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 \(p_1, p_2, \ldots, p_k\).

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 \(n, p_1, p_2, \ldots, p_k\). 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 \(i = 1, 2, \ldots , k\), the distribution of \(N_i\) is Poisson \((\mu p_i)\).

  • The counts \(N_1, N_2, \ldots , N_k\) 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, \(N_1, N_2, \ldots , N_k\) 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 \(N_1, N_2, \ldots , N_k\) 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

\(\big{(} \sum_{k=0}^2 e^{-3}\frac{3^k}{k!} \big{)}^6\)

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.