Frequency-Severity Models for Annual Aggregate Loss

Introduction

A catastrophe model may estimate the insured loss from each possible event, but an insurer’s annual loss depends on two things: how many events occur during the year and how much each event costs. A frequency-severity model combines these components to describe the total loss over a year. This is the annual aggregate loss, not the loss from a single occurrence.

The distinction matters when interpreting model output. An event-level loss distribution describes losses associated with one event. An annual aggregate distribution also allows for years with no events and years with multiple events. The frequency-severity framework is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Define the Annual Aggregate Loss

Definition: Let $N$ be the number of covered events in a year. Let $X_j$ be the covered loss from the $j$th event, for $j=1,2,\ldots,N$. The annual aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ with $S=0$ when $N=0$. The distribution of $N$ is the frequency model; the distribution of an event-level loss $X_j$ is the severity model.

The event loss $X_j$ should represent a consistent quantity. For example, it may be the insurer’s covered loss for one occurrence after applicable occurrence-level terms, rather than the underlying physical damage. Whether losses are gross or net of reinsurance must also be specified. An annual aggregate term, if relevant, is applied to the annual total; it does not change the definition of the event-level input.

A common baseline model assumes that, conditional on $N$, the event losses $X_1,\ldots,X_N$ are independent and identically distributed, and that their common distribution is independent of $N$. Write $E[X]$ and $\operatorname{Var}(X)$ for the mean and variance of one event loss, and $E[N]$ and $\operatorname{Var}(N)$ for the frequency moments. Under these assumptions,

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

These are moment formulas, not a complete description of the distribution of $S$. To obtain a full annual aggregate distribution, the frequency and severity distributions must be combined, as in the first worked example.

Why the Aggregate Moment Formulas Hold

The key is to condition on the event count. If $N=n$, the annual loss is the sum of $n$ independent event losses. Its conditional mean is $nE[X]$, and its conditional variance is $n\operatorname{Var}(X)$. This statement also gives zero conditional mean and variance when $n=0$.

Taking the expected value of the conditional mean gives

E[S]=E\bigl[E[S\mid N]\bigr] =E\bigl[NE[X]\bigr] =E[N]E[X].

For the variance, apply the law of total variance: the unconditional variance is the expected conditional variance plus the variance of the conditional mean. Therefore,

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

The first variance term reflects variation in event losses, given that events occur. The second reflects variation in the number of events. Both contribute to annual volatility. These formulas do not generally apply if event severities depend on the count, are not identically distributed, or have dependence between events. In those cases, the conditional mean and variance of the aggregate must be specified using the relevant model rather than these baseline expressions.

Building a Discrete Annual Loss Distribution

Worked Example: Combining Discrete Frequency and Severity

A hypothetical portfolio has annual event count $N$ with probabilities $P(N=0)=0.50$, $P(N=1)=0.40$, and $P(N=2)=0.10$. Assume event counts are independent of event severities, and event severities are independent and identically distributed: each event produces a covered loss of $2,000 or $8,000, each with probability 0.50. Construct the annual aggregate loss distribution and check its mean.

Method. For each possible count, enumerate the sums of the event losses. When $N=0$, the annual total is $0 with probability 0.50. When $N=1$, the annual total is $2,000 or $8,000, each with unconditional probability $0.40(0.50)=0.20$.

When $N=2$, there are four equally likely ordered pairs of event losses: $(2{,}000,2{,}000)$, $(2{,}000,8{,}000)$, $(8{,}000,2{,}000)$, and $(8{,}000,8{,}000)$. Each pair has conditional probability 0.25 given $N=2$. Thus the annual totals are $4,000 with conditional probability 0.25, $10,000 with conditional probability 0.50, and $16,000 with conditional probability 0.25. Multiplying by $P(N=2)=0.10$ gives unconditional probabilities 0.025, 0.050, and 0.025.

