\documentclass{article}
\include{macros}
\newtheorem{example}[theorem]{Example}
\begin{document}

% lecture number : lecture title : scribe
\lecture{26}{Poisson Point Processes}{Ben Hough}{jbhough@math.berkeley.edu}
In this lecture, we consider a measure space $(S,\cal{S},\mu)$, where $\mu$ is a $\sigma$-finite measure (i.e. $S$ may be written as a disjoint union of sets of finite $\mu$-measure).

\section{The Poisson Point Process}
\begin{definition}
A {\bf Poisson Point Process} (P.P.P.) with intensity $\mu$ is a collection of random variables $N(A,\omega)$, $A \in {\cal S}$, $\omega \in \Omega$ defined on a probability space $(\Omega, \F, \P)$ such that:
\begin{itemize}
\item[1.] $N(\cdot,\omega)$ is a counting measure on $(S, {\cal S})$ for each $\omega \in \Omega$.
\item[2.] $N(A, \cdot)$ is Poisson with mean $\mu(A)$:
\begin{center}
$\P(N(A)=k) = \frac{e^{-\mu(A)}(\mu(A))^k}{k!}$ all $A \in {\cal S}$.
\end{center}
\item[3.] If $A_1$, $A_2$, ... are disjoint sets then $N(A_1, \cdot)$, $N(A_2, \cdot)$, ... are independent random variables.
\end{itemize}
\end{definition}

\begin{theorem}
P.P.P.'s exist.
\end{theorem}
\begin{proofsketch}
It's not enough to quote Kolmogorov's Extension Theorem.  The only convincing argument is to give an explicit constuction from sequences of independent random variables.  We begin by considering the case $\mu(S)<\infty$.
\begin{itemize}
\item[1.]  Take $X_1$, $X_2$, ... to be i.i.d. random variables so that $\P(X_i \in A) = \frac{\mu(A)}{\mu(S)}$.
\item[2.]  Take $N(S)$ to be a Poisson random variable with mean $\mu(S)$, independent of the $X_i$'s.  Assume all random varibles are defined on the same probability space $(\Omega, \F, \P)$.
\item[3.]  Define $N(A) = \sum_{i = 1}^{N(S)} \1_{(X_i \in A)}$, for all $A \in {\cal S}$.
\end{itemize}
Now verify that this $N(A,\omega)$ is a P.P.P. with intensity $\mu$. If $\mu(S) = \infty$ we write $S=S_1 \bigcup_{i=1}^\infty S_i$ as a disjoint union where $\mu(S_i)<\infty$ and construct P.P.P.'s $N_i(\cdot)$ with intensity $\mu$ restricted to $S_i$.  Make the $N_i(\cdot)$ independent and define $N(A) = \sum_{i =1}^\infty N_i(A)$ for all $A \in {\cal S}$.  The superposition and thinning properties of Poisson random variables now imply that $N(\cdot)$ has the desired properties\footnote{For a reference, see {\it Poisson Processes}, Sir J.F.C. Kingman, Oxford University Press.}.
\end{proofsketch}

