Deriving Aggregate Loss Moments for Pricing

Why Aggregate Loss Moments Matter in Pricing

A prospective premium must account for the total claims generated over the period being priced, not just the cost of one claim. A frequency-severity model separates the number of claims from the size of each claim, then combines them into an aggregate loss. Its mean gives the expected claim cost; its variance describes process variability around that expectation. These are different quantities and support different pricing principles.

This tutorial derives the moments for a compound aggregate loss and applies them to hypothetical pricing situations. It focuses on process variation under a specified model; it does not include parameter uncertainty, expenses, or profit provisions. 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 exposure period. Let $X_1,X_2,\ldots$ be the claim severities, assumed independent and identically distributed, with each severity independent of $N$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where the sum is defined to be zero when $N=0$. Assume $E[N]$, $\operatorname{Var}(N)$, $E[X]$, and $\operatorname{Var}(X)$ are finite, where $X$ has the common severity distribution.

Here, $X$ is the amount paid for one claim, and $S$ is the total paid for all claims in the period. The independence assumptions matter: they rule out, for example, severity distributions that change systematically with the number of claims. If claims are dependent or the severity distribution depends on $N$, the formulas below need not apply.

Deriving the Aggregate Mean and Variance

Conditional on observing $N=n$ claims, the aggregate is the sum of $n$ independent severities. Its conditional mean is $nE[X]$, and its conditional variance is $n\operatorname{Var}(X)$. Applying the laws of total expectation and total variance gives the general compound-model formulas.

Condition on the claim count.
For $N=n$, add the means and variances of the $n$ independent claim severities.

Average the conditional mean.
Take the expectation of $E[S\mid N]=NE[X]$ over the distribution of $N$.

Decompose the aggregate variance.
Add the expected conditional variance to the variance of the conditional mean.

$$ E[S]=E[N]E[X] $$

\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)\{E[X]\}^2. \end{aligned}

The first variance term is the contribution from severity variation, averaged over the expected number of claims. The second is the contribution from the randomness in the claim count. The square on $E[X]$ is essential: variation in the claim count changes the total by approximately one average claim amount for each extra or missing claim.

Justification: Given $N=n$, independence of the severities implies $E[S\mid N=n]=nE[X]$ and $\operatorname{Var}(S\mid N=n)=n\operatorname{Var}(X)$. Therefore $E[S]=E[E[S\mid N]]=E[N]E[X]$. By the law of total variance, $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. Substituting the conditional expressions yields $E[N]\operatorname{Var}(X)+\operatorname{Var}(N)\{E[X]\}^2$.

An equivalent expression is useful when the severity second raw moment is supplied. Since $E[X^2]=\operatorname{Var}(X)+\{E[X]\}^2$, rearrangement gives:

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

For a Poisson claim count with parameter $\lambda$, $E[N]=\operatorname{Var}(N)=\lambda$. The second term in this last expression is then zero, so:

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

This Poisson simplification relies on the Poisson equality between the mean and variance of the count. It should not be used automatically for a binomial, fixed, or other non-Poisson frequency model.

Worked Applications

Worked Example: Poisson Frequency With Exponential Severity

A hypothetical annual policy portfolio has a Poisson claim count with parameter $\lambda=4$. Each claim amount $X$ is exponentially distributed with mean $500. A pricing analyst uses the variance principle $P=E[S]+\alpha\operatorname{Var}(S)$, with $\alpha=0.0001$ per dollar. Calculate the aggregate mean, aggregate variance, expected-cost premium, and variance-loaded premium.

Method and assumptions. The claim count is Poisson, the claim amounts are independent and identically distributed exponential variables, and severity is independent of claim count. The specified exponential mean is $500, so its variance is $(\$500)^2=\$250{,}000$. The period is one year.

For the exponential severity, the second raw moment is:

E[X^2]=\operatorname{Var}(X)+\{E[X]\}^2 =\$250{,}000+(\$500)^2 =\$500{,}000.

The compound Poisson formulas give:

E[S]=4(\$500)=\$2{,}000, \qquad \operatorname{Var}(S)=4(\$500{,}000)=\$2{,}000{,}000.

