Deriving the Compound Poisson Aggregate Loss Model

Why the Compound Poisson Model Matters

A catastrophe model may represent the number of loss-producing events in a year separately from the loss generated by each event. To obtain an annual aggregate-loss distribution, those two ingredients must be combined. The compound Poisson model gives a compact way to do this when event counts are Poisson and event severities follow a common distribution.

The model is useful because its probability-generating function summarizes the entire aggregate distribution, while its expected annual loss has a simple form. This tutorial derives both results and applies them to discrete severities. All examples are hypothetical, and dollar amounts are stated in thousands where indicated.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$ be the number of events in one year, where $\lambda>0$. Let $X_1,X_2,\ldots$ be independent and identically distributed event severities, independent of $N$, with $E[X_i]=\mu_X<\infty$. The annual aggregate loss is $S=\sum_{i=1}^{N}X_i$, with $S=0$ when $N=0$. When severities are nonnegative integers, their probability-generating function is $G_X(z)=E[z^X]$, and the aggregate probability-generating function is $G_S(z)=E[z^S]$.

A probability-generating function (PGF) applies to a nonnegative integer-valued random variable. For example, it is directly applicable when loss amounts are recorded in whole units such as thousands of dollars. If severity is continuous, a PGF is not the appropriate transform; the compound Poisson model still has expected loss $E[S]=\lambda E[X]$, but this PGF derivation is for integer-valued severities.

The key assumptions are that the annual event count is Poisson, severities have the same distribution and are independent of one another, and the count is independent of the severities. If event severities change with event count, or if events are dependent in a way that changes the aggregate model, the formulas below need not apply as written.

Deriving the Aggregate Probability-Generating Function

Condition first on the number of events. Given $N=n$, the aggregate is the sum of $n$ independent severities. Its PGF is therefore the product of the individual PGFs, $G_X(z)^n$. Averaging this conditional PGF over the Poisson distribution gives the aggregate PGF.

\begin{aligned} G_S(z) &=E[z^S]\\ &=\sum_{n=0}^{\infty}P(N=n)E[z^S\mid N=n]\\ &=\sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}G_X(z)^n\\ &=e^{-\lambda}\sum_{n=0}^{\infty}\frac{(\lambda G_X(z))^n}{n!}\\ &=\exp\!\left(\lambda(G_X(z)-1)\right). \end{aligned}

The fourth line uses the power-series identity $\sum_{n=0}^{\infty}a^n/n!=e^a$. This derivation explains why the Poisson assumption produces an exponential expression: the Poisson probabilities weight the conditional PGFs $G_X(z)^n$, and their sum is an exponential series.

Theorem: Under the stated compound Poisson assumptions, $G_S(z)=\exp(\lambda(G_X(z)-1))$. The probability $P(S=s)$ is the coefficient of $z^s$ in the power-series expansion of $G_S(z)$.

The PGF also gives the expected annual loss. For a nonnegative integer-valued variable with finite mean, $G'_S(1)=E[S]$. Differentiating the compound Poisson PGF and using $G_X(1)=1$ yields the result:

G'_S(z)=\lambda G'_X(z)\exp\!\left(\lambda(G_X(z)-1)\right), \qquad E[S]=G'_S(1)=\lambda G'_X(1)=\lambda E[X].

The same mean result follows directly by conditioning: $E[S\mid N=n]=n\mu_X$, so $E[S]=E[N]\mu_X=\lambda\mu_X$. Thus expected annual loss is expected event count multiplied by expected loss per event. This is an expectation, not a statement that the aggregate loss equals that amount in any particular year.

Applying the PGF to Discrete Severities

A practical calculation starts by writing the severity PGF from the possible severity amounts and their probabilities. Substitute it into the compound Poisson formula, then extract the coefficient or coefficients relevant to the requested aggregate loss. The following examples show how that process works.

