

\subsection{\bf Proof of Maximum Likelihood}
\subsection{\bf Theorem Part (a)---III}

\begin{slide}
\begin{itemize}
\item Note, this form for $\langle\alpha^*\rangle$ is only valid to $O(1/n)$.  
\[
\langle\alpha^* \rangle = \alpha_0 - \frac{\left\langle
\frac{\partial}{\partial\alpha} \ln P(x , \alpha_0 )
\right\rangle}{\left\langle \frac{\partial^2}{\partial\alpha^2} \ln P(x , 
\alpha_0 ) \right\rangle}
\]

\item Now, the numerator of formula for $\langle\alpha^* \rangle$ is
$\left\langle \frac{\partial}{\partial\alpha} \ln P(x, \alpha_0 )
\right\rangle$ which is by definition

\begin{eqnarray*}
\int\limits dx P(x, \alpha_0 ) \frac{\partial}{\partial\alpha} \ln P(x,
\alpha_0) & = & \int\limits dx \frac{\partial}{\partial\alpha} P(x, \alpha)
\big\vert_{\alpha=\alpha_0} \\
& = & \frac{\partial}{\partial\alpha} \left\lbrack \int\limits dx P(x,
\alpha) \right\rbrack \big\vert_{\alpha=\alpha_0} \\
& = & \frac{\partial}{\partial\alpha} \cdot 1 \\
& = & 0
\end{eqnarray*}
as $P$ is truely normalized.  

\item This proves part (a) that $\langle\alpha^* \rangle = \alpha_0$
with corrections from both of order $L_3/L_2$ and $L_1/L_2 \neq$ quite
0 terms which are both of order $1/n$.

\item These corrections have been calculated by Yellin.  
\end{itemize}
\end{slide}