The expected-cost premium is the aggregate expected loss, $2,000. Under the stated variance principle, the load is:

\alpha\operatorname{Var}(S) =0.0001\text{ per dollar}\times \$2{,}000{,}000 =\$200.

Thus the variance-loaded premium is $\$2{,}000+\$200=\$2{,}200$. The units are consistent: a coefficient measured per dollar times a variance measured in dollars squared produces a dollar amount.

Check. The expected number of claims is 4 and the mean claim amount is $500, so the expected aggregate loss is $2,000. The loaded premium exceeds expected cost by the calculated $200. The method avoids using the severity variance alone in the compound Poisson formula; the correct Poisson aggregate variance uses $E[X^2]$.

Worked Example: Binomial Frequency and Two-Point Severity

A hypothetical group of 10 units each has probability 0.20 of producing one claim during a year, independently of the other units. Thus $N\sim\operatorname{Binomial}(10,0.20)$. Each claim amount is independently $800 or $1,200, each with probability 0.50, and severities are independent of $N$. Find the aggregate mean and variance, identifying the contribution of each source of variability.

Method and assumptions. The count is binomial, so $E[N]=np$ and $\operatorname{Var}(N)=np(1-p)$. Claim severities are independent and identically distributed with the stated two-point distribution.

First calculate the frequency moments:

E[N]=10(0.20)=2, \qquad \operatorname{Var}(N)=10(0.20)(0.80)=1.6.

The severity mean and variance are:

E[X]=0.50(\$800)+0.50(\$1{,}200)=\$1{,}000,

\operatorname{Var}(X) =0.50(\$800-\$1{,}000)^2 +0.50(\$1{,}200-\$1{,}000)^2 =\$40{,}000.

The expected aggregate loss is:

E[S]=2(\$1{,}000)=\$2{,}000.

The severity-variation contribution is $E[N]\operatorname{Var}(X)=2(\$40{,}000)=\$80{,}000$. The frequency-variation contribution is $\operatorname{Var}(N)\{E[X]\}^2=1.6(\$1{,}000)^2=\$1{,}600{,}000$. Therefore:

\operatorname{Var}(S)=\$80{,}000+\$1{,}600{,}000 =\$1{,}680{,}000.

Check. The mean is 2 claims times $1,000 per claim, or $2,000. Most of the variance comes from uncertainty in the count, not from the relatively narrow two-point severity variation. The method avoids treating a binomial count as Poisson: its variance is 1.6, not 2.

Worked Example: Comparing Two Equal-Mean Portfolios

Two hypothetical annual portfolios each have expected aggregate losses of $2,000. Portfolio A has $N\sim\operatorname{Poisson}(2)$, with every claim amount fixed at $1,000. Portfolio B always has exactly two claims, with each claim amount independently equal to $800 or $1,200 with probability 0.50 each. Using $P=E[S]+\alpha\operatorname{Var}(S)$ with $\alpha=0.0001$ per dollar, calculate each premium and explain why they differ.

Method and assumptions. Within each portfolio, claim severities are independent and identically distributed and independent of frequency. The stated distributions apply to one year. Apply the general compound variance formula separately; equal expected losses do not imply equal aggregate variances.

For Portfolio A, $E[N]=\operatorname{Var}(N)=2$, $E[X]=\$1{,}000$, and $\operatorname{Var}(X)=0$. Thus:

E[S_A]=2(\$1{,}000)=\$2{,}000, \qquad \operatorname{Var}(S_A)=2(0)+2(\$1{,}000)^2 =\$2{,}000{,}000.

Its variance load is $0.0001\times \$2{,}000{,}000=\$200$, so its premium is $2,200.

For Portfolio B, $E[N]=2$ and $\operatorname{Var}(N)=0$, because the count is fixed. The two-point severity has mean $1,000 and variance $40,000, as calculated from deviations of $200 from the mean. Therefore:

E[S_B]=2(\$1{,}000)=\$2{,}000, \qquad \operatorname{Var}(S_B)=2(\$40{,}000)+0(\$1{,}000)^2 =\$80{,}000.

Its variance load is $0.0001\times \$80{,}000=\$8$, so its premium is $2,008.

