# Fokker-Planck equation for Ornstein-Uhlenbeck

I am trying to solve the following Ornstein-Uhlenbeck SDE

$dx = -\alpha x dt + \gamma dW$

where $x$ is a stochastic process, W is a standard Wiener process (Brownian motion) and $\alpha$ and $\gamma$ are both constants. I want to derive the probability density function $p(x,t)$ by using the Fokker-Planck equation

$\dfrac{\partial p(x,t)}{\partial t} = -\dfrac{\partial}{\partial x}[-\alpha x p(x,t)] + D\dfrac{\partial^2 p(x,t)}{\partial x^2}$

with $D = \gamma^2/2$ and initial condition $p(x,0) = \delta(x-x_0)$. I start by taking the Fourier transform with respect to $x$ of both sides, defined as

$\mathcal{F}\left\{f(x)\right\} = \dfrac{1}{\sqrt{2\pi}} \displaystyle\int_{-\infty}^{\infty} f(x) e^{-ikx}dx$

$\mathcal{F}^{-1}\left\{\hat{f}(k)\right\} = \dfrac{1}{\sqrt{2\pi}} \displaystyle\int_{-\infty}^{\infty} \hat{f}(k) e^{ikx}dk$

and using the identity

$\mathcal{F}\left\{\dfrac{\partial}{\partial x}[x p(x,t)] \right\} = k\dfrac{\partial\hat{p}(k,t)}{\partial k}$

This gives me the transformed PDE

$\dfrac{\partial \hat{p}(k,t)}{\partial t} = \alpha k \dfrac{\partial\hat{p}(k,t)}{\partial k} - k^2 D\hat{p}(k,t)$

With transformed initial condition

$\hat{p}(k,0) = \mathcal{F}\left\{p(x,0) \right\} = \dfrac{e^{-ix_0 k}} {\sqrt{2\pi}}$

At this point, I try using the method of characteristics to solve for $\hat{p}(k,t)$ and I always end up with the solution

$\hat{p}(k,t) = \dfrac{1}{\sqrt{2\pi}} \exp\left\{-\dfrac{D}{2\alpha}\left(e^{2\alpha t} - 1\right)k^2 - ix_0 e^{\alpha t}k \right\}$

The problem I have is that when I go ahead to inverse Fourier transform this function, I end up with a term $x_0 e^{\alpha t}$ for the mean of the process $x(t)$, but solving for the expectation of the SDE directly gives me $\mathbb{E}\left[x(t)\right] = x_0 e^{-\alpha t}$. I can't figure out why there is a sign difference in my exponent. Is there a mistake in one of my steps? And, would it be possible to obtain $p(x,t)$ from the Kolgomorov backward equation instead? It seems it would be better suited for this problem.

• I've never learned the definitions of forward equation versus backward equation unfortunately. What I do know is I would find the transition kernel in this case by solving $\frac{\partial f}{\partial t} - \Delta f + \alpha x \cdot \nabla f = 0$ using the Fourier transform and reading off the transition kernel from the solution. (Solution will be convolution of initial data with the kernel.) I was under the impression that you could also find the transition kernel using the adjoint equation (what you are attempting to do).
– user81375
May 24, 2018 at 14:34
• Since your mean has the time reversed, my best guess is there should be a minus sign in front of the $\frac{\partial}{\partial t}$ in your equation.
– user81375
May 24, 2018 at 14:35

