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Version 3 |
| \PMlinkescapeword{simple} |
\PMlinkescapeword{simple} |
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| A counting process $\lbrace X(t)\mid |
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 $\lambda$ if |
| t\in\mathbb{R}^{+}\cup\lbrace0\rbrace\rbrace$ is called a |
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| \emph{simple Poisson}, or simply a \emph{Poisson process} with |
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| parameter $\lambda$, also known as the \emph{intensity}, if |
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| \begin{enumerate} |
\begin{enumerate} |
| \item $X(0)=0$, |
\item $X(0)=0$, |
| \item $\lbrace X(t)\rbrace$ has stationary independent increments, |
\item $\lbrace X(t)\rbrace$ has stationary independent increments, |
| \item $P(X(t)=1)=\lambda t+o(t)$, |
\item $P(X(t)=1)=\lambda t+o(t)$, |
| \item $P(X(t)>1)=o(t)$, |
\item $P(X(t)>1)=o(t)$, |
| \end{enumerate} |
\end{enumerate} |
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| \begin{itemize} |
\begin{itemize} |
| \item The intensity $\lambda$ is assumed to be a constant in terms of $t$. |
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| \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 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, $\operatorname{E}(X(t))=\lambda t$. |
\item Therefore, $\operatorname{E}(X(t))=\lambda t$. |
| \end{itemize} |
\end{itemize} |
