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?
Best Answer
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.