P(X = k) = e^{-L}L^{k}/k!

The distribution has mean and variance L. The sum of two Poisson distributed variables with parameters L, M is itself Poisson distributed and has parameter L + M.

It turns out that the Poisson distribution is one of the distributions that tend to pop up naturally. What is the explanation for this?

Proposition:

Let X_{n} be a sequence of random variables, each binomially distibuted with parameters n, p, with L = np is constant. Then P(X_{n} = k) → e^{-L}L^{k}/k! as n → ∞.

Proof:

Using the definition of the binomial distribution we have

P(X_{n} = k) = (n!/k!(n-k)!)p^{k}(1-p)^{n-k} = (n!/(n-k)!n^{k})((np)^{k}/k!)(1-L/n)^{n}(1-L/n)^{-k} → e^{-L}L^{k}/k!

as n → ∞.

So the Poisson distribution can be considered as a limiting case of the binomial distribution. A few examples should illustrate the importance of this.

The distribution of misprints on the pages of a book can be modelled by saying that there are n = 1000 characters on each page, and each of these have an independent probablity of say p = 0,0001 of being a misprint. n is large, p is small but L = np = 0,1 is moderate in size. Therefore the Poisson distribution with parameter L = 0,1 will provide a good approximation to the number of misprints on each page.

Suppose that a radioactive sample has an average of L decays in an interval of time. To approximate the distribution of the number of decays in the interval we divide it into a large number n subintervals. For large n the probability probability of more than one decay in a subinterval is negligable. The probability that there is a decay in an interval is p = L/n. As n increases the approximation becomes more accurate, and since np = L remains constant the limiting case is the Poisson distribution with parameter L.