Binomial to Poisson

Fix the mean λ = np, then slide n. The blue bars are the binomial distribution: n trials, each a hit with chance p = λ/n, and the bar at k is the chance of exactly k hits. Every time you raise n the demo lowers p to keep the mean where it was. The orange dots are the Poisson distribution with the same mean, e^−λλ^k/k!, which does not depend on n at all. Slide n up and watch the bars converge onto the dots. The readout tracks the largest gap between them.

Mission. Set λ = 4. Find the smallest n for which no bar differs from its dot by more than 0.01. Then set λ = 10 and find it again. Which way did it move, and why?

Answer

For λ = 4 the largest gap drops under 0.01 at about n = 42, where p ≈ 0.1. For λ = 10 you need about n = 70, where p ≈ 0.14. The threshold moves up: a larger mean needs more trials to spread over. What governs the approximation is p = λ/n being small, since the error in each bar is roughly proportional to p, so for a fixed p the fit is about equally good whatever λ is. Extension: look at the variance line in the readout. The binomial variance np(1 − p) = λ(1 − p) is always a little under λ, and the gap is exactly λp.

mean λ = np 3 trials n 6 so p = λ/n =

What's going on

The binomial chance of k hits is C(n, k) p^k (1 − p)^n−k. Put p = λ/n and let n grow. The factor C(n, k)/n^k tends to 1/k!, because n(n − 1)⋯(n − k + 1) is nearly n^k when k is small compared with n. The factor (1 − λ/n)ⁿ tends to e^−λ, the same limit that defines e. And (1 − λ/n)^−k tends to 1. What is left is e^−λλ^k/k!, the Poisson distribution.

So the Poisson is what the binomial becomes when there are many chances for something to happen and each chance is slim: typos per page, calls to a help line per minute, radioactive decays per second, Prussian cavalry soldiers killed by horse kicks per year (the original 1898 data). You do not need to know n or p separately, only their product, and that is why the Poisson has a single parameter.

One consequence is visible in the readout: the Poisson has variance equal to its mean. The binomial variance np(1 − p) is slightly smaller, and the slack λp vanishes as p → 0. If you count rare events and the variance of your counts is much bigger than their mean, the events are not independent: something is making them cluster.

Sheet: The Poisson distribution. Related demo: the Galton board, where the binomial becomes normal instead.