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I need to find the steady state probability of an Ornstein-Uhlenbeck process. I've made a start, but I've got stuck at this point. First I start with the definition of the evolution of probability for the one variable Fokker-Planck equation:

$$\frac{\partial P}{\partial t}(x,t)= L_{FP} P(x,t)\\ L_{FP}=-\frac{\partial}{\partial x} D^{(1)}(x) + \frac{\partial^2}{\partial x^2} D^{(2)}(x)$$

where $P$ is the probability, and $D^{(1)}(x)$ and $D^{(2)}(x)$ are the drift and diffusion, respectively.

For an OU process, $$D^{(1)}(x)=-\gamma x, \ D^{(2)}(x)=D = \text{const},$$ where $\gamma$ is a constant. Substituting this into the Fokker-Planck equation gives $$\frac{\partial P}{\partial t}= \gamma\frac{\partial}{\partial x} (xP) + D\frac{\partial^2}{\partial x^2} P.$$

For a steady state solution, $\dfrac{\partial P}{\partial t}=0$, so the equation reduces to $$\frac{\partial^2 P}{\partial x^2}+\frac{\gamma}{D}\frac{\partial}{\partial x} (xP) =0.$$

I'm guessing that now I'd expand the second term to give

$$\frac{\partial^2 P}{\partial x^2}+\frac{\gamma}{D}\left(x\frac{\partial P}{\partial x} + P\right)=0,$$ but then I have no idea where to go from there. I'm also guessing there will be some initial condition like $$P(x,0) = \delta (x-x_0),$$ but I'm not sure. Can anyone help me please?

Ѕᴀᴀᴅ
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WilliamMorris
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1 Answers1

9

I found the answer in the end. Thought I should post it here in case anyone runs into the same difficulties.

Starting from the above steady state, where technically the derivatives become normal derivates

$$ \frac{d^2 P}{d x^2}+\frac{\gamma}{D}\frac{d}{d x} (xP) =0. $$

I rearrange then integrate

\begin{align} \frac{d}{d x} (xP) =& -\frac{D}{\gamma}\frac{d^2 P}{d x^2}\\ xP =& -\frac{D}{\gamma}\frac{d P}{d x} + C_1. \end{align}

Constant $C_1$ must be $0$ due to boundary conditions of $p,\frac{d p}{d x} \rightarrow 0$ as $x\rightarrow \pm\infty$.

Rearrange, integrate and rearrange:

$$ \int\frac{1}{P}\ dP = -\frac{\gamma}{D}\int x\ dx$$

$$ \ln P = -\frac{\gamma x^2}{2D} + C_2$$

$$ P = C_3 e^{-\frac{\gamma x^2}{2D}}.$$

Constant $C_3$ is found through the condition that $\int^{+\infty}_{-\infty}P(x)\ dx = 1$.

$$ C_3 \int^{+\infty}_{-\infty}e^{-\frac{\gamma x^2}{2D}}\ dx = 1 .$$

With a bit of change of variables $y^2=\frac{\gamma x^2}{2D}$

$$ C_3 \sqrt{\frac{2D}{\gamma}} \int^{+\infty}_{-\infty}e^{-y^2}\ dy = 1 .$$

The integral can be shown (Example: How do you integrate $e^{x^2}$?) to equal $\sqrt{\pi}$. So

$$ C_3 = \sqrt{\frac{\gamma}{2\pi D}}$$

and finally

$$ P = \sqrt{\frac{\gamma}{2\pi D}} e^{-\frac{\gamma x^2}{2D}}.$$

Andras Vanyolos
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WilliamMorris
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