

\subsection{\bf Gaussian Limit of Binomial}
\subsection{\bf and Poisson Distributions---III}

\begin{slide}
\begin{itemize}
\item We can do the same trick for the Poisson distribution.  

\item Fix $\lambda$ finite and let $x_i$ be independent with each $x_i$
as Poisson random variable with parameter $\lambda$.

\item Then from additivity $y=x_1\ldots + x_n$ has Poisson distribution
with parameter $n\lambda = \Lambda$.

\item Meanwhile, from central limit theorem, $y$ is Gaussian with mean
$n\mu$ and standard deviation $\sqrt{n} \sigma$.

\item Here again $\mu$ and $\sigma$ are mean and standard deviation of
individual distribution.

\item Here mean is given by
\begin{eqnarray*}
\mu & = & \sum \exp(-\lambda) k \frac{\lambda^k}{k!} \\
& = & \lambda e^{-\lambda} \frac{\partial}{\partial\lambda} \left(
\frac{\sum \lambda^k}{k!} \right) \\
& = & \lambda e^{-\lambda} \frac{\partial}{\partial\lambda} \left(
e^\lambda \right) \\
& = & \lambda
\end{eqnarray*}
\end{itemize}
\end{slide}

