

\subsection{\bf Proof of Maximum Likelihood}
\subsection{\bf Part (b):  Formula for Error---I}
\begin{slide}
\begin{itemize}
\item Now to prove part (b).  

\item First consider ``physics'' method.  

\item This takes standard deviation of $\alpha^*$ as
$\sqrt{-\frac{1}{L_2}}$ or rather this evaluated at $\alpha=\alpha^*$
not $\alpha=\alpha_0$ but these are same asymptotically.
\begin{eqnarray*}
\sigma^2_{\rm physics} & = & \frac{-1}{{\displaystyle \sum^n_{i=1}} \left\lbrack
\frac{\partial^2}{\partial\alpha^2} \ln P(x_i, \alpha_0) \right\rbrack} \\
\langle\sigma^2_{\rm physics}\rangle & = & \frac{-1}{n \left\langle
\frac{\partial^2}{\partial\alpha^2} \ln P(x_i, \alpha_0) \right\rangle}
\hspace{2em} \mbox{where} \\
\left\langle \frac{\partial^2}{\partial\alpha^2} \ln P(x,\alpha)
\right\rangle & = & \int\limits dx P \left\lbrace \frac{- (\partial P/ 
\partial\alpha)^2}{P^2} + \frac{\partial^2 P/\partial\alpha^2}{P}
\right\rbrace
\end{eqnarray*}

\item Now second term is just $\frac{\partial^2}{\partial\alpha^2}
{\displaystyle \int} P$ and is zero again because of normalization condition.
So,
\begin{equation}
\left\langle \sigma^2_{\rm physics} \right\rangle =
\frac{1}{\left\lbrace n {\displaystyle \int} dx \left( \frac{\partial
P}{\partial\alpha} \right)^2 / P \right\rbrace}
\end{equation}
\end{itemize}
\end{slide}

