Poisson Processes

Interspike Interval (ISI): time in-between spikes

Assume ISI is distributed $t_i \sim \mathrm{exp}(\lambda)$

The exponential distribution is memoryless

Proof: $\mathrm{Pr}\{T>t+s|T > t\} = \dfrac{e^{-\lambda(t+s)}}{e^{-\lambda t}} = e^{-\lambda s} = \mathrm{Pr}\{T>s\}$

Defining the Poisson Process

<t1-><t2-----><t3-><t4------->
|----|--------|----|----------|----------- t
0    T1       T2   T3         T4 

Let $t_1,\,t_2,\,…$ be independent exponentially distributed random variables with parameter $\lambda$.

$t_i \,\,\,\mathrm{i.i.d}\sim\mathrm{exp}(\lambda)$

Let $T_n =t_1 +t_2 +\, …\, +t_n$, for all $n \geq 1$, and let $T_0 = 0$.

So, $T_n$ is the absolute time of the $n$th spike.

We define $N(s) = \mathrm{max}\{n:\, T_n\leq s\}$.

In other words, $N(s)$ accepts an absolute time and returns the number of spikes that have occurred up to and including time s.

Why is $N(s)$ a Poisson Process and not an Exponential Process?

Property 1: $N(s)$ is a Poisson distributed random variable with mean $\lambda s$.

$\mathrm{Pr}\{N(s) =n\} = \dfrac{(\lambda s)^ne^{-\lambda s}}{n!}$

Proving Property 1:

$N(s) \iff T_{n-1} \leq s \leq T_n$

$\mathrm{Pr}\{N(s) = n\} = \mathrm{Pr}\{T_n \leq s,\, T_{n+1} > s\}$

$=\mathrm{Pr}\{T_n \leq s\} \cdot \mathrm{Pr}\{T_{n+1} > s | T_n \leq s\}$