Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
- $E$ be a $\mathbb R$-Banach space;
- $(X_t)_{t\ge0}$ be an $E$-valued right-continuous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with invariant distribution $\mu$;
- $f:E\to[0,\infty)$ be Borel measurable.
Assume that $X_0$ is already distributed according to $\mu$ (and hence $(X_t)_{t\ge0}$ is stationary). How can we calculate $$\operatorname E\left[\exp\left(-\int_0^tf(X_s)\:{\rm d}s\right)f(X_t)\right]\tag1?$$
I don't know how to proceed here. Since $X_0\sim\mu$, we know that $X_t\sim\mu$ for all $t\ge0$. But what does that mean for the distribution of the whole process $X$? Since the process is not independent (though there is a kind of independence from the Markov property which we may use instead), we cannot conclude that $$\left(X_{t_1},\ldots,X_{t_n}\right)\sim\mu^{\otimes n}\tag2$$ for all $0\le t_1<\cdots<t_n$ and $n\in\mathbb N$ ...