Compound Poisson Distribution¶

Definition¶

A compound Poisson loss is $\displaystyle S=\sum_{i=1}^NX_i$ with $N\sim\operatorname{Pois}(\lambda)$ $Pois (λ)$ , $\lambda\ge0$ $0$ , and i.i.d. nonnegative severities independent of $N$ . The empty sum is zero.

Interpretation¶

Poisson frequency makes aggregate losses particularly tractable. Even with continuous positive severities, aggregate loss includes a probability of no claims at all.

Bar chart of the probabilities of S = 2N with N Poisson of mean 3: mass at the even values 0, 2, 4 and so on, with the atom e^-3 at zero highlighted and no mass at odd values.

The figure plots the Example's loss $S=2N$ with $\lambda=3$ $3$ , so $\mathbb{P}(S=2k)=\mathrm{e}^{-3}3^k/k!$ $e^{- 3} 3^{k} / k!$ . The orange bar at zero is the no-claim probability $\mathbb{P}(S=0)=\mathrm{e}^{-\lambda}=\mathrm{e}^{-3}\approx0.05$ $0.05$ , present even though every claim is positive. The mass sits only on even values, and the mean $6=\lambda\mathbb{E}[X]$ and variance $12=\lambda\mathbb{E}[X^2]$ follow from the Key Results; a Poisson variable with mean 6 would have variance 6.

Move $\lambda$ and pick a claim-size law: $\mathbb E[S]=\lambda\mathbb E[X]$ and $\operatorname{Var}(S)=\lambda\mathbb E[X^2]$ $λ E [X^{2}]$ , and the spike at 0 always has height $e^{-\lambda}$ . With claims fixed at 2, $S=2N$ lives on the even numbers.

This specializes the compound risk distribution by choosing the count from the Poisson distribution. Since its count mean and variance both equal $\lambda$ , the general random-sum variance simplifies to $\lambda\mathbb{E}[X^2]$ .

Key Results¶

$\mathbb{E}[S]=\lambda\mathbb{E}[X]$ and $\operatorname{Var}(S)=\lambda\mathbb{E}[X^2]$ when finite.
The moment generating function (MGF) satisfies $M_S(t)=\exp(\lambda(M_X(t)-1))$ where finite.
With strictly positive severity, $\mathbb{P}(S=0)=\mathrm{e}^{-\lambda}$ $e^{- λ}$ .
Claim counts for disjoint severity categories are independent Poisson variables with rates $\lambda p_j$ (with $p_j$ the probability that a claim falls in category $j$ ); this uses the Poisson splitting result.

Examples¶

If each claim is 2 and $\lambda=3$ $3$ , then $S=2N$ has mean 6 and variance 12. Its support is $0,2,4,\ldots$ , so it is not an ordinary Poisson variable.

Prerequisites¶

Sources¶

Risk Insurance · Lecture 4

Aliases: compound Poisson aggregate loss, compound Poisson, compound Poisson random variable Keywords: Poisson frequency model, M_S(t) = exp(λ(M_X(t) - 1)), Var(S) = λ E[X^2], no-claim probability e^(-λ), Poisson splitting by claim category Symbols: lambda, sum

Last updated: 17 October 2026 21:53