Prove $E[f(X,Y)\mid Y]=E[f(X,Y)\mid Y,Z]$ if $X$ is independent of $Y$ and $Z$

Question

I would like to prove the following claim (which I think is true):

Suppose $\mathcal{F}_1 \subset \mathcal{F}_2$ are two $\sigma$-fields, $X,Y$ are random variables, and $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is a measurable function. If $X$ is independent of $\mathcal{F}_2$, $Y\in\mathcal{F}_1$, then $E[f(X,Y)\mid \mathcal{F}_1] = E[f(X,Y)\mid \mathcal{F}_2]$.

Intuitively, this is just saying $E[f(X,Y)\mid Y]=E[f(X,Y)\mid Y,Z]$ if $X$ is independent of the pair $(Y,Z)$, but I don't know how to prove the above claim rigorously.

Thanks in advance!

Answer 1

Let $F$ be an element of $\mathcal F_2$. Since $X$ is independent of $\mathcal F_2$, we have
$$\tag{*} \mathbb E\left[f\left(X,Y\right)\mathbf 1_F \right]=\int_{\mathbb R} \mathbb E\left[f\left(x,Y\right)\mathbf 1_F \right] \mathrm dP_X(x) $$
and for all real number $x$, $\mathbb E\left[f\left(x,Y\right)\mathbf 1_F \right] =\mathbb E\left[f\left(x,Y\right)\mathbb E\left[ \mathbf 1_F \mid \mathcal F_1 \right]\right] $ hence $$\mathbb E\left[f\left(X,Y\right)\mathbf 1_F \right]=\mathbb E\left[f\left(X,Y\right)\mathbb E\left[ \mathbf 1_F \mid \mathcal F_1 \right] \right] .$$ Let $Z:=f\left(X,Y\right)$. Since $$\mathbb E\left[Z\mathbb E\left[ \mathbf 1_F \mid \mathcal F_1 \right] \mid\mathcal F_1 \right]= \mathbb E\left[ \mathbf 1_F \mid \mathcal F_1 \right]\mathbb E\left[Z \mid\mathcal F_1 \right]= \mathbb E\left[ \mathbf 1_F \mathbb E\left[ Z \mid \mathcal F_1 \right] \mid\mathcal F_1 \right], $$ we have $$\mathbb E\left[f\left(X,Y\right)\mathbb E\left[ \mathbf 1_F \mid \mathcal F_1 \right] \right]=\mathbb E\left[\mathbb E\left[ f\left(X,Y\right)\mid\mathcal F_1 \right] \mathbf 1_F \right].$$ We showed that for all $F\in\mathcal F_2$,
$$ \mathbb E\left[f\left(X,Y\right)\mathbf 1_F \right]=\mathbb E\left[\mathbb E\left[ f\left(X,Y\right)\mid\mathcal F_1 \right] \mathbf 1_F \right], $$ which is what we wanted.

Proof of (*): we use the following fact: if $X$ is independent of a vector $(V_1,V_2)$, then for any function $g\colon \mathbb R^3\to\mathbb R$, $$\mathbb E\left[g \left(X,V_1,V_2\right) \right] =\int_{\mathbb R} \mathbb E\left[g\left(x,V_1,V_2\right) \right] \mathrm dP_X(x),$$ which follows from an application of Fubini's theorem and the fact that the law of $ \left(X,V_1,V_2\right)$ is the product measure between the laws of $X$ and $\left(V_1,V_2\right)$. We then apply this to $V_1=Y$, $V_2=\mathbf 1_F$ and $g \colon (x,v_1,v_2) \mapsto f(x,v_1)v_2$.