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Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.