4
$\begingroup$

What is the fundamental difference between Poisson Point Process and Binomial Point Process?

I am evaluating a solution in a Binomial Point Process setup. If I want to evaluate that in a Poisson Point Process setup, what all issues need to be considered?

$\endgroup$
3
$\begingroup$

This figure (copied from these notes) serves to illustrate the difference between a binomial and a Poisson point process. Shown are $n=100$ points randomly and independently placed in the unit square. This is a binomial point process, because the distribution of the number of points in an area $W$ inside the unit square is the binomial distribution with $n$ attempts and success probability $p=W$.

The same figure can also be seen as a realization of Poisson point process, given that there are $n=100$ points in the unit square. If this is not given, different realizations of the Poisson process will have different $n$'s, and the distribution of $n$ is the Poisson distribution. This is the only difference between the two processes, the binomial point process has a fixed number of points, while in the Poisson point process this number varies between different realizations.

If $n$ is large enough, the number of points in the unit square for the Poisson process will be peaked around the average of the Poisson distribution, so you might neglect the fluctuations around the average and just fix $n$ at the average value. In that approximation the Poisson and binomial point processes amount to the same thing.

$\endgroup$
  • $\begingroup$ thanks for your answer and nice explanation. However, can you make some comments on the intensity of the Poisson point process and its relation to $n$ of the Binomial point process? $\endgroup$ – Dimitrios Apr 10 '16 at 5:28
  • 1
    $\begingroup$ the intensity $\lambda$ of the Poisson process sets the average $\bar{n}=\lambda$ of $n$; so in the large-$\lambda$ limit the Poisson process becomes equivalent to a Binomial process with $n=\lambda$; fluctuations of $n$ around this value are smaller by a factor $1/\sqrt\lambda$. $\endgroup$ – Carlo Beenakker Apr 10 '16 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.