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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:

By 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

2 Answers 2

You are starting from $x$, not from $0$, so what you wrote about $E_x(\mathbf1(B)\circ \theta_t|\mathcal F^0_t)$ is a bit confusing.To clarify the expression, you'd better write it as $E(\mathbf1(B)\circ \theta_t|\mathcal F^x_t)$. And the equation can be established more clearly:Firstly,thanks to strong markov property, $$E(\mathbf1(B)\circ \theta_t|\mathcal F^x_t)=E(\mathbf1(B)\circ \theta_t| B^x_t) $$ which is the conditional probability of $B$ under the law of Brownian motion evaluated at time $t$ along with the starting point $x$.

And then you calculate the expectation of the above to achieve the probability, which is equal to

$$E(E(\mathbf1(B)\circ \theta_t| B^x_t))=\int P_y(B) P_t(x,dy)$$ The statement then holds. This question want you to confirm zero-one-law under any starting point of Brownian motion.

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The superscript zero in ${\cal F}_t^0$ has nothing to do with the starting point of the process. It simply denotes the "raw" $\sigma$-field ${\cal F}_t^0=\sigma(X_u,u\leq t)$. –  Byron Schmuland Jul 25 at 16:58
    
Yeah but I guess the superscript stands for the starting point if it exists, usually people don't put a superscript there. If the superscript doesn't count, then we can wrote the equation as $E_x(\mathbf 1(B)\circ \theta_t| \mathcal F_t)=E_x(\mathbf 1(B)\circ \theta_t| B_t) $ to express strong markov property in this case. –  Cjay Jul 26 at 2:24

Define the function $\phi(y)=E_y[\mathbf 1_B]$, that is, $\phi(y)=P_y(B)$. Then $E_{B_t}[\mathbf 1_B]$ is the random variable $\phi(B_t)$. The expected value of this random variable can be expressed in terms of the distribution $\nu$ of $B_t$ via $$E(\phi(B_t))=\int \phi(y)\, \nu(dy).\tag1$$ Under $E_x$, the distribution of $B_t$ is $\nu(dy)=P_t(x,dy)$, so substituting back into (1) gives $$ E_x(E_{B_t}[\mathbf 1_B])=E(\phi(B_t))=\int P_y(B)\, P_t(x,dy). $$

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