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So my professor asked me to prove this: $$ \mathbf Var(X) = \mathbb E (Var(X|Y))+ Var (\mathbb E (X|Y)) $$

(I apologize for the formating, new to MathJax)

I found this nifty proof here : http://www.macroeconomics.tu-berlin.de/fileadmin/fg124/financial_crises/exercise/Variances.pdf but I think I'm missing something.

If i'm getting this right, E(X) is the first moment and Var(X) is the second moment so how can the second moment (which is the second derivative I think) be so different. We also had to prove that E(X) = E (X|Y) so then the Var(X) = E (Var(X|Y)) + Var(E(X)).

This all seems very circuitous.

tl:dr how are Var(X) and E(X) related and what the heck is $E(X^2)$ and how is that different the $E(X)^2$

KennyWR
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  • I think the answer is here: http://math.stackexchange.com/questions/7040/conditional-and-total-variance?rq=1

    basically, they are like postion -> speed -> acceleration. Kind of. But i'm messing that up I think...

     – KennyWR Jan 28 '14 at 01:22

2 Answers2

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Provided $X$ has a variance (ie, $\mathbb{E}[X^2] < \infty$), you have that $$\operatorname{Var}(X) = \mathbb{E}\left[\left(X-\mathbb{E}[X]\right)^2\right]= \mathbb{E}[X^2] - \mathbb{E}[X]^2.$$ The first by definition, the second by expanding the square and massaging it a bit (using linearity of expectation).

Clement C.
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$Var [Y] = E[Y^2] -[E(Y)]^2 = E[E(Y^2 | X )] - {E[E(Y | X)]}^2 + zero \\ =E[E(Y^2 | X )] - {E[E(Y | X)]}^2 -E{[E(Y | X)]^2} + E{[E(Y | X)]^2}\\ =E[E(Y^2 | X )] - E{[E(Y | X)]^2} + E{[E(Y | X)]^2} - {E[E(Y | X)]}^2\\ = Var [E(Y | X)] + E[Var (Y | X)]$

That is :)

Community
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rodolfoaglez
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