Poisson Derivation 3

From Qwiki

Jump to: navigation, search

This derivation of the Poisson Distribution is based on the distribution of time intervals between events. This waiting-time distribution for the Poisson process is given by an exponential distribution. That is, the probability that the next event occurs in a time between Δ and Δ + dΔ from now is w_\lambda(\Delta) = \lambda e^{-\lambda \Delta}d\Delta.\, Also, the probability that no event occurs over a time interval Δ is just \exp(-\lambda\Delta)\,. The probability that exactly n events occur in a large time interval T is then given by the probabilities that n waiting times Δj, 1\leq j\leq n, add to less than T and a final interval \Delta_f = T-\Delta_1-\cdots-\Delta_n occurs with no more events. Mathematically, this probability is given by

P_\lambda(n,T) = \int d\Delta_f e^{-\lambda \Delta_f} \int d\Delta_1\cdots \int d\Delta_n\lambda^n e^{-\lambda(\Delta_1+\cdots\Delta_n)}\delta\left(T-\Delta_f-\cdots-\Delta_n\right)

But this is just a happy multiconvolution integral, so let's take a Laplace Transform with respect to T to find
\mathcal{L}\left[P_\lambda(n,T)\right](s) = \lambda^{n}\left\{\mathcal{L}\left[e^{-\lambda\Delta}\right](s)\right\}^{n+1}
=\frac{\lambda^n}{(s+\lambda)^{n+1}}.
One can just look up the inverse Laplace transform to find
P_\lambda(n,T) = \lambda^n \mathcal{L}^{-1}\left[\frac{1}{(s+\lambda)^{n+1}}\right] = \frac{e^{-\lambda T}(\lambda T)^n}{n!}.
Retrieved from "http://qwiki.stanford.edu/wiki/Poisson_Derivation_3"
Personal tools