% scribe: Matias Damian Cattaneo
% lastupdate: 12 November 2005
% lecture: 18
% references: Durrett, Section 2.6
% title: Poisson Processes: Part II
% keywords:  
% end

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\lecture{18}{Poisson Processes -- Part II}{Matias Damian Cattaneo}{cattaneo@econ.berkeley.edu}

\section{Compound Poisson Distribution}

We begin by recalling some things from last lecture.

Let $X_{1},X_{2},...$ be independent and identically distributed random
variables with distribution $F$ on $\R$; that is:%
\begin{equation*}
F\!\left( B\right) =\P \!\left[ X\in B\right]
\end{equation*}

Let $N_{\lambda }$ be a Poisson random variable with mean $\lambda $; that
is:%
\begin{equation*}
\P \!\left[ N_{\lambda }=k\right] =\frac{\lambda^{n}e^{-\lambda }}{k!},\text{ \ \ \ 
}n=0,1,2,\ldots.
\end{equation*}

\begin{theorem}
Let $S$ be the sum of a Poisson random number of i.i.d.\ random variables;
that is:%
\begin{equation*}
S=\sum_{i=1}^{N_{\lambda }}X_{i}.
\end{equation*}

Let $L\!\left( B\right) =\lambda F\!\left( B\right) $ for $B\in \borel$, where 
$L$ is a positive measure on $\R$. Then the distribution of $S$
is determined by its characteristic function:%
\begin{equation*}
\E\!\left[ e^{itS}\right] =\exp \!\left\{ \int_{\R%
}\left( e^{itx}-1\right) L\!\left( dx\right) \right\}.
\end{equation*}
\end{theorem}

\textbf{Remark: }
Call this distribution of $S$ compound Poisson with parameter $L$ 
(denoted by $CP\!\left(L\right)$) and observe that it is $\infty $-divisible 
since:%
\begin{equation*}
CP\!\left( L\right) \ast CP\!\left( M\right) =CP\!\left( L+M\right),
\end{equation*}%
so the convolution $n$-th root of $CP\!\left( L\right) $ is $CP\!\left( \frac{L}{%
n}\right) $.

\textbf{Interpretation of }$L$: Recall that Poisson point process $%
\longleftrightarrow $ counting measure, and we have%
\begin{equation*}
N\!\left( B\right) =\sum_{i=1}^{N_{\lambda }}1\!\left\{ X_{i}\in
B\right\}.
\end{equation*}%
That is, $N\!\left( B\right) $ is the number of values $1\leq i\leq N_{\lambda
}$ with $X_{i}\in B$. Observe%
\begin{equation*}
N\!\left( \R\right) =N_{\lambda }\thicksim Poisson\!\left( \lambda
\right)
\end{equation*}

What is the distribution of $N\!\left( B\right) $?  Apply the previous
theorem with $X_{i}$ replaced by $1\!\left\{ X_{i}\in B\right\} $. So we have%
\begin{equation*}
\E\!\left[ e^{itN\!\left( B\right) }\right] =\exp \!\left\{ \!\left(
e^{it}-1\right) L\!\left( B\right) \right\},
\end{equation*}%
so $N\!\left( B\right) \thicksim Poisson\!\left( L\!\left( B\right) \right) $%
.

More generally for $B_{1},B_{2},...,B_{m}$; $m$ disjoint sets we
can compute, by the same argument,%
\begin{equation*}
\E\!\left[ e^{i\sum_{k=1}^{m}t_{k}N\!\left( B_{k}\right) }%
\right] =\prod_{k=1}^{m}\E\!\left[ e^{it_{k}N\!\left(
B_{k}\right) }\right],
\end{equation*}%
and observe that the LHS is the multivariate characteristic function of the
vector $\left( N\!\left( B_{1}\right) ,N\!\left( B_{2}\right) ,...,N\!\left(
B_{m}\right) \right) $ at $\left( t_{1},t_{2},...,t_{m}\right) $, and the
RHS is the multivariate characteristic function of a collection of
independent random variables with a Poisson distribution. Consequently, by
the uniqueness theorem for multivariate characteristic function (see text)\
we conclude that $N\!\left( B_{1}\right) ,N\!\left( B_{2}\right) ,...,N\!\left(
B_{m}\right) $ are independent Poisson variables.

\bigskip

\section{Summary so far}
Now we summarize our work so far.

