Compound Poisson Distributions

From a Random Count to a Mixture of Convolutions

A sum of a fixed number of independent severities is described by repeated convolution. In many applications, however, the number of severities is itself random. A compound Poisson random variable combines these two ideas: a Poisson count chooses how many independent severity variables to add. Conditioning on that count reveals the full distribution as a weighted mixture, with one convolution term for each possible count.

Definition: Let \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda\geq 0\). Let \(X_1,X_2,\ldots\) be independent and identically distributed random variables, independent of \(N\), with common distribution \(G\). The compound Poisson random variable is \(S=\sum_{j=1}^{N}X_j\), with the empty sum defined to be \(0\) when \(N=0\). For \(n\geq 1\), write \(G^{*n}\) for the distribution of a sum of \(n\) independent variables with distribution \(G\), and set \(G^{*0}=\delta_0\), the point mass at zero.

The assumptions matter: the count is independent of the severities, and the severities are identically distributed and independent of one another. When \(N=n\), the sum has distribution \(G^{*n}\); when \(N=0\), it is exactly zero. The Poisson probability of each count supplies the mixture weight.

Theorem: For every Borel set \(A\subseteq\mathbb{R}\), the compound Poisson law is

P(S\in A)=\sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}G^{*n}(A).

Proof. The events \(\{N=n\}\), \(n=0,1,\ldots\), are disjoint and have total probability one. Conditioning on this count gives

P(S\in A) =\sum_{n=0}^{\infty}P(S\in A\mid N=n)P(N=n) =\sum_{n=0}^{\infty}G^{*n}(A)e^{-\lambda}\frac{\lambda^n}{n!}.

For \(n=0\), the conditional distribution is \(\delta_0\); for \(n\geq1\), it is the \(n\)-fold convolution \(G^{*n}\). This proves the formula. \(\square\)

This is a probability mixture, not just a formal series: its weights are nonnegative and sum to one, since \(\sum_{n=0}^{\infty}e^{-\lambda}\lambda^n/n!=1\). The zero-count term is \(e^{-\lambda}\delta_0\). Whether other terms also put probability at zero depends on the severity distribution.

Worked Example: A Constant Severity

Suppose \(N\sim\operatorname{Poisson}(2)\), and every severity equals \(3\), so \(X_j=3\) with probability one. Then \(S=3N\). Conditional on \(N=n\), all the mass is at \(3n\), giving

P(S=3n)=P(N=n)=e^{-2}\frac{2^n}{n!},\qquad n=0,1,2,\ldots.

In particular, \(P(S=0)=e^{-2}\), \(P(S=3)=2e^{-2}\), \(P(S=6)=2e^{-2}\), and \(P(S=9)=\frac{4}{3}e^{-2}\). The sum is supported on multiples of \(3\), so this compound Poisson law is discrete; it does not have a density. This example also shows why the general mixture formula must allow convolution laws that are not continuous.

Continuous Severities: The Atom and the Density

Now suppose each severity has a probability density function \(f\) with respect to Lebesgue measure. For \(n\geq1\), let \(f^{*n}\) denote the density of the sum of \(n\) independent severities. With \(f^{*1}=f\), the convolution recursion is

f^{*n}(x)=\int_{\mathbb{R}} f^{*(n-1)}(x-y)f(y)\,dy,\qquad n\geq2.

Each \(f^{*n}\) integrates to one. The zero-count term is different: it is a point mass, not a density. Separating it from the terms with \(n\geq1\) gives the continuous part of the law.

Theorem: If \(\lambda\geq0\) and the severities have a density \(f\), then \(S\) has an atom of size \(e^{-\lambda}\) at zero, and its remaining distribution has density

g(x)=\sum_{n=1}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}f^{*n}(x).

Proof. For any fixed \(a\in\mathbb{R}\) and \(n\geq1\), condition on \(X_2+\cdots+X_n\) (with this sum taken to be zero when \(n=1\)). Since \(X_1\) has a density, it assigns probability zero to every single point. Independence therefore gives \(P(X_1+\cdots+X_n=a)=0\). Thus every term with \(n\geq1\) is atomless, and only \(N=0\) contributes an atom. That event has probability \(e^{-\lambda}\) and gives \(S=0\).

For each \(n\geq1\), the conditional law has density \(f^{*n}\). The mixture of those laws has the displayed candidate density. To verify it, the summands are nonnegative, so Tonelli’s theorem allows integration term by term:

\int_{\mathbb{R}}g(x)\,dx =\sum_{n=1}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!} \int_{\mathbb{R}}f^{*n}(x)\,dx =\sum_{n=1}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!} =1-e^{-\lambda}.

