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This comes from page 2 of General Theory of Markov Processes by Michael Sharpe, with some changes in notation.

Suppose $$P(A_1\cap A_2\,|\, {\cal F}_{=t} )= P(A_1 \,|\, {\cal F}_{=t} )P( A_2\,|\, {\cal F}_{=t} )$$ for all $A_1\in {\cal F}_{\leq t}$ and $A_2\in {\cal F}_{\geq t}$. Using well known properties of conditional expectations, \begin{eqnarray*} P(A_1\cap A_2) &=&P(P(A_1\cap A_2\ |\ {\cal F}_{=t}))\cr &=&P\left( P(A_1\ |\ {\cal F}_{=t})\ P(A_2 \ | \ {\cal F}_{=t}) \right)\cr &=&P(P(A_2\ |\ {\cal F}_{=t}) ; A_1). \end{eqnarray*}

My question is by what property of conditional expectations, we can have $$ P\left( P(A_1\ |\ {\cal F}_{=t})\ P(A_2 \ | \ {\cal F}_{=t}) \right) = P(P(A_2\ |\ {\cal F}_{=t}) ; A_1)?$$

What is the general form for the property of conditional expectation?

Any reference?

Thanks in advance!

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@Byron: I still don't understand your reply there. – Ethan May 4 '11 at 5:12
up vote 3 down vote accepted

For every sigma-algebra G and every (suitably integrable) random variables X and Y such that Y is G-measurable, one has E(XY)=E(E(X|G)Y). This is one of the two conditions that characterize E(X|G), the other one being that E(X|G) is G-measurable. In particular, for every event A, E(Y;A)=E(P(A|G)Y).

Apply this to G your sigma-algebra $\mathcal{F}_{=t}$, A the event $A_1$ and Y the conditional probability of $A_2$ conditionally on G.

Note: in your post five Ps should be Es.

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Thank you! In definition of E(X|G), I remember it is E(X)=E(E(X|G); A) for any A in G. So is E(XY)=E(E(X|G)Y) for any G-measurable random variable Y really part of its definition? Why is E(XY)=E(E(X|G)Y) true? – Ethan May 4 '11 at 11:08
It is equivalent to ask that E(X;A)=E(E(X|G);A) for every A in G or to ask that E(XY)=E(E(X|G)Y) for every (bounded) G-measurable random variable Y. The latter implies logically the former but it is a standard exercise to prove that the two are in fact equivalent. – Did May 4 '11 at 11:42
From the former to the latter, is it proved by using simple functions to approximate Y when Y is nonnegative? – Ethan May 4 '11 at 11:56
Yes. This is called a functional monotone class theorem, see… – Did May 4 '11 at 12:45

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