Compound Poisson Aggregate Loss Components

Why Aggregate-Loss Components Matter

A risk’s aggregate loss combines two sources of randomness: how many claims occur and how much each claim costs. Even when the risk’s expected aggregate loss is known, the realized total can vary around that expectation. Across a population of risks, the expected totals can also differ from risk to risk. Separating these two kinds of variation is important when describing the Bühlmann model components for aggregate losses.

The key distinction is between process variation within a risk, measured by the aggregate-loss variance conditional on that risk’s parameters, and variation in risk-specific expected aggregate losses, measured across the risk population. The same aggregate-loss model supplies both components, but they answer different questions.

Definition: Let $\theta$ denote a risk’s characteristics. Conditional on $\theta$, let the number of claims $N$ have a Poisson distribution with mean $\lambda(\theta)$. Let claim severities $X_1,X_2,\ldots$ be independent and identically distributed conditional on $\theta$, independent of $N$, and have finite first and second moments. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Write $\mu_X(\theta)=E[X_j\mid\theta]$ and $m_2(\theta)=E[X_j^2\mid\theta]$.

Conditional Mean and Variance

For a fixed risk parameter $\theta$, the expected aggregate loss is the expected claim count multiplied by the expected severity. Its variance involves the second moment of severity, not just the severity variance.

E[S\mid\theta]=\lambda(\theta)\mu_X(\theta)

\operatorname{Var}(S\mid\theta)=\lambda(\theta)m_2(\theta) =\lambda(\theta)\left(\operatorname{Var}(X_j\mid\theta)+\mu_X(\theta)^2\right)

The second form makes clear why frequency variation contributes to aggregate variance: even if each claim had a fixed severity, a random number of claims would still make the total vary. The squared severity mean in $m_2$ captures that contribution.

Condition on the claim count.
For $N=n$, the total is a sum of $n$ independent severities, so $E[S\mid N=n,\theta]=n\mu_X(\theta)$ and $\operatorname{Var}(S\mid N=n,\theta)=n\operatorname{Var}(X_j\mid\theta)$.

Average the conditional mean over the count.
$E[S\mid\theta]=E[N\mid\theta]\mu_X(\theta)=\lambda(\theta)\mu_X(\theta)$.

Apply the variance identity over the count.
$\operatorname{Var}(S\mid\theta)=E[\operatorname{Var}(S\mid N,\theta)\mid\theta]+\operatorname{Var}(E[S\mid N,\theta]\mid\theta)$. The first term is $\lambda(\theta)\operatorname{Var}(X_j\mid\theta)$; the second is $\lambda(\theta)\mu_X(\theta)^2$, since a Poisson variable has variance equal to its mean.

Combine the terms.
The two terms sum to $\lambda(\theta)\{\operatorname{Var}(X_j\mid\theta)+\mu_X(\theta)^2\}=\lambda(\theta)m_2(\theta)$.

This derivation depends on the conditional independence and identical-distribution assumptions stated above. If claim severities are dependent, if the count is not conditionally Poisson, or if the severity moments do not exist, this formula need not apply. In particular, replacing the Poisson count with another frequency model generally changes the count-variance term.

Worked Example: Conditional Aggregate Loss With Exponential Severity

For a hypothetical risk, the annual claim count is Poisson with mean $\lambda=2.5$. Conditional on a claim, severity is exponential with mean $400. Assume claims are independent of the count and of one another. Find the conditional expected aggregate loss and variance, and check the variance using the aggregate second moment.

Method. Use the compound Poisson formulas with one fixed risk type. The exponential distribution with mean $400 has variance $400^2=\$160{,}000$, so its second moment is $160{,}000+400^2=\$320{,}000$.

The expected aggregate loss is $E[S]=2.5(400)=\$1{,}000$.

The variance is $\operatorname{Var}(S)=2.5(320{,}000)=\$800{,}000$. Equivalently, the two contributions are $2.5(160{,}000)=\$400{,}000$ from severity variation and $2.5(400^2)=\$400{,}000$ from count variation.

