26 Poisson Distributions

Example 26.1 Let \(X\) be the number of home runs hit (in total by both teams) in a randomly selected Major League Baseball game.

In what ways is this like the Binomial situation? (What is a trial? What is “success”?)
In what ways is this NOT like the Binomial situation?

The Binomial model has several restrictive assumptions that might not be satisfied in practice
Poisson models are often used to model the distribution of random variables that count the number of “relatively rare” events that occur over a certain interval of time/in a certain location

26.1 Poisson distributions

A discrete random variable \(X\) has a Poisson distribution with parameter \(\mu>0\) if its probability mass function \(p_X\) satisfies \[\begin{align*} p_X(x) & \propto \frac{\mu^x}{x!}, \;\qquad x=0,1,2,\ldots\\ & = \frac{e^{-\mu}\mu^x}{x!}, \quad x=0,1,2,\ldots \end{align*}\]
If \(X\) has a Poisson(\(\mu\)) distribution then \[\begin{align*} \text{E}(X) & = \mu\\ \text{Var}(X) & = \mu \end{align*}\]

Example 26.2 Let \(X\) and \(Y\) be the number of home runs hit by the home team and the away team, respectively, in a randomly selected baseball game. Assume that \(X\) and \(Y\) are independent, \(X\) has a Poisson(1.2) distribution and \(Y\) has a Poisson(1.1) distribution.

What does \(X+Y\) represent? What are the possible values of \(X+Y\)?
Compute and interpret \(\text{E}(X+Y)\).
Compute \(\text{Var}(X+Y)\) and \(\text{SD}(X+Y)\).
Compute and interpret \(\text{P}(X+Y=0)\).
Compute and interpret \(\text{P}(X+Y=1)\).
Compute \(\text{P}(X+Y=1)\). (Hint: what \((X, Y)\) pairs yield \(X+Y=1\)?)
Make an educated guess for the distribution of \(X+Y\).

Poisson aggregation. If \(X\) and \(Y\) are independent, \(X\) has a Poisson(\(\mu_X\)) distribution, and \(Y\) has a Poisson}(\(\mu_Y\)) distribution, then \(X+Y\) has a Poisson(\(\mu_X+\mu_Y\)) distribution.
- If component counts are independent and each has a Poisson distribution, then the total count also has a Poisson distribution.
Poisson disaggregation (a.k.a., splitting, a.k.a., thinning). If \(X\) and \(Y\) are independent, \(X\) has a Poisson(\(\mu_X\)) distribution, and \(Y\) has a Poisson(\(\mu_Y\)) distribution, then the conditional distribution of \(X\) given \(\{X+Y=n\}\) is Binomial(\(n\), \(\frac{\mu_X}{\mu_X+\mu_Y}\)).
- The total count of occurrences \(X+Y=n\) can be disaggregated into counts for occurrences of “type \(X\)” or occurrences of “type \(Y\)”.
- Given \(n\) occurrences in total, each of the \(n\) occurrences is classified as type \(X\) with probability proportional to the mean number of occurrences of type \(X\), \(\frac{\mu_X}{\mu_X+\mu_Y}\), and occurrences are classified independently of each other.

26.2 Poisson approximation

Example 26.3 Suppose that each page in the book contains exactly 2000 characters and that the probability that any single character is a typo is 0.00015, independently of all other characters. Let \(X\) be the number of characters on a randomly selected page that are typos.

Identify the distribution of \(X\) and its expected value and variance.
Compare the distribution of \(X\) to a Poisson distribution. (Which Poisson distribution do you think?) What do you notice?

Poisson approximation to Binomial. Consider \(n\) Bernoulli trials with probability of success on each trial equal to \(p_n\). Suppose that \(n\to\infty\) while \(p_n\to0\) and \(np_n\to\mu\), where \(0<\mu<\infty\). Then for \(x=0,1,2,\ldots\) \[ \lim_{n\to\infty} \binom{n}{x} p_n^x \left(1-p_n\right)^{n-x} = \frac{e^{-\mu}\mu^x}{x!} \]
That is, if \(n\) is “large” and \(p\) is “small” then a Binomial(\(n\), \(p\)) distribution is approximately a Poisson(\(np\)) distribution.

