This seems as an easy question, but however I can't handle it. In the following I need this fact:
If $X=(X_t)$ is a Markov process with transition semigroup $(K_t)$ and initial distribution $\mu$ then
$$E\left[\prod_{k=0}^m f_k(X_{t_k})\right]=\int_E \mu (dx_0)f_0(x_0)\int_E K_{t_1}(x_0,dx_1)f_1(x_1)\int_E K_{t_2-t_1}(x_1,dx_2)\dots\int_EK_{t_n-t_{n-1}}(x_{n-1},dx_n)f_n(x_n)$$ for all measurable functions $f_i\colon E \to [0,\infty)$ and $0=t_0<t_1<\dots<t_n$.
With $P_\mu$ I denote the distribution of $X$ under $P$ on $E':=E^{[0,\infty)}$ (space of all functions from $[0,\infty)\to E$). With $P_x$ we denote $P_{\delta_x}$, dirac measure at $x$. Furthermore let $\theta_t$ be a shift operator $E'$, i.e. $\theta_t(g)(s):=g(s+t)$. Now using the above fact, I want to verify this equation:
$$E_\mu [(Z\circ \theta_t) V]=E_\mu[E_{Y_t}[Z]V]$$
where $Y_t$ is the coordinate process on $E'$, i.e. $Y_t(f):=f(t)$. $E_\mu$ should denote the expectation with respect to the measure $P_\mu$. I want to show this fact for a special class of functions, i.e. $Z=\prod_{k=0}^m g_k(Y_{s_k})$ and $V=\prod_{i=0}^n f_i(Y_{t_i})$ with $t_n\le t$. What I did so far: $(Z\circ \theta_t)=\prod_{k=0}^m g_k(Y_{s_k+t})$.
Writing the right hand side down, leads to:
$$E_\mu\left[V \int_E \delta_{Y_t}(dx_0)g_0(x_0)\int_E K_{s_1}(x_0,dx_1)g_1(x_1)\dots \int_E K_{s_m-s_{m-1}}(x_{m-1},dx_m) g_m(x_m)\right].$$
Clearly $$\int_E \delta_{Y_t}(dx_0)g_0(x_0)\int_E K_{s_1}(x_0,dx_1)g_1(x_1)=g_0(Y_t)\int_E K_{s_1}(Y_t,dx_1)g_1(x_1).$$
So far, everything should be correct. But on the LHS, I stuck. First of all, why can I apply the fact above just to $(Z\circ\theta_t)$? Can I treat $V$ just like a constant? How is the correct form of the LHS?
