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I have a general system of two Ito SDEs and I want to convert that to a Fokker-Plack PDE.

The system of SDEs is:

$dX(t)=f_1(X)dt + \sigma g_1(X)dW(t)$

$dY(t)=f_2(X)dt + \sigma g_2(X)dW(t)$

What would be the corresponding Fokker-Plack? Moreover will the final PDE change if instead of $dW(t)$ in both we had $dW_1(t)$ and $dW_2(t)$ respectively?

My question is mainly motivated by whether there will be some second derivate term in the Fokker-Planck that includes $\frac{\partial^2}{\partial X \partial Y}$ depending on the Wiener increments.

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The drift part is more or less trivial: if you have a vector drift $b$ then the relevant formula contributes $-\nabla \cdot [ b p ]$ to the right side of the FPE.

The diffusion part depends very much on whether the BMs are the same or different. If they are the same, then the "pre-diffusion" matrix $\sigma$ looks like $\begin{bmatrix} s g_1 & 0 \\ s g_2 & 0 \end{bmatrix}$ (so that multiplication by $\begin{bmatrix} dW_1 \\ dW_2 \end{bmatrix}$ recovers the desired quantity). If they are different then you instead have $\begin{bmatrix} s g_1 & 0 \\ 0 & s g_2 \end{bmatrix}$. (My $s$ is your $\sigma$.) These drastically alter the properties of the PDE and of the process. In any case, the other term on the right side of the FPE will be $\frac{1}{2} \sum_{i=1}^2 \sum_{j=1}^2 \frac{\partial^2}{\partial x_i \partial x_j} [D_{ij} p]$ where $D=\sigma \sigma^T$. You can see that when they are distinct BMs there will be no cross terms because $D_{12}=D_{21}=0$.

See https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation for more details.

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    Thank you very much for your answer. This is exactly what I wanted to know. – 2017-02-15
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    @JimV If an answer helped you it is appreciated here if you upvote and more importantly accept it. – 2017-02-15