 Compound Poisson cumulants

Conditioning on the independent Poisson distribution count in a random sum gives $\mathbb{E}{e}^{tS}=\exp (\lambda (M_X(t)-1))$ 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.