How can the Poisson distribution be applied to a continuous frequency distribution?

Answer 1

Let L(t) be the instantaneous average rate of occurrences per unit time, at time t. So, for the ordinary Poisson distribution with parameter L, we just have L(t)=L for all t.

Let I be the integral of L(t) dt over a certain time interval [0,T], say.

Then, assuming that L(t) is continuous, or maybe just Riemann integrable, the total number of occurrences during [0,T] simply follows a Poisson distribution with parameter I. This is the simple answer one might expect.

To prove this (SKETCH: further estimates are needed to make this really rigorous): divide [0,T] into many small intervals [tj, tj+1). In each interval, the number of occurrences is approximately Poisson with parameter L(tj)(tj+1-tj).

The occurrences in each small interval are all independent of each other; hence the total number in [0,T], which is the sum of all these, follows a Poisson distribution with parameter the sum of L(tj)(tj+1-tj).

As you make the maximum size of the intervals shrink to zero, this sum tends towards I, the Riemann integral of L(t)dt over [0,T], as required.