= Compound Poisson cumulants
{title2=$K_S(t)=\lambda(M_X(t)-1),\quad\kappa_j(S)=\lambda\mathbb EX^j$}
Conditioning on the <independent> <Poisson distribution> count in a random sum gives $\mathbb Ee^{tS}=\exp(\lambda(M_X(t)-1))$ wherever finite. Differentiation at zero gives the displayed <cumulants> when the relevant derivatives exist. The aggregate <variance> is the count rate times the raw second claim <moment>, not just the claim <variance>. Positive <moment-generating function> arguments need an exponential <moment>; finite first and second claim <moments> suffice for the aggregate <mean> and <variance> by conditioning or right derivatives at zero of the <Laplace transform of a nonnegative random variable>.
Back to article page
