

\subsection{\bf Parameters of a}
\subsection{\bf Gaussian Distribution---II}

\begin{slide}
\begin{itemize}
\item {\em Variance}
\begin{eqnarray*}
& = & \frac{1}{\sqrt{2\pi} \sigma} \int\limits^{+\infty}_{-\infty}
(x-\mu)^2 p(x) dx \\
& = & \frac{\sigma^3}{\sqrt{2\pi} \sigma} \int\limits^{+\infty}_{-\infty} dy\
\frac{y^2}{\sigma^3}\ \exp \left\lbrack\frac{-y^2}{2\sigma^2}\right\rbrack
\end{eqnarray*}

\item Putting $y=x-\mu$ and multiplying, and dividing by $\sigma^3$.
\begin{eqnarray*}
Variance & = & \frac{\sigma^3}{\sqrt{2\pi} \sigma}\ 
\frac{\partial}{\partial\sigma} \left\lbrace \int\limits^{+\infty}_{-\infty} 
e^{-y^2/2\sigma^2} dy \right\rbrace \\
& = & \frac{\sigma^2}{\sqrt{2\pi}}\ \frac{\partial}{\partial\sigma} 
\left\lbrack \sqrt{2\pi} \sigma \right\rbrack \\
& = & \sigma^2
\end{eqnarray*}

\item so fundamental Gaussian parameter $\sigma$ is standard
deviation---as its nomenclature suggested.
\end{itemize}
\end{slide}

