$ E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y 1_A )$

Question

IF $Y \in \mathcal{L^1} (\Omega, \mathcal{F}, \Bbb{P})$

and $\mathcal{G_1} , \mathcal{G_2} ,\mathcal{G_3} $are $\sigma $ fields in $\mathcal{F} $

If we assume that $Y$ is $\mathcal{G_1}$ measurable and $\mathcal{G_3} $ is independent of $\mathcal{G_1}\bigvee\mathcal{G_2}$.

How can we prove that $ E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y 1_A )$ for every $A$ formed as $A= B \cap C$ , $B \in \mathcal{G_2} $,$ C \in \mathcal{G_3} $.

And then extend this to $\mathcal{G_2}\bigvee\mathcal{G_3}$ ( by using Dynkin's $π - λ $ theorem).

I found a similar exercise but I was unable to prove this problem .

Answer 1

By definition of $A$, $$ E [E (Y\mid\mathcal{G_2}) 1_A ] = E [E (Y\mid\mathcal{G_2}) 1_B1_C ].$$ Since $E (Y\mid\mathcal{G_2})$ is $\mathcal G_2$-measurable, so is $E (Y\mid\mathcal{G_2}) 1_B$ hence $1_C$ is independent of $Y\mid\mathcal{G_2}) 1_B$ and we get $$ E [E (Y\mid\mathcal{G_2}) 1_A ] = E [E (Y\mid\mathcal{G_2}) 1_B]\mathbb P(C).$$ Using the definition of conditional expectation gives $$ E [E (Y\mid\mathcal{G_2}) 1_A ] = E [Y 1_B]\mathbb P(C). $$ Conclude using the fact that $1_C$ is independent of $Y1_B$, a $\mathcal G_1\vee\mathcal G_2$-measurable random variable.

To conclude, let $\mathcal G$ be the collection of $\mathcal F$-measurables sets $A$ such that $$E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y 1_A ) a.s..$$ One can show that $\mathcal G$ is a $\lambda$-system containing the sets of the form $G_2\cap G_3$, $G_2\in\mathcal G_2$, $G_3\in\mathcal G_3$.

Answer 2

$E[E(Y|G_2)1_B1_C]=E[E(Y1_B|G_2)]E[1_C]=E[Y1_B]E[1_C]=E[Y1_A]$

I am unsure from the wording of your question whether you want help with the pi lambda extension.

In short, the set of A for which the equality holds is easily checked to be a Dynkin system D, sets which are an intersection of elements of $G_2$ and $G_3$ are clearly a pi system P. Hence, the equation is satisfied on $\sigma(P)\subset D$