Example 26.4 Recall the matching problem with a general \(n\): there are \(n\) objects that are shuffled and placed uniformly at random in \(n\) spots with one object per spot. Let \(X\) be the number of matches. We have seen:

The exact distribution of \(X\) when \(n=4\), via enumerating outcomes in the sample space
\(\text{E}(X)=1\) for any value of \(n\), via linearity of expected value

Now we’ll consider the distribution of \(X\) for general \(n\).

Use simulation to approximate the distribution of \(X\) for different values of \(n\). How does the approximate distribution of \(X\) change with \(n\)?
Does \(X\) have a Binomial distribution? Consider: What is a trial? What is success? Is the number of trials fixed? Is the probability of success the same on each trial? Are the trials independent?
If \(X\) has an approximate Poisson distribution, what would the parameter have to be? Compare this Poisson distribution with the simulation results; does it seem like a reasonable approximation?
For a general \(n\), approximate \(\text{P}(X=x)\) for \(x=0, 1, 2, \ldots\).
For a general value of \(n\), approximate the probability that there is at least one match. How does this depend on \(n\)?

Poisson models often provide good approximations for count data when the restrictive assumptions of Binomial models are not satisfied.
The following table summarizes the four distributions we have seen that are used to model counting random variables.
Note that Poisson distributions require the weakest assumptions.

Distribution	Number of trials	Number of successes	Independent trials?	Probability of success
Binomial	Fixed and known (\(n\))	Random (\(X\))	Yes	Fixed and known (\(p\)), same for each trial
Negative Binomial	Random (\(X\))	Fixed and known (\(r\))	Yes	Fixed and known (\(p\)), same for each trial
Hypergeometric	Fixed and known (\(n\))	Random (\(X\))	No	Fixed and known (\(p = \frac{N_1}{N_1+N_0}\)), same for each trial
Poisson	“Large” (could be random, could be unknown)	Random (\(X\))	“Not too dependent”	“Comparably small for all trials” (could vary between trials, could be unknown)

Example 26.5 Recall the birthday problem: in a group of \(n\) people what is the probability that at least two have the same birthday? (Ignore multiple births and February 29 and assume that the other 365 days are all equally likely.) We will investigate this problem using Poisson approximation. Imagine that we have a trial for each possible pair of people in the group, and let “success” indicate that the pair shares a birthday. Let \(X\) be the number of pairs that share a birthday.

Below, consider both a general \(n\) and \(n=30\).

How many trials are there?
Do the trials have the same probability of success? If so, what is it?
Are the trials independent?
If \(X\) has an approximate Poisson distribution, what would the parameter have to be? Use simulation to approximate the distribution of \(X\); does a Poisson distribution like a reasonable approximation?
Approximate the probability that at least two people share the same birthday. Compare to the theoretical value.

Poisson paradigm. Let \(A_1, A_2, \ldots, A_n\) be a collection of \(n\) events. Suppose event \(i\) occurs with marginal probability \(p_i=\text{P}(A_i)\). Let \(N = \text{I}_{A_i} + \text{I}_{A_2} + \cdots + \text{I}_{A_n}\) be the random variable which counts the number of the events in the collection which occur. Suppose

\(n\) is “large”,
\(p_1, \ldots, p_n\) are “comparably small”, and
the events \(A_1, \ldots, A_n\) are “not too dependent”,

Then \(N\) has an approximate Poisson distribution with parameter \(\text{E}(N) = \sum_{i=1}^n p_i\).

Example 26.6 Use Poisson approximation to approximate that probability that at least three people in a group of \(n\) people share a birthday. How large does \(n\) need to be for the probability to be greater than 0.5?