If we have two random variables X and Y which share a joint pdf and pmf (there is a discrete and a continuous scenario), how do we calculate:
Var[Var[X|Y]] - I looked at the Theory of Total Variance and it deals with Var[X|Y] but not this. An additional intuitive explanation will also be very much appreciated.
Additionally, does E[Var[X|Y]] = [E[X]]^2 * Var[Y] hold for continuous cases too?
Apologies in advance if the formatting is off. This is my first question on this site. Thanks in advance!
On
Use one of the Definitions for Variance (they are equivalent, so use whichever you find easiest): $$\begin{align}\mathsf{Var}(\mathsf {Var}(X\mid Y))&=\mathsf E(\mathsf{Var}(X\mid Y)^2)-\mathsf E(\mathsf {Var}(X\mid Y))^2\\[1ex] &=\mathsf E\big(\big(\mathsf{Var}(X\mid Y)-\mathsf E(\mathsf{Var}(X\mid Y))\big)^2\big)\end{align}$$
Since you say that you have the joint probability measure function to evaluate, you shall be able to calculate the conditional variance of $X$ under $Y$.
$$\begin{align}\mathsf{Var}(X\mid Y)&=\mathsf E(X^2\mid Y)-\mathsf E(X\mid Y)^2\\[1ex]&=\mathsf E((X-\mathsf E(X\mid Y))^2\mid Y)\end{align}$$
In the continuous case, for $k\in\Bbb N^+$:
$$\mathsf E(X^k\mid Y)= \dfrac{\displaystyle\int_\Bbb R x^kf_{X,Y}(x,Y)~\mathrm d x}{\displaystyle\int_\Bbb R f_{X,Y}(x,Y)~\mathrm d x}$$
And similarly, in the discrete case:
$$\mathsf E(X^k\mid Y)= \dfrac{\sum_{x} x^kf_{X,Y}(x,Y)}{\sum_{x} f_{X,Y}(x,Y)}$$
For any two random variables $X,Y$, we can define a random variable $X\vert Y$ which represents the value of $X$ given the value of $Y$. Thus, according to the standard definition, we have $$V(X\vert Y )= E(X^2\vert Y) - E^2 (X\vert Y)$$ The RHS is some function of $Y$ so $V(X\vert Y)$ would be a function of $Y$ which is a random variable and you can calculate it's variance in the usual manner. I don't see a way to provide a better answer, as it really depends on the random variables and their dependence.
For example, let $Y$ be some RV and $X\vert Y$ be either $Y$ or $-Y$ with equal probabilities. Then $E(X\vert Y)=0$ and $E(X^2\vert Y)=Y^2$, hence $Var(X\vert Y)=Y^2$. It follows that $$V(V(X\vert Y)) = V(Y^2)=E((Y^2)^2)-E^2(Y^2)=E(Y^4)-E^2(Y^2)$$ Choose your favorite $Y$, compute the appropriate expectations and you'll have your answer.