
\subsection{\bf Determination of Lifetime Example}
\subsection{\bf of Maximum Likelihood---II}
\begin{slide}
\begin{itemize}
\item The maximum of $L$ is given by:
\begin{eqnarray*}
\frac{dL}{d\tau} & = & 0 \hspace{1em} \mbox{at}\ \tau=\tau^* \ \
\mbox{or} \\
\frac{1}{\tau^{*^2}}\, {\displaystyle \sum^n_{i=1}} t_i - \frac{n}{\tau^*} 
& = & 0 \hspace{1em} \mbox{or} \\
\tau^* & = & \frac{1}{n}\, {\displaystyle \sum^n_{i=1}} t_i
\end{eqnarray*}
\item or the maximum likelihood estimate is just the mean: 

\item such a simple result is only true because of particular Poisson
distribution used in the example.
\end{itemize}
\end{slide}

