Frequency-Severity Aggregate Calculation Workshop

From Frequency and Severity to Aggregate Loss

An aggregate loss model combines two sources of uncertainty: how many claims occur and how large each claim is. A calculation can be wrong even when both component models are correct—for example, if a conditional probability is reported as an unconditional one, or if frequency variability is omitted from the aggregate variance. The reliable approach is to identify the count model and severity model, condition on the count when calculating probabilities, and then check moments independently.

Let $N$ be the number of claims, and let $X_1,X_2,\ldots$ be claim severities. In the standard independent and identically distributed (iid) frequency-severity model, $N$ is independent of the severities, and the $X_j$ are iid with finite mean and variance. The aggregate loss is $S=\sum_{j=1}^N X_j$, with $S=0$ when $N=0$. Unless stated otherwise, examples below use this model.

Definition: A frequency model specifies the distribution of $N$, and a severity model specifies the distribution of one claim amount $X$. The aggregate loss $S$ is their random sum. Write $\mu_X=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$. In the iid model with $N$ independent of severities, $E[S]=E[N]\mu_X$ and $\operatorname{Var}(S)=E[N]\sigma_X^2+\operatorname{Var}(N)\mu_X^2$, provided the required moments are finite.

Conditioning on the Claim Count

For an exact aggregate probability, split the calculation according to possible values of $N$. Given $N=n$, the aggregate is the sum of exactly $n$ severities. If the severity distribution is discrete, then

P(S=s)=\sum_{n=0}^{\infty}P(N=n)\, P(X_1+\cdots+X_n=s).

For $n=0$, the sum of severities is defined as zero, so its contribution is $P(N=0)$ when $s=0$, and zero otherwise. For a particular target $s$, many count values may be impossible and can be discarded. If severity is continuous, exact positive aggregate values generally have probability zero; conditioning is still valid, but probabilities of intervals or tail events are then more useful. An atom at zero may remain because no claim can occur.

Worked Example: A Binomial Count and Two Possible Severities

A hypothetical short-term coverage has $N\sim\operatorname{Binomial}(2,0.5)$. Given a claim, its amount $X$ is 2 or 4 dollars with probability $0.5$ each. Claim amounts are independent of each other and of $N$. Find $P(S=6)$, $E[S]$, and $\operatorname{Var}(S)$. Use count conditioning for the probability and check the moments by a second calculation.

Method and assumptions. The count can be 0, 1, or 2. An aggregate of 6 is possible only with two claims: one amount must be 2 and the other 4. The independence assumptions permit multiplication of the count probability and the conditional severity probability.

Probability. The probability of two claims is $P(N=2)=\binom{2}{2}(0.5)^2(0.5)^0=0.25$. Given two claims, the two possible orders for amounts 2 and 4 each have probability $0.5(0.5)=0.25$. Thus $P(S=6\mid N=2)=0.25+0.25=0.50$, and

P(S=6)=P(N=2)P(S=6\mid N=2)=0.25(0.50)=0.1250.

The value 0.25 is the probability of two claims, not the probability of an aggregate of 6. The conditional probability of the required severity pair must also be included.

Mean and variance. A single severity has $E[X]=0.5(2)+0.5(4)=3$ and $E[X^2]=0.5(2^2)+0.5(4^2)=10$. Therefore, $\operatorname{Var}(X)=10-3^2=1$. For the count, $E[N]=2(0.5)=1$ and $\operatorname{Var}(N)=2(0.5)(0.5)=0.5$. The random-sum identities give

E[S]=1(3)=3,\qquad \operatorname{Var}(S)=1(1)+0.5(3^2)=1+4.5=5.5.

Independent check. Enumerate the count cases for the mean. The probability of one claim is 0.5, and its expected aggregate is 3; the probability of two claims is 0.25, and their expected aggregate is 6. Thus $E[S]=0.5(3)+0.25(6)=3$, agreeing with the formula. The variance is 5.5, not 3.25: the frequency-variation term $0.5(3^2)=4.5$ must be included. The calculation also avoids reporting $P(S=6\mid N=2)=0.50$ as the unconditional answer.

