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\begin{document}


{\bf Question}

Let $X$ be a discrete random variable with probability generating
function $G(s).$  Show that
\begin{description}
\item[(i)] $G(1) = 1$
\item[(ii)] $G'(1) = E(X)$
\item[(iii)] $G''(1) = E(X(X-1))$
\item[(iv)] Var$(X) = G''(1) + G'(1) - (G(1))^2$
\end{description}


\vspace{.25in}

{\bf Answer}

Let (p$_n$) be the probability distribution of X

$\begin{array}{rclrcl} G(s) & =  & \ds\sum_{n=0}^{\infty} p_ns^n &
G(1) & = & \ds\sum p_n = 1 \\ G'(s) & =  & \ds\sum_{n=1}^{\infty}
np_ns^{n-1} & G'(1) & = & \ds\sum_{n=1}^\infty n p_n = E(X) \\
G''(s) & =  & \ds\sum_{n=2}^{\infty} n(n-1)p_ns^{n-2} & G''(1) & =
& \ds\sum_{n=2}^\infty n(n-1) p_n \\ & &  & & = & E(X(X-1))
\end{array} $


\begin{eqnarray*} Var(X) &  = & E(X^2) - E(X)^2 \\ & = & E(X(X-1))
+ E(X) - E(X)^2  \\ & = & G''(1) + G'(1) - (G'(1))^2
\end{eqnarray*}




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