Let $X_{1},X_{2},...$ be i.i.d.\ $F$. Let $N_{\lambda
}\thicksim Poisson\!\left( \lambda \right) $, independent of $X_{1},X_{2},...$%
. Let $N\!\left( B\right) =\sum_{i=1}^{N_{\lambda }}1\!\left\{ X_{i}\in
B\right\} $, the point process counting values in $B$ up to $N_{\lambda }$. Then $%
\left( N\!\left( B\right) ,B\in Borel\right) $ is a \textbf{Poisson random
measure} with mean measure $L$, meaning that if $B_{1},...,B_{m}$ are disjoint
Borel sets, $(N\!\left( B_{i}\right) $, $1\leq i\leq m)$ are independent 
with distributions $%
Poisson\!\left( L\!\left( B_{i}\right)\right)$  for $1\leq i\leq m$, 
respectively.

\begin{example}
(From previous lecture) Let $0<T_{1}<T_{2}<...$ be a sum of independent $%
Exponential\!\left( \lambda \right) $ variables. So $N_{t}=\sum%
_{i=1}^{\infty }1\!\left\{ T_{i}\leq t\right\} \thicksim
Poisson\!\left( \lambda t\right) $. Then we see that $\left( N_{t},0\leq t\leq
T\right) $ has the same distribution as $\left( N\!\left[ 0,t\right] ,0\leq
t\leq T\right) $ where%
\begin{equation*}
N\!\left[ 0,t\right] =\sum_{i=1}^{N_{\lambda }}1\!\left\{ X_{i}\leq
t\right\}
\end{equation*}%
for $X_{1},X_{2},...\thicksim \mathcal{U}\!\left[ 0,T\right] $.

This is an example of a famous connection between sums of exponentials and
uniform order statistics. Examples can be found in many texts, including
\cite{durrett}.
\end{example}

These are Poisson tricks!

\section{Computations with CP}
Now we discuss some computations with $CP\!\left( L\right) $. Think about
this: we have a Poisson scatter with mean intensity $L$, say $%
X_{1},X_{2},...,X_{n}$. Let $\lambda =L\!\left( \R\right) $. We have%
\begin{equation*}
S=\sum_{i=1}^{N_{\lambda }}X_{i}=\int xN\!\left( dx\right)
\end{equation*}%
and recall that%
\begin{equation*}
N\!\left( B\right) =\sum_{i=1}^{N_{\lambda }}1\!\left\{ X_{i}\in
B\right\} \thicksim Poisson\!\left( L\!\left( B\right) \right)
\end{equation*}%
and also%
\begin{eqnarray*}
N\!\left( \cdot \right) &=&\sum_{i=1}^{N_{\lambda }}\delta
_{X_{i}}\left( \cdot \right) \\
\int f\!\left( x\right) N\!\left( dx\right) &=&\sum_{i=1}^{N_{\lambda
}}f\!\left( X_{i}\right)
\end{eqnarray*}

Now we compute (You check details):%
\begin{eqnarray*}
\E\!\left[ S\right] &=&\E\!\left[ \int xN\!\left( dx\right) \right]
=\int x\E\!\left[ N\!\left( dx\right) \right] =\int xL\!\left( dx\right) \\
\mathbb{V}\!\left[ S\right] &=&\mathbb{V}\!\left[ \int xN\!\left( dx\right) \right]
=\mathbb{V}\!\left[ \sum ...\right] =...=\int x^{2}L\!\left( dx\right)
\end{eqnarray*}

\begin{example}
Consider%
\begin{eqnarray*}
L &=&\sum_{i}\lambda _{i}\delta _{X_i} \\
N\!\left( \cdot \right) &=&\sum_{i}N_i \delta_{X_{i}}\!\left( \cdot \right)
\end{eqnarray*}%
where $N_i \thicksim Poisson\!\left( \lambda _{i}\right) $
and as $i$ varies these are independent. Now we have:%
\begin{eqnarray*}
S &=&\int xN\!\left( dx\right) =\sum_{i}x_{i}N\!\left( x_{i}\right) \\
\E\!\left[ S\right] &=&\sum_{i}x_{i}\lambda _{i}=\int
xL\!\left( dx\right) \\
\mathbb{V}\!\left[ S\right] &=&\sum_{i}x_{i}^{2}\lambda _{i}=\int
x^{2}L\!\left( dx\right)
\end{eqnarray*}
\end{example}

\begin{theorem}
\textbf{(L-K)} Every $\infty $-divisible distribution on $\R$ is a
weak limit of shifted CP distributions.
\end{theorem}

Look at the characteristic function of a centered CP\ distribution to see
something new: take $S\thicksim CP\!\left( L\right) $ and look at $\left( S-%
\E\!\left[ S\right] \right) $.\ Assuming that $\int \left\vert
x\right\vert L\!\left( dx\right) <\infty $, we have%
\begin{eqnarray*}
\E\!\left[ e^{it\left( S-\E\!\left[ S\right] \right) }\right]
&=&\exp \!\left\{ -it\E\!\left[ S\right] \right\} \exp \!\left\{ \int
\left( e^{itx}-1\right) L\!\left( dx\right) \right\} \\
&=&\exp \!\left\{ \int \left( e^{itx}-1-itx\right) L\!\left( dx\right) \right\}
\end{eqnarray*}%
and%
\begin{equation*}
\E\!\left[ \left( S-\E\!\left[ S\right] \right) ^{2}\right]
=\int x^{2}L\!\left( dx\right)
\end{equation*}%
from before.

Observe that this formula defines a characteristic function for every
positive measure $L$ on $\R$ with $L\!\left( -1,1\right) ^{c}=0$ and $%
\int_{-1}^{1}x^{2}L\!\left( dx\right) <\infty $. You can easily check this; 
see texts such as \cite{durrett}.  This leads to the general L-K Formula.

