Independence of a Random Variable from a Sigma Field and Expectation

Question

The question is:

"Show that $X$ is independent of $\sigma(Y)$ if and only if for bounded and measurable $f,g$, $E[f(X)g(Y)]=E[f(X)]E[g(Y)]$."

I think I have managed to prove the forward statement by conditioning the expectation on $Y$, but I'm having difficulty doing the reverse. I assume we want to show that $E[X|\sigma(Y)]=E[X]$ but I'm stumped after that.

Answer 1

Proving that $X$ is independent of $\sigma\left(Y\right)$ is the same as proving that $X,Y$ are independent which on its turn is the same as proving that: $$P\left(X\in A,Y\in B\right)=P\left(X\in A\right)P\left(Y\in B\right)$$ for arbitrary Borel sets $A,B$.

Taking $f=\mathbf{1}_{A}$ and $g=\mathbf{1}_{B}$ we find: $$P\left(X\in A,Y\in B\right)=\mathbb{E}f\left(X\right)g\left(Y\right)=\mathbb{E}f\left(X\right)\mathbb{E}g\left(Y\right)=P\left(X\in A\right)P\left(Y\in B\right)$$

If conversely $X,Y$ are independent then so are $f\left(X\right)$ and $g\left(Y\right)$ for arbitrary Borel measurable functions.

This because: $$P\left(f\left(X\right)\in U,g\left(Y\right)\in V\right)=P\left(X\in f^{-1}\left(U\right),Y\in g^{-1}\left(V\right)\right)=$$$$P\left(X\in f^{-1}\left(U\right)\right)P\left(Y\in g^{-1}\left(V\right)\right)=P\left(f\left(X\right)\in U\right)P\left(g\left(Y\right)\in V\right)$$ for arbitrary Borel sets $U,V$.

If in that case moreover the functions $f$ and $g$ are bounded, so that the expectations of $f\left(X\right),g\left(Y\right)$ and $f\left(X\right)g\left(Y\right)$ exist then:$$\mathbb{E}\left[f\left(X\right)g\left(Y\right)\right]=\mathbb{E}f\left(X\right)\mathbb{E}g\left(Y\right)$$

Answer 2

First note that $\sigma(Y)$ consists of the sets $Y^{-1}(A)$ with $A$ Borel. Hence $X$ is independent if $\sigma(Y)$ iff $P(X^{-1}(A)\cap Y^{-1}(B))=P(X^{-1}(A))P(Y^{-1}(B))$ for all Borel sets $A$ and $B$.

If this condition holds then $Ef(X)g(Y)=Ef(X)Eg(Y)$ whenever $f$ and $h$ are indicator functions . This implies that the same holds when $f$ and $g$ are simple functions. Since bounded measurable functions are uniform limits of simple functions the equation holds for all such $f$ and $g$ (by Bounded Convergence Theorem).

The converse part follows by taking $f$ and $g$ to be indicator functions.