Expected value of conditional variance problem

Question

A dice is tossed untill 3 even numbers show up. Let Y be the number of times the number "1" showed up, and L the total amount of tosses. Compute E(V(Y|L)). I know L is distributed negative binominially, but besides that, i dont really understand how to approch this. I also dont quite understand the meaning of an expected value of variance, shouldnt E(V(Y|L))=V(Y|L)? Since an expected value of any value is just the value itself? Thanks in advance!

Answer 1

First let us work under condition $L=n$ where $n\in\{3,4,\dots\}$

Then among the $n$ tosses exactly $n-3$ did not show an even face, so had probability $\frac13$ to show face 1.

So we are dealing with $\mathsf{Bin}(n-3,\frac13)$ hence with variance $(n-3)\frac13\frac23=\frac29n-\frac23$.

We actually found that: $\mathsf{Var}(Y\mid L=n)=\frac29n-\frac23$.

That means that: $$\mathsf{Var}(Y\mid L)=\frac29L-\frac23$$

Note that $\mathsf{Var}(Y\mid L)$ is (in contrast with $\mathsf{Var}(Y\mid L=n)$) a random variable.

Now the job can be completed by finding:$$\mathsf E\left[\mathsf{Var}(Y\mid L)\right]=\mathsf E\left[\frac29L-\frac23\right]=\frac29\mathsf EL-\frac23$$

As you said $L$ has negative binomial distribution.