Splitting a Poisson Distribution

We consider a remarkable property of the Poisson distribution that has a connection to the multinomial distribution. We start with the following examples.

Example 1
Suppose that the arrivals of customers in a gift shop at an airport follow a Poisson distribution with a mean of \alpha=5 per 10 minutes. Furthermore, suppose that each arrival can be classified into one of three distinct types – type 1 (no purchase), type 2 (purchase under $20), and type 3 (purchase over $20). Records show that about 25% of the customers are of type 1. The percentages of type 2 and type 3 are 60% and 15%, respectively. What is the probability distribution of the number of customers per hour of each type?

Example 2
Roll a fair die N times where N is random and follows a Poisson distribution with parameter \alpha. For each i=1,2,3,4,5,6, let N_i be the number of times the upside of the die is i. What is the probability distribution of each N_i? What is the joint distribution of N_1,N_2,N_3,N_4,N_5,N_6?

In Example 1, the stream of customers arrive according to a Poisson distribution. It can be shown that the stream of each type of customers also has a Poisson distribution. One way to view this example is that we can split the Poisson distribution into three Poisson distributions.

Example 2 also describes a splitting process, i.e. splitting a Poisson variable into 6 different Poisson variables. We can also view Example 2 as a multinomial distribution where the number of trials is not fixed but is random and follows a Poisson distribution. If the number of rolls of the die is fixed in Example 2 (say 10), then each N_i would be a binomial distribution. Yet, with the number of trials being Poisson, each N_i has a Poisson distribution with mean \displaystyle \frac{\alpha}{6}. In this post, we describe this Poisson splitting process in terms of a “random” multinomial distribution (the view point of Example 2).

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Suppose we have a multinomial experiment with parameters N, r, p_1, \cdots, p_r, where

  • N is the number of multinomial trials,
  • r is the number of distinct possible outcomes in each trial (type 1 through type r),
  • the p_i are the probabilities of the r possible outcomes in each trial.

Suppose that N follows a Poisson distribution with parameter \alpha. For each i=1, \cdots, r, let N_i be the number of occurrences of the i^{th} type of outcomes in the N trials. Then N_1,N_2,\cdots,N_r are mutually independent Poisson random variables with parameters \alpha p_1,\alpha p_2,\cdots,\alpha p_r, respectively.

The variables N_1,N_2,\cdots,N_r have a multinomial distribution and their joint probability function is:

\displaystyle (1) \ \ \ \ P(N_1=n_1,N_2=n_2,\cdots,N_r=n_r)=\frac{N!}{n_1! n_2! \cdots n_r!} \ p_1^{n_1} p_2^{n_2} \cdots p_r^{n_r}

where n_i are nonnegative integers such that N=n_1+n_2+\cdots+n_r.

Since the total number of multinomial trials N is not fixed and is random, (1) is not the end of the story. The probability in (1) is only a conditional probability. The following is the joint probability function of N_1,N_2,\cdots,N_r:

\displaystyle (2) \ \ \ \ P(N_1=n_1,N_2=n_2,\cdots,N_r=n_r)

          \displaystyle \begin{aligned}&=P(N_1=n_1,N_2=n_2,\cdots,N_r=n_r \lvert N=\sum \limits_{k=0}^r n_k) \\&\ \ \ \ \ \times P(N=\sum \limits_{k=0}^r n_k) \\&\text{ } \\&=\frac{(\sum \limits_{k=0}^r n_k)!}{n_1! \ n_2! \ \cdots \ n_r!} \ p_1^{n_1} \ p_2^{n_2} \ \cdots \ p_r^{n_r} \ \times \frac{e^{-\alpha} \alpha^{\sum \limits_{k=0}^r n_k}}{(\sum \limits_{k=0}^r n_k!)} \\&\text{ } \\&=\frac{e^{-\alpha p_1} \ (\alpha p_1)^{n_1}}{n_1!} \ \frac{e^{-\alpha p_2} \ (\alpha p_2)^{n_2}}{n_2!} \ \cdots \ \frac{e^{-\alpha p_r} \ (\alpha p_r)^{n_r}}{n_r!} \end{aligned}

To obtain the marginal probability function of N_j, j=1,2,\cdots,r, we sum out the other variables N_k=n_k (k \ne j) in (2) and obtain the following:

\displaystyle (3) \ \ \ \ P(N_j=n_j)=\frac{e^{-\alpha p_j} \ (\alpha p_j)^{n_j}}{n_j!}

Thus we can conclude that N_j, j=1,2,\cdots,r, has a Poisson distribution with parameter \alpha p_j. Furrthermore, the joint probability function of N_1,N_2,\cdots,N_r is the product of the marginal probability functions. Thus we can conclude that N_1,N_2,\cdots,N_r are mutually independent.

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Example 1
Let N_1,N_2,N_3 be the number of customers per hour of type 1, type 2, and type 3, respectively. Here, we attempt to split a Poisson distribution with mean 30 per hour (based on 5 per 10 minutes). Thus N_1,N_2,N_3 are mutually independent Poisson variables with means 30 \times 0.25=7.5, 30 \times 0.60=18, 30 \times 0.15=4.5, respectively.

Example 2
As indicated earlier, each N_i, i=1,2,3,4,5,6, has a Poisson distribution with mean \frac{\alpha}{6}. According to (2), the joint probability function of N_1,N_2,N_3,N_4,N_5,N_6 is simply the product of the six marginal Poisson probability functions.

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2 thoughts on “Splitting a Poisson Distribution

  1. I think your expositions, those I looked at, are good.
    I have a problem which seems like it’s easy but my brain refuses to come through. Would you be willing to analyze a simple recursive algorithm that produces a distribution that seems to have order statistics of the Beta distribution; more or less?
    It might serve as an non-obvious example.

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