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?


1 Answer 1


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 $$ \begin{eqnarray*} &&\mathbb{E}_\mu[V\cdot\mathbb{E}_{Y_t}(Z)]\\ &=&\mathbb{E}_\mu[V\cdot\phi(Y_t)]\\ &=&\int\mu(dx_0)\cdots\int K_{t-t_n}(x_n,dz)\,\phi(z)\\ &=&\int\mu(dx_0)\cdots\int K_{t-t_n}(x_n,dz)\, g_0(z)\int K_{s_1}(z,dy_1)g_1(y_1)\cdots\int K_{s_m-s_{m-1}}(y_{m-1},dy_m)g_m(y_m)\\ &=&\mathbb{E}_\mu[f_0(Y_0)\cdots f_n(Y_{t_n})\,g_0(Y_t)g_1(Y_{t+s_1})\cdots g_m(Y_{t+s_m})]\\ &=&\mathbb{E}_\mu[V\cdot (Z\circ\theta_t)]. \end{eqnarray*} $$

  • $\begingroup$ 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! $\endgroup$
    – math
    Jun 24, 2012 at 8:18
  • $\begingroup$ 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 $$f_0(Y_{t_0})\,f_1(Y_{t_1})\,\cdots \,f_n(Y_{t_n})\,g_0(Y_{t})\,g_1(Y_{t+s_1})\,\cdots \,g_m(Y_{t+s_m}).$$ $\endgroup$
    – user940
    Jun 24, 2012 at 13:21
  • $\begingroup$ Now I got it! Thank you so much! You don't believe how much this helps! $\endgroup$
    – math
    Jun 25, 2012 at 8:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.