Feynman path integral is equivalent to Fokker–Planck equation. This is mentioned here, but it's not clear why.
This page says Schrodinger equation is also equivalent to Fokker–Planck equation which makes me even more confused.
Basically what I am asking is:
How to show that Fokker-Plank: $\frac{\partial}{\partial t} p(x, t) = -\frac{\partial}{\partial x}\left[\mu(x, t) p(x, t)\right] + \frac{\partial^2}{\partial x^2}\left[D(x, t) p(x, t)\right]$ is equivalent to path integral: $\psi(x,t)=\frac{1}{Z}\int_{\mathbf{x}(0)=x} e^{iS(\mathbf{x},\dot{\mathbf{x}})}\psi_0(\mathbf{x}(t))\, \mathcal{D}\mathbf{x}$?
Where $S(\mathbf{x},\dot{\mathbf{x}})= \int L( \mathbf{x}(t),\dot{\mathbf{x}}(t))dt$ is the action (integral over the Lagrangian) and $\mathcal{D}\mathbf{x}$ stands for the integration over all possible paths?