

\subsection{\bf Proof of Maximum Likelihood}
\subsection{\bf Part (b): Formula for Error---II}

\begin{slide}
\begin{itemize}
\item We can calculate the correct {\em statistical\/} standard
deviation in $\alpha^*$ by taking Equation~(17) and using our $F/G$ theorem
with $f_i = \frac{\partial}{\partial\alpha} \ln P(x_i, \alpha)$, $g_i =
\frac{\partial^2}{\partial\alpha^2} \ln P (x_i, \alpha)$.  Generally
\[
\sigma^2 = \frac{1}{n} \left\langle\left( \frac{f}{\langle g\rangle}
- \frac{g\langle f\rangle}{\langle g\rangle^2} \right)^2\right\rangle
\]
\item but $\langle f\rangle = 0$---proved above, so: $\sigma^2 = \frac{1}{n}
\frac{\langle f^2 \rangle}{\langle g\rangle^2}$.  

\item It is easy to see
that
\[
\left\langle f^2 \right\rangle = {\displaystyle\int\limits} \frac{\left( 
\frac{\partial P}{\partial\alpha} \right)^2}{P^2} \cdot P dx
\]
and this is $\langle g\rangle$.  
\end{itemize}
\end{slide}

