Prove that if the random variables $X$ and $Y$ are such that $EX^2$ and $EY^2$ are finite, then $$DX=E(D(X\mid Y))+D(E(X\mid Y))$$ where $D$ denotes the variance and $E$ denotes the expectation.
This semestar, we have been doing conditional expectation, characteristic functions, convergences of sequences of random variables,law of large numbers, central limit theorems, to name a few. But I cannot seem to get an idea on how to do this. Would very much appreciate if someone has done this before.
On
Hints: Write out the two variances on the right side in detail. You will then have fours terms. Two of these terms will cancel and the remaining two will combine to yield the variance of $X$. You will need to use the law of iterated expectations twice: $E(E(X^2|Y))=E(X^2)$ and $E(E(X|Y))=E(X)$.
$\mathrm{Var}X$ is equal to $\mathbf{E}X^2-(\mathbf{E}X)^2$. Therefore $$ \mathrm{Var}X=\mathbf{E}[\mathbf{E}[X^2|Y]]-(\mathbf{E}[\mathbf{E}[X|Y]])^2 $$ which is equal to $$ \mathrm{Var}X=(\mathbf{E}[\mathbf{E}[X^2|Y]]-\mathbf{E}[(\mathbf{E}[X|Y])^2])+(\mathbf{E}[(\mathbf{E}[X|Y])^2]-(\mathbf{E}[\mathbf{E}[X|Y]])^2). $$ Done!
Ps. The unique thing that you have to notice, apart from the definition of variance, is that the expected value of the conditional r.v. is equal to the expected value of the unconditional one. But this follows from the definition of conditional r.v., right?