Postscript version of this file

STAT 380 Week 7

Properties of Poisson Processes

1)

If $N_1$ and $N_2$ are independent Poisson processes with rates $\lambda_1$ and $\lambda_2$, respectively, then $N=N_1+N_2$ is a Poisson process with rate $\lambda_1+\lambda_2$.

2)

Suppose $N$ is a Poisson process with rate $\lambda$. Suppose each point is marked with a label, say one of $L_1,\ldots,L_r$, independently of all other occurences. Suppose $p_i$ is the probability that a given point receives label $L_i$. Let $N_i$ count the points with label $i$ (so that $N=N_1+\cdots+N_r$). Then $N_1,\ldots,N_r$ are independent Poisson processes with rates $p_i\lambda$.

3)

Suppose $U_1, U_2,\ldots$ independent rvs, each uniformly distributed on $[0,T]$. Suppose $M$ is a Poisson $(\lambda T)$ random variable independent of the $U$'s. Let

$$N(t) = \sum_1^M 1(U_i \le t)$$

Then $N$ is a Poisson process on $[0,T]$ with rate $\lambda$.

4)

Suppose $N$ is a Poisson process with rate $\lambda$. Let $S_1 < S_2 < \cdots$ be the times at which points arrive Given $N(T)=n$ $S_1,\ldots,S_n$ have the same distribution as the order statistics of a sample of size $n$ from the uniform distribution on $[0,T]$.

5)

Given $S_{n+1}=T$, $S_1,\ldots,S_n$ have the same distribution as the order statistics of a sample of size $n$ from the uniform distribution on $[0,T]$.

Indications of some proofs:

1) $N_1,\ldots,N_r$ independent Poisson processes rates $\lambda_i$, $N=\sum N_i$. Let $A_h$ be the event of 2 or more points in $N$ in the time interval $(t,t+h]$, $B_h$, the event of exactly one point in $N$ in the time interval $(t,t+h]$.

Let $A_{ih}$ and $B_{ih}$ be the corresponding events for $N_i$.

Let $H_t$ denote the history of the processes up to time $t$; we condition on $H_t$.

We are given:

$$P(A_{ih}\vert H_t) =o(h)$$

and

$$P(B_{ih}\vert H_t) = \lambda_i h+o(h)\, .$$

Note that

$$A_h \subset \bigcup_{i=1}^r A_{ih} \cup \bigcup_{i \neq j} \left(B_{ih}\cap B_{jh}\right)$$

Since

$$P(B_{ih}\cap B_{jh}\vert H_t) = P(B_{ih}\vert H_t) P(B_{jh}\vert H_t)$$

$$= (\lambda_i h +o(h))(\lambda_j h+o(h))$$

$$= O(h^2)$$

$$= o(h)$$

we have checked one of the two infinitesimal conditions for a Poisson process.

Next let $C_h$ be the event of no points in $N$ in the time interval $(t,t+h]$ and $C_{ih}$ the same for $N_i$. Then

$$P(C_h\vert H_t) = P(\cap C_{ih}\vert H_t)$$

$$= \prod P(C_{ih}\vert H_t)$$

$$= \prod (1-\lambda_ih +o(h))$$

$$= 1-(\sum \lambda_i)h + o(h)$$

shows

$$P(B_h\vert H_t) = 1-P(C_h\vert H_t) -P(A_h\vert H_t)$$

$$= (\sum \lambda_i) h +o(h)$$

Hence $N$ is a Poisson process with rate $\sum \lambda_i$.

2) The infinitesimal approach used for 1 can do part of this. See text for rest. Events defined as in 1): The event $B_{ih}$ that there is one point in $N_i$ in $(t,t+h]$ is the event, $B_h$ that there is exactly one point in any of the $r$ processes together with a subset of $A_h$ where there are two or more points in $N$ in $(t,t+h]$ but exactly one is labeled $i$. Since $P(A_h\vert H_t)=o(h)$

$$P(B_{ih}\vert H_t) = p_iP(B_{h}\vert H_t)+o(h)$$

$$= p_i (\lambda h+o(h)) + o(h)$$

$$= p_i \lambda h + o(h)$$

Similarly, $A_{ih}$ is a subset of $A_h$ so

$$P(A_{ih}\vert H_t) =o(h)$$

This shows each $N_i$ is Poisson with rate $\lambda p_i$. To get independence requires more work; see the text for the algebraic method which is easier.

3) Fix $s<t$. Let $N(s,t)$ be the number of points in $(s,t]$. Given $N=n$ the conditional distribution of $N(s,t)$ is Binomial$(n,p)$ with $p=(t-s)/T$. So

$$P(N(s ,t)=k)$$

$$= \sum_{n=k}^\infty P(N(s,t)=k,N=n)$$

$$= \sum_{n=k}^\infty P(N(s,t)=k\vert N=n) P(N=n)$$

$$= \sum_{n=k}^\infty \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \frac{(\lambda T)^n}{n!} e^{-\lambda T}$$

$$= \frac{e^{-\lambda T}}{k!} (\lambda T p)^k\sum_{n=k}^\infty \frac{(1-p)^{n-k} (\lambda T)^{n-k}}{(n-k)!}$$

$$= \frac{e^{-\lambda T}}{k!} (\lambda T p)^k \sum_{m=0}^\infty(1-p)^m(\lambda T)^m/m!$$

