Description of a compound binomial random variable

Question

Suppose $X$ is a binomial random variable with parameters $N$ and $p$. Suppose that $N$ itself is a random variable that has a binomial distribution with parameters $M$ and $\pi$. How can we describe $X$ as a compound random variable and its distribution?

Answer 1

$X \sim Bin(N, p)$ and $N \sim Bin(M, \pi)$.

\begin{align} P(X =x)&= \sum_{n=0}^M P(X=x|N=n)P(N=n) \\ &= \sum_{n=0}^M \binom{n}{x} \pi^x(1-\pi)^{n-x}\binom{M}{n}p^n(1-p)^{M-n} \end{align}

Answer 2

If X∼Bin(N,p) and N∼Bin(M,π), then S = Sum(Xi; i = 1 to N ) is Bin(M,πp). This can be demonstrated by summing on N = n for 0 to M, the terms of the compound distribution for S = k. Note that there are no terms for n < k. The terms need to be manipulated to give a sum from r = 0 to M-k that is binomial in y = π (1 - p)/(1 - πp) and 1 - y = (1 - π)/(1 - πp). The above result Bin(M,πp) then follows.

Answer 3

We can show that $X \sim \mathrm{Bin}(M,\pi p)$ using the characteristic function of the binomial.

Note that \begin{align*} \varphi_X(t) &= \mathbb{E}\!\left[e^{itX}\right] = \mathbb{E}\!\left[\mathbb{E}[e^{itX}|N]\right] = \mathbb{E}\!\left[(1-p+pe^{it})^N\right]. \end{align*}

Since $\varphi_N(t) = \mathbb{E}\!\left[e^{i t N}\right] = (1-\pi+\pi e^{it})^M$ and $\mathbb{E}\!\left[z^N\right] = (1-\pi+\pi z)^M$, then \begin{align*} \varphi_X(t) &= (1-\pi+\pi(1-p+pe^{it}))^M = (1-\pi p+\pi pe^{it})^M, \end{align*} which is the characteristic function of $\mathrm{Bin}(M,\pi p)$.