\section{More details on L\'evy Measure}

\begin{definition}
A measure $L$ on $\mathbb{R}$ is a L\'evy measure if it has 
the following properties: 
\begin{enumerate}
\item $L\{(-\varepsilon, \varepsilon)^c\}<+\infty$, for all $\varepsilon>0$. 
\item $L\{0\}=0$.
\item $\int_{-1}^{1} x^2L(dx)<+\infty$. 
\end{enumerate} 
\end{definition}

For such an $L$, $\sigma^2\geq 0$, $c\in \mathbb{R}$, define the 
L\'evy-Khinchine exponent in the following way: 
$$\Psi_{L, \sigma^2, c}(t)=\int \left(e^{itx}-1-it\tau(x)\right) L(dx) - 
\frac12\sigma^2t^2+itc,$$ where $\tau(x)$ is the truncation function 
defined by $\tau(x)=x\mbox{\bf 1}_{|x|\leq 1}+ \mbox{\bf 1}_{x> 1}- 
\mbox{\bf 1}_{x<1}$. 

\begin{theorem}
Two important results:
\begin{enumerate}
\item $e^{\Psi(t)}$ is an infinitely divisible characteristic function. 
\item $e^{\Psi(t)}$ determines $L, \sigma^2, c$ uniquely.
\end{enumerate}
\end{theorem}

Before we prove this theorem, we consider a few examples.

\begin{example}

\begin{enumerate}
\item Consider a point mass $\delta_c$ at $c$. Its characteristic function 
is $e^{itc}$, and we see that $itc= \Psi_{0,0,c}(t)$. 
\item Consider now a normal distribution $N(c,\sigma^2)$. Its characteristic 
function is $e^{itc-\sigma^2t^2/2}$ and it is easy to see that $\Psi(t) = 
itc-\sigma^2t^2/2$ corresponds to $(0,\sigma^2,c)$. 
\item Now, let $N$ be a Poisson random measure. For each $f\geq 0$, 
we have $$E\!\left(e^{-\theta\int fdN}\right) = \exp\!\left(\int 
\left(e^{-\theta f(x)} -1\right) \mu(dx)\right)\; 
\mbox{}$$ If $\mu$ is bounded measure, take $\theta=-it$, 
$$E\!\left(e^{it\int fdN}\right) = 
\exp\!\left( \int\left(e^{itf(x)}-1\right)\,\mu(dx)\right).$$ 
Let $L(dy)= \mu\{x: f(x)\in dy\}$ (restricted to $\{0\}^c$). For those who doesn't 
like to see $dy$'s outside the integral sign, the definition of $L$ could be
$L(B):=\mu(f^{-1}(B))$. 

Then $E\!\left(e^{it\int fdN}\right)=\exp\!\left(\int \left( 
e^{ity}-1\right) L(dy)\right)$. Here we can recognize the enemy from 
the beginning of the lecture, and the characteristic function of $\int fdN$ 
is $\exp(\Psi_{L,0,c})$ where $c=\int \tau(x)L(dx)$.
\end{enumerate}
\end{example}

\begin{proof}
First, we will prove that $e^{\Psi(t)}$ is a characteristic function, and 
the infinite divisibility is obvious ($n$-th root 
is $\Psi_{(L/n, \sigma^2/n, c/n)}$). Fix $t$. 
Observe that for $|x|<1$ we have 
%
\begin{equation}
e^{itx}-1-it\tau(x)=e^{itx}-1-itx\leq cx^2t^2
\label{lmproof1}
\end{equation}
%
for $|xt|$ small. Therefore, 
the integral converges because $\int_{-1}^1 x^2 L(dx)<+\infty$ and 
$L\{(-\varepsilon,\varepsilon)^c\}<+\infty$. Hence $\Psi(t)$ is a well-defined 
complex number for all $t\in \mathbb{R}$. 

Second, since the product of characteristic functions is also a characteristic 
function we may assume without loss of generality that $\sigma^2$ and $c$ 
are both $0$. Let $L_n$ be $L$ restricted to $\left(-\frac1n, 
\frac1n\right)^c$. Note that $\exp(\Psi_{L_n,0,0}(t))$ is a characteristic 
function: since $L_n$ is finite, $\exp(\Psi_{L_n,0,0}(t))$ is the 
characteristic function of a shifted compound Poisson variable with parameter 
$L_n$.  From \ref{lmproof1} and the dominated convergence theorem we see that  
%
$$\lim_{n\rightarrow \infty} \Psi_{L_n,0,0}(t)=\Psi_{L,0,0}(t).$$ 
%
Since $\exp$ is continuous function we immediately have that 
$\exp(\Psi_{L_n,0,0}(t))\rightarrow \exp(\Psi_{L,0,0}(t))$ and it only 
remains to prove that $\Psi(t)$ is continuous at $0$ (in order to apply the 
L\'evy continuity theorem). This is left as an 
exercise for the reader.  (The same dominated convergence theorem will work.) 
\end{proof}


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