If X is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$ measurable, why are $X,Y$ independent?

Question

Suppose there is a probability space $(\Omega, \mathcal{F}, P)$ and there exists a $\sigma$-algebra called $\mathcal{G}\subseteq \mathcal{F}$. Assume that $Y$ is a random variable which is $\mathcal{G}$-measurable and $X$ is independent of $\mathcal{G}$. My lecturer seems to think this means that $X,Y$ are independent?

I do not understand that this is true, since it seems to me that $\mathcal{G} \subseteq \sigma(Y)$, but there is not necessarily equality. My questions:

(1) Is the statement in the box above true?

(2) Can you prove it?

Answer 1

Let $\mathcal{H} = \sigma(X)$. Since $X$ and $\mathcal{G}$ are independent, for any $A_1 \in \mathcal{H}$ and $A_2 \in \mathcal{G}$,

$$\mathbb{P}(A_1\cap A_2) = \mathbb{P}(A_1)\mathbb{P}(A_2).$$

Now, let $B_1,B_2$ be measurable sets in the state spaces of $X$ and $Y$ respectively. Then, $\{X \in B_1\} \in \mathcal{H}$ and $\{Y \in B_2\} \in \sigma(Y)\subseteq \mathcal{G}$, so

$$\mathbb{P}(X\in B_1,Y\in B_2) = \mathbb{P}(X\in B_1)\mathbb{P}(Y\in B_2).$$

Answer 2

It's been many years since I took basic probability theory but I think the lecturer is right. I would try the proof below,

$X$ is independent of $\mathcal{G}$ means for all Borell sets $B_1$ and sets $A\in\mathcal{G}$, $$ P(X^{-1}(B_1)\cap A) = P(X^{-1}(B_1))P(A) $$

$Y$ measurable in $\mathcal{G}$ means that for all Borell sets $B_2$, $$ Y^{-1}(B_2) = A \in \mathcal{G} $$

Hence for all Borell sets $B_1,B_2$ $$ P(X^{-1}(B_1)\cap Y^{-1}(B_2)) = P(X^{-1}(B_1) \cap A) = P(X^{-1}(B_1))P(A) = P(X^{-1}(B_1))P(Y^{-1}(B_2)) $$ which shows that $X,Y$ is independent.

Answer 3

For sub-$\sigma$-algebras $\mathscr A$, $\mathscr B$, $\mathscr A_1$ of $(\Omega, \mathscr F, \mathbb P)$,

If $\mathscr A$ and $\mathscr B$ are independent, then $\mathscr A_1$ and $\mathscr B$ are independent $\forall \mathscr A_1 \subseteq \mathscr A$.

This is almost immediate from the definition of what it means for 2 $\sigma$-algebras to be independent: pick a ball from each box (pick an element of each $\sigma$-algebra). Then the two balls (elements) are independent. When we choose a sub-sub-$\sigma$-algebra, we're just restricting the choice of balls (elements) for one of the boxes (sub-$\sigma$-algebras), but the restriction doesn't matter since the definition holds for any element.