

\subsection{\bf Generation of Gaussian}
\subsection{\bf Random Numbers---II}
\begin{slide}
\begin{itemize}
\item You can prove this by taking polar coordinates in
\[
\exp \left\lbrack - \frac{1}{2} \left( g_1^2 + g_2^2 \right)
\right\rbrack dg_1\, dg_2
\]

\item with $g_1 = r\cos\theta$, $g_2 = r\sin\theta$.  

\item Then
\begin{eqnarray*}
&&\exp \left( - \frac{1}{2} \left[ g_1^2 + g_2^2 \right] \right) dg_1\, dg_2 \\
& = & \exp \left( - \frac{1}{2} r^2 \right) r\, dr\, d\theta \\
& = & \exp \left( - \frac{1}{2} r^2 \right) d\left(\frac{1}{2} r^2
\right) d\theta \\
& = & d \left( \exp \left( - \frac{1}{2} r^2 \right) \right) d\theta
\end{eqnarray*}

\item Therefore, the choice
\[
u_1 = \exp \left( - \frac{1}{2} r^2 \right) \hspace{2em} u_2 = \theta/2\pi
\]
as uniformly distributed in [0,1], implies that $g_1$, $g_2$ are Gaussian.
\end{itemize}
\end{slide}

