Expected value of the product of dependent geometric and binomial variables.

Question

Let $W$ be geometrically distributed with $p = a$.

Let Z be binomial with $n = 2W$ and $p = b$.

What is $E[W\cdot Z]$?

I can get a rough idea using simulation but would like to be able to derive the closed-form solution.

# R simulation:
a <- 0.05
b <- 0.01
mp <- function(a, b) {
  w <- rgeom(1, a)
  w * rbinom(1, w*2, b)
}
mean(replicate(1e6, mp(a, b))) # ~ 14.8

Answer 1

The result is

$$\tag{1}E(WX)=2b\dfrac{(1-a)(2-a)}{a^2}$$

which takes value $14.82$ for $a=0.05$ and $b=0.01$, in full agreement with the result of the $R$ simulation you have done (and that I have repeated as well on my side).

Before entering into the proof, it must be said that function rgeom in $R$ simulates a geometric distribution beginning at $0$, not at $1$.

Proof : Let us use the formula of conditional expectation under the form: $$E(WX)=E(E(WX|W))=E(E(wX|W=w))=E(wE(X|W=w))=$$

$$=E(wE(X|W=w))=E(w (2wb)|W=w)$$

(the last equality being an application of the classical value $np$ of the expectation of a binomial distribution $B(n,p)$). Thus:

$$E(WX)=E(2bw^2|W=w)=2bE(w^2|W=w)=2bE(W^2)=2b[E(W)^2+V(W)]$$

giving (take a look at the first column of the initial table in (https://en.wikipedia.org/wiki/Geometric_distribution)):

$$E(WX)=2b\left[\frac{(1-a)^2}{a^2}+\frac{1-a}{a^2}\right]$$

yielding (1).