For a check, the compound Poisson second moment is $E[S^2]=\operatorname{Var}(S)+E[S]^2=\$800{,}000+\$1{,}000{,}000=\$1{,}800{,}000$. Subtracting $E[S]^2$ returns a variance of $800,000. The common mistake is to use only $\lambda\operatorname{Var}(X)$, which would give $400,000 and omit the count-variation contribution.

Separating the Two Risk-Population Components

Now suppose the risk parameter $\theta$ varies across a population. For each risk, define its conditional expected aggregate loss as $m(\theta)=E[S\mid\theta]$. The average process variance is the population average of the within-risk conditional variance. The variance of hypothetical means is the variance across risks of their conditional expected aggregate losses.

Result: Under the compound Poisson assumptions, $E[\operatorname{Var}(S\mid\theta)]=E[\lambda(\theta)m_2(\theta)]$ is the average process variance, and $\operatorname{Var}(E[S\mid\theta])=\operatorname{Var}(\lambda(\theta)\mu_X(\theta))$ is the variance of risk-specific expected aggregate losses. Their sum is the unconditional variance of aggregate loss: $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid\theta)]+\operatorname{Var}(E[S\mid\theta])$.

The sum follows from the law of total variance: variation in $S$ consists of average variation around each risk’s own conditional mean plus variation in those conditional means around the population mean. The components have the same units, dollars squared, but a different interpretation. The first describes random outcomes for a risk with its parameters fixed; the second describes how expected totals differ across risks.

The risk parameter can affect frequency, severity, or both. Therefore, compute each risk’s aggregate mean as $\lambda(\theta)\mu_X(\theta)$ before finding its across-risk variance. Taking a variance of frequency parameters alone does not generally give the between-risk component.

Worked Example: Two Risk Types With a Shared Severity Distribution

A hypothetical portfolio contains Type A risks with probability 0.6 and Type B risks with probability 0.4. For Type A, annual claim count is Poisson with mean 1; for Type B, it is Poisson with mean 3. Both types have independent exponential claim severities with mean $1,000. Find the average process variance, the variance of risk-specific expected aggregate losses, and total aggregate-loss variance.

Method. For the exponential severity, the variance is $\$1{,}000{,}000$, and the second moment is $\$2{,}000{,}000$. Apply the conditional formulas separately to each risk type.

For Type A, $E[S\mid A]=1(1{,}000)=\$1{,}000$, and $\operatorname{Var}(S\mid A)=1(2{,}000{,}000)=\$2{,}000{,}000$. For Type B, $E[S\mid B]=3(1{,}000)=\$3{,}000$, and $\operatorname{Var}(S\mid B)=3(2{,}000{,}000)=\$6{,}000{,}000$.

The average process variance is $0.6(2{,}000{,}000)+0.4(6{,}000{,}000)=\$3{,}600{,}000$. The population mean aggregate loss is $0.6(1{,}000)+0.4(3{,}000)=\$1{,}800$.

The risk-specific expected losses are $1,000 and $3,000. Their variance is $0.6(1{,}000-1{,}800)^2+0.4(3{,}000-1{,}800)^2 =0.6(640{,}000)+0.4(1{,}440{,}000)=\$960{,}000$. Thus the total variance is $\$3{,}600{,}000+\$960{,}000=\$4{,}560{,}000$.

The calculation checks because the overall variance is the sum of the two nonnegative components. A common error is to treat all risks as if they had the portfolio-average frequency and then report only a process variance. That discards the $960,000 of variation in expected aggregate loss between risk types.

Worked Example: Frequency and Severity Both Differ by Risk

A hypothetical book consists of two equally likely risk types. Type A has a Poisson annual count with mean 2 and a fixed severity of $300 per claim. Type B has a Poisson annual count with mean 1 and exponential severity with mean $900. Assume conditional independence as in the compound Poisson model. Calculate both variance components and verify the total variance.

