Simulating Compound Frequency-Severity Losses

Why Simulate Aggregate Losses?

A frequency model describes how many claims occur, while a severity model describes the size of each claim. For many actuarial questions, the quantity of interest is the aggregate loss: the total amount paid across all claims in a period. A simulation of aggregate loss must therefore combine the two models in the correct order: generate a claim count, generate that many severities, and add them.

The zero-claim outcome is part of the model, not a special result to discard. If no claims occur, the aggregate loss is zero. A simulation that omits those trials changes the distribution being simulated. The procedure in this tutorial applies to a general discrete claim-count model; the count distribution need not be a particular named distribution.

Definition: Let $N$ be a nonnegative integer-valued claim count. Let $X_1,X_2,\ldots$ be claim severities, each with distribution $F_X$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where an empty sum is defined to be zero. Thus $S=0$ whenever $N=0$.

For the standard independent frequency-severity model, assume $N$ is independent of the severities, and that the severities $X_1,X_2,\ldots$ are independent and identically distributed. Each simulated aggregate loss uses one draw of $N$, followed by exactly $N$ severity draws. In applications, severities are usually nonnegative. If the model includes a payment per claim rather than a ground-up loss, $X_j$ should represent that payment amount consistently.

The Simulation Procedure

To simulate from a discrete frequency model, first use its cumulative probabilities to turn a uniform random number into a count. For a count distribution with probabilities $p_n=P(N=n)$, use the inverse cumulative rule: choose the smallest $n$ for which $\sum_{k=0}^{n}p_k$ is at least the uniform value. Then generate the corresponding number of severities. The earlier material on generating variates from uniforms gives the transformations for particular distributions.

A new, independent random input is required for each simulated quantity. In particular, the count draw and all severity draws in a trial must be independent under the model. For a simulation of several aggregate-loss trials, use a separate count draw for each trial. A trial with zero claims uses no severity draws; the next trial still needs its own count draw.

Generate the frequency.
Draw one uniform value and use the frequency distribution’s cumulative probabilities to obtain $N$.

Check for zero claims.
If $N=0$, record aggregate loss $S=0$ and proceed to the next trial.

Generate severities.
If $N>0$, independently generate $X_1,\ldots,X_N$ from the severity model.

Aggregate.
Add the $N$ severity values and record their sum as the trial’s aggregate loss.

Why This Procedure Produces the Compound Distribution

The aggregate loss has a different conditional distribution for each possible claim count. Conditional on $N=n$, it is the sum of $n$ independent severities. For $n=0$, that conditional distribution puts all probability at zero. The unconditional distribution combines these conditional cases, weighting each by the probability of its count.

Justification: For any set of possible aggregate values $A$, split the event $S\in A$ according to the mutually exclusive values of $N$. Independence ensures that, conditional on $N=n$, the severity draws still follow their specified severity model. Therefore $$ P(S\in A)=\sum_{n=0}^{\infty}P(N=n)\, P\left(\sum_{j=1}^{n}X_j\in A\right), $$ where the sum inside the probability is zero when $n=0$. Generating $N$ first selects a term of this mixture with probability $P(N=n)$; generating and summing $n$ severities then produces the corresponding conditional outcome. Repeating this process therefore samples from the aggregate-loss distribution, including its zero-claim probability.

This argument also explains why independently generating a fixed number of severities and then choosing a count afterward is not an equivalent procedure unless the simulation is carefully arranged to preserve the same joint model. The direct algorithm follows the model definition and makes the number of required severity values explicit.

Worked Example: Three Trials with a Discrete Severity Model

A hypothetical portfolio has claim count probabilities $P(N=0)=0.25$, $P(N=1)=0.45$, and $P(N=2)=0.30$. A claim severity is $100 with probability 0.50, $250 with probability 0.30, and $600 with probability 0.20. Use the supplied independent uniform values to simulate three aggregate-loss trials. For the count, use $U_N=0.14,\ 0.62,\ 0.91$, respectively. Use severity uniforms only when claims occur: $0.73$ for Trial 2 and $0.44,\ 0.88$ for Trial 3.

