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≥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,….
The counting process {N(t):t≥0} is said to be a Poisson process with mean rate λ 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 λt.
Therefore P(N(t)=n)=e−λt(λt)nn!
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)=e−λ(t−s)(λ(t−s))nn! and because of the properties of the Poisson distribution: E(N(t)−N(s))=V(N(t)−N(s))=λ(t−s).
Now consider the time at which arrivals occurs in a Poisson process. Let the first arrival occur at time A1, the second occur at time A1+A2 and so on. Thus A1,A2,… 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 {A1>t}={N(t)=0} and consequently P(A1>t)=P(N(t)=0)=e−λt Thus the probability that the first arrival will occur in [0,t] is given by P(A1≤t)=1−P(A1>t)=1−e−λt which is the cumulative density function of an exponential distribution with parameter λ. Hence, A1 is distributed exponentially with mean E(A1)=1/λ. It can also be shown that all inter-arrival times A1,A2,… are exponentially distributed and independent with mean 1/λ.
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: P(X>s+t|X>s)=P(X>s+t,X>s)P(X>s)=P(X>s+t)P(X>s)=1−P(X≤s+t)1−P(X≤s)=e−λ(s+t)e−λs=e−λt=P(X>t)