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)*:

- 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`\(t\geq 0\)`

,`\[P\{N(t+h)-N(t)=1\}=\lambda h+o(h),as\,h\rightarrow 0\]`

- The probability that 2 or more events occur in an interval of length
*h*is*o(h)*:`\[P\{N(t+h)-N(t)\geq 2\}=o(h),as\,h\rightarrow 0\]`

- 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:

- For
`\(A\subset E \)`

,`\[P\{N(A)=k\}=\begin{cases} \frac{e^{-\mu(A)}(\mu(A))^{k}}{k!} & if\,\mu(A)<\infty\\ 0 & if\,\mu(A)=\infty \end{cases} \]`

- 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 **. For another example, a Poisson process with Lebesgue measure in**

*unit-rate*`\(\mathbb{R}^{2}\)`

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

.

- Let
`\(E_{n}\)`

be an exponential random variable with parameter one, which is independent from all other random variables already generated. - Find the smallest
`\( \mu\geq 0\)`

for which`\[\int_{0}^{\mu}\lambda(s)ds=E_{1}+\cdots+E_{n}\]`

- Set n+1 to n.
- 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.

```
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")
```