## 7.1 Poisson Process

Consider random events such as the arrival of customer at a shop, the arrival of emails to a mail server or the arrival of calls to a call-center. These events can be described by a counting function $$N(t)$$ defined for all $$t\geq 0$$. This counting function represents the number of events that occurred in $$[0,t]$$. For each interval $$[0,t]$$ the value $$N(t)$$ is an observation of a random variable where the only possible values are the integers $$0,1,2,\dots$$.

The counting process $$\{N(t): t\geq 0\}$$ is said to be a Poisson process with mean rate $$\lambda$$ if the following assumptions are fulfilled:

• $$N(0) = 0$$;

• it has independent increments: that is the number of arrivals during non-overlapping time intervals are independent random variables.

• the number of events in any interval of length $$t$$ is a Poisson random variable with parameter $$\lambda t$$.

Therefore $P(N(t)=n)=\frac{e^{-\lambda t}(\lambda t)^n}{n!}$

The last assumption implies that the distribution of the number of arrivals between, say, $$t$$ and $$t+s$$ depends only on the length of the interval $$s$$ and not on the starting point $$t$$. This property is usually called stationarity. Consequently:

$P(N(t)-N(s)=n)=\frac{e^{-\lambda(t-s)}(\lambda(t-s))^n}{n!}$ and because of the properties of the Poisson distribution: $E(N(t)-N(s))=V(N(t)-N(s))=\lambda(t-s).$

Now consider the time at which arrivals occurs in a Poisson process. Let the first arrival occur at time $$A_1$$, the second occur at time $$A_1+ A_2$$ and so on. Thus $$A_1,A_2,\dots$$ are successive inter-arrival times. The first arrival occurs after time $$t$$ if and only if there are no arrivals in the interval $$[0,t]$$ so it is seen that $\{A_1>t\}=\{N(t)=0\}$ and consequently $P(A_1>t)=P(N(t)=0)=e^{-\lambda t}$ Thus the probability that the first arrival will occur in $$[0,t]$$ is given by $P(A_1\leq t)= 1- P(A_1>t)= 1- e^{-\lambda t}$ which is the cumulative density function of an exponential distribution with parameter $$\lambda$$. Hence, $$A_1$$ is distributed exponentially with mean $$E(A_1)=1/\lambda$$. It can also be shown that all inter-arrival times $$A_1,A_2,\dots$$ are exponentially distributed and independent with mean $$1/\lambda$$.

An alternative definition of a Poisson process is of a counting process whose inter-arrival times are distributed exponentially and independently.

Exponential distributions have the property of being memoryless, which is deeply connected to Poisson processes. For an exponential distribution $$X$$ it holds that $P(X> s+t|X>s)=P(X>t).$ Suppose $$X$$ represents the life of a light bulb. The above equation states that the probability that the light bulb lives for at least $$s+t$$ hours, given it has survived $$s$$ hours, is that same as the initial probability that it lives for at least $$t$$ hours. That is, the light bulb it does not remember that it has already been in use for a time $$s$$. Let’s show the equality is true: $\begin{eqnarray*} P(X>s+t | X>s) &=& \frac{P(X > s + t, X >s)}{P(X>s)}\\ &=& \frac{P(X > s + t)}{P(X>s)}\\ &=&\frac{1- P(X \leq s + t)}{1 - P(X \leq s)}\\ &=&\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}\\ &=& e^{-\lambda t}\\ &=& P(X>t) \end{eqnarray*}$