Poisson point processes

1 Basics

A Poisson process $\{\mathsf{n}\}\sim \mathrm{PoisP}(\lambda )$ is a stochastic process whose inter-occurrence times are identically and independently distributed such that ${\mathsf{t}}_i-{\mathsf{t}}_{i-1}\sim \mathrm{Exp}(\lambda )$ (rate parameterization). By convention, we set all ${\mathsf{t}}_i\ge 0$ and $\mathsf{n}(0)=0$ a.s. The counting process that increases for each event is one convenient representation for such a process. In that case, we think of the paths of the process as $\mathsf{n}:\mathbb{R}\mapsto {\mathbb{Z}}^{+}$ such that $\mathsf{n}(t)\equiv \sum _{i=1}^{\mathsf{n}}{\mathbb{I}}_{\{{\mathsf{t}}_i<t\}}$.

The following is a representation of such a creature based on Matt Motoki’s answer on stackoverflow.

set.seed(1234)
lambda <- 0.3
last_arrival <- 0
arrival_times <- c()
n <- 10
counts = 1:n
for (i in counts) {
  last_arrival <- rexp(1, rate = lambda) + last_arrival
  arrival_times <- c(arrival_times,last_arrival)
}

plot(arrival_times, counts, pch=16, ylim=c(0, n))
points(arrival_times, c(0, counts[-n]))
segments(
  x0 = c(0, arrival_times[-n]),
  y0 = c(0, counts[-n]),
  x1 = arrival_times,
  y1 = c(0, counts[-n])
)

It is easy to show that $\mathsf{n}(t)\sim \text{Pois}(\lambda t)$, where the pmf of a Poisson RV is

$$f(k;\lambda )=\frac{{\lambda }^n}{n!}{e}^{-\lambda }.$$

It is a standard result that the increments of such processes over disjoint intervals have a Poisson distribution also. For $t\ge s$:

$$\mathsf{n}(t)-\mathsf{n}(s)\sim \text{Pois}((t-s)\lambda ).$$

Note also the standard result that

$$\lambda :=\underset{h\to 0}{lim}\frac{\mathrm{E}(\mathsf{n}(t,t+h))}{h}.$$

We call $\lambda$ the rate.

2 Poisson distribution

3 Poisson bridge

Suppose we are given the value of a Poisson process at time $0$ and time $1$ and are concerned with some $t\in (0,1).$ We wish to know the conditional distribution $P(\mathsf{n}(t)\mid \mathsf{n}(1)=S)$, i.e. the Poisson bridge. Using the distributions of the increments and the independence property,

$$\begin{array}{rl}P[\mathsf{n}(t)=n\mid \mathsf{n}(1)=S]& =\frac{P[\mathsf{n}(t)=n\cap \mathsf{n}(1)=S]}{P[\mathsf{n}(1)=S]}\\ & =\frac{P[\mathsf{n}(t)=n\cap \tilde{\mathsf{n}}(1-t)=S-n]}{P[\mathsf{n}(1)=S]}\\ & =\frac{\frac{(\lambda t{)}^n}{n!}{e}^{-\lambda t}\frac{(\lambda (1-t{)}^{S-n})}{(S-n)!}{e}^{-\lambda (1-t)}}{\frac{{\lambda }^S}{S!}{e}^{-\lambda }}\\ & =\frac{(\lambda t{)}^n(\lambda (1-t){)}^{S-n}}{{\lambda }^S}\frac{e^{-\lambda t}{e}^{-\lambda (1-t)}}{e^{-\lambda }}\frac{S!}{(S-n)!n!}\\ & =(t{)}^n(1-t{)}^{S-n}\left(\begin{array}{c}S\\ n\end{array}\right)\\ & =\mathrm{Binom}(n;S,t).\end{array}$$ So we can simulate a point Poisson bridge at some $t<1$ by sampling a Binomial random variable.

4 Hitting time

Consider the hitting time of a level $\ell \in \mathbb{N}$ for a Poisson process $\mathsf{n}$.

$$\begin{array}{rl}\mathbb{P}[\mathsf{n}(t)<\ell ]& =\mathbb{P}[\underset{s}{inf}[\mathsf{n}(s)\ge \ell ]>t]\\ & =\mathbb{P}[{\mathsf{t}}_{\ell }>t]\\ & =\mathbb{P}[(\sum _{1<k\le \ell }{\mathsf{t}}_k-{\mathsf{t}}_{k-1})>t].\end{array}$$

But since ${\mathsf{t}}_k-{\mathsf{t}}_{k-1}\sim \mathrm{Exp}(\lambda ),$ and i.i.d. and the sum of i.i.d. Exponential variates is a Gamma variate, we have that

$${\mathsf{t}}_{\ell }\sim \mathrm{Gamma}(\ell ,\lambda )$$

and hence the density of this random variable is the Gamma distribution density,

$$f(t)=\frac{{\ell }^{\lambda }}{\mathrm{\Gamma }(\lambda )}{x}^{\lambda -1}{e}^{-\ell x}.$$

The Gamma process is a kind of dual to the Poisson; we can think of it as a distribution over specific hitting times of an imaginary latent Poisson process.

5 Taylor expansion

See Schoutens, Leuven, and Studer (2001).

6 References