= Solution
For the <Poisson distribution>, $G_N(z)=e^{\lambda(z-1)}$, so the aggregate has a <compound Poisson distribution> and
$$
\boxed{\kappa_S(t)=\lambda\bigl(M_{X_1}(t)-1\bigr).}
$$
Three differentiations immediately give
$$
\boxed{\kappa_{S,3}=\mathbb E(S-\mathbb ES)^3=\lambda m_3=\lambda\mathbb EX_1^3.}
$$
The first two <cumulants> are $\lambda m_1$ and $\lambda m_2$. If $\lambda>0$ and claim sizes are strictly positive, $m_3>0$, so \b[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.
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