Risk Measures for Compound Aggregate Losses

Introduction

An aggregate loss combines two sources of uncertainty: how many claims occur and how large each claim is. A risk measure for one claim does not, by itself, describe the total loss from a portfolio. The frequency and severity models must first be combined, after which the aggregate loss can be summarized by measures such as value-at-risk (VaR) and tail value-at-risk (TVaR).

This tutorial develops the aggregate mean and variance, states the compound Poisson distribution, and shows how these results support aggregate VaR and TVaR calculations. Normal approximations can be useful when an exact aggregate distribution is difficult to evaluate, but their limitations matter, particularly for skewed or heavy-tailed losses. The treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $N$ be the number of claims in a fixed period, and let $X_1,X_2,\ldots$ be independent and identically distributed claim severities, independent of $N$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where $S=0$ when $N=0$. Write $m=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$ for the severity mean and variance.

Aggregate Mean and Variance

The frequency distribution affects the aggregate variance as well as the aggregate mean. In particular, the variance of $N$ matters: even if claim sizes are modestly variable, an uncertain claim count can create substantial variation in total losses.

Theorem: If $N$ is independent of the iid severities $X_j$, and the moments below are finite, then $$ E[S]=E[N]E[X], \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2. $$ For $N\sim\operatorname{Poisson}(\lambda)$, this reduces to $$ E[S]=\lambda E[X], \qquad \operatorname{Var}(S)=\lambda E[X^2] =\lambda\left(\operatorname{Var}(X)+(E[X])^2\right). $$

The first pair of formulas applies to any independent frequency model with finite mean and variance. The Poisson formulas use the additional fact that $E[N]=\operatorname{Var}(N)=\lambda$. They do not apply unchanged to a fixed claim count or to a frequency model with different moments.

Why the Moment Formulas Hold

Condition on the observed claim count. Given $N=n$, the aggregate is a sum of $n$ independent severities, so its conditional mean is $nm$ and its conditional variance is $n\sigma_X^2$. Taking expectations gives $E[S]=E[N]m$. The law of total variance then separates uncertainty in claim sizes from uncertainty in claim count:

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N\sigma_X^2]+\operatorname{Var}(Nm)\\ &=E[N]\sigma_X^2+\operatorname{Var}(N)m^2. \end{aligned}

For Poisson frequency, substitute $E[N]=\operatorname{Var}(N)=\lambda$. The result is $\lambda(\sigma_X^2+m^2)$, and $\sigma_X^2+m^2=E[X^2]$. This explains why using only $\lambda\operatorname{Var}(X)$ omits the variation caused by the random count.

The Compound Poisson Distribution and Normal Approximation

If $N$ is Poisson with parameter $\lambda$, the aggregate distribution is a Poisson mixture of severity sums. Let $F_X^{*n}$ denote the distribution of the sum of $n$ independent severities; for $n=0$, that sum is zero. Then, for $s\geq 0$,

P(S\leq s)= \sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}F_X^{*n}(s).

This representation includes a probability mass $P(S=0)=e^{-\lambda}$ when severities are positive. Directly evaluating the mixture may be inconvenient, so a normal approximation is sometimes used. If the aggregate distribution is reasonably close to normal in the range of interest, use the aggregate mean and variance, not the severity mean and variance, as the normal parameters.

Approximation: Let $\mu_S=E[S]$, $\sigma_S=\sqrt{\operatorname{Var}(S)}$, and let $z_p$ satisfy $\Phi(z_p)=p$. Under a normal approximation to $S$, $$ \operatorname{VaR}_p(S)\approx \mu_S+z_p\sigma_S, \qquad \operatorname{TVaR}_p(S)\approx \mu_S+\sigma_S\frac{\phi(z_p)}{1-p}, $$ where $\Phi$ and $\phi$ are the standard normal distribution and density functions. The TVaR formula is for a continuous normal model; it is not an exact formula for an aggregate distribution with an atom at its quantile.

The approximation is not guaranteed to be accurate merely because the expected claim count is large. Strong skewness, very heavy severity tails, or a tail probability far beyond the region supported by the approximation can make estimated tail risk unreliable. When the aggregate distribution is discrete, its quantile and tail must also be handled with the atom-aware definition of TVaR.

