Exact Compound Poisson Loss With Exponential Severities

Introduction

A compound Poisson model combines a random number of claims with random claim amounts. When the claim count is Poisson and the individual severities are exponential, the aggregate distribution has a useful exact form: a probability atom at zero and a continuous density for positive losses. The density can be written using a modified Bessel function, giving an alternative to summing many severity convolutions.

The distinction between the atom and density matters. A positive aggregate loss has a density, not a point probability; zero loss has a point probability because it occurs exactly when there are no claims. The formulas below describe this model without a normal or other approximation.

Model and Exact Distribution

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, and let $X_1,X_2,\ldots$ be independent and identically distributed exponential claim severities with rate $\beta>0$, independent of $N$. Thus $f_X(x)=\beta e^{-\beta x}$ for $x>0$, and $E[X]=1/\beta$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$.

Here $\lambda$ is the expected claim count over the period being modeled, and $\beta$ is the severity rate in inverse loss units. For example, if losses are measured in dollars, $\beta$ is measured per dollar, and a density value has units per dollar.

Because every exponential severity is strictly positive, $S=0$ if and only if $N=0$. For $x>0$, conditioning on $N=n$ gives a gamma density with shape $n$ and rate $\beta$. Mixing those densities over the Poisson claim count yields the exact positive-loss density.

Theorem: Under the model above, $$ P(S=0)=e^{-\lambda}, $$ and, for $x>0$, $$ f_S(x)=e^{-(\lambda+\beta x)} \sqrt{\frac{\lambda\beta}{x}}\, I_1\!\left(2\sqrt{\lambda\beta x}\right), $$ where $I_1$ is the modified Bessel function of the first kind of order one. There is no other atom: the continuous component has total probability $1-e^{-\lambda}$.

Why the Bessel-Function Formula Holds

For a fixed positive integer $n$, the sum of $n$ independent exponential severities has density $f_{S\mid N=n}(x)=\beta^n x^{n-1}e^{-\beta x}/(n-1)!$ for $x>0$. The unconditional density at a positive $x$ is the Poisson-weighted sum of these conditional densities. The $n=0$ case contributes an atom at zero, not a density on the positive axis.

\begin{aligned} f_S(x) &=\sum_{n=1}^{\infty} e^{-\lambda}\frac{\lambda^n}{n!} \frac{\beta^n x^{n-1}e^{-\beta x}}{(n-1)!}\\ &=e^{-(\lambda+\beta x)} \sum_{k=0}^{\infty} \frac{(\lambda\beta)^{k+1}x^k}{k!(k+1)!}, \qquad x>0, \end{aligned}

The substitution $k=n-1$ gives the second line. The modified Bessel function of order one is defined by the convergent series $$ I_1(z)=\sum_{k=0}^{\infty} \frac{(z/2)^{2k+1}}{k!(k+1)!}. $$ Set $z=2\sqrt{\lambda\beta x}$. Multiplying this series by $\sqrt{\lambda\beta/x}$ gives exactly the sum in the density expression. This proves the stated Bessel representation.

The same mixture also verifies the total continuous probability. Each conditional density for $N=n\geq1$ integrates to one, so the positive part has total mass $$ \sum_{n=1}^{\infty}P(N=n)=1-e^{-\lambda}. $$ Adding the atom $e^{-\lambda}$ gives total probability one. This check depends on the severities being positive; if the severity model allowed zero-valued claims, zero aggregate loss would not be equivalent to $N=0$.

Evaluating the Density in Practice

For a given positive loss $x$, calculate the Bessel argument $2\sqrt{\lambda\beta x}$, the multiplier $\sqrt{\lambda\beta/x}$, and the exponential factor $e^{-(\lambda+\beta x)}$. Keep the units consistent throughout. The result is a density value, not the probability of observing exactly $x$.

The Bessel series is also useful for checking a numerical evaluation. Its terms are positive for $z>0$, and successive terms satisfy $$ \frac{a_{k+1}}{a_k} =\frac{(z/2)^2}{(k+1)(k+2)}, $$ where $a_k=(z/2)^{2k+1}/(k!(k+1)!)$. This ratio provides a direct way to see how quickly the series terms decrease.

Worked Example: Aggregate Atom and Density at a Positive Loss

A hypothetical portfolio has a Poisson claim count with mean $\lambda=2$. Claim severities are exponential with rate $\beta=1$ per loss unit. Find the probability of zero aggregate loss and the aggregate density at $x=2$ loss units.

Method. Use the zero-atom formula and then the exact positive-loss density. The Bessel argument is $2\sqrt{\lambda\beta x}$, so here it is $2\sqrt{2\cdot1\cdot2}=4$.

The atom at zero is $$ P(S=0)=e^{-2}=0.1353. $$ For the density, the multiplier is $\sqrt{2\cdot1/2}=1$, and the exponential factor is $e^{-(2+1\cdot2)}=e^{-4}$. The series gives $$ I_1(4)=2+4+2.666667+0.888889+0.177778+0.023704+0.002258+0.000161+0.000009+\cdots =9.759465\ldots. $$ Therefore, $$ f_S(2)=e^{-4}(1)(9.759465\ldots)=0.1788 $$ per loss unit, rounded to four decimal places. The total probability of a positive aggregate is $1-e^{-2}=0.8647$, which complements the zero atom. The density value $0.1788$ is not a point probability.

