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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 $Y$.

Lê Thành Đạt
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Qwertford
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    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

1 Answers1

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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 $$\mathbb{E} \text{Var}[X|Y] = \mathbb{E} \psi (Y) - \mathbb{E} \psi (Y)^2= \mathbb{E}(X^2)-\mathbb{E} \psi (Y)^2$$ $$\text{Var}\mathbb{E}[X | Y] = \text{Var} \phi(Y) = \mathbb{E} \psi (Y)^2 - (\mathbb{E}X)^2.$$

Sum this expressions and obtain the equality needed.

Kadmos
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Denis Korzhenkov
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