Worked Example: Compound Poisson Frequency with Exponential Severity

A hypothetical portfolio has a Poisson claim count with $\lambda=100$ per year. Claim severities are independent exponential random variables with mean $\$1{,}000$, independent of the count. Approximate the annual aggregate 99% VaR and TVaR using a normal model.

Method. Calculate the compound Poisson aggregate moments, then apply the normal VaR and TVaR formulas. For an exponential severity with mean $\theta$, $E[X]=\theta$, $\operatorname{Var}(X)=\theta^2$, and $E[X^2]=2\theta^2$. Here $\theta=\$1{,}000$, so $E[X^2]=\$2{,}000{,}000$.

The aggregate mean is $E[S]=100(\$1{,}000)=\$100{,}000$. The aggregate variance and standard deviation are

\operatorname{Var}(S)=100(\$2{,}000{,}000)=\$200{,}000{,}000, \qquad \sigma_S=\sqrt{\$200{,}000{,}000}=\$14{,}142.14.

For $p=0.99$, use $z_p=2.32635$ and $\phi(z_p)\approx0.0266521$. Thus

\operatorname{VaR}_{0.99}(S) \approx \$100{,}000+2.32635(\$14{,}142.14) =\$132{,}899.53.

The normal TVaR approximation is

\operatorname{TVaR}_{0.99}(S) \approx \$100{,}000+ \$14{,}142.14\left(\frac{0.0266521}{0.01}\right) =\$137{,}691.82.

Check. TVaR exceeds VaR, as expected for a nonconstant continuous loss distribution. These are normal approximations to the aggregate risk measures, not exact values from the compound Poisson mixture.

Common mistake avoided: Using $\lambda\operatorname{Var}(X)$ gives a variance of $\$100{,}000{,}000$ and a standard deviation of $\$10{,}000$. That omits frequency variation. For Poisson frequency the correct variance is $\lambda E[X^2]$, which here gives $\$200{,}000{,}000$.

Worked Example: A Non-Poisson Frequency Model

A hypothetical warranty portfolio has a claim count $N$ with $E[N]=40$ and $\operatorname{Var}(N)=60$. Claim severities are independent of $N$, with mean $\$500$ and standard deviation $\$500$. Approximate the 95% aggregate VaR and TVaR under a normal model.

Method. Use the general compound variance formula because the frequency is not specified as Poisson. The severity variance is $(\$500)^2=\$250{,}000$, and the squared severity mean is also $\$250{,}000$. Therefore,

E[S]=40(\$500)=\$20{,}000,

\operatorname{Var}(S) =40(\$250{,}000)+60(\$250{,}000) =\$25{,}000{,}000, \qquad \sigma_S=\$5{,}000.

For $p=0.95$, $z_p=1.64485$ and $\phi(z_p)\approx0.103136$. The approximations are

\operatorname{VaR}_{0.95}(S) \approx \$20{,}000+1.64485(\$5{,}000) =\$28{,}224.27,

\operatorname{TVaR}_{0.95}(S) \approx \$20{,}000+ \$5{,}000\left(\frac{0.103136}{0.05}\right) =\$30{,}313.60.

Check. The frequency contribution to variance is $60(\$500)^2=\$15{,}000{,}000$, while the severity contribution is $40(\$250{,}000)=\$10{,}000{,}000$; together they give the stated total. The TVaR approximation is above the VaR approximation.

Common mistake avoided: Treating this as Poisson frequency would impose $\operatorname{Var}(N)=E[N]=40$, contrary to the stated variance of 60. The general formula preserves the supplied frequency uncertainty.

TVaR When Aggregate Loss Has an Atom

A compound aggregate can be discrete even when the claim count is random. In particular, a fixed claim amount produces aggregate losses on a grid. If the VaR quantile itself has positive probability, TVaR includes only the needed fraction of the probability at that quantile. For $q_p=\operatorname{VaR}_p(S)$, the atom-aware expression is

\operatorname{TVaR}_p(S)= \frac{E[S\,\mathbf{1}_{\{S>q_p\}}]+q_p\bigl(F_S(q_p)-p\bigr)}{1-p}.

The first term includes all outcomes strictly above $q_p$. The second adds enough of the probability mass at $q_p$ to form an upper tail of probability $1-p$. This is consistent with the integral quantile definition of TVaR. Simply computing $E[S\mid S>q_p]$ discards the quantile mass and is generally not the same quantity.

