Are there any particular properties that \begin{align*} Var(Var(X\mid Y)) \end{align*} satisfies so that we can derive any upper and lower bounds on it. For example, if we replace $Var$ with expectation we have \begin{align*} E[E[X\mid Y]]=E[X] \end{align*}

This question is somewhat related to the one found here.

One way to proceed is to use \begin{align*} Var(Var(X\mid Y))&=E[Var^2(X\mid Y)]-(E[Var(X\mid Y)])^2\\ &=E[Var^2(X\mid Y)]-MMSE^2(X|Y)\\ \end{align*}

where we can bound $E[Var^2(X\mid Y)]$ as \begin{align*} E[Var^2(X\mid Y)]&=E[(E[(X-E[X|Y])^2|Y])^2] \\ &\le E[(E[(X-E[X|Y])^4|Y)]\\ &= E[(X-E[X|Y])^4] \end{align*} and we have a bound

\begin{align*} Var(Var(X\mid Y)) &\le E[(X-E[X|Y])^4]-MMSE^2(X|Y)\\ &=E[(X-E[X|Y])^4]-(E[(X-E[X|Y])^2])^2 \end{align*}

The question is can we do better and find a tighter bound?

If you need further assumption we can assume that $Y=X+Z$ where $X$ and $Z$ are finite variance and zero mean and independent.

I would be very grateful for any ideas you guys might have?


1 Answer 1


It is an old question so I don't know whether it makes sense to answer it now... Anyway, without assuming $Y=X+Z$ with $X$ independent of $Z$, your upper bound is sharp. To see this, note that it is reached if $$X= E(X|Y) + c(Y) e,$$ where $e$ is a Rademacher variable independent of $Y$. Then, $$E((X- E(X|Y))^2|Y)^2 = c(Y)^4 = E((X- E(X|Y))^4|Y)$$ and $E((X- E(X|Y))^2|Y)^2 \leq E((X- E(X|Y))^4|Y)$ is the only inequality you used (I think $X= E(X|Y) + c(Y) e$ is actually the only case where equality occurs).

Regarding the lower bound, it is actually 0: simply consider $X=E(X|Y)+ U$ with $U$ independent of $Y$. Note that this trivial lower bound can be reached even if $Y=X+Z$ with $X$ independent of $Z$. Take for instance $X$ and $Z$ normal, so that $(X,Y)$ is multivariate normal. Then $X-E(X|Y)$ is independent of $Y$ and $Var(Var(X|Y))=0$.


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