# On Poisson process

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

## Definitions

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.

###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.

library(ggplot2)
theme_set(theme_bw())
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){
time[i]=(3*sum(E_n[1:i]))^(1/3)
}
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")