Compound Frequency-Severity Models for Aggregate Loss

Why Combine Frequency and Severity?

An aggregate loss over a period depends on two distinct sources of uncertainty: how many claims occur and how large each claim is. A frequency model describes the claim count; a severity model describes the amount of an individual claim. Combining them gives a distribution for the total loss, which is the quantity needed for many risk-management calculations.

The construction begins by conditioning on the number of claims. If there are no claims, aggregate loss is zero. If there is one claim, aggregate loss has the individual-severity distribution. If there are two claims, aggregate loss has the distribution of the sum of two severities, and so on. This conditioning method works whether the frequency distribution is discrete or the severity distribution is discrete or continuous.

Definition: Let $N$ be the number of claims in a specified period, and let $X_j$ be the severity of claim $j$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where the sum is defined to be zero when $N=0$. A compound frequency-severity model specifies a distribution for $N$ and a severity distribution for the claim amounts, together with assumptions about how the two components are related.

Unless a problem states otherwise, the standard compound model assumes that, conditional on the claim count, claim severities are independent and identically distributed with distribution function $F_X(x)=P(X_j\le x)$, and that the count $N$ is independent of all severities. Independence means that learning the number of claims does not change the distribution of an individual claim. These are modeling assumptions, not consequences of choosing a frequency model and a severity model separately.

Condition on the Claim Count

Write $p_n=P(N=n)$. For $n\ge 1$, let $F_X^{*n}$ denote the distribution function of the sum of $n$ independent severities, each with distribution $F_X$. The notation $*n$ means that the severity distribution is convolved with itself $n$ times. For $n=0$, the sum is zero with probability one.

P(S\le s)=\sum_{n=0}^{\infty}P(N=n)P\left(\sum_{j=1}^{n}X_j\le s\mid N=n\right).

Under the standard independence and identical-severity assumptions, the conditional distribution of the severities given $N=n$ is the same as their unconditional joint distribution. Therefore,

F_S(s)=P(S\le s)=p_0\mathbf{1}_{\{s\ge 0\}}+\sum_{n=1}^{\infty}p_nF_X^{*n}(s).

Here $\mathbf{1}_{\{s\ge 0\}}$ equals 1 when $s\ge 0$ and 0 otherwise. If severities are nonnegative, $S$ is nonnegative as well. In particular, when individual claims are strictly positive, the probability of no aggregate loss is $P(S=0)=P(N=0)=p_0$. If zero severities are possible, positive claim counts can also contribute to $P(S=0)$.

Why the Conditional-Mixture Formula Holds

The possible values of $N$ form disjoint events: exactly zero claims, exactly one claim, and so on. The event $S\le s$ can be split across these events. Adding the probabilities of the disjoint cases gives

P(S\le s)=\sum_{n=0}^{\infty}P(S\le s,\ N=n) =\sum_{n=0}^{\infty}P(S\le s\mid N=n)P(N=n).

For a fixed $n\ge1$, the aggregate is the sum of $n$ severities, so its conditional distribution is the $n$-fold convolution when the severities are independent and identically distributed. For $n=0$, the aggregate is zero by definition. This proves the mixture representation. If severities change with the count, the conditional probability $P(S\le s\mid N=n)$ must instead use the severity model appropriate to that count; the unconditional severity distribution cannot simply be substituted.

Worked Examples

Worked Example: A Binomial Count With Two-Point Severity

Consider a hypothetical one-period portfolio with $N\sim\operatorname{Binomial}(2,0.5)$. Given $N$, the claim severities are independent, each equal to $1,000 or $3,000 with probability 0.5 each. The count and severities are independent. Find $P(S\le \$3{,}000)$ and give the aggregate-loss distribution.

Method. Condition on $N$, calculate each fixed-count sum distribution, and weight it by the probability of that count. This avoids treating an individual claim amount as though it were the total loss.

The count probabilities are $P(N=0)=0.25$, $P(N=1)=2(0.5)(0.5)=0.50$, and $P(N=2)=0.25$. For $N=0$, $S=0$. For $N=1$, $S$ is $1,000 or $3,000, each with conditional probability 0.5. The corresponding unconditional probabilities are $0.50(0.5)=0.25$ for each amount.

