Compound Poisson distribution ($S={\sum }_{i=1}^N{X}_i,N\sim \mathrm{Poisson}(\lambda )$)

The law of a sum of independent identically distributed claims with an independent Poisson distribution count of parameter $\lambda$. If the claim moment-generating function is $M_X$, the aggregate transform is $\exp (\lambda (M_X(r)-1))$ wherever finite. It has an atom at zero when the claims are positive. This is the fixed-time law of a Compound Poisson process.

If a Poisson distribution count $M$ of clusters has mean $\nu$, and each cluster independently contains a Poisson number of claims of mean $\lambda$, then the total count is a compound Poisson distribution with the displayed probability generating function. Conditioning gives $\mathbb{E}N=\nu \lambda$ and $\mathrm{Var}N=\nu \lambda (1+\lambda )$, so it is not Poisson for positive parameters. With independent severities of mean $\mu$ and variance ${\sigma }^2$, the aggregate payment has mean $\nu \lambda \mu$ and variance $\nu [\lambda ({\sigma }^2+{\mu }^2)+{\lambda }^2{\mu }^2]$. Independent validity thinning replaces the per-cluster count mean by $\lambda (1-p)$ and multiplies expected payment by $1-p$.

Let $N\sim \mathrm{Pois}(\lambda )$ be independent of independent and identically distributed random variables $(Y_i,Z_i)$ with finite second moments. Conditioning on $N$ gives covariance $N\mathrm{Cov}(Y_1,Z_1)$ and conditional expectations $N\mathbb{E}{Y}_1,N\mathbb{E}{Z}_1$. Law of total covariance and $\mathrm{Var}N=\lambda$ yield the formula. For per-claim excess of loss reinsurance payouts $Y=\min (X,M)$ and $Z=(X-M{)}_{+}$, it becomes $\lambda M\mathbb{E}[(X-M{)}_{+}]$. The insurer and reinsurer aggregates share their claim count and are generally dependent.

If zero-sized claims are allowed, a compound Poisson distribution count parameter need not be unique. Poisson thinning of zero claims gives the displayed positive-claim count parameter and the conditional claim size probability distribution given $X>0$, without changing the aggregate. If no positive claims occur, the aggregate is identically zero. This matters when representing a finite weighted sum of Poisson distribution counts whose coefficients include zero.

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.

For a compound Poisson distribution of count parameter $\lambda$ and severity moments $m_1,m_2$, the expected value is $\lambda m_1$ and the variance is $\lambda m_2$. As $\lambda \to \infty$ with the severity law fixed and $m_2<\infty$, the standardized aggregate converges in distribution to the standard normal distribution. Indeed, the logarithm of its characteristic function is $\lambda [{\phi }_X(t/\sqrt{\lambda m_2})-1]-it\lambda m_1/\sqrt{\lambda m_2}$; the second-order expansion of ${\phi }_X$ makes this tend to $-t^2/2$. The approximation loses the zero atom and assigns positive probability to negative losses.

Approximate a compound Poisson distribution of positive claims by a shifted gamma distribution whose first three cumulants agree. Here $m_j=\mathbb{E}{X}^j$ are raw severity moments, while the compound cumulants are $\lambda m_j$. Solve $k+\alpha /\nu =\lambda m_1$, $\alpha /{\nu }^2=\lambda m_2$, and $2\alpha /{\nu }^3=\lambda m_3$ to obtain the displayed parameters. The approximation captures positive skewness, but not the atom $e^{-\lambda }$ at zero. Its shift can be negative; with exponential claims of mean $\mu$ it is $-\lambda \mu /3$. Matching three cumulants alone supplies no general tail-error bound.

Write $m_k=\mathbb{E}{S}^k$, $m_0=1$, and ${\mu }_j=\mathbb{E}{X}^j$. Apply the Poisson size-bias identity with $h(s)=s^{k-1}$, and expand $(S+X{)}^{k-1}$ using the binomial theorem and independence. A finite $k$th severity moment suffices. In particular, the aggregate expected value, variance, and third central moment are $\lambda {\mu }_1$, $\lambda {\mu }_2$, and $\lambda {\mu }_3$, respectively; the latter two use raw severity moments.

Applying a measurable per-claim retention $g$ to independent claim sizes preserves the compound Poisson distribution form. The transformed severity is the pushforward measure of the original severity law. With count parameter $\lambda$, the retained aggregate has expected value $\lambda \mathbb{E}g(X)$ and variance $\lambda \mathbb{E}[g(X{)}^2]$. Zero retained marks may be removed by Poisson thinning.