For each given $p$, let $Z$ have a binomial distribution with parameters $p$ and $N$...

Question

For each given $p,$ let $Z$ have a binomial distribution with parameters $p$ and $N.$ Suppose that $N$ is itself binomially distributed with parameters $q$ and $M.$ Formulate $Z$ as a random sum and show that $Z$ has a binomial distribution with parameters $pq$ and $M.$

I'm unsure how to formulate $Z$ and show that it has such parameters.

Answer 1

Comments: Not quite finished. Maybe a useful start for you.

A random sum of Bernoulli random variables: Let $Z = \sum_{i=1}^N X_i,$ where $X_i \stackrel{indep}{\sim} \mathsf{Binom}(1, p) \equiv \mathsf{Bernoulli}(p)$ and (independently of the $X_i$) $N\sim \mathsf{Binom}(M, q).$

It is claimed that $Z \sim \mathsf{Binom}(M, pq),$ so that $E(Z) = Mpq$ and $Var(Z) = Mpq(1-pq).$

The usual formulas for a random sum of random variables give: $$E(Z) = E(N)E(X) = (Mq)(p) = Mpq,$$ as required.

Moreover, $$Var(Z) = E(N)Var(X) + Var(N)[E(X)]^2 \\= Mq[p(1-p)] + Mq(1-q)[p^2] = Mpq(1-pq),$$ also as required.

It remains to be shown that $Z$ is binomial with these parameters. I guess that the kind of argument using MGFs in the link in my previous Comment is sufficient to show that.

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Incidentally, in a simulation of a million realizations of $Z$ with $M = 10,\,p=.4$ and $p = .6.$ gives the following histogram, where the dots atop hisogram bars are exact values of $Binom(M = 10, pq=.24).$

B = 10^6;  z = numeric(B);  m = 10;  p = .4;  q = .6
for(i in 1:B) {
  n = rbinom(1, m, q);  z[i] = rbinom(1, n, p)}
mean(z)
## 2.400351

Answer 2

Let $Y_1,\dots,Y_M$ be IID $\mbox{Bern}(q)$. Set $N = \sum_{i\le M} Y_i$.

Let $X_1,\dots,X_M$ be IID $\mbox{Bern}(p)$, independent of $Y_i$'s.

Finally, let $Z=\sum_{i\le M} X_i Y_i$.

Then this is the representation you were looking for. Why ?

1. $Z\sim\mbox{Bin}(M,qp)$, because it's a sum of IID $\mbox{Bern}(pq)$. Indeed,$X_iY_i$ is either $0$ or $1$, and is equal to $1$ if and only if $X_i=Y_i=1$. This occurs with probability $pq$.

2. $P(Z=\cdot |N=l) \sim \mbox{Bin}(l,p)$ (what is the probability that among the $l$ $Y_i$'s equal to $1$, exactly $k$ from the corresponding $X_i$'s are $1$ ? $\binom{l}{k} p^k(1-p)^{l-k}$).