= Solution
Write $M(t)=\mathbb Ee^{tX}$, with $M(0)=1$, and differentiate $\kappa(t)=\log M(t)$:
$$
\begin{aligned}
\kappa'(t)&=\frac{M'(t)}{M(t)},\\
\kappa''(t)&=\frac{M''(t)}{M(t)}-\frac{M'(t)^2}{M(t)^2},\\
\kappa'''(t)&=\frac{M'''(t)}{M(t)}-\frac{3M'(t)M''(t)}{M(t)^2}
+\frac{2M'(t)^3}{M(t)^3}.
\end{aligned}
$$
Using $M^{(j)}(0)=\mathbb EX^j$ and $\mu=\mathbb EX$ gives
$$
\boxed{\kappa_1=\mu,\qquad\kappa_2=\mathbb EX^2-\mu^2=\operatorname{Var}X,\qquad
\kappa_3=\mathbb EX^3-3\mu\mathbb EX^2+2\mu^3=\mathbb E(X-\mu)^3.}
$$
Thus 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> gives
$$
M_S(t)=\exp\{\lambda(M_{X_1}(t)-1)\},\qquad
\kappa_S(t)=\lambda(M_{X_1}(t)-1).
$$
Consequently the <compound Poisson distribution> has
$$
\kappa_{S,1}=\lambda m_1,\qquad
\kappa_{S,2}=\lambda m_2,\qquad
\kappa_{S,3}=\lambda m_3.
$$
These are raw severity <moments>, not severity <cumulants>.
The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ gives
$$
M_Y(t)=\left(\frac\nu{\nu-t}\right)^\alpha\quad(t<\nu),\qquad
\kappa_V(t)=kt-\alpha\log(1-t/\nu).
$$
Therefore the required matching equations for the <shifted gamma distribution> are
$$
k+\frac\alpha\nu=\lambda m_1,\qquad
\frac\alpha{\nu^2}=\lambda m_2,\qquad
\frac{2\alpha}{\nu^3}=\lambda m_3.
$$
For $\lambda>0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha,\nu$ and the unique solution
$$
\boxed{\nu=\frac{2m_2}{m_3},\qquad
\alpha=\frac{4\lambda m_2^3}{m_3^2},\qquad
k=\lambda m_1-\frac{2\lambda m_2^2}{m_3}.}
$$
This 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> is
$$
\boxed{S\approx\mathcal N(\lambda m_1,\lambda m_2),\qquad
\mathbb P(S\leq x)\approx\Phi\!\left(\frac{x-\lambda m_1}{\sqrt{\lambda m_2}}\right).}
$$
To justify its large-count limit, let $\varphi_X$ be the claim <characteristic function>. For fixed severity law with $m_2<\infty$, its expansion is $\varphi_X(u)=1+im_1u-m_2u^2/2+o(u^2)$. The <characteristic function> of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ is
$$
\exp\left\{\lambda\left[\varphi_X\left(\frac t{\sqrt{\lambda m_2}}\right)-1\right]
-\frac{it\lambda m_1}{\sqrt{\lambda m_2}}\right\}
\longrightarrow e^{-t^2/2}.
$$
Thus 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 \b[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.
Back to article page
