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I have a Markov process $X_t$ then I know that $E[f(X_t)|X_s=x]=g_1(x)$ for any measurable defined everywhere $f$ and some $g_1$, for $t>s$. Is that also true that $E[f(X_t, X_q)|X_s=x]=g_2(x)$ where $q>t>s$ for some $g_2$? If so are $g_1$ and $g_2$ the same? thanks!

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Where is $f$ defined? –  Davide Giraudo Apr 18 '12 at 18:54
    
it a measurable function defined everywhere. say it is $f(x)=e^x$ and $e^{x+y}$. The point of the question that my function depends basically on two random variables $X_t$ and $X_q$ but by the Markov property the expectation of each depends only on the value at s time, but I am confused what is the expectation of both t the same time. –  Medan Apr 18 '12 at 19:29

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In the first case, the result $g_1(x)$ really depends on $x$, $t-s$ and $f$ (and the transition kernel of the Markov process), but nothing else. In the second case, the result $g_2(x)$ really depends on $x$, $q-t$, $t-s$ and $f$ (and the transition kernel of the Markov process), but nothing else. If the transition kernel is $(P_t)_{t\geqslant0}$, the result in the first case is $$ g_1(x)=\int f(y)P_{t-s}(x,\mathrm dy), $$ and the result in the second case is $$ g_2(x)=\iint f(y,z)P_{t-s}(x,\mathrm dy)P_{q-t}(y,\mathrm dz). $$

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Thank You very much. Now I understand what are those functions and see how they are formed. –  Medan Apr 22 '12 at 18:59

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