5 min read

On Poisson process

This is my study note for Math 605, UW-Madison. See more details at here.


Almost every entry level statistical course introduce the Poisson distribution formula (we call this as the first formulation): \[f(k;\mu)=Pr(X=\mu)=\frac{\mu^{k}e^{-\mu}}{k!}\] where \(X\) is a discrete random variable, \(\mu\) is the mean measure. When you need to calculate the probability of observing k events, you can just put k back to the formula. One important property of Poisson distribution is that its mean equals with its variance, i.e. \(\mu=E(X)=Var(X)\).

Now let’s consider continuous process. Suppose that we start at time 0 to count events (earthquakes, car accidents, number of death, etc.). For each time t, we obtain a number N(t), which is the total number of events that has occurred up to time t. We then make the following modeling assumptions on the process N(t):

  1. For some \(\lambda>0\), the probability of exactly one event occurring in a given time interval of length h is equal to \(\lambda h+o(h)\). That is, for any \(t\geq 0\), \[P\{N(t+h)-N(t)=1\}=\lambda h+o(h),as\,h\rightarrow 0\]
  2. The probability that 2 or more events occur in an interval of length h is o(h): \[P\{N(t+h)-N(t)\geq 2\}=o(h),as\,h\rightarrow 0\]
  3. The random variables \(N(t_{1}) - N(s_{1})\) and \(N(t_{2}) - N(s_{2})\) are independent for any choice of \( s_{1}\leq t_{1}\leq s_{2}\leq t_{2} \). This is usually termed an independent interval assumption.

We then sat that N(t) is a homogeneous Poisson process with intensity, propensity, or rate \(\lambda\).

Proposition 1: Let N(t) be a Poisson process satisfying the three assumptions above. Then for any \( t\geq s\geq0\) and \(k\in\{0,1,2,...\}\), we can prove that \[P\{N(t)-N(s)\}=e^{-\lambda(t-s)}\frac{(\lambda(t-s))^{k}}{k!} \] If we choose \(s=0\), then we get \[P\{N(t)-N(s)\}=e^{-(\lambda t)}\frac{(\lambda t)^{k}}{k!} \]

Letting \(S_{1}\) denote the time of the first increase in the process (i.e. the first event occurred), then according to Proposition 1, \[ P\{S_{1}>t\}=P\{N(t)=0\}=e^{-\lambda t} \] Therefore, the distribution of the first event time is exponentially distributed with a parameter of \(\lambda\). Because of the independent increments assumption (assumption 3), we can see that the distribution of \(S_{2}-S_{1}\) is also exponentially distributed with a parameter of \(\lambda\). Therefore, we see that *N(t)* is simply the counting process of a renewal process with inter-event times determined by *exponential random variables*(we call this as the second formulation).

We now move to the third formulation of a one dimensional Poisson process. We say the N is a Poisson process with intensity \(\lambda\) if for any \( A\subset\mathbb{R}_{\geq 0} \) and \( k\geq 0 \), we have that \[P\{N(A)=k\}=e^{-\lambda|A|}\frac{(\lambda|A|)^{k}}{k!} \] where \(|A|\) is the Lebesgue measure of A and if \(N(A_{1}),...,N(A_{k}) \) are independent random variables whenever \(A_{1},...,A_{k} \) are disjoint subsets of state space \( E\).

Definition 1. Let N be a point process with state space \( E\in\mathbb{R}^{d}\), and let \(\mu\) be a measure on \(\mathbb{R}^{d} \). We say that N is a Poisson process with mean measure \(\mu \), or a Poisson random measure, if the following two conditions hold:

  1. For \(A\subset E \), \[P\{N(A)=k\}=\begin{cases} \frac{e^{-\mu(A)}(\mu(A))^{k}}{k!} & if\,\mu(A)<\infty\\ 0 & if\,\mu(A)=\infty \end{cases} \]
  2. If \(A_{1},...,A_{k} \) are disjoint subsets of state space \( E\), then \(N(A_{1}),...,N(A_{k}) \) are independent random variables.

Note that the mean measure of a Poisson process, \( \mu(A) \), completely determines the process. One choice of the mean measure would be a multiple of Lebesgue measure, which gives length in \( \mathbb{R}^{1}\), area in \( \mathbb{R}^{2}\), volume in \( \mathbb{R}^{3}\), etc. That is, if \( \mu((a,b])=\lambda(b-a)\), for \(a,b\in\mathbb{R}\), then \(\mu\) is said to be Lebesgue measure with rate, or intensity, \(\lambda\). If \(\lambda=1\), then the measure is said to be unit-rate. For another example, a Poisson process with Lebesgue measure in \(\mathbb{R}^{2}\) satisfies \(\mu(A)=Area(A) \). When the mean measure is a multiple of Lebesgue measure, we call the process homogeneous (\(\lambda>1\)).

If \(\wedge\) is a non-decreasing, absolutely continuous function, and over an open interval (a,b), then mean measure \(\mu\) for a Poisson process is \[\mu((a,b))=\wedge(b)-\wedge(a)\] If \(\wedge\) has density \(\lambda\) (i.e. it is differentiable), then \[\mu((a,b))=\wedge(b)-\wedge(a)=\int_{a}^{b}\lambda(s)ds \] As a result, for any \( A\subset \mathbb{R}\), \[P\{N(A)=k\}=e^{-\int_{a}^{b}\lambda(s)ds}\frac{(\int_{a}^{b}\lambda(s)ds)^{k}}{k!}\]

The three formulations above are all equivalent.

Some definitions: >1. Renewal process: it is used to model occurrences of events happening at random time, where gaps between points (inter-event time) are i.i.d. random variables. >1. Point process: it is used to model a random distribution of points in space. It is a renewal process which distributes points so gaps are i.i.d. exponential random variables. e.g. locations of diseased deer in a given region (space); the breakdown times of certain part of a car (time). The simplest and most ubiquitous example of a point process is the Poisson point process. >1. Lebesgue measure: length, area, volume, etc.

Transformations of Poisson processes

  • If the position of the points of a homogeneous process of rate \(\lambda>0\) are multiplied by \(\lambda\), then the resulting point process is also Poisson, and it is, in fact, a homogeneous process of rate 1.
  • Likewise, we could start with a unit-rate process and divide the position of each point by \(\lambda\) to get a homogeneous process with rate \(\lambda\).
  • Also, move the points around via an one-to-one function, or transformation, resulted in another Poisson process.

Simulating non-homogeneous Poisson process

Set \(t_{0}=0,n=1 \).

  1. Let \(E_{n}\) be an exponential random variable with parameter one, which is independent from all other random variables already generated.
  2. Find the smallest \( \mu\geq 0\) for which \[\int_{0}^{\mu}\lambda(s)ds=E_{1}+\cdots+E_{n}\]
  3. Set n+1 to n.
  4. Return to step 1 or break.

###Example Let N be a non-homogeneous Poisson process with local intensity \(\lambda(t)=t^2\). Write a code that simulates this process until 500 jumps have taken place.

N=500 # number of jumps
time=vector() # to hold times of jumps
E_n=rexp(N) # N exp random values
for (i in 1:N){
ggplot(data=NULL)+geom_step(aes(x=c(0,time), y=c(0:500)))+
  geom_line(aes(x=c(0,time), y=c(0,time)^3/3), color="blue")