Fokker Planck equation in the Stratonovich approach

Question

I'm a physics master student and I have difficulties understanding how to derive the Fokker Planck equation from the Stratonovich SDE.

With the Ito SDE it is simple since the noise is independent of $X_t$ :

• With a physics notation :

$\dot{X} = \mu(X,t)+\sigma(X,t)\xi(t) $ with $\langle\xi(t)\rangle=0$ and $\langle\xi(t)\xi(s)\rangle = 2B\delta(t-s) $

Which can be written rigorously : $X_t = \mu(X_t,t)dt+\sigma(X_t,t)\sqrt{2B}dB_t$

• Then one takes an arbitrary function f with the right properties ($\mathcal{C}^2$, compactly supported etc.) and evaluates the expected value of $f(X_t)$ to derive the F.-P. equation :

$\frac{d\langle f(X(t)\rangle}{dt} = \langle f'(X)\dot{X}+B\sigma(X,t)^2f''(X)\rangle$

$\frac{d\langle f(X(t)\rangle}{dt} = \langle f'(X)\mu(X,t)+B\sigma(X,t)^2f''(X)\rangle+0$ since $\langle f(X)\sigma(X,t)\xi(t)\rangle=0$

• Finally a simple integration by part enables you to get :

$\int f(x) \partial_t p(x,t) dx = \int f(x) \partial_x\left( -\mu(x,t)p(x,t) + B\partial_x(\sigma(x,t)^2p(x,t))\right) \implies \partial_t p(x,t)=\partial_x\left( -\mu(x,t)p(x,t) + B\partial_x(\sigma(x,t)^2p(x,t))\right)$

If we consider the same problem with a Stratonovich approach:

$\frac{d\langle f(X(t)\rangle }{dt} = \langle f'(X)\dot{X}\rangle= \langle f'(X)\left( \mu(X,t)+\sigma(X,t)\xi(t) \right)\rangle$

But $\langle f'(X)\sigma(X,t)\xi(t)\rangle \neq 0$

How do we get from here to the Stratonovich Fokker-Planck equation ?

Answer 1

Using notes from here Lecture 5 - Mathematical Foundations of Stochastic Processes and Lecture 10: Forward and Backward equations for SDEs .

For the Stratonovich SDE

$$dX=\mu_{1}(X)dt+\sigma_{1}(X)\circ dW_{t}$$

the corresponding Stratonovich-Fokker-Plank equation for the transition density $\rho$ is

$$\partial_{t}\rho=\partial_{x}((-\mu_{1}\rho+\frac{1}{2}\sigma_{1}\partial_{x}(\sigma_{1}\rho))).$$

So since the corresponding derivation for the Itô-formulation is clear, lets first convert to it

$$f\circ dW_{t}=\frac{1}{2}f\partial_{x}fdt+fdW_{t}.$$

So we study

$$dX=\mu_{2}(X)dt+\sigma_{2}(X) dW_{t}$$

for $\mu_{2}=\mu_{1}+\frac{1}{2}\sigma_{1}\partial_{x}\sigma_{1}$ and $\sigma_{2}=\sigma_{1}.$

Here as done in eq 3-4 Lecture 10: Forward and Backward equations for SDEs we simply apply Itô-formula to some $f\in C^{2}$ to get

$$df(X)=(\mu_{2} \partial_{x}f+\frac{1}{2}\sigma_{2}^{2}\partial_{xx}f)dt+(\partial_{x}f)\sigma dW_{t}.$$

From here we just integrate from s to t and take the conditional expectation with $X_{s}=y$. As you mentioned the Itô-integral term has expectation zero:

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$.

In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

For a proof see theorem 5 in Martingales and Elementary Integrals .

So now are left with the generator

$$\mathcal{L}=\mu_{2} \partial_{x}+\frac{1}{2}\sigma_{2}^{2}\partial_{xx}.$$

From here we proceed as in eq. 10,11 to compute the adjoint and revert to $\mu_{1},\sigma_{1}$

$$\mathcal{L}^{*}f=- \partial_{x}(\mu_{2}f)+\frac{1}{2}\partial_{xx}(f\sigma_{2}^{2})=\partial_{x}[- \mu_{2}f+\frac{1}{2}\partial_{x}(f\sigma_{2}^{2})]$$

$$=\partial_{x}[- \mu_{1}f+\frac{1}{2}\sigma_{1}\partial_{x}(f\sigma_{1})].$$