Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 37 / 1 / b

Put $q=1-p$. The count has the zero-based geometric distribution, whose probability generating function is $G_N(z)=p/(1-qz)$. Hence the aggregate has a compound geometric distribution, withThis transform requires $q{M}_{X_1}(t)<1$; it holds for every $t\le 0$. To differentiate explicitly, let $g(t)=M_{X_1}(t)$. ThenandAt zero, $g=1$ and $g^{(j)}=m_j$, givingEvery term is positive for $0<p<1$ and strictly positive claim sizes. Therefore the unstandardized skewness is positive, even if the claim-size third central moment itself is negative. The zero-based convention matters: there is no factor $M_{X_1}(t)$ in the numerator of this aggregate transform.