If $X$ and $Y$ are both independent from $\mathscr {F}$, is $f(X,Y)$ also independent from $\mathscr{F}$?

Question

Let $X$ and $Y$ are random variables and they are both independent from the $\sigma$-algebra $\mathscr F$. Given a Borel measurable function $f(x,y)$, is $f(X,Y)$ also independent from $\mathscr F$?

I guess it is right, but actually not quite sure. Here is what I do:

Let $A$ be a Borel set, then $B = \{ f(X,Y) \in A \}$ is also a Borel set. I consider the case where $B = B_1 \times B_2$ is a rectangular area. Then given $E \in \mathscr{F}$, $P(\{f(X,Y)\in A \}\cap E) = P((\{ X \in B_1\} \cap E) \cap (\{Y \in B_2\} \cap E))$. It seems I cannot proceed any more. And I even get skeptical whether it is right.

Any help is appreciated!

Answer 1

If $\sigma(X),\sigma(Y)$ and $\mathscr {F}$ are independent, then $f(X,Y)$ is independent from $\mathscr{F}$.

In this respect, your question is similar to asking about pairwise independence. If $\sigma(X)$ and $\mathscr {F}$ are independent and $\sigma(Y)$ and $\mathscr {F}$ are independent, then it is not sufficient to conclude $\sigma(X,Y)$ is independent from $\mathscr{F}$. Of course $\sigma(f(X,Y))$ merely $\subseteq \sigma(X,Y)$. That's gotta be some $f$ to achieve independence.

If $\sigma(X,Y) = \sigma(\sigma(X) \cup \sigma(Y))$ and $\mathscr {F}$ are independent, then $f(X,Y)$ is independent from $\mathscr{F}$.

In this respect, your question is similar to asking if the conclusion holds simply for $\sigma(X) \cup \sigma(Y)$ and $\mathscr {F}$ being independent: $\sigma(X) \cup \sigma(Y)$ isn't even a $\sigma$-algebra!