Vtyjkyuk

My problem is the following:
$X$ and $Y$ are two random variables and $mathcalF$ is a $sigma$-algebra. Given that $X$ and $Y$ are independent, and that $X$ is independent of $mathcalF$, can I affirm that
$$mathbbE(XY mid mathcalF) = mathbbE(X) cdot mathbbE(Y mid mathcalF)?$$
My intuition is yes but I did not manage to find a formal proof. Any idea ?

Thanks a lot.

No, in general the statement is not correct.

Example Consider sets $A_1,A_2,A_3$ such that $mathbbP(A_1 cap A_2 cap A_3) = 0$ and $mathbbP(A_i)>0$, $mathbbP(A_i cap A_j) = mathbbP(A_i) cdot mathbbP(A_j)$ for any two $i,j in 1,2,3$. Set $X:=1_A_1$, $Y=1_A_2$, $mathcalF := sigma(1_A_3)$. Then $X$ and $Y$ are independent, as well as $X$ and $mathcalF$, so the assumption on the independence is satisfied. For $F:= A_3$ we have

$$int_F X cdot Y , dmathbbP = int 1_A_1 cdot 1_A_2 cdot 1_A_3 , dmathbbP = mathbbP(A_1 cap A_2 cap A_3) = 0 tag1 $$

On the other hand,

$$int_F mathbbE(X) cdot mathbbE(Y mid mathcalF), dmathbbP = mathbbP(A_1) cdot underbracemathbbE(Y)_mathbbP(A_2) cdot mathbbP(A_3)>0 tag2$$

where we used that $Y$ and $mathcalF$ are independent (hence $mathbbE(Y mid mathcalF)=mathbbE(Y)$). Since by the definition of conditional expectation

$$0 stackrel(1)= int_F X cdot Y , dmathbbP stackrel!= int_F mathbbE(X cdot Y mid mathcalF) , dmathbbP$$

we conclude from $(2)$ that

$$mathbbE(X cdot Y mid mathcalF) neq mathbbE(X) cdot mathbbE(Y mid mathcalF)$$

But: If $sigma(X)$ is independent of $sigma(Y,mathcalF)$ (i.e. the smallest $sigma$-algebra containing $mathcalF$ and $sigma(Y)$), then the claim holds as the following proof shows:

Denote by $mathcalG := sigma(mathcalF,sigma(Y))$ the $sigma$-algebra generated by $mathcalF$ and $Y$. Since $sigma(X)$ and $mathcalG$ are independent, we have

$$mathbbE(X mid mathcalG) = mathbbE(X)$$

Consequently, by the tower property

$$mathbbE(X cdot Y mid mathcalF) = mathbbE(Y cdot underbracemathbbE(X mid mathcalG)_mathbbE(X) mid mathcalF) = mathbbE(X) cdot mathbbE(Y mid mathcalF)$$