Point processes are random scattering of points, randomly scattered with respect to some law of distribution. Poisson processes are a fundamental example of a point process that are used to model a variety of different phenomena and serve as the basis of more complicated processes. For example we may be interested in a model for the arrival times or total of number of customers \(N(t)\) that have visited a high street shop up to an arbitrary time \(t\). Poisson processes are often used to develop probabilistic models for \(N(t)\); in this context, the actual arrival times of the customers form a Poisson point process.

We will consider a few different ways of defining and constructing Poisson processes, all of which give a process with the same law, but each description will have certain advantages over the other,

Some review

We say that \(X\) is Poisson random variable with mean \(\mu\) if it has probability mass function given by \[ \mathbb{P}(X = n) = \frac{ e^{-\mu} \mu ^n}{n!} \ \text{ for $n = 0,1, 2, \ldots$}.\] We say that \(Z\) is an exponential random variable with mean \(\mu\) (rate \(\lambda = 1/\mu\)) if it is continuous random variable with probability density function given by \[ z \mapsto \tfrac{1}{\mu} e^{-\tfrac{ z} {\mu}} \ \text{ for $z \geq 0$}.\]

Exercises: Poisson and exponential random variables

  • Check that the mean of a Poisson random variable with parameter \(\mu\) does indeed have mean \(\mu\). Show that the variance is also \(\mu\).

  • Show that the minimal of two independent exponential random variables is again an exponential random variable.

  • What is the memoryless property? Prove that exponential random variables have this property.

  • Let \(n>0\) and \(\mu >0\). Let \(W\) be a Poisson random variable with mean \(\mu\). Let \((X_i)_{i=1}^n\) be independent Bernoulli random variables with parameter \(p=\lambda/n\). Let \(S_n = X_1 + \cdots + X_n\). Show for all nonnegative integers \(k\), we have \(\lim_{n \to \infty} \mathbb{P}(S_n =k) = \mathbb{P}(W=k)\).

  • Let \(\mu >0\), \(p \in (0,1)\), and \(W\) be a Poisson random variable with mean \(\mu\). Define the random variable \(X\) in the following way: if \(W=n\), then we let \(X\) be the sum of \(n\) independent Bernoulli random variables with parameter \(p\). Show that \(X\) is a Poisson random variable with mean \(p\lambda\).

  • Show that the sum of independent Poisson random variables is again a Poisson random variable.

Generating a Poisson process as exponential inter-arrivals

We will start with the following construction which is also useful for simulations.

We say that \(N\) is a Poisson process on \([0, \infty)\) with rate \(\lambda\). We call the set of random arrival times \(\{T_n: n \in \mathbb{Z}^{+}\}\) the corresponding Poisson point process on \([0, \infty)\) of intensity \(\lambda\).

With R it is easy to simulate a Poisson process:

inter = rexp(15, 1) 
arr = cumsum(inter)
one = rep(1, times = length(arr))
plot(arr, one, yaxt = 'n', ann=FALSE)


We can make use of R and simulate many realizations of Poisson point processes. In particular, pay attention to the following points:

I think the last point is very important in applications; everything has units in the real world, and it can be quite revealing to make sure all the units work out.

Whatโ€™s Poisson got to do with it?

It is not obvious what Poisson random variables have to do with Poisson processes from its construction as exponential inter-arrival times. In the following more mathematical description, the word Poisson appears from the outset.

Let \(N\) be a Poisson process on \([0, \infty)\) with rate \(\lambda\). It is possible to show that:

These two properties also characterize a Poisson process and can be used to define spatial Poisson processes in higher dimensions. We will prove that our construction using exponential random variables satisfies a weak version of the first property.

Lemma: Let \(N\) be a Poisson process constructed with exponential inter-arrival times of rate \(\lambda>0\). Then \(N(t)\) is a Poisson random variable with mean \(t \lambda\).


