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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?

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up vote 4 down vote accepted

Note that both $(Z\circ \theta_t)\, V$ and $ \mathbb{E}_{Y_t}[Z]\, V$ are product functions whose expectation can be found using your main equation. Define $$\phi(z)=\mathbb{E}_z(Z)=g_0(z)\int K_{s_1}(z,dy_1)g_1(y_1)\cdots\int K_{s_m-s_{m-1}}(y_{m-1},dy_m)g_m(y_m).$$ Then we calculate $$ \begin{eqnarray*} &&\mathbb{E}_\mu[V\cdot\mathbb{E}_{Y_t}(Z)]\\ &=&\mathbb{E}_\mu[V\cdot\phi(Y_t)]\\ &=&\int\mu(dx_0)\cdots\int K_{t-t_n}(x_n,dz)\,\phi(z)\\ &=&\int\mu(dx_0)\cdots\int K_{t-t_n}(x_n,dz)\, g_0(z)\int K_{s_1}(z,dy_1)g_1(y_1)\cdots\int K_{s_m-s_{m-1}}(y_{m-1},dy_m)g_m(y_m)\\ &=&\mathbb{E}_\mu[f_0(Y_0)\cdots f_n(Y_{t_n})\,g_0(Y_t)g_1(Y_{t+s_1})\cdots g_m(Y_{t+s_m})]\\ &=&\mathbb{E}_\mu[V\cdot (Z\circ\theta_t)]. \end{eqnarray*} $$

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I'm very thankful for you answer. This problem is bothering me for quite a while. I understand the definition and form of $\phi(z)$. Though, I do not understand the second equation, i.e. $E_\mu[V\phi (Y_t)]=\int\mu (dx_0)\dots \int K_{t-t_n}(x_n,dz)\phi(z)$. Where is the function $V$? Maybe I do not understand the general form of $E_\mu[f]$ for a suitable $f$. Also I do not understand your fourth equality. It would be appreciated a lot, if you could explain these things. As I said, this problem is bothering me for quite a while. Anyway, I'm very thankful for your help and patience! – math Jun 24 '12 at 8:18
It may be helpful if you write out my answer showing more of the factors. There is a lot of stuff implied by the dots $\cdots$. Both of the equations that are giving you trouble are your equation; first applied to the product $$f_0(Y_{t_0})\,f_1(Y_{t_1})\,\cdots \,f_n(Y_{t_n})\,\phi(Y_t)$$ and secondly (in my fourth equation) to the product $$f_0(Y_{t_0})\,f_1(Y_{t_1})\,\cdots \,f_n(Y_{t_n})\,g_0(Y_{t})\,g_1(Y_{t+s_1})\,\cdots \,g_m(Y_{t+s_m}).$$ – Byron Schmuland Jun 24 '12 at 13:21
Now I got it! Thank you so much! You don't believe how much this helps! – math Jun 25 '12 at 8:07

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