Check. Both portfolios have expected cost $2,000, but Portfolio A has a random claim count and Portfolio B does not. The much larger frequency variation in A produces the larger aggregate variance and, under the specified pricing rule, the larger premium. The method avoids comparing portfolios on expected cost alone when the stated pricing principle also responds to process variance.

Using a Variance Load Carefully

The expected-cost principle sets the premium equal to $E[S]$, before any expenses, profit provisions, or other adjustments. A variance principle adds a nonnegative amount proportional to aggregate variance:

Variance principle: For aggregate loss $S$, define $P=E[S]+\alpha\operatorname{Var}(S)$, where $\alpha\geq 0$ is a specified loading coefficient. If $S$ is measured in dollars, $\alpha$ must have units of inverse dollars so that the loading is measured in dollars. The principle is a pricing rule, not a claim that this premium is uniquely correct for every purpose.

A larger variance increases the indicated premium when $\alpha>0$, even if expected losses are unchanged. The coefficient must be supplied or otherwise justified; it cannot be inferred from the moment formulas alone. Because the load depends on variance in squared monetary units, the coefficient's units and the loss measurement scale must remain consistent. Changing from dollars to thousands of dollars requires converting the coefficient as well.

This calculation measures variability under the assumed frequency and severity distributions. It does not automatically account for uncertainty in estimated parameters, changing conditions, dependence among claims, or a particular capital target. If the model assumptions fail, a precise calculation can still give a misleading pricing indication.

Common Mistake: Omitting Frequency Variation

For the Poisson model in the first example, a tempting but incorrect calculation is $\operatorname{Var}(S)=\lambda\operatorname{Var}(X)$. It would give $4(\$250{,}000)=\$1{,}000{,}000$, only half the correct variance. The correction is to use $\lambda E[X^2]$:

\lambda E[X^2] =4\bigl(\operatorname{Var}(X)+\{E[X]\}^2\bigr) =4(\$250{,}000+\$250{,}000) =\$2{,}000{,}000.

The omitted amount, $\lambda\{E[X]\}^2$, reflects randomness in the number of claims. More generally, a second common error is to use $\operatorname{Var}(N)\operatorname{Var}(X)$ for the frequency term. The correct term is $\operatorname{Var}(N)\{E[X]\}^2$: count fluctuations change the number of claims, each with an average size, rather than multiplying the two variances.

Key takeaway: Under independent, identically distributed severities independent of frequency, aggregate expected loss is $E[N]E[X]$, and aggregate variance is $E[N]\operatorname{Var}(X)+\operatorname{Var}(N)\{E[X]\}^2$. Match the frequency moments to the actual count model before making a pricing comparison.

Check Your Understanding

For each hypothetical annual model, calculate the requested aggregate moments or premium. Use dollars for monetary quantities.

$N\sim\operatorname{Poisson}(3)$, and every claim costs $400. Find $E[S]$ and $\operatorname{Var}(S)$. (Answer: $E[S]=\$1{,}200$, $\operatorname{Var}(S)=\$480{,}000$.)
$N\sim\operatorname{Binomial}(8,0.25)$, and every claim costs $600. Find $E[S]$ and $\operatorname{Var}(S)$. (Answer: $E[S]=\$1{,}200$, $\operatorname{Var}(S)=\$540{,}000$.)
A frequency model has $E[N]=5$ and $\operatorname{Var}(N)=4$. Independent severities have mean $200 and variance $\$900$. Find the aggregate mean and variance. (Answer: $E[S]=\$1{,}000$, $\operatorname{Var}(S)=\$164{,}500$.)
$N\sim\operatorname{Poisson}(1.5)$. Independent severities have mean $2,000 and variance $\$1{,}000{,}000$. Find the aggregate mean and variance. (Answer: $E[S]=\$3{,}000$, $\operatorname{Var}(S)=\$7{,}500{,}000$.)
Portfolio A has $E[N]=\operatorname{Var}(N)=3$, and severity mean $500 and variance $\$25{,}000$. Portfolio B has exactly three claims and the same severity distribution. Find each variance-principle premium for $\alpha=0.0002$ per dollar. (Answer: Portfolio A, $1,665; Portfolio B, $1,515.)

What You'll Learn