Method. Apply inverse cumulative probabilities separately to the count and severity draws, then sum the severities within each trial. The count cumulative probabilities are $0.25,\ 0.70,\ 1.00$ for counts 0, 1, and 2. The severity cumulative probabilities are $0.50,\ 0.80,\ 1.00$ for $100, $250, and $600.

Trial 1. Since $0.14\leq0.25$, the simulated count is $N=0$. There are no severity draws, so $S=0$.

Trial 2. Since $0.25<0.62\leq0.70$, the simulated count is $N=1$. The severity uniform value $0.73$ lies above $0.50$ and at or below $0.80$, so the one severity is $250. Thus $S=250$.

Trial 3. Since $0.70<0.91\leq1.00$, the simulated count is $N=2$. The first severity uniform value, $0.44$, selects $100 because it is at or below $0.50$. The second value, $0.88$, selects $600 because it is above $0.80$. Therefore $S=100+600=700$.

The three simulated aggregate losses are $0, $250, and $700. The total across these three trials is $0+250+700=$$950. This check confirms that each nonzero total equals the sum of exactly the number of severities specified by its simulated count.

Common mistake avoided: Do not generate a severity for Trial 1 simply to keep a fixed number of random draws per trial. A zero count requires an empty severity sum, not a severity that is ignored after generation.

Using a Continuous Severity Transformation

The same count-then-sum structure applies when severity is continuous. For example, if $X$ has an exponential distribution with mean $\theta>0$, its cumulative distribution function is $F_X(x)=1-e^{-x/\theta}$ for $x\geq0$. Setting $F_X(x)=U$ and solving for $x$ gives $X=-\theta\ln(1-U)$. Each independent uniform value produces one severity; the generated amounts are then added.

Worked Example: Two Exponential Severities

For a hypothetical coverage, the claim count has probabilities $P(N=0)=0.10$, $P(N=1)=0.40$, and $P(N=2)=0.50$. Each claim severity is exponentially distributed with mean $\theta=$$80, independently of the count and other severities. Use count uniform $U_N=0.86$, then severity uniform values $0.25$ and $0.75$. Calculate the simulated aggregate loss to the nearest cent.

Method. The count cumulative probabilities are $0.10,\ 0.50,\ 1.00$. Since $0.50<0.86\leq1.00$, the count is $N=2$, so two independent exponential severities are required.

For the first severity, $$ X_1=-80\ln(1-0.25)=-80\ln(0.75) =80(0.2876821\ldots)=23.0146\ldots. $$ Thus $X_1$ is $23.01 to the nearest cent.

For the second severity, $$ X_2=-80\ln(1-0.75)=-80\ln(0.25) =80(1.3862944\ldots)=110.9036\ldots. $$ Thus $X_2$ is $110.90 to the nearest cent. Add the unrounded amounts before rounding the aggregate: $$ S=23.0146\ldots+110.9036\ldots=133.9181\ldots. $$ The simulated aggregate loss is $133.92.

Check. The simulation used two severity draws because the count was two. Both severities are nonnegative, and their sum rounds to $133.92. Rounding each severity for display did not affect the aggregate calculation.

Common mistake avoided: The inverse transformation is $-\theta\ln(1-U)$, not $-\theta\ln(U)$ under this uniform convention. For $U=0.25$, the correct severity is based on $\ln(0.75)$.

Careful Handling of a Random-Number Stream

In a practical simulation, random values may be supplied as a stream rather than as a separately labelled list for each claim. The algorithm must still preserve the roles of the values: take one value for the count, then take one value per claim for severity. When the count is zero, take no severity values for that trial. Otherwise, values can become misaligned—for example, a value intended for the next trial’s count might accidentally be used as a severity.

A simulation record is easier to audit if it stores the count, the generated severities, and their aggregate for each trial. For a single trial, the record should satisfy the identity $S=\sum_{j=1}^{N}X_j$, with $S=0$ for $N=0$. This simple bookkeeping check can reveal omitted claims, extra severity draws, or a sum assigned to the wrong trial.

Worked Example: Sampling a Heavy-Tailed Severity

A hypothetical portfolio has $P(N=0)=0.15$, $P(N=1)=0.35$, and $P(N=2)=0.50$. Each severity follows a Pareto Type I distribution with minimum value $x_m=$$80 and shape parameter $\alpha=2$, so $F_X(x)=1-(80/x)^2$ for $x\geq80$. The count uniform is $0.80$, and the severity uniforms are $0.36$ and $0.75$. Find the aggregate loss to the nearest cent.

Method. The cumulative count probabilities are $0.15,\ 0.50,\ 1.00$. Because $0.50<0.80\leq1.00$, the simulated count is $N=2$.

For a Pareto Type I severity, solving $U=1-(80/x)^2$ for $x$ gives $x=80/(1-U)^{1/2}$. With $U=0.36$, $$ X_1=\frac{80}{\sqrt{1-0.36}} =\frac{80}{\sqrt{0.64}} =\frac{80}{0.8}=100. $$ With $U=0.75$, $$ X_2=\frac{80}{\sqrt{1-0.75}} =\frac{80}{\sqrt{0.25}} =\frac{80}{0.5}=160. $$ Therefore $S=100+160=$$260.00.

Check. Both generated severities are at least the minimum $80, as required by the model, and their sum is $260.00. The count of two corresponds to exactly two severity values.

Common mistake avoided: The Pareto inverse uses the square root because the shape parameter is 2. Using $80/(1-U)$ would incorrectly omit the exponent $1/\alpha$.

Common Simulation Traps

Dropping zero-count trials. A zero count produces an aggregate of zero and must remain in the simulated sample. Removing it changes the modeled outcome distribution.
Generating the wrong number of severities. The count determines the number of severity draws within that trial. Do not use a fixed number of severities for every trial unless the model itself specifies a fixed count.
Reusing a uniform value. Under the independent model, count and severity values must be generated independently. Reusing the count uniform as a severity uniform introduces dependence.
Adding across trials instead of within a trial. Each aggregate is the sum of the severities for one count draw. A portfolio of simulated trials is a collection of those separate totals.
Rounding too early. Keep full precision in transformed severities and round the aggregate only at the requested reporting step.

The procedure is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus because it requires translating frequency and severity assumptions into a simulated aggregate outcome. The central discipline is to preserve the model’s structure: one count, that count’s number of independent severities, and one total—including zero when there are no claims.

Check Your Understanding

For each problem, use the stated inverse-cumulative convention and retain full precision until the final answer.

A count has probabilities $P(N=0)=0.30$, $P(N=1)=0.50$, and $P(N=2)=0.20$. The count uniform is $0.27$. What aggregate loss is recorded, regardless of the severity distribution? (Answer: $0.00)
A count has probabilities $0.20,\ 0.30,\ 0.50$ for $N=0,1,2$. Severity is $40 with probability 0.40 and $100 with probability 0.60. The count uniform is $0.44$, and the two severity uniforms are $0.20$ and $0.90$. Find the aggregate. (Answer: $140.00)
A count has probabilities $0.10,\ 0.60,\ 0.30$ for $N=0,1,2$. Severities are exponential with mean $50. The count uniform is $0.75$, and severity uniforms are $0.20$ and $0.60$. Find the aggregate to the nearest cent. (Answer: $56.97)
A count has probabilities $0.15,\ 0.35,\ 0.50$ for $N=0,1,2$. Severities are Pareto Type I with $x_m=$$80 and $\alpha=2$. The count uniform is $0.80$, and the severity uniforms are $0.36$ and $0.84$. Find the aggregate. (Answer: $300.00)
For one trial, the count uniform selects $N=1$. A second trial’s count uniform selects $N=0$. How many severity draws are required across the two trials, and what is the second trial’s aggregate? (Answer: One severity draw; the second trial’s aggregate is $0.00)

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