Aggregate Variance By Total Variance

Why Aggregate Variance Depends on Both Frequency and Severity

An aggregate loss varies for two reasons: the number of claims can change, and the amount of each claim can change. A calculation that accounts for only one source of variation will generally understate or misstate the uncertainty in total loss. The law of total variance gives a direct way to include both sources using the mean and variance of the claim count and of an individual claim.

Let $N$ be the number of claims in a fixed period, let $X_1,X_2,\ldots$ be the claim severities, and define the aggregate loss by $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. This is the random-sum model introduced in the earlier material on aggregate loss as a random sum.

Definition: In the standard compound aggregate model, $N$ is a nonnegative integer-valued random variable. The severities $X_1,X_2,\ldots$ are independent and identically distributed copies of a nonnegative random variable $X$, and the severity sequence is independent of $N$. Assume $E[N]$, $\operatorname{Var}(N)$, $E[X]$, and $\operatorname{Var}(X)$ are finite. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with an empty sum of zero.

Write $\mu=E[X]$ and $\sigma^2=\operatorname{Var}(X)$. Under these assumptions, aggregate variance is determined by the first two moments of frequency and severity:

Theorem: For the standard compound aggregate model, $$ \operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)\big(E[X]\big)^2. $$

Derivation by the Law of Total Variance

The law of total variance states that, for a random variable $S$ and a conditioning variable $N$, $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. The first term averages the variability remaining in aggregate loss after the count is known. The second measures how the conditional mean changes as the count changes.

Condition on a fixed count.
If $N=n$, then $S$ is the sum of $n$ independent severities. Thus $E[S\mid N=n]=n\mu$ and $\operatorname{Var}(S\mid N=n)=n\sigma^2$. For $n=0$, both quantities are zero, so these expressions also hold.

Average the conditional variance.
Since $\operatorname{Var}(S\mid N)=N\sigma^2$, taking expectations gives $E[\operatorname{Var}(S\mid N)]=E[N]\sigma^2$.

Find the variance of the conditional mean.
Because $E[S\mid N]=N\mu$, its variance is $\operatorname{Var}(N\mu)=\mu^2\operatorname{Var}(N)$.

Combine the two terms.
Substitution into the law of total variance yields $\operatorname{Var}(S)=E[N]\sigma^2+\mu^2\operatorname{Var}(N)$, the stated formula.

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

This derivation also explains the formula: conditional severity variation contributes $E[N]\operatorname{Var}(X)$, while variation in the number of claims contributes $\operatorname{Var}(N)(E[X])^2$. The latter uses the square of the severity mean, not the severity variance.

Applying the Formula to Common Frequency Models

The aggregate-variance formula stays the same when the frequency model changes. Only $E[N]$ and $\operatorname{Var}(N)$ change. For a binomial count, use $q=1-p$. For a negative binomial count below, $N$ means the number of failures before the $r$th success in independent Bernoulli trials with success probability $p$; this parameterization matters because other conventions count successes instead.

Frequency model	$E[N]$	$\operatorname{Var}(N)$	Resulting aggregate variance
Poisson with parameter $\lambda$	$\lambda$	$\lambda$	$\lambda\operatorname{Var}(X)+\lambda(E[X])^2$
Binomial with parameters $n,p$, $q=1-p$	$np$	$npq$	$np\operatorname{Var}(X)+npq(E[X])^2$
Negative binomial: failures before $r$ successes, probability $p$	$r(1-p)/p$	$r(1-p)/p^2$	$\frac{r(1-p)}{p}\operatorname{Var}(X)+\frac{r(1-p)}{p^2}(E[X])^2$

For a Poisson count, the mean and variance of $N$ are both $\lambda$. Therefore the result can also be written as $\lambda E[X^2]$, since $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. This simplification depends on the Poisson frequency model; it is not the general compound-model formula.

Worked Example: Poisson Frequency With Two Possible Claim Amounts

A hypothetical portfolio has a Poisson number of claims with parameter $\lambda=4$ during a year. Independently, each claim is either $100 or $200, each with probability $0.5$. Find the variance of annual aggregate loss.

Method. Calculate the severity mean and variance, then use the compound Poisson specialization of the total-variance formula. This avoids treating the Poisson simplification as valid for every frequency model.

For an individual claim, $E[X]=0.5(100)+0.5(200)=50+100=\$150$. Its second moment is $E[X^2]=0.5(100^2)+0.5(200^2)=0.5(10{,}000)+0.5(40{,}000)=25{,}000$. Therefore $\operatorname{Var}(X)=E[X^2]-(E[X])^2=25{,}000-150^2=25{,}000-22{,}500=2{,}500$ dollars squared.

For $N\sim\operatorname{Poisson}(4)$, $E[N]=4$ and $\operatorname{Var}(N)=4$. Hence $\operatorname{Var}(S)=4(2{,}500)+4(150^2)=10{,}000+4(22{,}500)=10{,}000+90{,}000=100{,}000$ dollars squared.

Check. The compound Poisson form gives $4E[X^2]=4(25{,}000)=100{,}000$, agreeing with the result. The common mistake of using only $E[N]\operatorname{Var}(X)$ would give $10{,}000$, omitting the substantial contribution from variation in claim count.

Worked Example: Binomial Frequency With Fixed Exposure Count

A hypothetical program has five eligible exposures. Each independently produces one claim with probability $p=0.4$, so $N\sim\operatorname{Binomial}(5,0.4)$. Each claim is independently $800 or $1,200 with equal probability. Find the aggregate-loss variance.

Method. Find the binomial count moments and the two severity moments, then substitute both frequency contributions. The binomial variance is not equal to its mean unless the parameters happen to make them equal.

