7.3. Poissonizing the Multinomial

This is just an extension of Poissonizing the binomial.

Suppose you have a sequence of i.i.d. multinomial trials. For example, suppose you are drawing at random with replacement from a population that has $k$ classes of indviduals in proportions $p_1,p_2,\dots ,p_k$.

If you draw a fixed number of times $n$, then the joint distribution of the counts of the classes in the sample is multinomial with parameter $n,p_1,p_2,\dots ,p_k$. For each fixed $i$, the distribution of the count in Class $i$ is binomial $(n,p_i)$.

If you replace the fixed number $n$ of trials by a Poisson $(\mu )$ random number of trials, then the multinomial gets Poissonized as follows:

• For each $i=1,2,\dots ,k$, the distribution of $N_i$ is Poisson $(\mu p_i)$.

• The counts $N_1,N_2,\dots ,N_k$ in the $k$ different categories are mutually independent.

We won’t go through the proof which is a straightforward extension of the proof in the case $k=2$ given in an earlier section. Rather, we will look at why the result matters.

When the number of trials is fixed, $N_1,N_2,\dots ,N_k$ are all dependent on each other in complicated ways according to the multinomial distribution. But when you let the sample size be a Poisson random variable, then the independence of the counts $N_1,N_2,\dots ,N_k$ lets you quickly calculate the chance of any particular configuration of classes in the sample.

For example, suppose that in your population the distribution of classes is as follows:

• Class 1: 20%

• Class 2: 30%

• Class 3: 50%

Now suppose you draw $N$ independent times where $N$ has the Poisson $(20)$ distribution. Then in the sample,

• the number of Class A individuals has the Poisson distribution with parameter $20$% of $20$, that is Poisson $(4)$,

• the number of Class B individuals has the Poisson $(6)$ distribution,

• the number of Class C indidviduals has the Poisson $(10)$ distribution,

• and these three counts are independent.

Note that the Poisson parameters of the three Class counts must add up to the original Poisson parameter of the distribution of $N$. The independence makes it easy to find the chance of any specified configuration of the three classes in the sample: just multiply the three individual Poisson probabilities.

For example, the chance that you will get at least 3 individuals in Class A, at least 5 in Class B, and at least 8 in Class C is about 42.5%.

(1 - stats.poisson.cdf(2, 4))*(1-stats.poisson.cdf(4, 6))*(1-stats.poisson.cdf(7, 10))

0.42475602042528027

The number of factors in the answer is equal to the number of classes, unlike the inclusion-exclusion formula in which the amount of work increases much more with each additional class, as you have seen in exercises.

Quick Check

Suppose I roll a Poisson $(18)$ number of dice. What is the chance that each face appears at most twice?

Answer

$(\sum _{k=0}^2{e}^{-3}\frac{3^k}{k!}{)}^6$

Poissonization helps data scientists tackle questions like, “How many times must I sample so that my chance of seeing at least one individual of each class exceeds a given threshold?” The answer depends on the distribution of classes in the population, of course, but allowing the sample size be a Poisson random variable can make calculations much more tractable. For applications, see for example the Abstract and References of this paper.