

\subsection{\bf Addition of Independent}
\subsection{\bf Random Variables}

\begin{slide}
\begin{itemize}
\item Suppose $y=\frac{1}{n} {\displaystyle \sum^n_{i=1}} x_i$ 

\item where the $x_i$ are independent and they all have the same
distribution with mean $\mu$ and standard deviation $\sigma$.

\item Then clearly $\langle y\rangle$ is also $\mu$.  

\item Further, to find standard deviation, calculate
\begin{eqnarray*}
\left\langle (y-\mu )^2 \right\rangle & = & \frac{1}{n^2} \sum_{i,j} 
\left\langle \left( x_i - \mu \right) \left( x_j - \mu \right) \right\rangle \\
& = & \frac{\sigma^2}{n}
\end{eqnarray*}
as all off diagonal terms vanish.

\item So it is precisely true that $y$ has mean $\mu$ and standard deviation
$\frac{\sigma}{\sqrt{n}}$.  

\item Later, we will show that as $n\rightarrow\infty$
(i.e., not exactly), all other moments of $y$ vanish faster than $1/n$ 

\item and so $y$ becomes Gaussian whatever the original distributions
of $x$.
\end{itemize}
\end{slide}

