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<record version="5" id="6733">
 <title>Poisson process</title>
 <name>PoissonProcess</name>
 <created>2005-02-09 20:22:19</created>
 <modified>2006-10-04 09:49:54</modified>
 <type>Definition</type>
 <creator id="3771" name="CWoo"/>
 <author id="13753" name="Mathprof"/>
 <author id="3771" name="CWoo"/>
 <classification>
	<category scheme="msc" code="60G51"/>
 </classification>
 <defines>
	<concept>simple Poisson process</concept>
	<concept>intensity</concept>
 </defines>
 <synonyms>
	<synonym concept="Poisson process" alias="homogeneous Poisson process"/>
 </synonyms>
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A counting process $\lbrace X(t)\mid
t\in\mathbb{R}^{+}\cup\lbrace0\rbrace\rbrace$ is called a
\emph{simple Poisson}, or simply a \emph{Poisson process} with
parameter $\lambda$, also known as the \emph{intensity}, if
\begin{enumerate}
\item $X(0)=0$,
\item $\lbrace X(t)\rbrace$ has stationary independent increments,
\item $P(X(t)=1)=\lambda t+o(t)$,
\item $P(X(t)&gt;1)=o(t)$,
\end{enumerate}
where $o(t)$ is the O notation.

\textbf{Remarks}.
\begin{itemize}
\item The intensity $\lambda$ is assumed to be a constant in terms of $t$.
\item Condition 3 above says that the \emph{rate} in which the an event occurs once in time interval $t$, as $t$ approaches 0, is $\lambda$. Condition 4 says that the event occurs more than once is very unlikely (the rate approaches zero as the time interval shrinks to zero).
\item It can be shown that $X(t)$ has a Poisson distribution (hence the name of the stochastic process) with parameter $\lambda t$: $$P(X(t)=n)=e^{-\lambda t}\frac{{(\lambda t)}^n}{n!}.$$
\item Therefore, $\operatorname{E}[X(t)]=\lambda t$.
\end{itemize}</content>
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