How is the Ornstein-Uhlenbeck process stationary if the mean and variance are time dependent? I'm just beginning to study stochastic processes (not rigorously, this is in terms of Langevin dynamics) and I stumbled upon a problem. I think I'm misunderstanding some concepts, so I thought I could get some clarification here.
The Ornstein-Uhlenbeck process is stationary. This means that the mean, variance, etc. do not depend on time.
Yet, when I solve the appropriate Fokker-Planck equation for the conditional pdf (with a delta initial condition and an absorbing boundary at infinity), the answer I get is a normal distribution with mean and variance explicitly time dependent! For example, for an initial condition at $v_0$, the mean is $v_0 e^{-\gamma t}$.
The Fokker-Planck equation is for the pdf of the velocity of a particle in a gas of colliding particles. Let $p \equiv p(v,t|v_0,0)$, then:
$$\frac{\partial p}{\partial t} = \gamma \frac{\partial }{\partial v} \left( v p\right) + \frac{\gamma kT}{m} \frac{\partial ^2 p}{\partial v^2}.$$
This corresponds to the following SDE (Langevin equation):
$$ \frac{dv}{dt} = -\gamma v + \sqrt{\frac{2\gamma kT}{m}} \eta(t),$$ 
where $\eta(t)$ is a unit delta-correlated Gaussian white noise. This is clearly an Ornstein-Uhlenbeck process, right?
Back to the Fokker-Planck equation: assuming $p(v,0|v_0 ,0) = \delta (v-v_0)$ and a zero probability current at $|v| \to \infty$, the solution comes out to be:
$$p(v,t|v_0,0) = \left( \frac{m}{2\pi kT (1-e^{-2\gamma t})} \right) ^\frac{1}{2} \exp \left( -\frac{m(v-v_0 e^{-\gamma t})^2}{2kT(1-e^{-2\gamma t})} \right).$$
This is obviously a normal distribution with a time dependent mean and variance. How do I reconcile this with the statement that the process is stationary? Or is this process actually not Ornstein-Uhlenbeck? 
 A: The O-U process with a delta initial condition is not stationary in this sense.  But that's the wrong initial condition.
The O-U process is a Markov process which admits a stationary distribution, so if you want a stationary process, you should start it in the stationary distribution.  Here it's a Gaussian distribution, and it isn't hard to work out what the mean and variance of that distribution ought to be.
This corresponds to a time-independent solution of the Fokker-Planck equation, which you can easily verify is of the form
$p(v) = e^{-x^2/2 \sigma^2}$, and you can work out the right value for $\sigma$ in terms of your parameters.
A: The process should be mean-reverting (e.g. for long time $t \to \infty$ is should approach the mean value). Here your mean value is 0. Stationary may be a bit misleading here. This is a diffusion equation and of course it depends on time.
A: A stochastic process is called stationary (in this general sense, without giving details about initial conditions) when the propagator between two times (or Green's function, or two-times conditional probability) depends ONLY on the difference between these two times (i.e., time homogeneity, or time translational invariant when both times are shifted). You see this is your case if instead of taking initial time equal to 0, you take it as $t_0$ arbitrary. Then in your result both times will appear only in the combination $t-t_0$. This is formally why the process is called stationary. Another, very different stuff, if if the system has relaxed to stationary state (in terms of one-time PDF) or not. This terminology is common in physics. I'm a physicist, not a mathematician, that's why I'm kind of used to hear it in this way.