$$= \frac{e^{-\lambda T}}{k!} (\lambda T p)^ke^{\lambda T(1-p)}$$

$$= \frac{e^{-\lambda(t-s)}(\lambda(t-s))^k}{k!}$$

4): Fix $s_i,h_i$ for $i=1,\ldots, n$ such that

$$0 < s_1 < s_1+h_1 < s_2 < \cdots < s_n<s_n+h_n< T$$

Given $N(T)=n$ we compute the probability of the event

$$A=\bigcap_{i=1}^n \{s_i < S_i < s_i+h_i\}$$

Intersection of $A$, $N(T)=n$ is ($s_0=h_0=0$):

$$\begin{multline*} B \equiv \\ \bigcap_{i=1}^n \{N(s_{i-1}+h_{i-1},s_i]=0,N(s_i,s_i+h_i]=1\} \\ \cap \{N(s_n+h_n,T]=0\} \end{multline*}$$

whose probability is

$$\left(\prod \lambda h_i\right) e^{-\lambda T}$$

So

$$P(A\vert N(t)=n) = \frac{P(A,N(T)=n)}{P(N(T)=n)}$$

$$= \frac{ \lambda^n e^{-\lambda T} \prod h_i}{(\lambda T)^n e^{-\lambda T} / n!}$$

$$= \frac{n!\prod h_i}{T^n}$$

Divide by $\prod h_i$ and let all $h_i$ go to 0 to get joint density of $S_1,\ldots,S_n$ is

$$\frac{n!}{T^n} 1(0 < s_1 < \cdots < s_n <T)$$

which is the density of order statistics from a Uniform$[0,T]$ sample of size $n$.

5) Replace the event $S_{n+1}=T$ with $T < S_{n+1}< T+h$. With $A$ as before we want

$$\begin{multline*} P(A\vert T < S_{n+1}< T+h) \\ = \frac{ P(B,N(T,T+h] \ge 1)}{P(T < S_{n+1}< T+h)} \end{multline*}$$

Note that $B$ is independent of $\{N(T,T+h] \ge 1\}$ and that we have already found the limit

$$\frac{P(B)}{ \prod h_i} \to \lambda^n e^{-\lambda T}$$

We are left to compute the limit of

$$\frac{P(N(T,T+h] \ge 1)}{P(T < S_{n+1}< T+h)}$$

The denominator is

$$\begin{multline*} \sum_{k=0}^n P(N(0,T]=k,N(T,T+h]=n+1-k) \\ +o(h) = P(N(0,T]=n)\lambda h+o(h) \end{multline*}$$

Thus

$$\frac{P(N(T,T+h] \ge 1)}{P(T < S_{n+1}< T+h)} = \frac{\lambda h+o(h)}{\frac{(\lambda T)^n}{n!} e^{-\lambda T} \lambda h+o(h)}$$

$$\to \frac{n!}{(\lambda T)^ne^{-\lambda T}}$$

This gives the conditional density of $S_1,\ldots,S_n$ given $S_{n+1}=T$ as in 4).

Inhomogeneous Poisson Processes

The idea of hazard rate can be used to extend the notion of Poisson Process. Suppose $\lambda(t) \ge 0$ is a function of $t$. Suppose $N$ is a counting process such that

$$P(N(t+h)=k+1\vert N(t)=k,H_t)= \lambda(t) h +o(h)$$

and

$$P(N(t+h) \ge k+2\vert N(t)=k,H_t)= o(h)$$

Then $N$ has independent increments and $N(t+s)-N(t)$ has a Poisson distribution with mean

$$\int_t^{t+s} \lambda (u) du$$

If we put

$$\Lambda(t) = \int_0^t \lambda(u) du$$

then mean of $N(t+s)-N(t)$ is $\Lambda(t+s)-\Lambda(t)$.

Jargon: $\lambda$ is the intensity or instaneous intensity and $\Lambda$ the cumulative intensity.

Can use the model with $\Lambda$ any non-decreasing right continuous function, possibly without a derivative. This allows ties.

Compound Poisson Processes

Imagine insurance claims arise at times of a Poisson process, $N(t)$, (more likely for an inhomogeneous process).

Let $Y_i$ be the value of the $i$th claim associated with the point whose time is $S_i$.

Assume that the $Y$'s are independent of each other and of $N$.

Let

$$\mu = {\rm E}(Y_i)$$    and $$\sigma^2 = {\rm var}(Y_i)$$

Let

$$X(t) = \sum_{i=1}^{N(t)} Y_i$$

be the total claim up to time $t$. We call $X$ a compound Poisson Process.

Useful properties:

$${\rm E}\left\{X(t)\vert N(t)\right\} = N(t) \mu$$

$${\rm Var}\left\{X(t)\vert N(t)\right\} = N(t)\sigma^2$$

$${\rm E}\left\{X(t)\right\} = \mu {\rm E}\left\{N(t)\right\}$$

$$= \mu \lambda t$$

$${\rm Var}\left\{X(t)\right\} = {\rm Var}\left[{\rm E}\left\{X(t)\vert N(t)\right\}\right]$$

$$\qquad +{\rm E}\left[{\rm Var}\left\{X(t)\vert N(t)\right\}\right]$$

$$= \lambda t \mu^2 + \lambda t \sigma^2$$

(Look at all familiar? See homework.)