11  Poisson Distributions

Table 11.1: Poisson(1) distribution, the approximate distribution of \(X\), the number of matches in the matching problem for general \(n\) and uniformly random placement of objects in spots.
x P(X=x)
0 0.36788
1 0.36788
2 0.18394
3 0.06131
4 0.01533
5 0.00307
6 0.00051
7 0.00007

Example 11.1 Recall the matching problem from Example 6.3, Exercise 7.3, Exercise 8.2, and Exercise 9.2. There are \(n\) distinct objects that are shuffled and placed uniformly at random in \(n\) distinct spots with one object per spot, and \(X\) is the number of correct matches of object with 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., \(n \le 6\)), \(\text{Var}(X)=1\) and the approximate distribution of \(X\) is given by Table 11.1.

Starting with \(x=0\), describe the distribution of \(X\) in terms of relative likelihoods:

  1. 1 is [blank] times as likely as 0



  2. 2 is [blank] times as likely as 1



  3. 3 is [blank] times as likely as 2



  4. 4 is [blank] times as likely as 3



  5. 5 is [blank] times as likely as 4



  6. 6 is [blank] times as likely as 5



  7. In general, \(x\) is [blank] times as likely as \(x-1\), for \(x=1, 2, 3, \ldots\)



Example 11.2 Let \(X\) be the number of home runs hit (in total by both teams) in a randomly selected Major League Baseball game. Technically, there is no fixed upper bound on what \(X\) can be, so mathematically it is convenient to consider \(0, 1, 2, \ldots\) as the possible values of \(X\). Assume that \(X\) has a Poisson distribution with parameter 2.4.

  1. Interpret 2.4 in context.





  2. Identify what the Poisson(2.4) patterns implies in context.





  3. Sketch a plot of the distribution of \(X\).





  4. Assuming that the probability that \(X\) is greater than 9 is negligible, make a table representing the distribution of \(X\).





  5. The pmf of \(X\) is \[ \text{P}(X = x) = e^{-2.4}\frac{2.4^x}{x!}, \quad x = 0, 1, 2, \ldots \] Use the pmf to compute \(\text{P}(X = 0)\).





  6. Use the pmf to compute \(\text{P}(X = 1)\).





  7. Use the pmf to compute \(\text{P}(X = 2)\).





  8. Use the pmf to construct a table and plot representing the distribution of \(X\). Does it match your previous work?





  9. Use the pmf to compute \(\text{E}(X)\) and verify that it is 2.4.





  10. Compute and interpret \(\text{P}(X \le 2)\).





Example 11.3 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.

  1. In what ways is this like the Binomial situation?




  2. In what ways is this NOT like the Binomial situation?




  3. In what sense are home runs “relatively rare”?




11.1 Exercises

Exercise 11.1 Continuing Example 11.1. Explain why \(X\) in the matching problem has a Poisson distribution when \(n\) is large.




Exercise 11.2 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”).

  1. Make an educated guess for \(\text{E}(X)\).




  2. Compute \(\text{P}(X = 0)\).




  3. Compute \(\text{P}(X = 1)\).




  4. How could you use simulation to approximate the distribution of \(X\)?




  5. Use simulation to approximate the distribution of \(X\). Does the approximate distribution follow (approximately) the Poisson pattern?




  6. 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.

  1. Recall the series expansion \(e^{\mu} = \sum_{x=0}^\infty \frac{\mu^x}{x!}\)↩︎