1
$\begingroup$

I'm trying to prove some basic Covariance relationships, specifically this: $$ \operatorname{cov}(E(Y\mid X),Y-E(Y\mid X))=0 $$ I managed to show that: $$ \operatorname{cov}(E(Y\mid X),Y-E(Y\mid X))=E\left[Y\cdot E(Y\mid X)-(E(Y\mid X))^2\right] $$ but I don't know what to do now. The main theorem that I'm using for this is the law of iterated expectations: $$E\left(E\left(X\mid Y\right)\right)=E(X)$$ and I don't know how to apply it in this situation. Can anyone give me some hints or ideas?

Thanks for helping!!! :D

$\endgroup$
1
$\begingroup$

What you have to use is: $$ E(Y\cdot E(Y\mid X))=E\left(E(Y\cdot E(Y\mid X)\mid X))\right)=E(E(Y\mid X)E(Y\mid X)) =E((E(Y\mid X))^2) $$ where the first equality is given by the law of iterated expectations, and the second by noticing that $E(Y\mid X)$ is already a function of $X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.