Show that $Var(X)=Var(\Bbb{E}(X|Y))+\Bbb{E}(Var(X|Y))$

Question

Let $X\in L^2(\Omega,\mathcal{F},\Bbb{P})$. Let $Y$ be a random variable, we define

$$Var(X|Y)=\Bbb{E}((X-\Bbb{E}(X|Y))^2|Y))$$

Show that $Var(X)=Var(\Bbb{E}(X|Y))+\Bbb{E}(Var(X|Y))$

Ok, so

$\Bbb{E}(Var(X|Y))=\Bbb{E}((X-\Bbb{E}(X|Y))^2)=\Bbb{E}(X^2)-2\Bbb{E}(X\Bbb{E}(X|Y))+\Bbb{E}(\Bbb{E}(X|Y)^2)$

While

$Var(\Bbb{E}(X|Y))=\Bbb{E}((\Bbb{E}(X|Y))^2)-[\Bbb{E}(\Bbb{E}(X|Y))]^2=\Bbb{E}((\Bbb{E}(X|Y))^2)-[\Bbb{E}(X)]^2$

Adding both, I obtain

$Var(\Bbb{E}(X|Y))+\Bbb{E}(Var(X|Y))=Var(X) + 2(\Bbb{E}(\Bbb{E}(X|Y)^2)-\Bbb{E}(X\Bbb{E}(X|Y)))$

But I don't know how to deal with $\Bbb{E}(X\Bbb{E}(X|Y))$ nor with $\Bbb{E}(\Bbb{E}(X|Y)^2)$. What can I do with those terms?

Answer 1

Note that \begin{align*} \Bbb{E}(Var(X \mid Y))&=\Bbb{E}(X^2)-2\Bbb{E}(X\Bbb{E}(X \mid Y))+\Bbb{E}((\Bbb{E}(X \mid Y))^2)\\ &=\Bbb{E}(X^2)-2\Bbb{E}(E(X\Bbb{E}(X \mid Y) \mid Y ))+\Bbb{E}((\Bbb{E}(X \mid Y))^2)\\ &= \Bbb{E}(X^2) - \Bbb{E}((\Bbb{E}(X \mid Y))^2). \end{align*} The remaining is now straightforward.

Answer 2

I got to

$Var(\Bbb{E}(X|Y))+\Bbb{E}(Var(X|Y))=Var(X) + 2(\Bbb{E}(\Bbb{E}(X|Y)^2)-\Bbb{E}(X\Bbb{E}(X|Y)))$

So a possibility is to show

$\Bbb{E}(\Bbb{E}(X|Y)^2)-\Bbb{E}(X\Bbb{E}(X|Y))=0$

So I can proceed in the following way:

$\Bbb{E}(X\Bbb{E}(X|Y))=\Bbb{E}(\Bbb{E}(X\Bbb{E}(X|Y))|Y)$ Since $\Bbb{E}(X|Y)=g(Y)$ for some $g$, we have that $\Bbb{E}(\Bbb{E}(X\Bbb{E}(X|Y))|Y)=\Bbb{E}(\Bbb{E}(Xg(Y))|Y)=\Bbb{E}(g(Y)\Bbb{E}(X|Y))=\Bbb{E}(\Bbb{E}(X|Y)^2)$

And therefore, $\Bbb{E}(\Bbb{E}(X|Y)^2)-\Bbb{E}(X\Bbb{E}(X|Y))=0$ concluding that:

$Var(\Bbb{E}(X|Y))+\Bbb{E}(Var(X|Y))=Var(X)$