# Why is $Var(X)= EVar(X|Y)+Var(E(X|Y))$

I See this identity but it does not make sense to me.

$Var(X)= EVar(X|Y)+Var(E(X|Y))$

is this something that should intuitively make sense? Is it saying the variance of X is equal to the expected variance of X given Y plus the variance of Y? I say the second part because since E(X|Y) is a function with Y as variable Var(E(X|Y)) would capture the variance of Y.

• Intuitively to me, the $E[\text{Var}(X\mid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $\text{Var}(E[X\mid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[X\mid Y]=Y$ – Henry Oct 20 '16 at 15:02

The main thing here is this one: $\mathbb{E}(\mathbb{E}[f(X) | Y]) = \mathbb{E} f(X)$.
Just denote $\psi (Y) = \mathbb{E}[X^2 | Y]$, $\phi(Y) = \mathbb{E}[X | Y]$.
Then $$\mathbb{E} \text{Var}[X|Y] = \mathbb{E} \psi (Y) - \mathbb{E} \psi (Y)^2,$$ $$\text{Var}\mathbb{E}[X | Y] = \text{Var} \phi(Y) = \mathbb{E} \psi (Y)^2 - (\mathbb{E}X)^2.$$