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QUESTION
 Write down th mgf of a Poisson distribution with mean $\mu$.
  Hence write down the mgf for the standardized Poisson variable.
  Show that as $\mu \rightarrow \infty$ this latter mgf tends to
  $E^{\frac{1}{2}t^2}$, and hence confirm that a Poisson variable
  can be approximated by a normal distribution when $\mu$ is
  large.

ANSWER
\begin{eqnarray*}
   M(t)&amp;=&amp;\sum_o^\infty e^{xt}e^{-\mu}\frac{\mu^x}{x!}\ \
   \textrm{for}\phi(\mu)\\
   &amp;=&amp;e^{-\mu}\sum_0^\infty\frac{(\mu e^t)^x}{x!}\\
   &amp;=&amp;e^{-\mu}e{\mu e^t}
  \end{eqnarray*}
  For $\phi(\mu)$ mean $=\sigma^2=\mu$. Standardized variable
  $Y=\frac{x-\mu}{\sqrt{\mu}}$

  \begin{eqnarray*}
  M_Y(t)&amp;=&amp;e^{-\frac{\mu}{\sigma}t}M_x(\frac{t}{6})\\
  &amp;=&amp;e^{-\sqrt{\mu}t}e^{-\mu}e^{\mu e^\frac{t}{\sqrt{\mu}}}\\
  &amp;=&amp;e^{-\mu-\sqrt{\mu}t}e^{\mu(1+\frac{t}{\sqrt{\mu}}+\frac{t^2}{2\mu}+\frac{t^3}{6\mu\sqrt{\mu}}+\ldots)}\\
  &amp;=&amp;e^{\frac{t^2}{2}+\frac{t^3}{6\sqrt{\mu}}}\rightarrow e^{\frac{1}{2}t^2}
  \textrm{ as }\mu \rightarrow \infty
  \end{eqnarray*}
  which is the mgf of N(0,1) hence by uniqueness.

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