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. $\mu =E\left(X\right)=Var\left(X\right)$.

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 $\lambda h+o\left(h\right)$. That is, for any $t\ge 0$, $$P\left\{N\right(t+h)-N(t)=1\}=\lambda h+o\left(h\right),ash\to 0$$
2. The probability that 2 or more events occur in an interval of length h is o(h): $$P\left\{N\right(t+h)-N(t)\ge 2\}=o\left(h\right),ash\to 0$$
3. The random variables $N\left(t_1\right)-N\left(s_1\right)$ and $N\left(t_2\right)-N\left(s_2\right)$ are independent for any choice of $s_1\le t_1\le s_2\le t_2$. 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 $t\ge s\ge 0$ and $k\in \{0,1,2,...\}$, we can prove that $$P\left\{N\right(t)-N(s\left)\right\}=e^{-\lambda (t-s)}\frac{\left(\lambda \right(t-s){)}^k}{k!}$$ If we choose $s=0$, then we get $$P\left\{N\right(t)-N(s\left)\right\}=e^{-\left(\lambda t\right)}\frac{(\lambda t{)}^k}{k!}$$

Letting $S_1$ denote the time of the first increase in the process (i.e. the first event occurred), then according to Proposition 1, $$P\{S_1>t\}=P\left\{N\right(t)=0\}=e^{-\lambda t}$$ 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 $A\subset R_{\ge 0}$ and $k\ge 0$, we have that $$P\left\{N\right(A)=k\}=e^{-\lambda |A|}\frac{(\lambda |A|{)}^k}{k!}$$ where $|A|$ is the Lebesgue measure of A and if $N\left(A_1\right),...,N\left(A_k\right)$ 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 $E\in R^d$, and let $\mu$ be a measure on $R^d$. 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 $A\subset E$, $$P\left\{N\right(A)=k\}=\{\begin{array}{cc}\frac{e^{-\mu \left(A\right)}\left(\mu \right(A){)}^k}{k!}& if\mu \left(A\right)<\infty \\ 0& if\mu \left(A\right)=\infty \end{array}$$
2. If $A_1,...,A_k$ are disjoint subsets of state space $E$, then $N\left(A_1\right),...,N\left(A_k\right)$ are independent random variables.

Note that the mean measure of a Poisson process, $\mu \left(A\right)$, completely determines the process. One choice of the mean measure would be a multiple of Lebesgue measure, which gives length in $R^1$, area in $R^2$, volume in $R^3$, etc. That is, if $\mu \left(\right(a,b\left]\right)=\lambda (b-a)$, for $a,b\in R$, 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 $R^2$ satisfies $\mu \left(A\right)=Area\left(A\right)$. 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 $$\mu \left(\right(a,b\left)\right)=\wedge \left(b\right)-\wedge \left(a\right)$$ If $\wedge$ has density $\lambda$ (i.e. it is differentiable), then $$\mu \left(\right(a,b\left)\right)=\wedge \left(b\right)-\wedge \left(a\right)={\int }_a^b\lambda \left(s\right)ds$$ As a result, for any $A\subset R$, $$P\left\{N\right(A)=k\}=e^{-{\int }_a^b\lambda \left(s\right)ds}\frac{\left({\int }_a^b\lambda \right(s)ds{)}^k}{k!}$$

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 $\mu \ge 0$ for which $${\int }_0^{\mu }\lambda \left(s\right)ds=E_1+\cdots +E_n$$
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 $\lambda \left(t\right)=t^2$. 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")