Why the Aggregate Moment Formulas Hold

The moment identities follow by first fixing the count. When $N=n$, the aggregate is a sum of $n$ independent severities, so its conditional mean is $n\mu_X$ and its conditional variance is $n\sigma_X^2$. Taking expectations gives the mean. For the variance, use the law of total variance: average the conditional variance, then add the variance of the conditional mean.

E[S]=E[E[S\mid N]]=E[N\mu_X]=E[N]\mu_X,

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

This separates two sources of aggregate variance. The first term is variation in severity among claims, accumulated over the expected number of claims. The second is variation in the count, scaled by the squared mean severity. The formula requires independent claim severities with a common distribution and a count independent of those severities. If severities are dependent, or their distribution depends on the count, the conditional variance and mean must be recalculated for that model rather than using this identity unchanged.

Compound Poisson Calculations and a Cross-Check

For $N\sim\operatorname{Poisson}(\lambda)$, $E[N]=\operatorname{Var}(N)=\lambda$. Consequently, the general moment formula simplifies to $\operatorname{Var}(S)=\lambda(\sigma_X^2+\mu_X^2)=\lambda E[X^2]$. This gives a useful independent check: calculate the variance either by the general frequency-severity identity or directly from the second severity moment. The two results must agree.

Worked Example: An Exact Probability in a Compound Poisson Model

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.5)$. Each claim amount is independently 1 or 2 dollars, each with probability 0.5, and claim amounts are independent of $N$. Find $P(S=2)$, $E[S]$, and $\operatorname{Var}(S)$. Round the probability to four decimal places.

Method. Condition on the claim count. Since each claim is at least 1 dollar, counts greater than 2 cannot produce an aggregate of 2. Count zero also cannot produce 2. There are two possible contributions:

For $N=1$, the severity must be 2. The contribution is $P(N=1)P(X_1=2)=e^{-0.5}(0.5)(0.5)=0.1516327$.
For $N=2$, both severities must be 1. The contribution is $P(N=2)P(X_1=1,X_2=1)=e^{-0.5}(0.5^2/2)(0.5^2)=0.0189541$.

Adding the disjoint cases gives $P(S=2)=0.1516327+0.0189541=0.1705868$, or 0.1706 to four decimal places. In particular, the one-claim contribution is about 0.1516 and the two-claim contribution is about 0.01895; the Poisson probability for the count and the conditional severity probability are both required.

Moments. The severity moments are $E[X]=0.5(1)+0.5(2)=1.5$, $E[X^2]=0.5(1^2)+0.5(2^2)=2.5$, and $\operatorname{Var}(X)=2.5-1.5^2=0.25$. The aggregate mean is $0.5(1.5)=0.75$. The general variance formula gives

\operatorname{Var}(S)=0.5(0.25)+0.5(1.5^2) =0.125+1.125=1.25.

Independent check. The compound Poisson shortcut gives $\operatorname{Var}(S)=\lambda E[X^2]=0.5(2.5)=1.25$, matching the first calculation. A common mistake is to use only $\lambda\operatorname{Var}(X)=0.125$, which ignores variability in the number of claims. The omitted count-variation term is $1.125$.

Frequency Models Beyond Poisson

The conditioning method applies to any frequency distribution, not just binomial or Poisson counts. For example, let $N$ be the number of failures before the first success in independent Bernoulli trials with success probability $p$. Then $P(N=n)=(1-p)^n p$ for $n=0,1,\ldots$, $E[N]=(1-p)/p$, and $\operatorname{Var}(N)=(1-p)/p^2$. This convention starts the count at zero; it is not the distribution of the number of trials required for the first success.