Method. Determine each type’s severity second moment, then calculate its conditional aggregate mean and variance.

For Type A, $E[X\mid A]=\$300$ and $E[X^2\mid A]=\$300^2=\$90{,}000$. Therefore $E[S\mid A]=2(300)=\$600$ and $\operatorname{Var}(S\mid A)=2(90{,}000)=\$180{,}000$.

For Type B, the exponential severity has mean $900, variance $900^2=\$810{,}000$, and second moment $\$810{,}000+\$810{,}000=\$1{,}620{,}000$. Hence $E[S\mid B]=1(900)=\$900$ and $\operatorname{Var}(S\mid B)=1(1{,}620{,}000)=\$1{,}620{,}000$.

The average process variance is $0.5(180{,}000)+0.5(1{,}620{,}000)=\$900{,}000$. The mean across risk types is $0.5(600)+0.5(900)=\$750$. The variance of the two risk-specific means is $0.5(600-750)^2+0.5(900-750)^2=\$22{,}500$. The total variance is therefore $\$922{,}500$.

The answer is checked by adding the two components. The common mistake is to apply the Type B severity moments to Type A, or to pool the two severity distributions before computing the conditional quantities. Conditional moments must be calculated using the severity model belonging to each risk type.

Interpretation and Exam Traps

The conditional variance formula is compact, but the second moment can be easy to overlook. Since $E[X^2]=\operatorname{Var}(X)+E[X]^2$, using only severity variance ignores the uncertainty created by having a random number of claims. For deterministic severity $c$, for example, the conditional aggregate variance is still $\lambda c^2$, not zero.

Likewise, the across-risk component concerns variation in expected aggregate losses, not variation in conditional variances. Two risk types could have identical expected aggregate losses and very different process variances; their between-risk component would be zero even though their process variance might be substantial. Conversely, risk types with different means produce a positive between-risk component, even if each risk’s process variance is small.

Use the severity second moment. For compound Poisson aggregate loss, $\operatorname{Var}(S\mid\theta)=\lambda(\theta)E[X^2\mid\theta]$, not $\lambda(\theta)\operatorname{Var}(X\mid\theta)$.
Condition before averaging. Calculate the mean and variance for each risk type using its own frequency and severity assumptions, then average or take variance across risks as appropriate.
Keep the components distinct. Average process variance is an expectation of conditional variances; variance of hypothetical means is a variance of conditional means.
Check the model assumptions. The stated formulas presume a Poisson count and independent, identically distributed severities conditional on the risk. Do not transfer them automatically to other frequency models or dependent claims.

Key takeaway: For a compound Poisson risk, the conditional aggregate mean is $\lambda\mu_X$ and the conditional variance is $\lambda E[X^2]$. Across risks, average the conditional variance for the process component and take the variance of the conditional mean for the risk-to-risk component.

Check Your Understanding

For each problem, identify the conditional or across-risk component requested and keep monetary variances in dollars squared.

A hypothetical risk has Poisson claim count with mean 0.8 and deterministic severity of $500. Find its conditional expected aggregate loss and variance. (Answer: mean $400; variance $200,000.)
A risk has Poisson claim count with mean 2 and exponential severity with mean $200. Find its conditional aggregate mean and variance. (Answer: mean $400; variance $160,000.)
A portfolio has Type A probability 0.25 and Type B probability 0.75. Counts are Poisson with means 1 and 3, respectively, and every claim has deterministic severity $100. Find the average process variance and variance of risk-specific expected aggregate losses. (Answer: average process variance $25,000; variance of means $7,500.)
Two equally likely risk types have Poisson count means 2 and 4, respectively, and both have exponential severity with mean $50. Find the total aggregate-loss variance by calculating both components. (Answer: average process variance $15,000; variance of means $2,500; total variance $17,500.)
A hypothetical risk has a Poisson count with mean 1.5 and Gamma severity with shape 2 and scale $300. Find the conditional aggregate mean and variance. (Answer: mean $900; variance $810,000.)

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