0$ case has drift, the $bX_tdt$ term is called the drift. And we say that with $a\ne0$ we add a nonzero mean reversion level to the problem, see Vasicek model $\endgroup$ Dec 31, 2023 at 17:46

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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 $X_0\sim N(0, -\frac{1}{2b})$.

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  • $\begingroup$ 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, 2020 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 $t\to\infty$. $\endgroup$
    – null
    May 7, 2020 at 17:42

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