\noindent
The most common way to construct a P.P.P. is to define 
\begin{equation}
N(A) = \sum_i \1_{(T_i \in A)}
\end{equation}
for some sequence of random variables $T_i$ which are called the points of the process.  
\begin{example}
Let $\mu$ be Lebesgue measure on $[0,\infty)$.  Define random variables $T_i$ so that $0<T_1<T_2<T_3<...$ with $T_1 = W_1$, and $T_r = W_1 + ... + W_r$ where the $W_i$ are i.i.d. exponential($\lambda)$ random variables. That is, $\P(W_i>t) = e^{-\lambda t}$ and if we define $N_t = N\left( [0,t) \right)$ we find that
\begin{equation}
\P(N_t = j) = \frac{e^{-\lambda t}(\lambda t)^j}{j!}
\end{equation}
\end{example}
\section{An Application}
We now describe a general method for constructing a process with independent increments from a P.P.P.  In particular, we wish to construct a process $(X_t; t \geq 0)$ of the form
\begin{equation}
	X_t = \sum_{0<s \leq t} \Delta X_s
\end{equation}
where the above sum has only countably many non-zero terms, and the collection $\{(s,\Delta X_s):s>0,\Delta X_s \neq 0\}$ is the set of points of a P.P.P. on $(0,\infty) \times \R-\{0\}$ with intensity measure $m \times L$ for some Levy-measure $L$ on $\R-\{0\}$ ($m$ denotes Lebesgue measure).
\begin{example}
A generalized Poisson Process may be constructed as follows.  Let $N_1$, $N_2$, ..., $N_m$ be independent P.P.P.'s with rate $\lambda_i$.  Define
\begin{equation}
X_t = \sum_{i= 1}^m a_i N_i(t)
\end{equation}
so that $X_t$ jumps by $a_i$ whenever $N_i$ jumps by 1.  In this case $\Delta X_t = \sum_i a_i \Delta N_i(t)$ and the Levy-measure is 
\begin{equation}
L(A) = \sum_{i=1}^m \lambda_i 1_{(a_i \in A)}.
\end{equation}
The familiar Poisson Process with parameter $\lambda$ is obtained by letting $m=1$, $\lambda_1 = \lambda$ and $a_1 =1$.
\end{example}
This example illustrates the concept for a discrete Levy-measure $L$.  From the previous lecture, we can handle a general finite measure $L$ by setting 
\begin{equation}
X_t = \sum_{i=1}^\infty Y_i 1_{(T_i \leq t)}
\end{equation}
where the $T_i$ are the points of jumps of a standard Poisson Process with rate $L(\R)$ and the $Y_i$ are i.i.d. with $\P(Y_i \in A) = \frac{L(A)}{L(\R)}$.  If $L$ is supported on $(0,\infty)$, i.e. we only allow positive jumps, we can compute:
\begin{eqnarray}
E\left[e^{-\theta X_t}\right] &=& \sum_{i=1}^\infty \P(N_t = i)(E\left[e^{-\theta Y}\right])^i \\
	&=& \sum_{i=1}^\infty e^{-\lambda t} \frac{(\lambda t)^i}{i!} \left( E\left[e^{-\theta Y} \right] \right)^i \\
&=& e^{-\lambda t} exp \left[\lambda t E\left[e^{-\theta Y}\right]\right] \\
&=& e^{-\lambda t} exp\left[t \int_0^\infty e^{-\theta x} L(dx)\right] \\
E\left[e^{-\theta X_t}\right] &=& exp\left[t \int_0^\infty \left(e^{-\theta x} -1 \right) L(dx) \right].
\end{eqnarray}
To summarize:  we have shown that if $(X_t, t \geq 0)$ is a process with independent increments where the set of jumps $\left( (s,\Delta X_s), s>0 \right)$ is the set of points of a P.P.P. on $(0,\infty) \times \R$ with intensity distribution $L(dx)$, then $X_t$ has distribution specified by the Laplace Transform above (26.11).  This formula is an instance of the Levy-Khintchine equation.

\begin{example}
Consider two independent processes $X_t$ and $Y_t$ which correspond to P.P.P.'s with measures $dt \times L(dx)$ and $dt \times M(dx)$ respectively.  Then $X_t + Y_t$ is a P.P.P. with measure $dt \times (M+L)(dx)$ as one may check via direct computation using equation (26.11).
\end{example}

These results may also be extended to suitable infinite measures.  Suppose we can write $L = \sum_i L_i$ where the $L_i$'s are finite and $\int_0^\infty x L(dx)<\infty$.  Then we may define P.P.P.'s $X_i$ according to the measures $dt \times L_i(dx)$ and sum them to obtain a process $X$ with jumps according to the measure $dt \times L(dx)$.

\noindent
Finally, it is of interest to ask for which measures $L$ equation (26.11) give the Laplace Transform of a distribution on $[0,\infty]$?  One may show that the following three conditions are equivalent:
\begin{itemize}
\item[1.] Equation (26.11) is the Laplace Transform of a distribution on $[0,\infty]$
\item[2.]$\int_0^\infty \left( 1-e^{-\theta x} \right) L(dx)<\infty$ for all $\theta<\infty$
\item[3.]$\int_0^\infty \left( 1-e^{-\theta x} \right) L(dx)<\infty$ for some $\theta<\infty$.
\end{itemize}

\begin{exercise}
Check that if you make this Poisson construction of jumps with $L(dx) = x^{-1}e^{-x} dx$ then $X_t=$ gamma($t$).  That is:
\begin{equation}
	\P(X_t \in dx) = \frac{1}{\Gamma(t)} e^{-x} x^{t-1} dx.
\end{equation}
\end{exercise}



\end{document}