$E(Y|Χ_1,Χ_2)=E(Y|Χ_1) $

Question

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

and random Vectors $X_1$ , $X_2$ such that $\sigma(X_2)$ independent of $\sigma(Y,X_1)$ .

I want to show that $E(Y|Χ_1,Χ_2)=E(Y|Χ_1) $ a.s.

$$$$ Thought :

We could maybe use the below results and if we define $\mathcal{G_1}, \mathcal{G_2} ,\mathcal{G_3}$ in an appropriate way . We may be able to show the desired result.

$$$$ Thus, $\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 $\mathcal{G_1}\bigvee\mathcal{G_2}$.

$ 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} $.

I should define $\mathcal{G_1} \mathcal{G_2} \mathcal{G_3}$ in an apropriate way and we will be able to show that

$E(Y|X_1,X_2)=E(Y|X_1) $ a.s.

Answer 1

Extending $E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y 1_A )$ to the sets $A$ of $\mathcal G_2\vee\mathcal G_3$, we derive that $$ \mathbb E\left[\left(\mathbb E\left[Y\mid\mathcal G_2\right]-Y\right)\mid \mathcal G_2\vee\mathcal G_3\right]=0 $$ and since $\mathcal G_2\subset \mathcal G_2\vee\mathcal G_3$, this can be written as $$ \mathbb E\left[Y\mid\mathcal G_2\right]=\mathbb E\left[Y\mid\mathcal G_2\vee\mathcal G_3\right]. $$ Now let $\mathcal G_1=\sigma(Y)$, $\mathcal G_2=\sigma(X_1)$ and $\mathcal G_3=\sigma(X_2)$.