Mean And Variance Of A Compound Aggregate

Why Frequency and Severity Both Matter

An aggregate loss is a random sum: the number of claims is uncertain, and the amount of each claim is uncertain. Its expected value depends on the average number of claims and the average claim amount. Its variance reflects two sources of uncertainty: variation in claim amounts and variation in the number of claims. Keeping those sources distinct is essential when calculating aggregate risk.

This tutorial develops the mean and variance formulas for a compound aggregate and applies them to Poisson, binomial, and negative binomial claim counts. The calculations concern undiscounted losses; no interest or mortality assumption is involved. Each example describes a hypothetical portfolio.

Definition: Let $N$ be a nonnegative integer-valued claim count. Let $X_1,X_2,\ldots$ be independent and identically distributed claim amounts, independent of $N$, with $E[X]=\mu$ and $\operatorname{Var}(X)=\sigma^2$. The aggregate loss is $S=\sum_{i=1}^{N}X_i$, with $S=0$ when $N=0$. Assume $E[N]$, $\operatorname{Var}(N)$, and the severity second moment are finite.

The Compound Mean and Variance

Write $m=E[N]$ and $v_N=\operatorname{Var}(N)$. The mean aggregate is the expected count multiplied by the expected severity. The variance has one term for severity variation at the expected count and another term for the variation in the count itself.

Theorem: Under the independence and moment assumptions above,

E[S]=E[N]E[X]=m\mu, \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2 =m\sigma^2+v_N\mu^2.

A useful equivalent expression uses the severity second raw moment $E[X^2]=\sigma^2+\mu^2$:

\operatorname{Var}(S)=E[N]E[X^2]+\bigl(\operatorname{Var}(N)-E[N]\bigr)(E[X])^2.

The equivalent form is especially useful when a problem gives $E[X^2]$ directly. Do not substitute $E[X^2]$ for $\operatorname{Var}(X)$ in the main formula: they differ by $\mu^2$.

Why the Formulas Hold

Condition on the claim count. Given $N=n$, the aggregate is a sum of $n$ independent severities, so its conditional mean is $n\mu$ and its conditional variance is $n\sigma^2$. The law of total expectation gives the mean. The law of total variance separates the average conditional variance from the variance of the conditional mean:

Condition on $N$.
For $N=n$, $S$ is the sum of $n$ independent copies of $X$. Therefore $E[S\mid N]=N\mu$ and $\operatorname{Var}(S\mid N)=N\sigma^2$.

Take the expectation.
$E[S]=E[E[S\mid N]]=E[N\mu]=E[N]\mu$.

Separate the two variance sources.
$\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])=E[N]\sigma^2+\operatorname{Var}(N\mu)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$.

The first variance term arises because severities can differ even if the count were fixed. The second arises because the conditional expected aggregate, $N\mu$, changes when the count changes. The derivation requires severity amounts to be independent of the count and independent across claims. If severity depends on the count or claims are dependent, these formulas need not apply in this form.

Applying the Formulas to Common Count Models

Once the count model is specified, obtain its mean and variance, then substitute them together with the severity moments. For a binomial count, the probability parameter below is the probability of a claim for each of $n$ opportunities. For a negative binomial count, we use the convention that $N$ counts failures before the $r$th success.

Claim count model	$E[N]$	$\operatorname{Var}(N)$	Compound aggregate variance
$N\sim\operatorname{Poisson}(\lambda)$	$\lambda$	$\lambda$	$\lambda(\sigma^2+\mu^2)=\lambda E[X^2]$
$N\sim\operatorname{Binomial}(n,q)$	$nq$	$nq(1-q)$	$nq\sigma^2+nq(1-q)\mu^2$
$N$ is failures before $r$ successes, success probability $p$	$\dfrac{r(1-p)}{p}$	$\dfrac{r(1-p)}{p^2}$	$\dfrac{r(1-p)}{p}\sigma^2+\dfrac{r(1-p)}{p^2}\mu^2$

For Poisson frequency, $\operatorname{Var}(N)=E[N]$, so the compound variance simplifies to $\lambda E[X^2]$. That simplification is specific to Poisson counts. It does not mean that the aggregate variance is generally the expected count times the severity variance.

