Compound Poisson Aggregate Loss

Building a Compound Poisson Model

A Poisson frequency model describes how many claims occur, while a severity model describes the amount of each claim. A compound Poisson model combines them to describe total loss over a fixed period. Its formulas depend on keeping two quantities distinct: the number of claims and the amount paid per claim. A claim can occur but produce no payment, for example if the model includes zero severities; in that case, a positive claim count does not necessarily mean positive aggregate loss.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$ is the expected number of claims in a specified period. Let $X_1,X_2,\ldots$ be independent, identically distributed, nonnegative claim amounts, independent of $N$, with $E[X_1]<\infty$ and $E[X_1^2]<\infty$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where a sum with $N=0$ is defined to be zero. This is a compound Poisson aggregate.

The severity distribution can be discrete or continuous. The probability generating function is directly useful when severities are nonnegative integer amounts (or have been represented on an integer monetary grid). For general severity distributions, the moment generating function provides the corresponding transform when it exists near zero. The rate $\lambda$ refers to the full period being modeled; it is not the expected loss amount.

Generating Functions and Moments

Write $G_X(z)=E[z^X]$ for the severity probability generating function when $X$ is nonnegative integer-valued. Conditioning on the claim count gives a direct derivation of the aggregate generating function. Given $N=n$, independence of severities makes the generating function of their sum $G_X(z)^n$. Averaging over the Poisson probabilities then yields:

G_S(z)=E[z^S] =\sum_{n=0}^{\infty} e^{-\lambda}\frac{\lambda^n}{n!}G_X(z)^n =e^{-\lambda}\exp\{\lambda G_X(z)\} =\exp\{\lambda(G_X(z)-1)\}.

The same conditioning argument applies to moment generating functions. If $M_X(t)=E[e^{tX}]$ is finite for $t$ in a neighborhood of zero, then:

M_S(t)=\exp\{\lambda(M_X(t)-1)\}.

To obtain the mean and variance, take the logarithm, which is the cumulant generating function:

K_S(t)=\log M_S(t)=\lambda(M_X(t)-1).

Differentiating at zero gives $K_S'(0)=\lambda E[X]$ and $K_S''(0)=\lambda E[X^2]$. The first cumulant is the mean and the second is the variance, so the compound Poisson formulas are:

Theorem: Under the model assumptions above, $$ E[S]=\lambda E[X], \qquad \operatorname{Var}(S)=\lambda E[X^2] =\lambda\big(\operatorname{Var}(X)+(E[X])^2\big). $$

The second moment $E[X^2]$, not just the severity variance, appears in the aggregate variance. This also follows from conditioning: $E[S\mid N]=NE[X]$ and $\operatorname{Var}(S\mid N)=N\operatorname{Var}(X)$. Applying the law of total variance gives

\operatorname{Var}(S) =E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2 =\lambda\operatorname{Var}(X)+\lambda(E[X])^2 =\lambda E[X^2],

because a Poisson random variable has both mean and variance $\lambda$. The second term reflects variation in the number of claims, even if every claim amount were fixed.

Zero Aggregate Loss and the Meaning of the Rate

When every claim amount is strictly positive, $S=0$ exactly when $N=0$. Thus $P(S=0)=e^{-\lambda}$. If a claim amount can itself be zero, the aggregate can also be zero when claims occur but all their amounts are zero. Let $p_0=P(X=0)$. Conditional on $N=n$, the chance that all $n$ amounts are zero is $p_0^n$, including $p_0^0=1$. Therefore:

P(S=0)=E[p_0^N] =\exp\{\lambda(p_0-1)\} =\exp\{-\lambda(1-p_0)\}.

The effective rate of claims with positive amounts is $\lambda(1-p_0)$. This explains why the probability of zero aggregate loss is not always the same as the probability of zero claims. In particular, if the severity is strictly positive, then $p_0=0$ and the formula reduces to $e^{-\lambda}$.

The rate parameter is tied to the exposure period and population defined in the model. If independent, identical risks each have a Poisson count with rate $r$ over one unit of exposure, then the total count across $m$ such risks has rate $mr$. More generally, a rate per policy-year must be multiplied by policy-years to obtain the expected count for the period. This scaling changes the aggregate distribution even when the severity distribution stays fixed.

Worked Applications

Worked Example: Discrete Severity With Zero Payments

A hypothetical benefit portfolio has a Poisson claim count with mean $\lambda=2$ per quarter. Each counted claim has amount $X=0$ with probability $0.2$, $X=\$1{,}000$ with probability $0.5$, and $X=\$3{,}000$ with probability $0.3$. Find the aggregate probability generating function, mean, variance, and probability of zero aggregate payment.

Method and setup. The count and claim amounts are independent, and the claim amounts are identically distributed. In units of dollars, the severity PGF is $G_X(z)=0.2+0.5z^{1000}+0.3z^{3000}$. The compound Poisson formula gives:

G_S(z)=\exp\{2(G_X(z)-1)\} =\exp\{-1.6+z^{1000}+0.6z^{3000}\}.

Mean. $E[X]=0(0.2)+1{,}000(0.5)+3{,}000(0.3)=0+500+900=\$1{,}400$. Hence $E[S]=2(1{,}400)=\$2{,}800$.

Variance. $E[X^2]=0^2(0.2)+1{,}000^2(0.5)+3{,}000^2(0.3)=0+500{,}000+2{,}700{,}000=\$3{,}200{,}000$. Thus $\operatorname{Var}(S)=2(3{,}200{,}000)=\$6{,}400{,}000$ squared. As a check, $E[X^2]-(E[X])^2=3{,}200{,}000-1{,}960{,}000=1{,}240{,}000$, and $2(1{,}240{,}000+1{,}960{,}000)=\$6{,}400{,}000$.

