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I'm solving exercise 2.28 in Revuz/Yor. I was able to prove 1). Unfortunately at 2) I got stuck. I have to show:

Let $B$ be a d-dimensional Brownian motion and $A\in \mathcal{A}:=\cap_t \mathcal{A}_t$, where $\mathcal{A}_t:=\sigma(B_s;s\ge t)$. For any fixed $t$, there is an event $B\in \mathcal{A}$ such that $\mathbf1_{A}=\mathbf1_B\circ \theta_t$, ($\theta$ is the shift operator), then $$P_x[A]=\int P_y(B)P_t(x,dy)$$ and conclude that either $P_\cdot[A]=0$ or $P_\cdot[A]=1$.

I did the following:

Bye Proposition 1.7 (Markov Property) I get:

$$E_x[\mathbf1_B\circ \theta_t|\mathcal{F}_t^0] =E_{B_t}[\mathbf1_B]$$

Well integrating the LHS, we get:

$$E_x[E_x[\mathbf1_B\circ \theta_t|\mathcal{F}_t^0]]=E_x[\mathbf1_A]=P_x[A]$$

Therefore I'm done with the first part, if

$$E_x[E_{B_t}[\mathbf1_B]]=\int P_y(B)P_t(x,dy)$$

My problem is, I do not see why this is true. It would be very nice if someone could tell me, why this equation is true. I run always in trouble using regular conditional probability. Also some hints for the second statement would be appreciated. I'm very thankful for your help.

cheers

math

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@ math : isn't this Chapman-Kolmogorov property of markovian kernel ? –  TheBridge Mar 28 '12 at 18:56
    
@ TheBridge: You mean the equation $ E_x[E_{B_t}[\mathbf1_B]]=\int P_y(B)P_t(x,dy) $ ? Could you make this a little bit more precise in an answer, then I can accept it. If I read the article on Wikipedia to Chapman-Kolmogorov, I do not see why this should follow. –  math Mar 28 '12 at 19:01
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