Worked Example: Exact Loss from One or Two Events

A hypothetical portfolio has a Poisson event count with mean $\lambda=1.5$ per year. Each event produces a loss of either $\$1{,}000$ with probability $0.60$ or $\$2{,}000$ with probability $0.40$, independently. Find the PGF, the probability that annual aggregate loss is exactly $\$2{,}000$, and expected annual loss.

Method. Measure loss in thousands of dollars, so the possible severities are 1 and 2. The severity PGF is $G_X(z)=0.60z+0.40z^2$. Substituting $\lambda=1.5$ gives

G_S(z)=\exp\!\left(1.5(0.60z+0.40z^2-1)\right) =e^{-1.5}\exp(0.9z+0.6z^2).

To find the probability of exactly $\$2{,}000$, take the coefficient of $z^2$. In the exponential expansion, $z^2$ can arise from one $0.6z^2$ term or from two $0.9z$ terms. Therefore,

P(S=2)=e^{-1.5}\left(0.6+\frac{0.9^2}{2!}\right) =e^{-1.5}(0.6+0.405) =e^{-1.5}(1.005) =0.2242.

As a check, the same loss arises from one event of severity 2 or two events each of severity 1. Their probabilities are $e^{-1.5}(1.5)(0.40)=0.133878$ and $e^{-1.5}(1.5^2/2)(0.60^2)=0.090368$, respectively. Their sum is $0.224246$, which rounds to $0.2242$. The mean severity is $0.60(1)+0.40(2)=1.4$ thousand dollars, so expected annual loss is $1.5(1.4)=2.1$ thousand dollars, or $\$2{,}100$.

Common mistake avoided: Counting only one $\$2{,}000$ event omits two $\$1{,}000$ events, which also produce the requested aggregate. The coefficient method includes both possibilities.

Worked Example: A Cumulative Aggregate-Loss Probability

A hypothetical book has $N\sim\operatorname{Poisson}(0.8)$. Each event produces $\$2{,}000$ with probability $0.75$ and $\$5{,}000$ with probability $0.25$. Calculate the probability that annual aggregate loss is at most $\$4{,}000$, and calculate expected annual loss.

Method. In thousands of dollars, $G_X(z)=0.75z^2+0.25z^5$, and

G_S(z)=\exp\!\left(0.8(0.75z^2+0.25z^5-1)\right) =e^{-0.8}\exp(0.6z^2+0.2z^5).

To have aggregate loss at most 4, no $\$5{,}000$ event can occur. The possible aggregate amounts are 0, 2, and 4 thousand dollars. Their probabilities are the coefficients of $z^0,z^2,z^4$:

\begin{aligned} P(S=0)&=e^{-0.8},\\ P(S=2)&=e^{-0.8}(0.6),\\ P(S=4)&=e^{-0.8}\frac{0.6^2}{2!}=e^{-0.8}(0.18). \end{aligned}

Hence $P(S\leq4)=e^{-0.8}(1+0.6+0.18)=e^{-0.8}(1.78)=0.7998$, rounded to four decimal places. The mean severity is $0.75(2)+0.25(5)=2.75$ thousand dollars. Expected annual loss is $0.8(2.75)=2.2$ thousand dollars, or $\$2{,}200$.

Common mistake avoided: Using $P(S=4)$ as the answer would omit the outcomes 0 and 2. “At most $\$4{,}000$” requires adding the probabilities for every possible aggregate amount up to and including 4.

Worked Example: Several Event Counts Produce the Same Total

A hypothetical model has $N\sim\operatorname{Poisson}(2)$. Each event produces either $\$1{,}000$ or $\$3{,}000$, each with probability $0.50$. Find the probability that annual aggregate loss is exactly $\$6{,}000$, and find expected annual loss.

Method. With losses measured in thousands, $G_X(z)=0.5z+0.5z^3$, so

G_S(z)=\exp\!\left(2(0.5z+0.5z^3-1)\right) =e^{-2}\exp(z+z^3).