For the frequency, $q=1-p=0.6$, so $E[N]=np=5(0.4)=2$ and $\operatorname{Var}(N)=npq=5(0.4)(0.6)=1.2$.

For a claim, $E[X]=0.5(800)+0.5(1{,}200)=400+600=\$1{,}000$. Also, $E[X^2]=0.5(800^2)+0.5(1{,}200^2)=0.5(640{,}000)+0.5(1{,}440{,}000)=1{,}040{,}000$. Thus $\operatorname{Var}(X)=1{,}040{,}000-(1{,}000)^2=40{,}000$ dollars squared.

Now $\operatorname{Var}(S)=2(40{,}000)+1.2(1{,}000)^2=80{,}000+1.2(1{,}000{,}000)=80{,}000+1{,}200{,}000=1{,}280{,}000$ dollars squared.

Check. The count can take values only from zero through five, and its variance is $1.2$, as calculated from $npq$. The aggregate variance is greater than the severity-only term $80{,}000$, because the claim count is also random.

Worked Example: Negative Binomial Frequency and Parameterization

A hypothetical claims process continues until three successes in independent trials, where each trial is a success with probability $p=0.5$. Let $N$ count failures before the third success. Each claim is independently $10 or $30 with equal probability. Find $\operatorname{Var}(S)$.

Method. Use the negative binomial moments for the stated failures-count convention. A common mistake is to use a formula for the number of trials through the final success, which defines a different count.

Here $r=3$ and $p=0.5$. The failure count has $E[N]=r(1-p)/p=3(0.5)/0.5=3$, and $\operatorname{Var}(N)=r(1-p)/p^2=3(0.5)/(0.5)^2=1.5/0.25=6$.

For severity, $E[X]=0.5(10)+0.5(30)=5+15=\$20$. Also, $E[X^2]=0.5(10^2)+0.5(30^2)=0.5(100)+0.5(900)=500$, so $\operatorname{Var}(X)=500-20^2=500-400=100$ dollars squared.

Therefore $\operatorname{Var}(S)=3(100)+6(20^2)=300+6(400)=300+2{,}400=2{,}700$ dollars squared.

Check. The frequency moments used are for failures before the third success: mean $3$, variance $6$. Substituting them into the general formula gives $2{,}700$; using a count of trials instead would not describe the stated aggregate model.

Assumptions, Interpretation, and Exam Traps

The formula applies when the claim severities are identically distributed and independent of one another and of $N$, and when the required moments are finite. If severity depends on the count, or claim amounts are dependent, then the conditional mean or variance given $N$ may not be $N E[X]$ or $N\operatorname{Var}(X)$. In that case, the law of total variance still holds, but its conditional terms must be calculated for the model actually specified.

The two terms have distinct roles. $E[N]\operatorname{Var}(X)$ captures the variation in claim amounts across the expected number of claims. $\operatorname{Var}(N)(E[X])^2$ captures count fluctuation, scaled by the square of the average claim amount. If severity is deterministic, $\operatorname{Var}(X)=0$, but aggregate loss can still vary whenever $\operatorname{Var}(N)>0$.

Leaving out the count term. Using only $E[N]\operatorname{Var}(X)$ ignores variability from the number of claims.
Using the wrong severity quantity. The count term uses $(E[X])^2$, not $\operatorname{Var}(X)$.
Assuming every frequency model is Poisson. Only for Poisson frequency are $E[N]$ and $\operatorname{Var}(N)$ both $\lambda$.
Misreading the negative binomial convention. Check whether the random variable counts failures, successes, or trials before selecting its moments.
Dropping units too early. If severities are in dollars, aggregate variance is in dollars squared.

A reliable exam calculation is to write down $E[N]$, $\operatorname{Var}(N)$, $E[X]$, and $\operatorname{Var}(X)$ separately before substitution. This makes the frequency convention visible and reduces the chance of omitting one source of variation. The method and applications here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the frequency moments and severity moments before applying the total-variance formula.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2.5)$. Each claim has mean severity 4 thousand dollars and severity variance 9 thousand-dollars squared. Find the aggregate variance in thousand-dollars squared. (Answer: 62.5 thousand-dollars squared)
A program has $N\sim\operatorname{Binomial}(8,0.25)$. Each claim has mean severity 3 units and variance 4 units squared. Find $\operatorname{Var}(S)$. (Answer: 21.5 units squared)
Let $N$ be the number of failures before two successes, with success probability $p=0.4$. Each claim has mean severity 5 and variance 6. Find the aggregate variance. (Answer: 205.5)
A frequency model has $E[N]=4$ and $\operatorname{Var}(N)=4.5$. Independent claim severities have mean 12 and variance 16. Calculate $\operatorname{Var}(S)$. (Answer: 712)
Six independent exposures each generate a claim with probability $0.2$. Every claim is exactly $50. Calculate the aggregate-loss variance in dollars squared. (Answer: 2,400 dollars squared)

Frequency model	\(E[N]\)	\(\operatorname{Var}(N)\)	Resulting aggregate variance
Poisson with parameter \(\lambda\)	\(\lambda\)	\(\lambda\)	\(\lambda\operatorname{Var}(X)+\lambda(E[X])^2\)
Binomial with parameters \(n,p\), \(q=1-p\)	\(np\)	\(npq\)	\(np\operatorname{Var}(X)+npq(E[X])^2\)
Negative binomial: failures before \(r\) successes, probability \(p\)	\(r(1-p)/p\)	\(r(1-p)/p^2\)	\(\frac{r(1-p)}{p}\operatorname{Var}(X)+\frac{r(1-p)}{p^2}(E[X])^2\)

What You'll Learn