Independence with single random variables implies the independence of collection.

Question

On probability space $\left(\Omega,\mathscr{F},P\right)$, assume a sub $\sigma$-algebra $\mathscr{F}_{0}$ and random variable $X,Y$ are independent with $\mathscr{F}_{0}$. Do we have $\left(X,Y\right)$ is independent with $\mathscr{F}_{0}$?

Since $\sigma\left(X,Y\right)=\sigma\left(\mathscr{G}\right)$where $$\mathscr{G}=\left\{ \omega\in\Omega:X^{-1}\left(\omega\right)\in A,Y^{-1}\left(\omega\right)\in B,A,B\in\mathscr{B}\left(\mathbb{R}\right)\right\}$$ Since $\mathscr{G}$ is a $\pi$-system, we just need to show that whether $\mathscr{F}_{0}$ and $\mathscr{G}$ are independent. But how do we show this? (We may also assume the independence of $X$ and $Y$)

Answer 1

Let $X,Y$ be two i.i.d. random variables, with $\Pr(X=1)=\Pr(X=-1)=1/2$. Let $\mathcal F_0$ be the $\sigma$-algebra generated by the random variable $XY$. Then:

• $X$ is independent of $Y$ by definition;
• $X$ is independent of $\mathcal F_0$ since for $\epsilon,\epsilon'\in\{-1,1\}$, $$\Pr(X=\epsilon;XY=\epsilon')=P(X=\epsilon,Y=\epsilon\epsilon')=1/4=\Pr(X=\epsilon )\Pr( XY=\epsilon') ;$$
• for the same reason, $Y$ is independent of $\mathcal F_0$.
• But $(X,Y)$ is not independent of $\mathcal F_0$, otherwise, a measurable function of $(X,Y)$ would be independent of $\mathcal F_0$, in particular, $XY$, which is not possible since this random variable is not degenerated.