 Cumulant composition for a random sum

For a random sum of independent claims, conditioning on the count gives $M_S(t)=\mathbb{E}[M_X(t{)}^N]$ and therefore the displayed cumulant-generating function. Write $a_j={\kappa }_{N,j}$ and $b_j={\kappa }_{X,j}$. The chain rule gives ${\kappa }_{S,1}=a_1{b}_1$, ${\kappa }_{S,2}=a_1{b}_2+a_2{b}_1^2$, and ${\kappa }_{S,3}=a_1{b}_3+3a_2{b}_1{b}_2+a_3{b}_1^3$. In raw severity moments $m_j=\mathbb{E}{X}^j$, the third cumulant is $a_1{m}_3+3(a_2-a_1)m_1{m}_2+(a_3-3a_2+2a_1)m_1^3$. Positive exponential moments justify two-sided differentiation near zero; finite ordinary moments alone suffice for the corresponding left derivatives when claims are positive.