Worked Example: Poisson Frequency With Two Possible Claim Amounts

A hypothetical equipment portfolio has a Poisson claim count with parameter $\lambda=2.5$ for the year. Each claim amount $X$, in dollars, is independently \$100 with probability 0.6 or \$250 with probability 0.4, independently of the count. Find the aggregate mean and variance.

Method. Find the severity mean and variance, then use the compound Poisson moments. The count has $E[N]=\operatorname{Var}(N)=2.5$.

First calculate the severity mean:

\mu=E[X]=0.6(100)+0.4(250)=60+100=160.

Next calculate the second raw moment and severity variance:

E[X^2]=0.6(100^2)+0.4(250^2) =0.6(10{,}000)+0.4(62{,}500) =6{,}000+25{,}000=31{,}000, $$ $$ \sigma^2=E[X^2]-\mu^2=31{,}000-160^2 =31{,}000-25{,}600=5{,}400.

The aggregate mean is:

E[S]=\lambda\mu=2.5(160)=\$400.

The aggregate variance is:

\operatorname{Var}(S)=\lambda\sigma^2+\lambda\mu^2 =2.5(5{,}400)+2.5(25{,}600) =13{,}500+64{,}000=77{,}500\text{ dollars}^2.

Check using the Poisson simplification: $2.5E[X^2]=2.5(31{,}000)=77{,}500$ dollars squared, matching the result. The common mistake avoided here is using only $2.5\sigma^2$, which omits the variability due to the random claim count.

Worked Example: Binomial Frequency and the Count-Variance Term

A hypothetical warranty covers six independent devices, each having probability $q=0.2$ of producing one claim during the coverage period. Thus $N\sim\operatorname{Binomial}(6,0.2)$. Claim amounts are independent of the count, with mean $\mu=\$50$ and variance $\sigma^2=225$ dollars squared. Find the aggregate mean and variance.

Method. Calculate the binomial count moments and substitute them separately into the two terms of the compound formulas.

E[N]=nq=6(0.2)=1.2, \qquad \operatorname{Var}(N)=nq(1-q)=6(0.2)(0.8)=0.96.

The aggregate mean is:

E[S]=1.2(50)=\$60.

For the variance, the severity-variation term is $1.2(225)=270$ dollars squared. The count-variation term is $0.96(50^2)=0.96(2{,}500)=2{,}400$ dollars squared. Thus:

\operatorname{Var}(S)=270+2{,}400=2{,}670\text{ dollars}^2.

The result is checked directly from the formula: $E[N]\sigma^2+\operatorname{Var}(N)\mu^2=1.2(225)+0.96(2{,}500)=2{,}670$. A common mistake is to set $\operatorname{Var}(N)=E[N]$, which is true for Poisson counts but not for this binomial count.

Worked Example: Negative Binomial Frequency With Failures-First Parameterization

A hypothetical repair program records the number of unsuccessful service calls before the third successful resolution. Let $N$ be the number of failures before $r=3$ successes, with success probability $p=0.5$ on each independent call. Claim amounts are independent of $N$, with mean $\mu=\$100$ and variance $\sigma^2=1{,}000$ dollars squared. Find the aggregate mean and variance.

Method. Use the failures-before-$r$-successes moments, not the alternative negative binomial convention that counts total trials. Here $1-p=0.5$.

E[N]=\frac{r(1-p)}{p} =\frac{3(0.5)}{0.5}=3, \qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2} =\frac{3(0.5)}{0.5^2} =\frac{1.5}{0.25}=6.