Check and mistake avoided. The atom and continuous mass sum to $0.1353+0.8647=1.0000$, to four decimal places. A common error is to report $f_S(2)$ as $P(S=2)$; for a continuous component, the probability at any single positive point is zero.

Worked Example: Evaluating a Small Density in Dollars

A hypothetical book has $N\sim\operatorname{Poisson}(1)$, and each claim amount is exponential with mean $\$250$. Calculate the aggregate density at a loss of $\$1{,}000$, and state the zero-loss probability.

Method. The exponential rate is the reciprocal of the mean severity: $\beta=1/250=0.004$ per dollar. At $x=1{,}000$, $\lambda\beta x=1(0.004)(1{,}000)=4$, so the Bessel argument is $2\sqrt{4}=4$. The multiplier is $\sqrt{\lambda\beta/x}=\sqrt{0.004/1{,}000}=0.002$ per dollar, and the exponential factor is $e^{-(1+4)}=e^{-5}$.

Using $I_1(4)=9.759465\ldots$, $$ \begin{aligned} f_S(1{,}000) &=0.002e^{-5}I_1(4)\\ &=0.002(0.006737947)(9.759465\ldots)\\ &=0.0001315\ \text{per dollar}. \end{aligned} $$ The zero-loss probability is $P(S=0)=e^{-1}=0.3679$. The density is rounded to seven decimal places per dollar, and the probability is rounded to four decimal places.

Check and mistake avoided. The factor $0.002$ is essential: omitting it would produce a number 500 times too large. The density’s units are per dollar because its multiplier contains $1/\text{dollar}$; $I_1$ and the exponential factor are dimensionless.

Worked Example: Recovering the Rate From the Mean Severity

For a hypothetical portfolio, the claim count is Poisson with mean $1$, and the mean severity is $\$100$. Find the aggregate density at $\$100$, and compare it with the probability of no claims.

Method. The exponential rate is $\beta=1/100=0.01$ per dollar. At $x=100$, $\lambda\beta x=1$, so the Bessel argument is $2$, and the multiplier is $\sqrt{1(0.01)/100}=0.01$ per dollar. The exponential factor is $e^{-(1+0.01\cdot100)}=e^{-2}$.

From the defining series, $$ I_1(2)=1+0.5+0.083333+0.006944+0.000347+0.000012+\cdots =1.590637\ldots. $$ It follows that $$ f_S(100)=0.01e^{-2}I_1(2) =0.002153\ \text{per dollar}, $$ rounded to six decimal places. Separately, $P(S=0)=e^{-1}=0.3679$, while $P(S>0)=1-e^{-1}=0.6321$.

Check and mistake avoided. The positive aggregate mass is $0.6321$, not the density value at $\$100$. A density has inverse-dollar units; probabilities are dimensionless. Also, the rate is $1/100$, not $100$: the supplied $\$100$ is a mean, not an exponential rate.

Assumptions, Checks, and Common Traps

The exact formula is useful when the Poisson and exponential assumptions are appropriate and a continuous aggregate density is needed. It does not apply unchanged to a different claim-count distribution, a different severity distribution, or a model with a separate probability of zero severity. In particular, changing the frequency model changes the mixture weights, while changing the severity model changes the conditional convolution densities.

Separate the atom from the density. The probability at zero is $e^{-\lambda}$; the Bessel formula applies only for $x>0$.
Use the severity rate, not its mean, as $\beta$. If the exponential mean is $m$, then $\beta=1/m$.
Keep units consistent. If $x$ is in dollars, $\beta$ must be per dollar and the density is per dollar.
Do not interpret a density as a point probability. For a small interval of width $\Delta x$ near a positive $x$, its probability is approximately $f_S(x)\Delta x$, provided the density does not change much over that interval.
Do not omit the $n=0$ term conceptually. It is not part of the positive density sum, but it is required to describe the full aggregate distribution.

For an exam calculation aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus, a reliable sequence is to identify $\lambda$, convert the severity mean to a rate if necessary, determine whether the question asks for the atom or a positive-loss density, and only then substitute into the appropriate formula. A final unit check often exposes a lost multiplier or a rate-versus-mean error.

Check Your Understanding

For each problem, identify whether the requested quantity is an atom, a density, or both before calculating.

A hypothetical model has a Poisson claim count with mean $0.8$ and strictly positive exponential severities. Find the probability of zero aggregate loss and the probability of a positive aggregate loss. (Answer: $P(S=0)=0.4493$; $P(S>0)=0.5507$.)
Let $\lambda=1$, with exponential claim severities of mean $\$100$. Find the aggregate density at $\$100$ using $I_1(2)=1.590637\ldots$. (Answer: $0.002153$ per dollar, to six decimal places.)
A portfolio has $\lambda=1$ and exponential severity rate $0.004$ per dollar. For a loss of $\$1{,}000$, state the Bessel argument and the multiplier before the exponential factor, then evaluate the density using $I_1(4)=9.759465\ldots$. (Answer: Bessel argument $4$; multiplier $0.002$ per dollar; density $0.0001315$ per dollar, to seven decimal places.)
A student uses $e^{-\lambda}$ as the density at every positive loss. Explain the error for the model with $\lambda=2$, $\beta=1$, and $x=2$, and give the correct density. (Answer: $e^{-\lambda}$ is the zero-loss atom, not a positive-loss density; the correct density is $e^{-4}I_1(4)=0.1788$ per loss unit, to four decimal places.)

What You'll Learn