For $N=2$, there are four equally likely ordered severity pairs: ($1,000, $1,000), ($1,000, $3,000), ($3,000, $1,000), and ($3,000, $3,000). Their aggregate amounts are $2,000, $4,000, $4,000, and $6,000. Thus the conditional probabilities are 0.25, 0.50, and 0.25 at those amounts. Weighting by $P(N=2)=0.25$ gives unconditional probabilities 0.0625, 0.1250, and 0.0625.

Combine the count cases: $P(S=0)=0.2500$, $P(S=\$1{,}000)=0.2500$, $P(S=\$2{,}000)=0.0625$, $P(S=\$3{,}000)=0.2500$, $P(S=\$4{,}000)=0.1250$, and $P(S=\$6{,}000)=0.0625$. Therefore, $P(S\le \$3{,}000)=0.2500+0.2500+0.0625+0.2500=0.8125$.

Check. The aggregate probabilities sum to $0.2500+0.2500+0.0625+0.2500+0.1250+0.0625=1.0000$. The main mistake avoided is omitting the two-claim case or counting its two different ways to produce $4,000 as only one outcome.

Worked Example: A Poisson Count and a Discrete Severity Model

A hypothetical protection pool has $N\sim\operatorname{Poisson}(1)$ claims in a year. Independently of $N$, severities are independent and equal to $100 with probability 0.75 or $300 with probability 0.25. Find the probability that annual aggregate loss is no more than $200.

Method. Identify which claim counts and severity combinations can satisfy the threshold, then add their probabilities. This avoids including possible outcomes whose minimum loss already exceeds the threshold.

With zero claims, aggregate loss is zero, contributing $P(N=0)=e^{-1}=0.367879\ldots$. With one claim, the loss is no more than $200 only when its severity is $100. This contributes $P(N=1)(0.75)=e^{-1}(0.75)=0.275909\ldots$. With two claims, the loss is no more than $200 only if both severities are $100. This contributes $P(N=2)(0.75)^2=(e^{-1}/2)(0.5625)=0.103466\ldots$.

Three or more claims cannot meet the threshold because every claim is at least $100, so three claims have aggregate loss of at least $300. The total is therefore $P(S\le \$200)=e^{-1}[1+0.75+0.5625/2]$. The bracket equals $2.03125$, giving $P(S\le \$200)=0.7473$ to four decimal places.

Check. The allowed cases are exactly zero claims, one $100 claim, or two $100 claims. The probability is less than 1 because other possible outcomes exceed $200. The common trap is to treat $N=2$ as automatically producing $200; both severities must actually be $100.

Worked Example: When Severity Depends on Claim Count

Consider a hypothetical portfolio where $P(N=0)=0.50$, $P(N=1)=0.30$, and $P(N=2)=0.20$. If $N=1$, the single claim is $100 or $400 with equal probability. If $N=2$, the two severities are conditionally independent, and each is $100 with probability 0.80 or $400 with probability 0.20. Find $P(S\le \$300)$.

Method. Use the conditional-mixture formula with a severity distribution specific to each count. This is necessary because the severity distribution is different when $N=1$ and $N=2$.

For $N=0$, $S=0$, so the contribution is $0.50$. For $N=1$, only the $100 severity satisfies the threshold; the contribution is $0.30(0.50)=0.15$. For $N=2$, the sum is at most $300 only when both claims are $100. A $100 and a $400 claim already total $500. Conditional independence within the two-claim case gives contribution $0.20(0.80)(0.80)=0.128$.

Adding the three disjoint cases gives $P(S\le \$300)=0.50+0.15+0.128=0.7780$.

Check and trap. The conditional probabilities sum to 1: the zero-, one-, and two-claim cases have probabilities 0.50, 0.30, and 0.20. A mistaken approach would first average the low-severity rate across claims and then assume that rate applies independently of $N$. The expected count is $0.30+2(0.20)=0.70$, and the low-severity rate among claims under that averaging is $[0.30(1)(0.50)+0.20(2)(0.80)]/0.70=0.6714$ to four decimals. Substituting this rate as if it were valid in both count cases gives approximately $0.50+0.30(0.6714)+0.20(0.6714)^2=0.7916$, not the correct $0.7780$. The error is the unsupported independence assumption between claim count and severity.