Worked Example: A Geometric Count with Fixed Claim Amounts

A hypothetical contract has a claim count $N$ equal to the number of failures before the first success, with success probability $p=0.5$. Every claim has a fixed amount of 2 dollars, independently of the count. Find $P(S\leq2)$, $E[S]$, and $\operatorname{Var}(S)$.

Method: translate the aggregate event into a count event. Since each claim is exactly 2 dollars, $S=2N$. Therefore $S\leq2$ is equivalent to $N\leq1$, not just $N=0$. The relevant count probabilities are $P(N=0)=0.5$ and $P(N=1)=(0.5)(0.5)=0.25$. Hence

P(S\leq2)=P(N=0)+P(N=1)=0.5+0.25=0.7500.

The count has mean $(1-0.5)/0.5=1$ and variance $(1-0.5)/(0.5)^2=2$. With fixed severity $X=2$, $E[X]=2$ and $\operatorname{Var}(X)=0$. Thus

E[S]=1(2)=2,\qquad \operatorname{Var}(S)=1(0)+2(2^2)=8.

Check. Because $S=2N$, scaling a random variable by 2 multiplies its mean by 2 and its variance by $2^2=4$. The results are therefore $E[S]=2E[N]=2$ and $\operatorname{Var}(S)=4\operatorname{Var}(N)=8$. The event $S\leq2$ includes both zero claims and one claim; using only $P(N=0)$ would give an incorrect probability of 0.5000.

A Reliable Calculation Sequence

For an integrated frequency-severity problem, a short sequence of checks helps keep the probability and moment calculations separate. State the frequency model’s parameterization, since names such as “geometric” and “negative binomial” can use different count conventions. Then state the severity support and independence assumptions before applying a formula.

Define the random sum.
Write $S=\sum_{j=1}^N X_j$, including the convention $S=0$ when $N=0$.

Find possible counts for the event.
Use the severity support to identify which values of $N$ can produce the target aggregate or interval.

Calculate conditional probabilities.
For each possible count, find the conditional probability of the target severity sum, then multiply by $P(N=n)$.

Calculate moments separately.
Find $E[X]$, $\operatorname{Var}(X)$, $E[N]$, and $\operatorname{Var}(N)$, then apply the random-sum identities.

Cross-check.
Use a direct count enumeration for a small model, or a second valid moment identity such as the compound Poisson variance formula.

The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In practice, keeping the probability calculation and the moment calculation as distinct steps is especially useful: exact probabilities require the possible severity combinations, while the moment identities require only the first two moments of frequency and severity.

Check Your Understanding

For each problem, assume the count and severities are independent and the severities are iid unless the statement specifies fixed amounts.

$N\sim\operatorname{Binomial}(1,0.4)$, and every claim amount is 5 dollars. Find $P(S=0)$, $E[S]$, and $\operatorname{Var}(S)$. (Answer: $0.6000,\ \$2,\ \$6$.)
$N\sim\operatorname{Binomial}(2,0.5)$. Each severity is 1 or 3 dollars with equal probability. Find $P(S=4)$, $E[S]$, and $\operatorname{Var}(S)$. (Answer: $0.1250,\ \$2,\ \$3$.)
$N\sim\operatorname{Poisson}(2)$, and each claim amount is fixed at 3 dollars. Find $P(S=0)$, $E[S]$, and $\operatorname{Var}(S)$. (Answer: $0.1353,\ \$6,\ \$18$.)
$N$ is the number of failures before the first success, with success probability $0.25$, and every claim amount is fixed at 4 dollars. Find $P(S\leq4)$, $E[S]$, and $\operatorname{Var}(S)$. (Answer: $0.4375,\ \$12,\ \$192$.)
$N\sim\operatorname{Poisson}(1)$, and each claim amount is 0 or 2 dollars with equal probability. Find $P(S=0)$, $E[S]$, and $\operatorname{Var}(S)$. (Answer: $0.6065,\ \$1,\ \$2$.)

What You'll Learn