Annual aggregate loss $S$	Probability
$0	0.500
$2,000	0.200
$4,000	0.025
$8,000	0.200
$10,000	0.050
$16,000	0.025

The probabilities sum to $0.500+0.200+0.025+0.200+0.050+0.025=1.000$. The expected annual loss from the distribution is $$ E[S]=2{,}000(0.200)+4{,}000(0.025)+8{,}000(0.200)+10{,}000(0.050)+16{,}000(0.025) =$3{,}000. $$

As a check, $E[N]=0(0.50)+1(0.40)+2(0.10)=0.60$, while $E[X]=0.50(2{,}000)+0.50(8{,}000)=$5{,}000$. The moment formula gives $E[S]=0.60($5{,}000)=$3{,}000$, agreeing with the distribution. A single event cannot produce a $0 loss under this severity model, yet the annual total is $0 with probability 0.50 because no event occurs. That is one important difference between severity and annual aggregate loss.

Using Frequency and Severity Moments

Often the requested quantities are the expected annual loss and its variance rather than every value of the aggregate distribution. The following examples show how the frequency distribution affects those moments.

Worked Example: Binomial Frequency and Two-Point Severity

For a hypothetical portfolio, the number of events in a year follows a binomial distribution with two trials and probability 0.40 per trial. Thus $P(N=0)=0.36$, $P(N=1)=0.48$, and $P(N=2)=0.16$. Assume the count is independent of the event severities. Each event loss is $20,000 or $60,000 with probability 0.50 each. Calculate the annual mean, variance, and standard deviation.

Frequency moments. The mean count is $E[N]=2(0.40)=0.80$, and the variance is $\operatorname{Var}(N)=2(0.40)(0.60)=0.48$.

Severity moments. The event mean is $E[X]=0.50($20{,}000)+0.50($60{,}000)=$40{,}000$. Each value is $20,000 from the mean, so $$ \operatorname{Var}(X) =0.50(20{,}000-40{,}000)^2+0.50(60{,}000-40{,}000)^2 =0.50(400{,}000{,}000)+0.50(400{,}000{,}000) =$400{,}000{,}000. $$

The expected annual loss is $E[S]=0.80($40{,}000)=$32{,}000$. The annual variance is $$ \begin{aligned} \operatorname{Var}(S) &=0.80($400{,}000{,}000)+0.48($40{,}000)^2\\ &=$320{,}000{,}000+$768{,}000{,}000\\ &=$1{,}088{,}000{,}000. \end{aligned} $$

The standard deviation is $\sqrt{$1{,}088{,}000{,}000}=$32{,}984.845\ldots$, or $32,984.85 rounded to the nearest cent. The calculation uses the event-loss variance in the first term and the squared event mean in the second term.

Common mistake and correction. A tempting error is to calculate the severity variance as $800,000,000 by adding the two squared deviations without correctly weighting them. Each squared deviation is $400,000,000, and each has probability 0.50; their weighted average is $400,000,000, not $800,000,000. Using the incorrect severity variance would give $0.80($800{,}000{,}000)+$768{,}000{,}000=$1{,}408{,}000{,}000$. The correct aggregate variance is $1,088,000,000.

Worked Example: Same Expected Count, Different Frequency Variance

A second hypothetical portfolio has $N$ following a binomial distribution with four trials and probability 0.25 per trial. Its event count is independent of event severity. Each event loss is $4,000 or $16,000 with probability 0.50 each. Calculate the annual mean and variance, and identify how the frequency model contributes to the result.

The frequency mean and variance are $E[N]=4(0.25)=1.00$ and $\operatorname{Var}(N)=4(0.25)(0.75)=0.75$. For the event loss, $E[X]=0.50($4{,}000)+0.50($16{,}000)=$10{,}000$. The deviations from this mean are both $6,000 in absolute value, so $\operatorname{Var}(X)=0.50($36{,}000{,}000)+0.50($36{,}000{,}000)=$36{,}000{,}000$.