Probability-Generating Functions for Discrete Models

When both claim counts and severities take nonnegative integer values, probability-generating functions provide a compact representation of the compound model. For a nonnegative integer-valued random variable $W$, its probability-generating function is $G_W(z)=E[z^W]$, where $z$ is a formal variable (or a value for which the expectation exists).

Theorem: If $N$ is independent of the independent, identically distributed integer-valued severities $X_1,X_2,\ldots$, then the aggregate-loss probability-generating function is $$ G_S(z)=G_N(G_X(z)). $$ This composition requires the stated independence and integer-valued models. It is not a direct probability-generating-function formula for continuously distributed severities.

To justify the result, first condition on $N=n$. Independence of the severities gives $E[z^S\mid N=n]=E[z^{X_1+\cdots+X_n}]=G_X(z)^n$, because $z^{X_1+\cdots+X_n}=\prod_{j=1}^n z^{X_j}$. Averaging over the count then gives $G_S(z)=\sum_{n=0}^{\infty}P(N=n)G_X(z)^n=G_N(G_X(z))$. For $n=0$, the product is 1, consistent with aggregate loss zero.

For example, if $N\sim\operatorname{Poisson}(\lambda)$ and each severity is one unit with probability one, then $G_X(z)=z$. Consequently $G_S(z)=G_N(z)=\exp(\lambda(z-1))$, so the aggregate loss has the same distribution as the claim count measured in units of loss. More generally, coefficients of $G_S(z)$ give the probabilities of the possible integer aggregate amounts.

Modeling Checks and Common Errors

Before calculating an aggregate distribution, make the modeling choices explicit. A clear specification should say what period the count covers, the units of severity, the frequency distribution and its parameters, the severity distribution and its parameters, and whether counts and severities are independent. If there are different claim types, clarify whether their severities share one distribution or require separate conditional models.

Confusing count and total. $N$ counts claims; $S$ is the sum of their amounts. A frequency probability is not itself an aggregate-loss probability.
Forgetting the zero-claim case. When $N=0$, $S=0$. This usually creates an atom at zero even when individual severities are continuous.
Using a single severity distribution without justification. The convolution formula $F_X^{*n}$ presumes the same severity distribution for all claims in the fixed-count case. If the distribution changes with $N$, condition explicitly as in the third example.
Assuming independence because separate models were supplied. A frequency model and a severity model do not establish independence. If claim severity tends to be larger in periods with more claims, the conditional severity model must reflect that relationship.
Dropping combinations in discrete calculations. Different claim orderings can yield the same aggregate amount. Count every valid combination, or use convolution or a probability-generating function.

The compound model is useful because it separates two questions while preserving their combined effect: how often losses occur and how much each loss costs. Its results are only as appropriate as its assumptions. Independence and identical severity may be reasonable approximations for some portfolios, but should be questioned when common events, changing claim mix, or other mechanisms link claim count and claim size.

Check Your Understanding

For each problem, use the stated count and severity assumptions; probabilities in the answers are rounded to four decimal places.

A hypothetical policy group has $N\sim\operatorname{Bernoulli}(0.20)$. If a claim occurs, its severity is $50 or $150 with equal probability, independently of $N$. Find $P(S\le \$100)$. (Answer: 0.9000)
A hypothetical two-exposure portfolio has $N\sim\operatorname{Binomial}(2,0.40)$, and each claim has fixed severity $100. Find $P(S\le \$100)$, including the zero-claim case. (Answer: 0.8400)
For a hypothetical portfolio, $N\sim\operatorname{Poisson}(0.5)$ and every claim has fixed severity $200. Find the probability aggregate loss is at most $200. (Answer: 0.9098)
A hypothetical pool has $N\sim\operatorname{Poisson}(1)$. Independent severities are $100 with probability 0.80 and $300 with probability 0.20. Find $P(S=\$300)$, accounting for every way the total can equal $300. (Answer: 0.1050)
A hypothetical model has $P(N=0)=0.50$, $P(N=1)=0.30$, and $P(N=2)=0.20$. Given $N=1$, severity is $100 with probability 0.50; given $N=2$, the two independent severities are each $100 with probability 0.80. All other severities are $400. Find $P(S\le \$300)$. (Answer: 0.7780)

What You'll Learn