 Solution

For the Poisson distribution, $G_N(z)=e^{\lambda (z-1)}$, so the aggregate has a compound Poisson distribution andThree differentiations immediately giveThe first two cumulants are $\lambda m_1$ and $\lambda m_2$. If $\lambda >0$ and claim sizes are strictly positive, $m_3>0$, so the paper's unstandardized skewness is strictly positive. For the degenerate count $\lambda =0$, the aggregate is zero and the third central moment is zero; strict positivity presumes a positive claim intensity.