

\subsection{\bf The Likelihood Maximizes the}
\subsection{\bf Quivering Rod---III}

\begin{slide}
\begin{itemize}
\item So as $L= Pr \left( \alpha \vert x_i , T \right)$

\item we say that maximum likelihood estimate of $\alpha$ is $\alpha^*
= \frac{1}{n} {\displaystyle \sum^n_{i=1}} x_i$

\item and its standard deviation is $\sigma/\sqrt{n}$.

\item Note that whatever a priori $P(\alpha\vert T)$ you put in,

\begin{description}
\item{$\bullet$} the shape of $L \cdot P(\alpha\vert T)$ is essentially
independent of $P(\alpha\vert T)$ as $n\rightarrow\infty$
\item{$\bullet$} and we get a Gaussian peaked at $\alpha^*$ riding on
top of the constant value $P(\alpha^* \vert T)$.
\end{description}
\end{itemize}
\end{slide}