Zero payment. Here $p_0=0.2$, so $P(S=0)=e^{-2(0.8)}=e^{-1.6}=0.2019$, to four decimal places. This includes quarters with no claims and quarters in which all counted claims have zero amounts. The method avoids treating every counted claim as a positive payment.

Worked Example: Exponential Claim Amounts

For a hypothetical equipment cover, the quarterly claim count is Poisson with mean $\lambda=3$. Given a claim, its amount $X$ has an exponential distribution with mean $\$500$. Find the aggregate MGF, mean, variance, and zero-loss probability.

Method and setup. The exponential severity has rate $1/500$ per dollar, so $M_X(t)=1/(1-500t)$ for $t<1/500$. The aggregate MGF is therefore:

M_S(t)=\exp\left\{3\left(\frac{1}{1-500t}-1\right)\right\},\qquad t<\frac{1}{500}.

For an exponential variable with mean $\theta=500$, $E[X]=500$ and $E[X^2]=2\theta^2=2(500^2)=500{,}000$. Thus $E[S]=3(500)=\$1{,}500$, and $\operatorname{Var}(S)=3(500{,}000)=\$1{,}500{,}000$ squared. Because an exponential amount is strictly positive with probability one, $P(S=0)=e^{-3}=0.0498$, to four decimal places. The result is for one quarter, the period to which the stated Poisson mean applies.

Check. The variance is larger than the expected claim count times the severity variance alone: the random number of claims contributes additional variation. This calculation uses the second raw moment $500{,}000$, not the severity variance $250{,}000$.

Worked Example: Aggregating Exposure Across Risks

A hypothetical group covers 8 identical one-year risks. Each risk has an independent Poisson claim count with mean $0.25$ per year. Claim amounts are independent of counts and equal to $\$400$ with probability $0.75$ or $\$1{,}600$ with probability $0.25$. Find the group’s annual aggregate mean, variance, and probability of no loss.

Method and rate conversion. The total annual count is Poisson with rate $\lambda=8(0.25)=2$. For one claim, $E[X]=400(0.75)+1{,}600(0.25)=300+400=\$700$, and $E[X^2]=400^2(0.75)+1{,}600^2(0.25)=120{,}000+640{,}000=\$760{,}000$.

The group aggregate mean is $E[S]=2(700)=\$1{,}400$. Its variance is $\operatorname{Var}(S)=2(760{,}000)=\$1{,}520{,}000$ squared. Both severity amounts are positive, so the probability of no loss is $P(S=0)=e^{-2}=0.1353$, to four decimal places.

Check and trap avoided. The count rate doubles from $0.25$ to $2$ because eight independent exposures are combined. Using $0.25$ for the whole group would give an incorrect mean of $\$175$ and would describe one risk, not the group.

Common Exam Trap: Using Severity Variance Alone

A frequent error is to write $\operatorname{Var}(S)=\lambda\operatorname{Var}(X)$. That expression omits the aggregate variation caused by the random claim count. In the first worked example, the incorrect calculation would be $2(1{,}240{,}000)=\$2{,}480{,}000$ squared. The correct calculation is $2E[X^2]=2(3{,}200{,}000)=\$6{,}400{,}000$ squared. The missing amount is $\lambda(E[X])^2=2(1{,}960{,}000)=\$3{,}920{,}000$ squared. Use severity variance alone only as one component in the total-variance identity, not as the final compound Poisson variance.

Key takeaway: For a compound Poisson aggregate, the frequency rate multiplies the severity mean to give the aggregate mean and the severity second raw moment to give the aggregate variance. The zero-loss probability is $e^{-\lambda(1-P(X=0))}$, so it equals $e^{-\lambda}$ only when a claim amount is strictly positive.

Check Your Understanding

For each problem, use the stated period and distinguish counted claims from positive amounts where necessary.

A quarterly claim count is Poisson with mean $1.2$. Claim amount is zero with probability $0.1$ and $\$200$ with probability $0.9$. Find the aggregate mean and the probability of zero aggregate payment. (Answer: Mean $=\$216$; $P(S=0)=e^{-1.08}=0.3396$.)
A Poisson count has mean $2.5$ per year, and every claim amount is exactly $\$80$. Find the aggregate variance and probability of no loss. (Answer: $\operatorname{Var}(S)=\$16{,}000$ squared; $P(S=0)=e^{-2.5}=0.0821$.)
A monthly count is Poisson with mean $1.5$. Claim amounts are $\$100$ with probability $0.4$ and $\$300$ with probability $0.6$. Find aggregate mean, variance, and zero-loss probability. (Answer: Mean $=\$330$; variance $=\$87{,}000$ squared; $P(S=0)=e^{-1.5}=0.2231$.)
Each of 12 independent policies has a Poisson count with mean $0.1$ per year. Claim amounts are exponential with mean $\$2{,}500$. Find the portfolio’s annual aggregate mean, variance, and zero-loss probability. (Answer: Mean $=\$3{,}000$; variance $=\$15{,}000{,}000$ squared; $P(S=0)=e^{-1.2}=0.3010$.)
A quarterly Poisson count has mean $4$. Each counted claim has amount zero with probability $0.25$ and $\$1{,}000$ with probability $0.75$. Find the aggregate mean, variance, and probability of zero payment. (Answer: Mean $=\$3{,}000$; variance $=\$3{,}000{,}000$ squared; $P(S=0)=e^{-3}=0.0498$.)

This treatment of aggregate loss models is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

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