

\subsection{\bf Number of Points Needed}
\subsection{\bf in Joint Monte Carlo}

\begin{slide}
\begin{itemize}
\item Then assuming
special case written above where integrand has product form
$f\,g$ and $f$ and $g$ are independent, we need
$$
n = n_k = 10^4 \left( {\sigma_{fg}\over{\mu_f\,\mu_g}} \right)
^2 \hbox{points}
$$
$$
\sigma^2_{fg} = \sigma^2_f\,\sigma^2_g + \mu^2_f\,\sigma^2_g +
\mu^2_g\,\sigma^2_f
$$
as can easily be proven.
$$
{\sigma^2_{fg}\over{\mu^2_f\,\mu^2_g}} \sim
{\sigma^2_f\over{\mu^2_f}} + {\sigma^2_g\over{\mu^2_g}}.
$$
\item We see that we need NOT $n_k=n_f \cdot n_g$ points, but rather
$n_k=n_f + n_g$ points!

\item The exact calculation requires one to relax the unrealistic
assumption that $f$ and $g$ are independent, but the point is general
and the estimate of $n_k$ is reasonably accurate!
\end{itemize}
\end{slide}

