

\subsection{\bf Proof of Maximum Likelihood}
\subsection{\bf Theorem Part (a)---I}
\begin{slide}
\begin{itemize}
\item Can be found in Cramer, p.~479 or Hoel p.~379.  

\item For (a) and (b), write
\[
\ln L(\alpha) = \sum^n_{i=1} \ln P(x_i , \alpha)
\]

\item Expand $L$ about true theoretical value.
\end{itemize}

\[
\ln L(\alpha) = L_0 + (\alpha - \alpha_0 ) L_1 + \frac{(\alpha -
\alpha_0)^2}{2} L_2 + \frac{(\alpha - \alpha_0)^3}{6} L_3 + \ldots
\]
\begin{eqnarray*}
L_0 & = & \sum^n_{i=1} \ln P(x_i , \alpha_0 ) \\
L_1 & = & \sum^n_{i=1} \frac{\partial}{\partial\alpha} \ln P(x_i ,
\alpha_0 ) \\
L_2 & = & \sum^n_{i=1} \frac{\partial^2}{\partial\alpha^2} \ln P(x_i ,
\alpha_0 ) , \hspace{1em} \mbox{etc.}
\end{eqnarray*}
\begin{itemize}
\item We get maximum likelihood estimate, by calculating $\partial\ln
L/\partial\alpha^* = 0$ or
\begin{eqnarray}
L_1 & + & (\alpha^* - \alpha_0 ) L_2 = 0 \hspace{1em} \mbox{i.e.,}
\nonumber \\
&& \alpha^* = \alpha_0 - \frac{L_1}{L_2}
\end{eqnarray}
\end{itemize}
\end{slide}

