The Poisson distribution
Tonight's idea
Misprints on a page, calls to a switchboard in a minute, atoms decaying in a second. Each is a binomial count: \(n\) chances, each with a small probability \(p\), and a mean \(\lambda = np\) (lambda) of ordinary size. Fix \(\lambda\), let \(n\) grow, so \(p = \lambda/n\) shrinks. The chance of exactly \(k\) events is
\[\binom{n}{k}\Big(\frac{\lambda}{n}\Big)^{k}\Big(1-\frac{\lambda}{n}\Big)^{n-k} = \underbrace{\frac{n(n-1)\cdots(n-k+1)}{n^k}}_{\to\,1}\cdot\frac{\lambda^k}{k!}\cdot\underbrace{\Big(1-\frac{\lambda}{n}\Big)^{n}}_{\to\,e^{-\lambda}}\cdot\underbrace{\Big(1-\frac{\lambda}{n}\Big)^{-k}}_{\to\,1}.\]
Keep \(k\) fixed and send \(n \to \infty\). The first factor is \(k\) fractions each tending to 1; so does the last. The middle one is the compound-interest limit from the \(e\) sheet, with \(-\lambda\) for 1. What survives is
\(P(k) = \dfrac{e^{-\lambda}\lambda^k}{k!}, \qquad k = 0, 1, 2, \ldots\)
the Poisson distribution. Two parameters have become one: only the product \(np\) matters, so you never need to know how many "trials" a page of print contains.
The mean is \(np = \lambda\). The binomial variance \(np(1-p) = \lambda(1 - \lambda/n)\) tends to \(\lambda\) too. So mean = variance: a quick test of whether a count is Poisson.
Watch one: horse kicks
The classic data set is a count of Prussian cavalrymen killed by horse kicks, published by Ladislaus Bortkiewicz in 1898: 200 corps-years, 122 deaths, so \(\lambda = 0.61\) per corps per year. Poisson predicts \(200\, e^{-0.61}\, 0.61^k/k!\) corps-years with \(k\) deaths:
| deaths \(k\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| observed | 109 | 65 | 22 | 3 | 1 |
| Poisson | 108.7 | 66.3 | 20.2 | 4.1 | 0.6 |
Thousands of soldiers at tiny risk, no record of how many: only \(\lambda\) matters.
Slide \(n\) up with \(np\) fixed and watch the binomial bars settle onto the Poisson curve: mathdetours.com/demos/poisson.html.
Your turn
- For \(\lambda = 2\), find \(P(0)\), \(P(1)\), \(P(2)\) and \(P(3 \text{ or more})\) to three decimal places. Why are \(P(1)\) and \(P(2)\) equal?
- A typesetter averages 0.5 misprints per page. (a) What is the probability a page has at least one? (b) In a 300-page book, how many clean pages do you expect? (c) What fraction of pages have two or more?
- Prove mean = variance from the formula, not from the binomial. (a) Show \(\sum_{k \ge 0} k\,P(k) = \lambda\) by cancelling \(k\) into \(k!\) and taking out a factor \(\lambda\). (b) In the same way show \(\sum k(k-1)P(k) = \lambda^2\). (c) Deduce the variance, \(E[X^2] - \lambda^2\), is \(\lambda\).
- A baker stirs raisins into dough for 100 cookies; each raisin lands in a cookie at random. (a) What are \(n\), \(p\) and \(\lambda\) for the number of raisins in one cookie? (b) How many raisins are needed so that at most one cookie in a hundred has none? (c) With that many, what fraction of cookies have exactly one raisin?
Stretch: the birthday problem, Poisson style
Among 23 people there are \(\binom{23}{2} = 253\) pairs, each sharing a birthday with probability \(1/365\). Treat the number of matching pairs as Poisson. What is \(\lambda\)? Estimate the chance of no shared birthday, and compare with the exact \(0.4927\) from the grade 8 birthday sheet. Why does \(\ln 2\) appear?
Answers
- \(e^{-2} = 0.135\); \(2e^{-2} = 0.271\); \(2e^{-2} = 0.271\); \(1 - 5e^{-2} = 0.323\). \(P(2) = P(1) \cdot \lambda/2\), and \(\lambda/2 = 1\).
- (a) \(1 - e^{-0.5} = 0.393\). (b) \(300\,e^{-0.5} = 182\) clean pages. (c) \(1 - e^{-0.5}(1 + 0.5) = 0.090\), about one page in eleven.
- (a) \(\sum_{k\ge1} \dfrac{e^{-\lambda}\lambda^k}{(k-1)!} = \lambda e^{-\lambda}\sum_{j \ge 0}\dfrac{\lambda^j}{j!} = \lambda e^{-\lambda}e^{\lambda} = \lambda\). (b) The \(k = 0, 1\) terms vanish; cancel \(k(k-1)\) into \(k!\) and take out \(\lambda^2\): \(\lambda^2 e^{-\lambda}e^{\lambda} = \lambda^2\). (c) \(E[X^2] = E[X(X-1)] + E[X] = \lambda^2 + \lambda\), so the variance is \(\lambda^2 + \lambda - \lambda^2 = \lambda\).
- (a) \(n\) = number of raisins, \(p = 1/100\), \(\lambda = n/100\). (b) Need \(e^{-\lambda} \le 0.01\), so \(\lambda \ge \ln 100 = 4.605\): about 461 raisins. (c) \(\lambda e^{-\lambda} = 4.605 \times 0.01 = 0.046\), under 5%. Most cookies get 3 to 6.
Stretch. \(\lambda = 253/365 = 0.6932\), so \(P(\text{no match}) \approx e^{-0.6932} = 0.500\). The exact answer is 0.4927; the approximation is slightly off because the pairs are not quite independent (if A matches B and B matches C, then A matches C). The number of pairs that makes the chance one half is the one with \(\lambda = \ln 2 = 0.6931\), that is \(365 \ln 2 = 253\) pairs, and 23 people give exactly 253. That is why 23 is the famous answer.