What is the distribution of a sum of binomial distributions with the same parameter q but with the sample sizes following a Poisson distribution?

Question

Let $\{a_1,a_2,\ldots,,a_n\}$ be a random sample of a Poisson distribution. Consider the following random variables $X_1=\mathrm{Binomial}(a_1,q), ~X_2=\mathrm{Binomial}(a_2,q),\ldots,~X_n=\mathrm{Binomial}(a_n,q)$.

Is there a method to find the distribution of $X=\sum_{i=1}^nX_i$?

Answer 1

The answer to this particular problem is immediate if you think in terms of how these distributions usually arise.

Consider a homogeneous Poisson process of rate $\mu.$ In the first interval of time you observe $a_1$ events, then in the second interval you observe $a_2 $ events, and so on. The Binomial probabilities reflect the results of selecting randomly, with equal probabilities $q,$ from each set of $a_i$ events. The result is a Poisson process that has been "thinned" by keeping only a fraction $q$ of all events. Thus, its rate must be $q\mu$ and it has run for $n$ time intervals. Consequently, writing $\lambda=n\mu$ for the overall rate during these time intervals, the answer must be the Poisson probability,

$$\Pr(X=k) = e^{-q\lambda} \frac{(q\lambda)^k}{k!}.$$

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General solution

Let's now address the general question where the $a_i$ arise independently according to any distribution.

Let $A=a_1+a_2+\cdots+a_n$ be the random variable equal to this sum. By conceiving of a Binomial$(a,q)$ distribution as being that of the sum of $a$ independent Bernoulli$(q)$ distributions, it is evident that the sum of the $X_i$ is distributed as the sum of all $a_1+a_2+\cdots + a_n$ Bernoulli variables. Let the probability function for $A$ be $p,$ so that for $a=0, 1, 2, \ldots,$

$$p_a = \Pr(a_1+a_2+\cdots+a_n=a).$$

Conditional on $A=a,$ the Binomial probability law is

$$\Pr(k) = \binom{a}{k}q^k(1-q)^{a-k}.$$

The event $X=k$ is the disjoint union of the events $(A,X)=(0,k),$ $(A,X)=(1,k),$ $(A,X)=(2,k),$ and so on. Therefore its probability is the sum of the probabilities of these events, each of which can be computed in terms of conditional probabilities as

$$\Pr(A,X)=(a,k) = \Pr(X=k\mid A=a)\Pr(A=a) = \binom{a}{k}q^k(1-q)^{a-k}\, p_a.$$

Their sum is

$$\Pr(X=k) = \sum_{a=0}^\infty \Pr(a,k) = \sum_{a=0}^\infty\binom{a}{k}q^k(1-q)^{a-k}\, p_a.$$

Notice that $X=k$ is impossible unless $a \ge k.$ Let us therefore reindex the sum in terms of $a-k=0, 1, 2, \ldots,$ writing $n=a-k:$

$$\Pr(X=k) = \sum_{n=0}^\infty\binom{n+k}{k}q^k(1-q)^{n}\, p_{k+n} = q^k\sum_{n=0}^\infty\binom{n+k}{k}(1-q)^{n}\, p_{k+n}.$$

This the general solution. It was obtained using only the basic axioms of probability (and therefore many explanations can be found in textbooks, if you wish to do further research).

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Worked example

When the $a_i$ have Poisson distributions of rates $\lambda_i,$ their sum has a Poisson distribution of rate $\lambda = \sum_i\lambda_i,$ whence

$$p_a = e^{-\lambda} \frac{\lambda^a}{a!}.$$

Plugging this into the foregoing formula gives

$$\Pr(X=k) = q^k\sum_{n=0}^\infty\binom{n+k}{k}(1-q)^{n}\, e^{-\lambda} \frac{\lambda^{k+n}}{(k+n)!} = e^{-\lambda}\frac{(\lambda q)^k}{k!}\sum_{n=0}^\infty\frac{(\lambda(1-q))^{n}}{n!}.$$

This last sum is the series for $\exp(\lambda(1-q)),$ simplifying the result to

$$\Pr(X=k) = e^{-\lambda}\frac{(\lambda q)^k}{k!}e^{\lambda(1-q)} = e^{-q\lambda}\frac{(\lambda q)^k}{k!},$$

which is the Poisson probability for $k$ with rate $\lambda q$ we obtained at the outset (using no calculation at all).