Show $\operatorname{Var}(X)=E(\operatorname{Var}(X\vert \mathcal{F}))+\operatorname{Var}(E(X\vert \mathcal{F}))$

Question

Show $\operatorname{Var}(X)=E(\operatorname{Var}(X\vert \mathcal{F}))+\operatorname{Var}(E(X\vert \mathcal{F}))$

I think the fact that:

$\operatorname{Var}(X\vert \mathcal{F})=E[X^2\vert \mathcal{F}]-E[X\vert \mathcal{F}]^2(*)$

may help me:

using $(*)$

$E(E[X^2\vert \mathcal{F}]+E[X\vert \mathcal{F}]^2)+E[E(X\vert \mathcal{F})^2\vert \mathcal{F}]-E[E(X\vert \mathcal{F})\vert \mathcal{F}]^2$

I know that via the tower property $E[E[X\vert \mathcal{F}]]=E[X]$ BUT does this also hold for $E[E[X\vert \mathcal{F}]^2]$, i.e. is $E[E[X\vert \mathcal{F}]^2]=E[X]?$ Any why?

Any ideas on how to solve the underlying problem

Answer 1

The exercise is nothing more than just writing out the right hand side and seeing that the two terms that you don't know what to do with cancel.

$$ E(Var(X|\mathcal{F}))=E (X^2)-E((E(X|\mathcal{F}))^2), $$ and the last term is sort impossible to say something intelligent about.

Fortunately, $$ Var(E(X|\mathcal{F}))=E((E(X|\mathcal{F}))^2)-(EX)^2, $$ where we've used that $E(E(X|\mathcal{F}))=EX$ by definition.

Adding these together, we get

$$ E(Var(X|\mathcal{F}))+Var(E(X|\mathcal{F}))=E(X^2)-(EX)^2=Var(X) $$

Answer 2

Partial answer. A valid proof has already been given. I would like to point out why $E((E(X|\mathcal F)^{2}) \neq EX$ in general. If $X$ is independent of $\mathcal F$ the this equation becomes $(EX)^{2}=EX$ or $EX=0$ or $1$. Also, if $X$ is measurable w.r.t. $\mathcal F$ it becomes $EX^{2}=EX$. This is false for any mean $0$ random variable which is not $0$ almost surely.

Answer 3

$\def\Var{\operatorname{\sf Var}}\def\E{\operatorname{\sf E}}\def\F{\mathcal F}$You wish to show that: $\Var(X)=\E(\Var(X\mid\F))+\Var(\E(X\mid\F))$

By definition of variance

$$\begin{align}\Var(X)&=\E(X^2)-\E(X)^2\\[2ex]\E\big(\Var(X\mid\F)\big)&=\E\big(\E(X^2\mid\F)-\E(X\mid\F)^2\big)\\&=\E\big(\E(X^2\mid\F)\big)-\E\big(\E(X\mid\F)^2\big)\\[2ex]\Var\big(\E(X\mid\F)\big)&=\E\big(\E(X\mid\F)^2\big)-\E\big(\E(X\mid\F)\big)^2\end{align}$$

So...

$~~$

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PS: Don't forget to use the Law of Total Expectation, $\E(g(X))=\E\big(\E(g(X)\mid\F)\big)$

I know that via the tower property $E[E[X\vert \mathcal{F}]]=E[X]$ BUT does this also hold for $E[E[X\vert \mathcal{F}]^2]$, i.e. is $E[E[X\vert \mathcal{F}]^2]=E[X]?$ Any why?

It doesn't also hold. Fortunately it does not have to.

You just need $\E(\E(X\mid\F))=\E(X)$ and $\E(\E(X^2\mid F))=\E(X^2)$