DES - Definition of a 'Poisson Process'

Summary

Given a set of independent, events all described by the same exponential distribution, we have that the probability of exactly $n$ events occurring during the interval $[t,t+T]$ is:

$$P\left(N_e(t,t+T)=n\right)=\frac{(\lambda T{)}^n}{n!}{e}^{-\lambda T},n=0,1,2,\dots$$

Where $P\left(N_e(t,t+T)=n\right)$ can be seen as the probability mass function of the Poisson distribution with parameter $\lambda T$. (From here the formula).

While the average number of said events occurring in an interval $[t,t+T]$ is:

$$E\left[N_e(t,t+T)\right]=\lambda T\text{. }$$

NOTE: Both formulas do NOT depend on time.

---

Poisson Process:

A Poisson process is the counting process of an event which is always possible, and characterized by interval times (interval between one event and the next) that are:

• Independent
• Identically distributed with an exponential distribution.

---

Observations:

The rate of the exponential distribution is also called the rate of the Poisson process.

A Poisson process can be viewed as a stochastic timed automaton with Poisson clock structure: $\left(\mathcal{E},\mathcal{X},\Gamma ,f,x_0,F\right)$

• Where $\mathcal{E}=\{e\},$
• $\mathcal{X}=\{0,1,2,3,\dots \}$, the state counts the number of occurrences of event $e$
• The transition diagram is given by:
• and $F=\left\{F_e\right\}$ with $F_e(t)=1-e^{-\lambda t},t⩾0,\lambda >0$.

NOTE: The transition diagram and the state $\mathcal{X}$ are infinite.

---

Observation:

Lets declare now a random variable $N_e(t,t+T)$ as:

$N_e(t,t+T)\triangleq$ number of occurrences of event $e$ over the interval $(t,t+T]$

$N_e(t,t+T)$ is a discrete random variable taking values in $\{0,1,2,\dots \}$. It can be shown that the pmf of $Ne(t,t+T)$ has the following form:

$$P\left(N_e(t,t+T)=n\right)=\frac{(\lambda T{)}^n}{n!}{e}^{-\lambda T},n=0,1,2,\dots$$

Which is also the pmf of Poisson Distribution with parameter $\lambda T$

$P[N_e(t,t+T)=n]$ : Probability that exactly $n$ events $e$ occurs in the interval $[t,t+T]$

• Demonstration for $n=0$ so: $P\left(N_e(t,t+T)=0\right)$

---

Expected Value of Poisson Process

The expected value of $N_e(t,t+T)$ is:

$$E\left[N_e(t,t+T)\right]=\lambda T\text{. }$$

Notice that the above formula does not depend on the time instant $t$, but only on the length T of the interval. $\Rightarrow$ It is a consequence of the memory-less property of the exponential distribution

---

~Example:

---