Therefore:

E[S]=3(100)=\$300, $$ $$ \operatorname{Var}(S)=3(1{,}000)+6(100^2) =3{,}000+6(10{,}000) =63{,}000\text{ dollars}^2.

As a check, the count has variance twice its mean in this parameterization because $r=3$ and $p=0.5$ give $E[N]=3$ and $\operatorname{Var}(N)=6$. The common mistake avoided is substituting the count mean for its variance, or using a different negative binomial convention without adjusting the count moments.

Common Exam Traps and Checks

The formulas are short, but the substitutions are easy to misread. In particular, severity variance, severity second moment, count mean, and count variance play different roles. Check each quantity against the model before combining them.

Using the wrong severity moment. The general variance formula uses $\sigma^2$ in its first term and $\mu^2$ in its second. If using $E[X^2]$, use the equivalent form with the adjustment $\operatorname{Var}(N)-E[N]$.
Assuming every count has Poisson variance. The equality $\operatorname{Var}(N)=E[N]$ is not valid for binomial or negative binomial counts in general.
Mixing negative binomial conventions. State whether $N$ counts failures or total trials. The moments in the table apply to failures before $r$ successes.
Forgetting the squared severity mean. The count-variance contribution is $\operatorname{Var}(N)\mu^2$, not $\operatorname{Var}(N)\mu$.
Ignoring independence. The stated formulas assume severities are identically distributed, independent of one another, and independent of the count.

A reliable calculation records $E[N]$, $\operatorname{Var}(N)$, $E[X]$, and $\operatorname{Var}(X)$ in separate lines before substituting. This makes both the model parameterization and the units easier to check. The aggregate mean has currency units; the aggregate variance has squared-currency units.

Key takeaway: For an independent-severity random sum, $E[S]=E[N]E[X]$, while $\operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$. Determine both count moments from the specified model before using the formula.

Check Your Understanding

For each hypothetical aggregate, identify the count moments before calculating. Answers are shown at the end of each question.

A Poisson count has parameter $\lambda=1.5$. Independent severities have mean $\$80$ and variance $400$ dollars squared. Find the aggregate mean and variance. (Answer: mean $\$120$; variance $10{,}200$ dollars squared.)
A count is binomial with $n=5$ and claim probability $q=0.2$. Severity mean is $\$50$, and severity variance is $225$ dollars squared. Find the aggregate mean and variance. (Answer: mean $\$50$; variance $2{,}225$ dollars squared.)
$N$ counts failures before $r=2$ successes, with $p=0.5$. Severity mean is $\$30$, and severity variance is $100$ dollars squared. Find the aggregate mean and variance. (Answer: mean $\$60$; variance $3{,}800$ dollars squared.)
A Poisson count has parameter $\lambda=3$. Severity mean is $\$40$, and severity variance is $900$ dollars squared. Find the aggregate mean and variance. (Answer: mean $\$120$; variance $7{,}500$ dollars squared.)
A count is binomial with $n=8$ and $q=0.25$. Severity mean is $\$120$, and severity variance is $1{,}600$ dollars squared. Find the aggregate mean and variance. (Answer: mean $\$240$; variance $24{,}800$ dollars squared.)

Claim count model	\(E[N]\)	\(\operatorname{Var}(N)\)	Compound aggregate variance
\(N\sim\operatorname{Poisson}(\lambda)\)	\(\lambda\)	\(\lambda\)	\(\lambda(\sigma^2+\mu^2)=\lambda E[X^2]\)
\(N\sim\operatorname{Binomial}(n,q)\)	\(nq\)	\(nq(1-q)\)	\(nq\sigma^2+nq(1-q)\mu^2\)
\(N\) is failures before \(r\) successes, success probability \(p\)	\(\dfrac{r(1-p)}{p}\)	\(\dfrac{r(1-p)}{p^2}\)	\(\dfrac{r(1-p)}{p}\sigma^2+\dfrac{r(1-p)}{p^2}\mu^2\)

What You'll Learn