Poisson Process

The Poisson process N(t) is an example of a counting process, that is N(t) is the number of times something has happened up to time t.

Example: Consider an ATM machine and let N(t) be the number of customers served by the ATM machine at time t.

Because of the way it is defined every counting process has the following properties:
1) N(t)≥0
2) N(t) is an integer
3) If s<t then N(s)≤N(t)
4) If s<t then N(t)-N(s) is the number of events that have occured in the interval (s,t].

Definition

Let {N(t);t≥0} be a counting process. Then
a) {N(t);t≥0} is said to have independent increments if the number of events that occur in disjoint intervals are independent.
b) {N(t);t≥0} is said to have stationary increments if the distribution of the number of events that occur in any interval of time depend only on the length of the interval.

Example: The process of our ATM machine probably has independent increments but not stationary increments. Why?

The most important example of a counting process is the Poisson process. To define it we need the following notation, called Landau's o symbol:
a function f is said to be o(h) if limh0f(h)/h = 0

Example: f(x)=x2 is o(h) but f(x)=x is not.

Definition

A counting process {N(t);t≥0} is said to be a Poisson process with rate l>0 if
1) N(0)=0
2) N(t) has stationary and independent increments
3) P(N(h)=1) = lh + o(h)
3) P(N(h)≥2) = o(h)

Notice that this implies that during a short time period the probability of an event occuring is proportional to the length of the interval and the probability of a second (or even more) events occuring is very small.

Proposition

Let {N(t);t≥0} be a Poisson process, then
N(t+s)-N(s) ~ Pois(t)

Proof Let pn = P(N(t)=n). Then

where the last equation follows from (P(N(h)=0) = 1-P(N(h)≥1) = 1-P(N(h)=1) - P(N(h)≥2)
Now
The same basic idea works for the case pn(t) as well to finish the proof

Remark It is intuitively clear why the definition above should lead to the Poisson distribution. Take the interval (0,t] and subdivide it into k equal size intervals (0,t/k], (t.k, 2t/k) .. ((k-1)t/k,t]. The probability of 2 or more events in any one interval goes to 0 as k goes to ∞ because
P(2 or more events in any subinterval)
≤∑P(2 or more events in the kth subinterval)
= ko(t/k)
t×o(t/k)/(t/k) 0 as k ∞
Hence N(t) will (with probability going to 1) just equal the number of subintervals in which an event occurs. However, by independent and stationary increments this number will have a binomial distribution with parameters k and p=lt+o(t/k). hence by the Poisson approximation to the binomial we see that N(t) will have a Poisson distribution with rate lt.

Definition

Let T1 be the time when the first event occurs, T2 the time from the first event until the second event etc. The sequence T1, T2, .. is called the sequence of interarrival times.

Proposition

Let {N(t);t≥0} be a Poisson process, and {Ti;i≥1} be the interarrival times. Then T1, T2, .. are iid Exp(l)

Proof Note that {T1>t} is equivalent to {no events occured in (0,t]} and so

and we see that T1 ~ Exp(l). But

because of independent and stationary increments. So we find that T2 ~ Exp(l) and that T1 T2. By induction it is clear that the sequence {Tn,n=1,2,..} is an iid sequence of exponential r.v. with mean 1/l.

Remark This result should not come as a surprise because the assumption of independent and stationay increments means that the process from any moment on is independent of all that occured before and also has the same distribution as the process started at 0. In other words the process is memoryless, and we have previously shown that any continuous rv on (0,∞) with the memoryless property has to have an exponential distribution.

Definition

Let Sn be the arrival time of the nth event. (This is also often called the waiting time until the nth event).

Proposition

Let {N(t);t≥0} be a Poisson process, and {Sn;n≥1} be the waiting times. Then Sn~ G(n,l)

Proof Clearly Sn = Sni=1Ti, and so we find Sn ~ G(n,l).

Example: Up to yesterday a store has 999,856 customers. They are planning to hold a little party when the 1,000,000th customer comes into the store. From experience they know that a customer arrives about every 4 minutes, and the store is open from 9am to 6pm. What is the probability that they will have the party today?
They will have the party if at least 144 customers come into the store today. Let's assume that the customers arrive according to a Poisson process with rate 4min (?) then we want the probability P(S144 < 9× 60). Now S144 ~G(144,4) and
P(S144 < 540) = 0.23.

Here is another proof of the last proposition. We use the fact that the nth event occurs at or before time t if and only if the number of events occuring by time t is at least n. So
N(t)≥n Sn≤t

This is a very useful equivalence, and much more general than just for the Poisson process, so here is an illustration:

With this we find