Worked Example: Exact TVaR for a Discrete Compound Poisson Loss

A hypothetical book has $N\sim\operatorname{Poisson}(2)$ claims, each with a fixed severity of $\$1{,}000$. Find the exact 90% VaR and TVaR of $S=\$1{,}000N$.

Method. Since the severity is fixed, the aggregate is determined by the Poisson count. The needed probabilities are $P(N=0)=0.135335$, $P(N=1)=0.270671$, $P(N=2)=0.270671$, $P(N=3)=0.180447$, and $P(N=4)=0.090224$. Consequently, $P(N\leq3)=0.857123$ and $P(N\leq4)=0.947347$. The smallest aggregate amount whose cumulative probability is at least 0.90 is $\$4{,}000$, so $\operatorname{VaR}_{0.90}(S)=\$4{,}000$.

For the TVaR numerator, first calculate the expected loss strictly above $\$4{,}000$. This corresponds to $N\geq5$. For a Poisson variable with mean 2, the identity $E[N\mathbf{1}_{\{N\geq5\}}]=2P(N\geq4)$ follows by shifting the Poisson probability index. Therefore,

E[S\mathbf{1}_{\{S>\$4{,}000\}}] =\$1{,}000\left(2P(N\geq4)\right) =\$1{,}000\left(2(1-0.857123)\right) =\$285.75.

The probability mass to include at the quantile is $F_S(\$4{,}000)-0.90=0.947347-0.90=0.047347$. Its contribution is $\$4{,}000(0.047347)=\$189.39$. Thus,

\operatorname{TVaR}_{0.90}(S) =\frac{\$285.75+\$189.39}{0.10} =\$4{,}751.41.

Check. The upper 10% consists of all outcomes above $\$4{,}000$, whose probability is $1-0.947347=0.052653$, and a further 0.047347 of probability at $\$4{,}000$. These total 0.10, as required.

Common mistake avoided: $E[S\mid S>\$4{,}000]$ is not the 90% TVaR here: it excludes the quantile mass. The correction is to include the required fraction at $\$4{,}000$ using the atom-aware formula.

Practical Interpretation and Exam Traps

Aggregate risk measures depend on both frequency and severity. The aggregate mean is the expected total loss, while aggregate VaR identifies a quantile of that total and TVaR summarizes the upper tail. They answer different questions: losses can have the same mean but different dispersion or tail behavior, and hence different VaR and TVaR.

Check which frequency formula applies. Use the general formula when $E[N]$ and $\operatorname{Var}(N)$ are given. Use $\lambda E[X^2]$ for variance only when $N$ is Poisson and independent of severities.
Keep severity variance and second moment distinct. $E[X^2]=\operatorname{Var}(X)+(E[X])^2$; confusing them understates compound Poisson variance.
Use aggregate, not individual, moments in the normal approximation. First calculate $E[S]$ and $\operatorname{Var}(S)$, then apply the normal quantile or tail formula.
Check for a quantile atom. If $P(S=q_p)>0$, do not replace TVaR by the conditional mean strictly above $q_p$.
State whether a result is exact or approximate. Moment matching does not make a normal model identical to the compound distribution, especially in a far tail.

Key takeaway: For independent frequency and severity, $E[S]=E[N]E[X]$ and $\operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$. These moments support normal approximations to aggregate VaR and TVaR, while exact discrete tails require correct treatment of probability at the quantile.

Check Your Understanding

For each problem, identify the appropriate aggregate model before calculating. Treat requested normal-model risk measures as approximations.

A hypothetical portfolio has Poisson frequency with $\lambda=25$, and independent severities with mean $\$200$ and variance $\$40{,}000$. Find the aggregate mean, variance, and approximate 95% VaR. (Answer: mean $\$5{,}000$; variance $\$2{,}000{,}000$; VaR approximately $\$7{,}326.21$)
A count has mean 10 and variance 15. Independent claim severities have mean $\$300$ and variance $\$160{,}000$. Find the aggregate variance and approximate 99% VaR. (Answer: variance $\$2{,}950{,}000$; VaR approximately $\$6{,}995.29$)
A hypothetical aggregate is $S=\$500N$, where $N\sim\operatorname{Poisson}(1)$. Find the 80% VaR. (Answer: $\$1{,}000$)
A hypothetical portfolio has Poisson frequency with $\lambda=9$ and independent exponential severities with mean $\$100$. Using a normal approximation, find the 95% VaR. (Answer: approximately $\$1{,}597.89$)

What You'll Learn