10
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ the OU process admits a "steady state" probability distribution: math.stackexchange.com/q/2671880/532409 .... IF you take this distribution as the initial condition, then it is "stationary". $\endgroup$
    – Quillo
    Commented Apr 17, 2022 at 15:22

3 Answers 3

7
$\begingroup$

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 $p(v) = e^{-v^2/2 \sigma^2}/\sqrt{2\pi\sigma^2}$, and you can work out the right value for $\sigma$ in terms of your parameters.

$\endgroup$
7
  • 3
    $\begingroup$ Thanks. So - a stationary process is one for which there exists a stationary distribution. If that distribution is chosen to be the initial distribution, then nothing happens (in terms of dynamics), and the mean and all the moments are constant. But if some other initial distribution is chosen, the system will evolve in such a way that on a suitably long time scale, the distribution will tend to the stationary distribution. Is this correct? $\endgroup$ Commented May 22, 2017 at 16:31
  • 3
    $\begingroup$ @SpineFeast: Basically. There is an ambiguity in the meaning of the word "process". A "Markov process" is really a whole family of processes: one for each possible initial distribution. To say that "a Markov process has a stationary distribution" is to say that one process in that family is a stationary process. Under appropriate conditions (look up "ergodicity"), every other process in the family will have its distribution converge to that same stationary distribution. $\endgroup$ Commented May 22, 2017 at 16:56
  • $\begingroup$ I see. So, the evolution of a Markov process is essentially described by a PDE + a boundary condition, where the PDE is for conditional probabilities. And to this I must add a further input (initial condition) to know the joint pdf and have a full description of a particular problem (realization of the process). Right? One further question about "stationary processes": when it's said that they are invariant under a time shift (the conditional pdf is a function of only $t-t'$), this is only true for that one particular process with the right initial value, not the entire family, correct? $\endgroup$ Commented May 22, 2017 at 17:23
  • 1
    $\begingroup$ @SpineFeast: First part right. For the second part, the conditional pdf is the same for all processes in the family, so if one has this property (often called "time homogeneity") then they all do. $\endgroup$ Commented May 22, 2017 at 17:32
  • 1
    $\begingroup$ @SpineFeast: I think what you're calling "Green's function" is what I mean by conditional pdf: the pdf at time $t$ conditioned on a delta mass at $v_0$ at time $t_0$. When you integrate against an initial condition you get an unconditional pdf. I'm afraid this is probably the last comment I have time for today. $\endgroup$ Commented May 22, 2017 at 17:47
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Yes, certainly in the $\gamma t > > 1$ regime it thermalizes and tends to the Maxwell distribution, losing memory of the initial velocity, but it should still be stationary? The Fokker-Planck equation corresponds to the SDE: $dv/dt = -\gamma v + \sqrt{2\gamma kT/m} \eta(t)$ with $\eta$ a unit delta correlated Gaussian white noise. $\endgroup$ Commented May 22, 2017 at 15:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .