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: \[\begin{eqnarray*} P(X>s+t | X>s) &=& \frac{P(X > s + t, X >s)}{P(X>s)}\\ &=& \frac{P(X > s + t)}{P(X>s)}\\ &=&\frac{1- P(X \leq s + t)}{1 - P(X \leq s)}\\ &=&\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}\\ &=& e^{-\lambda t}\\ &=& P(X>t) \end{eqnarray*}\]