# How do I interpret the following $E^{X_S}[f(X_t)]$?

Given a Markov family $$X=\{X_t, \mathcal{F}_t, t\ge 0\}$$ on some $$(\Omega,\mathcal{F})$$,together with a family of probability measures $$\{P^x\}_{x \in \mathbb{R}^d}$$ on $$(\Omega,\mathcal{F})$$ and a nice function $$f$$ so that $$f(X_t)$$ is always integrable w.r.t to the entire spectrum of probability measures $$\{P^x\}_{x \in \mathbb{R}^d}$$, and S is a finite optional time. How do I interpret the following function?

$$E^{X_S}[f(X_t)]$$

Can I think of think of it as $$\omega \mapsto E^{X_{S(\omega)}(\omega)}[f(X_t)]$$ This interpretation seems very strange to me since we first fix $$\omega$$ to compute $$X_S(\omega)(\omega)=x$$ and then compute the expectation of the random variable $$\omega \mapsto X_t(\omega)$$ on $$\Omega$$ w.r.t to the probability measure $$P^x$$.

This is bothering me since I cant follow statement for example in $$e'$$ in Proposition 6.7 shown below in Karatzas and Shreve

• Note that $P_t f(x) := \mathbb{E}^x f(X_t)$ is the semigroup associated with the Markov process $(X_t)_{t \geq 0}$. The expression $\mathbb{E}^{X_s} f(X_t)$ hence equals $P_t f(X_s)$. Nothing too strange about that.... – saz Dec 3 '18 at 15:48