I have a problem:
Let $X$ and $Y$ in $\mathcal L^{2}$, and put $Y' = E(Y |X)$. Show that:
- $Cov(Y, X) = Cov(Y', X)$
- $Corr(Y, X) ≤ Corr(Y' , X)$
I have tried to solve 1 it by deriving the definition of covarience:
$E(XY) - E(X)E(Y) = E(XY')-E(X)E(Y')$
and implacing $Y' = E(Y|X) = \int Y *f_{Y|X}dy$, but can't see the further way.
Also, to solve 2, it's just needed to proove, that $\sigma _{Y} > \sigma _{Y'} $, which is intuativly understandble, but i can't show it neither.
You are down the correct path. You just have to compute the $E(Y^{'})$ realizing that it is a function of $X$, $$ E(Y^{'}) = \int Y^{'} f_x(x) dx = \int y \ f_{y|x}(y|x) \ f_x(x) \ dx dy = \int y \ f_{x,y}(x,y) \ dx dy = E(Y) $$ Which follows from the definition of conditional pdf. An alternate approach is to directly use the law of total expectation: $$ E(Y) = E ( E(Y|X) ) = E (Y') $$ Follow the same approach on the variance. Follow the same approach on $E(Y' \ X)$ to get that $E(Y'X) = E(YX)$. For the second part we apply the law of total variance: $$ Var(Y) = E ( Var(Y|X) ) + Var( E(Y|X) ) = E ( Var(Y|X) ) + Var( Y' ) > Var( Y' ) $$ Since $ E ( Var(Y|X) )>0$.