The coefficient of $z^6$ in $\exp(z)\exp(z^3)$ includes three possibilities: six events of severity 1; four events, consisting of three severity-1 events and one severity-3 event; or two events, both of severity 3. From the exponential expansions, their coefficients are $1/6!$, $1/3!$, and $1/2!$, respectively. Thus

P(S=6)=e^{-2}\left(\frac{1}{720}+\frac{1}{6}+\frac{1}{2}\right) =e^{-2}\frac{481}{720} =0.0904.

The mean severity is $0.5(1)+0.5(3)=2$ thousand dollars. Therefore, expected annual loss is $2(2)=4$ thousand dollars, or $\$4{,}000$. The probability is rounded to four decimal places. The three coefficient terms account for all ways the event severities can sum to 6; no other combination of 1s and 3s does so.

Common mistake avoided: Treating $\$6{,}000$ as necessarily two events of $\$3{,}000$ misses the four-event and six-event combinations. Extracting a coefficient from the PGF systematically includes each allowable event count and severity combination.

Interpretation and Common Traps

The aggregate PGF is a compact distributional result, but its use depends on matching the model to the question. In a discrete model, a coefficient gives an exact-loss probability; a cumulative probability requires summing all relevant coefficients. Expected annual loss is obtained from frequency times mean severity and does not require calculating the full aggregate distribution.

A particularly important distinction is between “no events” and “no aggregate loss.” If all event severities are strictly positive, then $S=0$ exactly when $N=0$, and $P(S=0)=e^{-\lambda}$. If a counted event can have zero severity, however, aggregate loss can be zero even when one or more events occur. In that case $P(S=0)=G_S(0)=\exp(\lambda(G_X(0)-1))$, which need not equal $e^{-\lambda}$.

Wrong calculation: Setting expected annual loss equal to the event rate, $E[S]=\lambda$. This ignores the severity amount and can also mismatch units.
Correction: Calculate the mean severity in the stated units, then use $E[S]=\lambda E[X]$. For example, if frequency is in events per year and mean severity is in thousands of dollars per event, the result is thousands of dollars per year.
Wrong probability shortcut: Using only the one-event case when several event counts can yield the target aggregate.
Correction: Expand the PGF and include every contribution to the requested coefficient, or enumerate the contributing event-count and severity combinations.

These methods are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. The main discipline is to state the frequency-severity assumptions, keep severity units consistent, and translate the requested aggregate event into the appropriate PGF coefficient or sum of coefficients.

Check Your Understanding

For each hypothetical model, identify the required distributional quantity before calculating. Loss amounts are in thousands of dollars unless stated otherwise.

A portfolio has $N\sim\operatorname{Poisson}(0.5)$, and every event produces $\$4{,}000$. Find expected annual loss and the probability of zero aggregate loss. (Answer: $E[S]=\$2{,}000$; $P(S=0)=0.6065$.)
A portfolio has $N\sim\operatorname{Poisson}(1)$; each event produces 1 or 3, each with probability 0.50. Write the aggregate PGF and find expected annual loss. (Answer: $G_S(z)=\exp(0.5z+0.5z^3-1)$; $E[S]=\$2{,}000$.)
A portfolio has $N\sim\operatorname{Poisson}(0.4)$; event severity is 2 or 6, each with probability 0.50. Find $P(S=2)$ and expected annual loss. (Answer: $P(S=2)=e^{-0.4}(0.2)=0.1341$; $E[S]=\$1{,}600$.)
A portfolio has $N\sim\operatorname{Poisson}(1)$, with each event producing 1 or 2, each with probability 0.50. Find $P(S=3)$ and expected annual loss. (Answer: $P(S=3)=e^{-1}(0.5^3/3!+0.5^2/2!)=0.0996$; $E[S]=\$1{,}500$.)
A portfolio has $N\sim\operatorname{Poisson}(2)$; each counted event has severity 0 with probability 0.25 and severity 2 with probability 0.75. Find the aggregate PGF, expected annual loss, and probability of zero aggregate loss. (Answer: $G_S(z)=\exp(1.5(z^2-1))$; $E[S]=\$3{,}000$; $P(S=0)=e^{-1.5}=0.2231$.)

What You'll Learn