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{\bf Question}

Let $X$ be a Poisson random variable with parameter $\lambda.$
Show that the p.g.f. for $X$ is $G(s) = e^{\lambda s - \lambda}.$

Use this to find the mean and variance of $X.$



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{\bf Answer}

For a Poisson distribution with parameters $\lambda,$ $$p_n =
\frac{e^{-\lambda}\lambda^n}{n!}$$ So $$ G(s)  =  e^{-\lambda}
\sum_{n=0}^\infty \frac{\lambda^ns^n}{n!} = e^{(\lambda s -
\lambda)}$$ $$\begin{array}{rclrcl}  G'(s) & = & \lambda
e^{(\lambda s - \lambda)} & G'(1) & = & \lambda = E(X) \\ G''(s) &
= &  \lambda^2e^{(\lambda s - \lambda)} & G''(1) & =& \lambda^2
\end{array}$$ $$Var(X) = \lambda^2 + \lambda - \lambda^2 =
\lambda$$



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