so I've been trying to prove the following relation for quite a while now but haven't really figured it out yet, maybe I'm missing something really obvious.
To prove:
$\frac{Var(E\left[X|Y\right])}{Var(X)} = \frac{Var(E[Y|X])}{Var(Y)}$
This problem seems to be related to the law of total Variance: $Var(X) = Var(E\left[X|Y\right]) + E\left[Var(X|Y)\right])$
So if you look at it in terms of the explained and unexplained parts the equation above basically states that the percentage of the Variance of X explained by Y is the same as the percentage of the Variance of Y explained by X which seems a bit odd to me.
Does anyone have an idea how to tackle this problem? Or is there already a similar or related relation which can be used to proof this equation?
Many thanks in advance!
The equality is not true in general and here is a counterexample. Let $(X,Y)$ has joint pmf $$ \mathbb P(X=0,Y=0)=\frac14, \quad \mathbb P(X=0,Y=1)=\frac14, \quad \mathbb P(X=1,Y=-1)=\frac12. $$ Then $\mathbb E(Y\mid X)=\frac12 \mathbb 1_{\{X=0\}}-\mathbb 1_{\{X=1\}}$, $\text{Var}(\mathbb E(Y\mid X))=\frac{9}{16}$, $\text{Var}(Y)=\frac{11}{16}$ and $$ \frac{\text{Var}(\mathbb E(Y\mid X))}{\text{Var}(Y)} = \frac{9}{11}.$$ And $\mathbb E(X\mid Y)=\mathbb 1_{\{Y=-1\}}$, $\text{Var}(\mathbb E(X\mid Y))=\frac{1}{4}$, $\text{Var}(X)=\frac{1}{4}$ and $$ \frac{\text{Var}(\mathbb E(X\mid Y))}{\text{Var}(X)} = 1.$$