In what way are Fokker-Planck equations and Feynman path integrals equivalent? 
*

*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?
 A: I will start by apologising for a physicist's level of rigor in the following derivations but I think they give good insight into the connection between what I would call stochastic physics' Trinity of Langevin, Fokker-Planck and Path integral.
Fokker-Planck from Langevin equation
Given a Langevin equation $dZ_t=b(Z_t)dt+\sigma(Z_t)dW_t$ (the most common form found in physics) we can derive the Fokker-Planck equation from the Chapman-Kolmogorov equation in a quick and dirty way:
$$
p(y,t|x)=\int dz\, p(y,t|z)p(z,t'|x)
$$
We suppose that $t = \delta t$ and thus we can write $$p(y,t|z) = \left\langle \delta (y-z-h)\right\rangle_h$$ where $h=\delta Z$ is defined by the associated Langevin equation. Now we have 
$$
p(y,t|x)=\int dz\, \left\langle \delta (y-z-h)\right\rangle_hp(z,t'|x)\,.\tag{$\ast$}
$$
We can then Taylor expand the delta function as
$$
\left(1+\left\langle h\right\rangle \frac{\partial}{\partial y} +\frac{1}{2}\left\langle h^{2}\right\rangle \frac{\partial^2}{\partial y^2} +\ldots\right)\delta(y-z) = (1+\mathscr{L})\delta(y-z)\,,
$$
then integrating by parts and the delta function we are left with
$$
p(y,t|x)=(1+\mathscr{L^\dagger})p(y,t'|x)\,.
$$
Expanding $p(y,t|x)= p(y,t'|x) + \frac{\partial}{\partial t}p(y,t'|x) + \ldots$ in the limit $t \rightarrow 0$ we arrive at the Fokker-Planck equation
$$
\frac{\partial p(y,t|x)}{\partial t}=\mathscr{L^\dagger}p(y,t|x)
$$
relabelling $t'$ as $t$.
Path integral from Langevin
We apply the Chapman-Kolmogorov equation many times:
$$
p(y,t|x)=\int\prod_{i=1}^{N}dz_{i}\, p(y,t_{N}|z_{N})\ldots p(z_{2},t_{2}|z_1)p(z_{1},t_{1}|x)
$$
And then use $(\ast)$ to replace each  along with the identity $\delta(x) = \int dk \exp(ikx)$ we get a sequence of averages over complex exponentials:
$$
p(y,t|x)=\int Dz Dk\, \left\langle e^{ik_N(z_N-z_{N-1}-h_N)}\right\rangle\ldots \left\langle e^{ik_2(z_2-z_{1}-h_2)}\right\rangle \left\langle e^{ik_1(z_1-x-h_1)}\right\rangle
$$
however we can still do better. Knowing the probability distribution of $h$ (usually Gaussian in physics) we can write 
$$
\left\langle O \right\rangle = \int dh P(h) O
$$
thus after replacing every averaged over exponential we get
$$
p(y,t|x)=\int Dz Dk Dh\, e^{i\int dt \,k\dot{z}} P[h_t]
$$
and when you perform the integrations over the paths of $k$ and $h$ you will get your path integral.
Path integral from Fokker-Planck
I am less sure about how to go about directly showing the equivalence without the presence of a Langevin equation. However if we create a Lagrangian by multiplying the F-P equation by an auxiliary field and then integrate over all paths of our probability density and our auxiliary field, I have a feeling this will work but I am not certain so I will leave my answer here.
The main point of this is to show that the path integral and the FP equation are essentially two different representations of the Chapman-Kolmogorov equation. As far as I can tell they are basically interchangeable and I hope someone more knowledgeable than myself can come in and explain why you would use one over the other in certain situations.
I hope this has been useful!