Therefore, $$ E[S]=1.00($10{,}000)=$10{,}000, $$ and $$ \begin{aligned} \operatorname{Var}(S) &=1.00($36{,}000{,}000)+0.75($10{,}000)^2\\ &=$36{,}000{,}000+$75{,}000{,}000\\ &=$111{,}000{,}000. \end{aligned} $$

The first variance term is the contribution from event severity. The second is the contribution from variation in the count. A frequency model with the same expected count but a different variance would leave the mean unchanged while changing this second term. Thus the expected number of events alone does not determine annual aggregate volatility.

Assumptions, Interpretation, and Exam Traps

Before applying a frequency-severity formula, state what counts as an event, what loss is assigned to each event, and what period the count covers. If the event loss is already net of an occurrence deductible and limit, the aggregate is a sum of those net event losses under the stated model. If an annual aggregate cap or other annual term is also being considered, first form the annual total and then apply that annual term to the total.

The independence assumption is substantive. For example, a severe event could increase the chance of further events, or event losses could share common conditions that make them dependent. In such a model, the conditional distribution of event losses given $N$ may differ from the unconditional severity distribution. The standard formula should not be used without checking that its assumptions fit the stated model.

Do not substitute a single-event quantity for an annual quantity. $E[X]$ is the mean loss per event; $E[S]$ is the mean total loss over a year.
Do not omit zero-event years. The aggregate is zero when the count is zero, even if every event severity is strictly positive.
Do not use only the expected count to calculate variance. The frequency variance also enters, multiplied by the squared event mean.
Do not forget the square. The second variance term is $\operatorname{Var}(N)(E[X])^2$, not $\operatorname{Var}(N)E[X]$.
Keep the model’s basis consistent. Distinguish gross from net loss and event-level terms from annual aggregate terms.

Key takeaway: An annual aggregate model combines a count of events with the losses from those events. Under independent, identically distributed event severities that are independent of the count, the annual mean is $E[N]E[X]$, and the annual variance has separate severity and frequency contributions.

Check Your Understanding

For each problem, treat all event losses as independent and identically distributed and independent of the count unless the problem says otherwise.

A hypothetical portfolio has $P(N=0)=0.70$, $P(N=1)=0.20$, and $P(N=2)=0.10$. Each event loss is exactly $5,000. Find $E[N]$, $\operatorname{Var}(N)$, and the annual aggregate mean and variance. (Answer: $E[N]=0.40$; $\operatorname{Var}(N)=0.44$; $E[S]=$2,000$; $\operatorname{Var}(S)=$11,000,000$.)
A hypothetical annual count has mean 2 and variance 1. Event losses have mean $3,000 and variance $4,000,000. Find the annual aggregate mean and variance. (Answer: $E[S]=$6,000$; $\operatorname{Var}(S)=$17,000,000$.)
The annual count is Poisson with mean 0.40. Each event loss is $10,000 or $30,000 with probability 0.50 each. Find the annual mean and variance. (Answer: $E[S]=$8,000$; $\operatorname{Var}(S)=$200,000,000$.)
The annual count is binomial with four trials and probability 0.25 per trial. Each event loss is $2,000 or $6,000 with probability 0.50 each. Find the annual mean and variance. (Answer: $E[S]=$4,000$; $\operatorname{Var}(S)=$16,000,000$.)
In a hypothetical model, $P(N=0)=0.50$, $P(N=1)=0.30$, and $P(N=2)=0.20$. Conditional on $N=1$, the sole event loss is $10,000; conditional on $N=2$, each of the two event losses is $20,000. Find the annual mean and variance directly, and state whether the independent, identically distributed severity formula applies. (Answer: $E[S]=$11,000$; $\operatorname{Var}(S)=$229,000,000$; the formula does not apply because the event-loss distribution depends on the count.)

What You'll Learn