

\subsection{\bf (i) Example from Histogramming}
\subsection{\bf (Physics) Events---III}

\begin{slide}
{\Smallsize
\begin{itemize}
\item Also, $\langle g^2 \rangle = \langle g\rangle$
\item And so error of estimate is $\sqrt{p(1-p)}/\sqrt{N\langle
g\rangle}$
\item where we can estimate
\item $\sqrt{N\langle g\rangle} = N_{\rm hist} - $ the number of
events in histogram.
\item Actually, we can ``derive'' this result more simply using just the
single indicator variable $\theta$ on the events that fall in the
histogram.  
\item This is a little flaky because the value $N_{\rm hist}$ is
a random variable (with an error $\pm \sqrt{N_{\rm hist}}$) as it
comes from selecting events from the larger sample $N$.  
\item The tortuous
argument above can convince you that it doesn't matter that $N_{\rm
hist}$ is a random variable!
\item The naive argument which gives correct answer is
\begin{eqnarray*}
\langle\theta\rangle & = & \frac{1}{N_{\rm hist}} \sum^{N_{\rm
hist}}_{i=1} \theta \left( x_i \right) \\
\left\langle \theta^2 \right\rangle & = & \frac{1}{N_{\rm hist}}
\sum^{N_{\rm hist}}_{i=1} \theta^2 \left( x_i \right) =
\langle\theta\rangle \\
\sigma & = & \left\langle \theta^2 \right\rangle - \langle
\theta\rangle^2 = p - p^2
\end{eqnarray*}
\item and so the error is $\sigma/\sqrt{N}$ in agreement with above
($\sigma$ is standard deviation of $\theta$).
\end{itemize}
}
\end{slide}

