

\subsection{\bf Bartlett's Function for $\chi^2$}

\begin{slide}
\begin{itemize}
\item From Equation~(35) we have
\[
\frac{\partial}{\partial\lambda} \ln L = - \frac{2\alpha(\lambda -
\lambda_0 )}{2} \hspace{2em} \mbox{for}\ \ L = \exp \left( -
\frac{\chi^2}{2} \right) 
\]

\item and using our earlier asymptotic result
\begin{eqnarray*}
\alpha & = & I(\lambda) \hspace{1em} \mbox{we find} \\
S(\lambda ) & = & \sqrt{I(\lambda )} (\lambda_0 - \lambda ) \\
& = & \sqrt{\alpha} (\lambda_0 - \lambda )
\end{eqnarray*}

\item Therefore, $\lambda = \lambda_0 \pm one$.  Standard Deviation is
just $S(\lambda) = \mp 1$.  

\item In fact, for any number of parameters $n$,

\item $S(\lambda )$ has mean 0 and variance 1, and so the rule $S(\lambda )
= \pm 1 $ ($\pm l$ for $l$ Standard Deviations) is an appropriate
generalization of the rule that one increases $\chi^2$ by $1(l)$.
\end{itemize}
\end{slide}