More generally, integrating \(g\) over any Borel set \(A\) gives the mixture probability from the terms \(n\geq1\). Together with the mass \(e^{-\lambda}\) at zero, this accounts for the entire distribution. The density is defined up to changes on sets of Lebesgue measure zero. \(\square\)

When severities are nonnegative and \(P(X_j=0)=0\), a sum with at least one term is positive with probability one. In that case the continuous part is concentrated on \((0,\infty)\). Its topological support, however, need not exclude zero: for example, positive exponential severities have density arbitrarily close to zero, so zero belongs to the support of their continuous component.

Worked Example: Exponential Severities

Let \(N\sim\operatorname{Poisson}(3)\), independently of severities \(X_j\) with exponential density \(f(x)=2e^{-2x}\) for \(x>0\) and \(f(x)=0\) otherwise. The atom at zero is

P(S=0)=e^{-3}\approx0.0498.

For \(x>0\), the \(n\)-fold convolution is

f^{*n}(x)=\frac{2^n x^{n-1}e^{-2x}}{(n-1)!}.

This follows by induction. It holds for \(n=1\). If it holds for \(n-1\), the convolution recursion, with the integral restricted to \(0<y<x\), gives

f^{*n}(x) =\int_0^x\frac{2^{n-1}(x-y)^{n-2}e^{-2(x-y)}}{(n-2)!}\,2e^{-2y}\,dy =\frac{2^ne^{-2x}}{(n-2)!}\int_0^x(x-y)^{n-2}\,dy =\frac{2^nx^{n-1}e^{-2x}}{(n-1)!}.

Substitution in the mixture formula gives

g(x)=e^{-3}\sum_{n=1}^{\infty}\frac{3^n}{n!} \frac{2^nx^{n-1}e^{-2x}}{(n-1)!},\qquad x>0.

For example, at \(x=1\), the \(n=1,2,3\) contributions are respectively \(6e^{-5}\), \(18e^{-5}\), and \(18e^{-5}\). Their partial sum is \(42e^{-5}\); it is not the full density value, because all \(n\geq4\) terms must also be included.

Worked Example: Uniform Severities on a Bounded Interval

Let \(N\sim\operatorname{Poisson}(2)\) and let the severities be independent \(\operatorname{Uniform}(0,1)\) variables. For \(0<x<1\), the \(n\)-fold convolution has density \(f^{*n}(x)=x^{n-1}/(n-1)!\). Indeed, it holds for \(n=1\); if it holds for \(n\), then for \(0<x<1\),

f^{*(n+1)}(x) =\int_0^x \frac{(x-y)^{n-1}}{(n-1)!}\,dy =\frac{x^n}{n!}.

Therefore the compound density on this interval is

g(x)=e^{-2}\sum_{n=1}^{\infty}\frac{2^n}{n!}\frac{x^{n-1}}{(n-1)!},\qquad 0<x<1.

At \(x=\frac12\), the contributions for \(n=1,2,3,4\), before the common factor \(e^{-2}\), are \(2\), \(1\), \(\frac16\), and \(\frac1{72}\). For instance, the \(n=4\) contribution is \(\frac{2^4}{4!}\frac{(1/2)^3}{3!}=\frac1{72}\). These are only the first four terms; higher counts also contribute. The separate atom at zero is \(e^{-2}\), so the density terms do not replace or include that atom.

Common Mistakes and Why the Mixture Form Matters

Calling the whole law a density. With continuous severities, \(S\) still has probability \(e^{-\lambda}\) at zero. The density \(g\) describes only the part arising from \(N\geq1\), and integrates to \(1-e^{-\lambda}\), not one.
Dropping the zero-count case. The sum has an explicitly defined empty-sum value, \(S=0\), when \(N=0\). Omitting this case loses a genuine atom.
Calling the sum Poisson merely because the count is Poisson. The law of \(S\) depends on the severity distribution. Constant severities produce a lattice law; continuous severities produce a continuous component. Neither should be mistaken for a Poisson count law without further argument.
Claiming the continuous component’s support lies strictly above zero. If severities are positive, the component is concentrated on positive values. But its topological support may include zero when positive sums can approach zero.

The count-weighted convolution representation is useful because it keeps the roles of the two ingredients visible: \(\lambda\) determines the weights on possible counts, while \(G\) determines the convolution law at each count. It also makes mixed distributions straightforward to describe: one can report the atom and the continuous density separately without forcing either into the wrong representation.

Check Your Understanding

Use the mixture formula and the conditions on the severity distribution to answer these questions.

In the general compound Poisson law, what distribution corresponds to the term \(n=0\), and what is its mixture weight?
If severities have a density, why can no term with \(n\geq1\) contribute an atom at a specified point?
Why does the continuous density in the theorem integrate to \(1-e^{-\lambda}\), rather than to one?
If severities are positive with probability one, what can be said about where the continuous component is concentrated? Why does this not necessarily exclude zero from its topological support?
For constant severity \(c>0\), describe the possible values of \(S\) and give the probability that \(S=0\).

What You'll Learn