\begin{align} \sigma_S^2 &= E \left[ \left( S - E[S] \right)^2 \right] \\ &= E \left[ S^2 - 2E[S]S + \left( E[S] \right)^2 \right] \\ &= E \left[ S^2 \right] - 2 \left( E[S] \right)^2 + \left( E[S] \right)^2 \\ &= E \left[ S^2 \right] - \left( E[S] \right)^2 \end{align}
\begin{align} \frac{dM_X(t)}{dt} &= \lambda e^t \exp \left( \lambda (e^t-1) \right) \\ \frac{d^2 M_X(t)}{dt^2} &= \lambda e^t \left(\lambda e^t + 1 \right) \exp \left( \lambda (e^t-1) \right) \end{align} より、 \begin{align} \mu_X &= \frac{dM_X(0)}{dt} \\ &= \lambda \\ E \left[ X^2 \right] &= \frac{d^2 M_X(0)}{dt^2} \\ &= \lambda (\lambda + 1) \\ \sigma_X^2 &= E \left[ X^2 \right] - \mu_X^2 \\ &= \lambda \end{align} を得る。
\begin{align} M_Y(t) &= \sum_{y=1}^\infty e^{yt} \cdot (1-q) q^{y-1} \\ &= \frac{1-q}{q} \sum_{y=1}^\infty \left( q e^t \right)^y \\ &= \frac{1-q}{q} \cdot q e^t \cdot \frac{1}{1-qe^t} \\ &= \frac{(1-q)e^t}{1-qe^t} \end{align}
$X,Y$ が独立なので $E[XY]=E[X]E[Y]$ が成り立つことを考慮して、 \begin{align} M_Z(t) &= E \left[ e^{tZ} \right] \\ &= E \left[ e^{t(X+Y)} \right] \\ &= E \left[ e^{tX} e^{tY} \right] \\ &= E \left[ e^{tX} \right] E \left[ e^{tY} \right] \\ &= M_X(t) M_Y(t) \end{align} を得る。
\begin{align} M_Z(t) &= M_X(t) M_Y(t) \\ &= e^t \exp \left( e^t-1 \right) \cdot \frac{e^t}{2-e^t} \\ &= \frac{\exp \left( e^t-1 \right)}{2e^{-t}-1} \end{align}
\begin{align} \frac{dM_Z(t)}{dt} &= \frac{\exp \left( e^t-1 \right) \left( 2 - e^t + 2e^{-t} \right)}{\left( 2e^{-t}-1 \right)^2} \end{align}
\begin{align} \mu_Z &= \left. \frac{dM_Z(t)}{dt} \right|_{t=0} \\ &= 3 \end{align}