So I'm looking at a diffusion process with killing with a state- and time-dependent killing rate. This is described in Oksendal's Stochastic differential equations pages 143-145 "The Feynman-Kac Formula. Killing". Basically, you have a generator
$$
L f = -\sum\_i \frac{\partial }{\partial x\_i} A\_i(x,t) f(x) + \frac{1}{2} \sum\_{i,j} \frac{\partial^2}{\partial x\_i \partial x\_j} B\_{i,j}(x,t) f(x) - c(x,t)f(x),
$$
and this corresponds to a process with drift $A$, diffusion $B$ and a killing rate given by $c(x,t)$.


However, no one mentions if the killing rate can be time-dependent. The demonstration uses the stochastic process
$$
Z\_t = exp(-\int\_0^t c(X\_s)) ds,
$$
with $dZ\_t$ given by
$$
dZ\_t = -Z\_t c(X\_t) dt.
$$
However, if $c$ is a function of time too, there would be an additional term in the differential, right? Wikipedia's [Feynman-Kac formula page](http://en.wikipedia.org/wiki/Feynman%25E2%2580%2593Kac_formula) states the problem with a time-dependent potential $V(x,t)$ but then goes on to drop this dependence throughout the page.
Can i go on to use the Feynman-Kac formula if the potential is time-dependent? Do I have to include some additional terms somewhere? My hunch is yes, but I'm not sure how to derive the correct formulation!


Thanks to everyone! Man, stochastic calculus can sure put knots in your brain!



Is the Feynman-Kac formula valid for a time-dependent potential
I was probably a bit confused because of the operators used. The operator shown in the question is the Kolmogorov forward operator. The backwards operator is the adjoint of $L$, given by
$$
L^\* f = \sum\_i A\_i(x,t) \frac{\partial f}{\partial x\_i} + \sum\_{i,j}\frac{B\_{i,j}(x,t)}{2}\frac{\partial^2 f}{\partial x\_i \partial x\_j} - c(x,t) f.
$$
Furthermore, the operator only helps to define the process $X\_t$, which will lead to the Ito formula and then to the Feynman-Kac formula.


