Feynman-Kac proof

Question

I have seen numerous proofs, showing that if $u\left(x,t\right)$ satisfies a PDE of the form:
$$\frac{\partial u}{\partial t}+\mu\left(x,t\right) \frac{\partial u}{\partial x}+\frac{1}{2}{\left(\sigma\left(x,t\right)\right)}^2 \frac{\partial^2 u}{{\partial x}^2}-V\left(x,t\right) u+f\left(x,t\right)=0$$
then, $M_s$ defined as:
$$M_s=e^{-\int\limits_{t}^{s}{V\left(\tau,X_{\tau}\right)}d\tau}u\left(s,X_s\right)+\int\limits_{t}^{s}{e^{-\int\limits_{t}^{r}{V\left(\tau,X_{\tau}\right)d\tau}}f\left(r,X_r\right)dr}$$
where $X_t$ is a random process, with evolution defined as:
$$dX\left(x,t\right)=\mu\left(x,t\right) dt+\sigma\left(x,t\right) dW$$
is a martingale, and as such, its average at multiple realizations at an arbitrary point in time (i.e. terminating time), could be used as an estimator for the solution. However, I could not find any proof that such a solution actually satisfies the original equation. Any good reference or tip would be greatly appreciated. Thanks in advance!

Answer 1

It looks like you mean $$ dX_s=\mu(X_s,s)\,ds+\sigma(X_s,s)\,dW_s\,,\quad s\ge t\,,\quad X_t=x\,. $$ Because $M_s$ is a martingale, $\mathbb E[M_s]=M_t$ which is seen to collapse to $u(t,X_t)=u(t,x)\,.$

Apart from that: The book Brownian Motion and Stochastic Calculus by Karatzas & Shreve is the best reference on Feynman-Kac that I know.