Let \(N\) be given by exponential inter-arrival times, so that \(X_i\) are the inter-arrival times, and \(T_k\) is the time of the \(k\) arrival. Observe that \[\mathbb{P}(N(t) = 0) = \mathbb{P}( X_1 >t) = e^{-\lambda t}.\] For \(k >0\), we have \[ \begin{eqnarray*} \mathbb{P}(N(t) = k) &=& \mathbb{P}( T_k \leq t, T_{k+1} >t) \\ &=& \mathbb{P}( T_k \leq t, T_{k} + X_{k+1} >t) \\ &=& \int_0 ^t \int_{t-y} ^{\infty} g(y) f(x) dx dy, \end{eqnarray*} \] where \(g\) is the pdf for \(T_k\) and \(f\) is the pdf for \(X_{k+1}\); here we need to appeal the independence of these two random variables. Thus \[\mathbb{P}(N(t) = k) = \int_0 ^t g(y) e^{-(t-y) \lambda } dy\] We recall that the sum of independent exponentials has the law of a gamma distribution; specifically \[ g(y) = \frac{\lambda^k}{\Gamma(k)} y^{k-1} e^{-\lambda y}.\] Thus \[ \begin{eqnarray*} \mathbb{P}(N(t) = k) &=& \frac{\lambda^k e^{-t \lambda} }{\Gamma(k)} \int_0 ^ t y^{k-1} dy \\ &=& \frac{\lambda^k e^{-t \lambda} t^k}{k\Gamma(k) } \\ &=& \frac{(\lambda t)^k e^{-t \lambda} }{k! }, \end{eqnarray*} \] as promised.


We demonstrate the last exercise by simulations in the following way. We simulate Poisson processes, and record the position of the first arrival if it occurs in \([0,1]\) and the second arrival occurs outside this interval. We plot a histogram of the recorded occurrences

    record <- function(){
 inter = rexp(2,1)
 arr = cumsum(inter)
 r = 2
 if ( (arr[1] < 1) & (arr[2] >1)){
 r <- arr[1]

x = replicate(10000, record())
y = setdiff(x, 2)
hist(y, prob=T)

A characterization via modelling assumptions

Poisson processes are good models for arrivals only if it is reasonable to assume that one also has the memoryless property that comes with the exponential distribution. In addition, if an arrival process \(N(t)\) satisfies the following mild conditions, then it can be shown that it is a Poisson process of intensity \(\lambda\).

We can sketch a proof of our previous lemma which connected Poisson random variables with Poisson processes. Let \(t >0\), and partition the interval \([0,t]\) into \(n\) intervals of size \(t/n\), where \(n\) is large. By orderliness condition with stationarity, we can assume that in each interval there is at most one arrival. Let \(p = \mathbb{P}( N(t/n) =1))\). By the independent increments and rate conditions, we have that probability that there are \(k\) arrives is given by \[\mathbb{P}(N(t) = k) = {n \choose k }p^k (1-p)^{n-k} = \frac{n!}{(n-k)! k!}\frac{(\lambda t)^k}{n^k}\Big(1- \frac{\lambda t}{n}\Big)^{n-k} + g(n),\] where \(g(n) \to 0\) as \(n \to \infty\). Thus the desired limit follows.


Poisson processes as uniform random variables

Another way to simulate Poisson point processes is given by the following characterization on finite volumes. To simulate a Poisson point process of intensity \(\lambda\) on an interval \([s,t]\):

This characterization is my favourite and also has the advantage that it can easily be generalized to higher dimensions and other spaces.

The following code outputs the a sequence of arrivals of a Poisson process of intensity \(2\) in the interval \([0,1]\)

M = rpois(1, 2) 
P = runif(M) 
sort(P, decreasing=F)
## [1] 0.1836958


Other spaces and spatial Poisson point processes

In more general spaces, it is more natural to consider the point process (the location of the arrivals) rather than the counting process (the total number arrivals). Here we think of the point process \(\Pi\) as a random subset of \(\mathbb{R}^d\), and we let \(\Pi(A)\) to be the number of \(\Pi\)-point in \(A \subset \mathbb{R}^d\). We say that \(\Pi\) is a Poisson point process on \(\mathbb{R}^d\) with intensity \(\lambda\) if:

This characterization is also lends itself to easy simulations. To simulate a Poisson point process \(\Pi\) of intensity \(\lambda\) on a set \(A \subset \mathbb{R}^d\):

The following R code gives a Poisson point process on a square.

M= rpois(1,10)
x = runif(M)
y = runif(M)
plot(x,y, xlim=c(0,1), ylim=c(0,1))

Some other nice properties



Version: 30 October 2022