Prove $\operatorname{Corr}(X,\mathbb{E}(X\mid Y)) > 0$

Question

Well, the exercise tell me:

Be $X$ and $Y$ r.v with $\mathbb{E}(X) = \mathbb{E}(Y) = 0$. Prove that $\operatorname{Corr}(X,\mathbb{E}(X\mid\sigma(Y))) \geq 0$

Well my idea is the following:

$\operatorname{Corr}(X,\mathbb{E}(X\mid\sigma(Y))) = \mathbb{E}(X \cdot \mathbb{E}(X\mid\sigma(Y))) - \mathbb{E}(X)\mathbb{E}(\mathbb{E}(X\mid\sigma(Y))) = \mathbb{E}(X \cdot \mathbb{E}(X \mid \sigma(Y)))$ because $\mathbb{E}(X) = 0$.

From here I don't know how can I prove what they ask to me. My idea to continue my solution is to assume that $X$ is $\sigma(Y)$ measurable so $\mathbb{E}(X \mid \sigma(Y)) = X$ and then we will have that $\operatorname{Corr}(X,\mathbb{E}(X\mid\sigma(Y))) = \mathbb{V}(X) \geq 0$.

My problem is that I don't know if I can assume that $X$ is $\sigma (Y)$ measurable.

Answer 1

I think you might rely on the fact that $E[X|Y]$ is $\sigma(Y)$ measurable, so: $$\textrm{Cov}[X,E[X|Y]]=E[E[X|Y]X]=E[E[E[X|Y]X|Y]]=E[E[X|Y]^2]\geq 0$$