| x | P(X=x) |
|---|---|
| 0 | 0.3679 |
| 1 | 0.3679 |
| 2 | 0.1839 |
| 3 | 0.0613 |
| 4 | 0.0153 |
| 5 | 0.0031 |
| 6 | 0.0005 |
| 7 | 0.0001 |
19 Poisson Distributions
- Binomial and Negative Binomial models have several restrictive assumptions that might not be satisfied in practice
- Poisson models are more flexible and 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
Example 19.1 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. Note: \(n\) is a fixed and known number, but we are considering different values of it.
We have seen that \(\text{E}(X)=1\) for any \(n\) and that unless \(n\) is small (e.g., \(n \le 6\)) the approximate distribution of \(X\) is the Poisson(1) distribution.
- Starting with \(x=0\), describe the distribution of \(X\) in terms of relative likelihoods:
- 1 is [blank] times as likely as 0
- 2 is [blank] times as likely as 1
- 3 is [blank] times as likely as 2
- 4 is [blank] times as likely as 3
- 5 is [blank] times as likely as 4
- 6 is [blank] times as likely as 5
- In general, \(x\) is [blank] times as likely as \(x-1\), for \(x=1, 2, 3, \ldots\)
- Specify the pmf of \(X\).
- Use the pmf to compute \(\text{E}(X)\).
- Use the pmf to compute \(\text{Var}(X)\).
- A random variable \(X\) has a Poisson distribution with parameter \(\mu>0\) if the possible values are \(0, 1, 2, \ldots\) and the pmf follows the \[ \text{Poisson(}\mu\text{) pattern:}\quad x \text{ is } \frac{\mu}{x} \text{ as likely as } x-1 \quad \text{ (for } x = 1, 2, 3, \ldots) \]
- As a function of \(x=0, 1, 2, \ldots\), a Poisson(\(\mu\)) pmf increases for \(x<\mu\) and then decreases for \(x >\mu\)
- If \(\mu\) is a whole number, then the Poisson pmf assigns the same probability to the values \(\mu\) and \(\mu-1\).
- If \(\mu<1\) then \(x=0\) has the highest probability and the pmf decreases from there.
- A discrete random variable \(X\) has a Poisson distribution with parameter \(\mu>0\) if and only if its probability mass function \(p_X\) satisfies \[
p_X(x) = \frac{e^{-\mu}\mu^x}{x!}, \quad x=0,1,2,\ldots
\]
- The Poisson pattern implies that the shape of a Poisson pmf as a function of \(x\) is given by \(\mu^x/x!\). \[ p(x) = p(0)\frac{\mu^x}{x!}, \qquad x = 0, 1, 2, \ldots \]
- The constant \(p(0) = e^{-\mu}\) simply renormalizes the pmf so that the probabilities sum1 to 1.
- If \(X\) has a Poisson(\(\mu\)) distribution then \[\begin{align*} \text{E}(X) & = \mu\\ \text{Var}(X) & = \mu \end{align*}\]
Example 19.2 Let \(X\) be the number of home runs hit (in total by both teams) in a randomly selected Major League Baseball game. Assume that \(X\) has a Poisson(2.4) distribution.
- Interpret the parameter 2.4 in context.
- Compute \(\text{P}(X = 2.4)\).
- Identify the most likely value of \(X\).
- Specify the pmf of \(X\).
- Compute \(\text{P}(X = 0)\) and interpret the value in context.
- Construct a table and spinner corresponding to the distribution of \(X\).
Example 19.3 Suppose that the number of fatal crashes in SLO County in a week follows, approximately, a Poisson(0.53) distribution, independently from week to week. Let \(X\) be the number of fatal crashes in a period of 4 consecutive weeks (close enough to a “month”).
- What is \(\text{E}(X)\)? You should be able to compute it without knowing the distribution of \(X\).
- Compute \(\text{P}(X = 0)\).
- Compute \(\text{P}(X = 1)\).
- How could you use simulation to approximate the distribution of \(X\)?
- Use simulation to approximate the distribution of \(X\). Does the approximate distribution follow (approximately) the Poisson pattern?
- Use a Poisson pmf to compute \(\text{P}(X = 5)\).
- 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.
Recall the series expansion \(e^{\mu} = \sum_{x=0}^\infty \frac{\mu^x}{x!}\)↩︎