Derive Poisson Distribution

Let's say we are observing an event that to us seems random. E.g. The number of cars passing through a road.

Suppose we learn that on average, there are m cars passing the road within 1 hour.

Goal: Can we make a model to suggest what is the probability of seeing exactly r cars passing the road in an hour?

Let's represent one hour as a line, and on the line there are marks to represent the time on which a car passes. Let's suppose there are m marks, to represent the average case of seeing m cars on the road in an hour.

Let's divide the 1 hour line into n uniform time segments, n large enough such that in one time segment there is at most 1 car. Out of the n time segments, m of them contains a car. Then it is reasonable to say that the probability of a time segment to contain a car is m/n. Let's analyse as n goes to infinity.

We can say P(seeing exactly r cars in an hour) =  P(exactly r cars contained in n segments) = \( \binom{n}{r} \left( \frac{m}{n} \right)^r \left( 1 - \frac{m}{n} \right)^{(n-r)} \).

After rearranging algebraicly, we see that it is then product of the following terms:

\( \frac{m^r}{r!} \) which is constant w.r.t n,

\( \left( 1 - \frac{m}{n} \right)^n \) which converges to \( e^{-m} \) as n grows.

\( \frac{n!}{(n-r)!(n-m)^r} \) which converges to 1 as n grows.

So when n is large, P(seeing exactly r cars in an hour) = \( \frac{m^r}{r!}e^{-m} \) where m is the average in an hour.

And this is actually Poisson distribution. We have just derived Poisson distribution from a thought experiment.