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QUESTION A random variable has pgf $\frac{(1+2z)^4}{81}$. Find
$p(2),\
  p(4),\ \mu$ and $\sigma^2$.

ANSWER
 \begin{eqnarray*}
  G(z)&=&\frac{(1+2z)^4}{81}\\
  G^{(1)}(z)&=&\frac{8(1+2z)^3}{81}\\
  G^{(2)}(z)&=&\frac{16(1+2z)^2}{27}\\
  2!p(2)&=&G^{(2)}(0)=\frac{16}{27}\textrm{ therefore
  }p(2)=\frac{8}{27}\\
  G^{(3)}(z)&=&\frac{64(1+2z)}{27}\\
  G^{(4)}(z)&=&\frac{128}{27}\\
  4!p(4)&=&G^{(4)}(0)=\frac{128}{27}\textrm{ therefore
  }p(4)=\frac{16}{81}
  \end{eqnarray*}

  $\mu=G^{(1)}(1)=\frac{8 \times 3^3}{81}=\frac{8}{3}\\
  E(X(X-1))=G^{(")}(1)=\frac{16 \times 9}{27}=\frac{16}{3}$
  therefore
  $\sigma^2=\frac{16}{3}+{8}{3}-(\frac{8}{3})^2=\frac{8}{9}$\\
  Note that $G(z)=(\frac{1}{3}+\frac{2}{3}z)^4$. This corresponds
  to B(4,$\frac{2}{3}$) Hence $p(2)=\left(\begin{array}{c}4\\
  2\end{array}\right)(\frac{2}{3})^2(\frac{1}{3})^2=\frac{8}{27}$,
  $p(4)=(\frac{2}{3})^4=\frac{16}{81}$.\\
  $\mu=4 \times \frac{2}{3}=\frac{8}{3},\ \sigma^2= 4 \times
  \frac{2}{3} \times \frac{1}{3}=\frac{8}{9}$


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