Somewhat embarrassingly I have never familiarized myself with the "forward/backward equation" nomenclature so I won't comment on that directly here. Instead, observe that if $$(X_{t})_{t \geq 0}$$ satisfies $$\begin{equation*} dX_{t} = - \alpha X_{t} dt + dB_{t}, \end{equation*}$$ then, given a nice function $$f : \mathbb{R} \to \mathbb{R}$$, the process $$(U_{t})_{t \geq 0} = (f(X_{t})_{t \geq 0}$$ satisfies $$\begin{equation*} dU_{t} = f'(X_{t}) dB_{t} + \left(\frac{1}{2} f''(X_{t}) - \alpha X_{t} f'(X_{t})\right) dt. \end{equation*}$$ Thus, we deduce that the function $$u(x,t) = \mathbb{E}^{x}(f(X_{t}))$$ satisfies $$\begin{equation*} \frac{\partial u}{\partial t} = \frac{1}{2} \frac{\partial^{2}u}{\partial x^{2}} - \alpha x \frac{\partial u}{\partial x}. \end{equation*}$$ Note that the adjoint of the generator $$L = \frac{1}{2} \frac{\partial^{2}}{\partial x^{2}} - \alpha x \frac{\partial}{\partial x}$$ (acting on $$L^{2}(\mathbb{R},dx)$$) is given by $$L^{*}v = \frac{1}{2} \frac{\partial^{2}v}{\partial x^{2}} + \frac{\partial}{\partial x}(\alpha x v)$$, as can be verified using integration by parts.

Now if $$p$$ is the transition kernel, then we know that $$u(x,t) = \int_{\mathbb{R}} f(y) p(t,x,y) \, dy$$. Therefore, one way to find $$p$$ is to solve for $$u$$ using the Fourier transform, which will give an integral for $$u$$ in terms of $$f$$. Also note that the equation solved by $$u$$ and the integral representation formula implies $$\frac{\partial}{\partial t} p(t,x,y) = \frac{1}{2} \frac{\partial^{2}}{\partial x^{2}} p(t,x,y) - \alpha x \frac{\partial}{\partial x} p(t,x,y)$$.

Alternatively, suppose $$\varphi$$ is a nice function and $$v$$ solves the adjoint equation $$\begin{equation*} \left\{ \begin{array}{r l} - \frac{\partial v}{\partial t} = L^{*}v & \text{in} \, \, (0,T) \times \mathbb{R}^{d} \\ v(x,T) = \varphi(x) \end{array} \right. \end{equation*}$$ Observe that \begin{align*} \frac{d}{dt} \left\{ \int_{\mathbb{R}} u(x,s) v(x,s) \, dx \right\} &= \int_{\mathbb{R}} \left(\frac{\partial u}{\partial t} v + u \frac{\partial v}{\partial t} \right) \, dx \\ &= 0. \end{align*} and, thus, $$\begin{equation*} \int_{\mathbb{R}} u(x,T) \varphi(x) \, dx = \int_{\mathbb{R}} f(x) v(x,0) \, dx. \end{equation*}$$ This implies $$\begin{equation*} \int_{\mathbb{R}}f(x) v(x,0) \, dx = \int_{\mathbb{R}} \int_{\mathbb{R}} f(y) p(T,x,y) \varphi(x) \, dx dy. \end{equation*}$$ Since $$f$$ was arbitrary, we deduce $$v(x,0) = \int_{\mathbb{R}} \varphi(x) p(T,x,y) \, dy$$. Since there's nothing special about $$T$$, we see that $$v(x,s) = \int_{\mathbb{R}} \varphi(x) p(T - s, x,y) \, dy$$. Since this is true independently of the choice of $$\varphi$$, we see that $$p(t,x,y)$$ solves $$\begin{equation*} -\frac{\partial}{\partial t} p(t,x,y) = \frac{1}{2} \frac{\partial^{2}}{\partial y^{2}} p(t,x,y) + \frac{\partial}{\partial y}\left\{\alpha x p(t,x,y)\right\}. \end{equation*}$$

Long story short, the equation you were solving for $$p$$ was off by a minus sign. If you wanted, you could have used the easier-to-remember equation with $$L$$ instead of the equation with $$L^{*}$$.

I noticed my mistake was in the definition of

$\mathcal{F}\left\{\dfrac{\partial}{\partial x}[x p(x,t)] \right\} = -k \dfrac{\partial \hat{p}(k,t)}{\partial k}$

I had the wrong sign there for some reason, which was the cause of all my problems. Thank you all for your help.