Let $X$ and $Y$ be two random variables; I want to show now two things that should actually be not too hard to prove:
If $\mathbb{E}(X\mid Y)=\mathbb{E}(X)$, then Cov$(X,Y)=0$
If $\mathbb{E}(X\mid Y)=a+bY$, then b=Cov$(X,Y)/\sigma_X^2$, where $\sigma_X^2$ is the variance of $X$ and we assume that it exists and is non-zero.
Now for both things I should use the law of iterated expectations:
$$\mathbb{E}(\mathbb{E}(A\mid B))=\mathbb E (A)$$However I don't really see how to do it - can someone give me a hint?
Hints:
For each question, use the fact that:
$\mathsf {Cov}(X,Y)~{=\mathsf E(XY)-\mathsf E(X)\,\mathsf E(Y)\\=\mathsf E(\mathsf E(X\mid Y)Y)-\mathsf E(X)\,\mathsf E(Y)}$
For (b) also use that: $\mathsf E(X)=\mathsf E(\mathsf E(X\mid Y))$ and $\sigma_X^2= \mathsf E(\mathsf E(X^2\mid Y))-\mathsf E(\mathsf E(X\mid Y))^2$