Simulation of Aggregate Loss from Fitted Models

Why Simulate Aggregate Losses?

A fitted frequency–severity model gives a probability distribution for aggregate loss, but that distribution may be difficult to calculate directly. Simulation offers a flexible alternative: repeatedly generate a claim count, generate that many claim severities, and add them. The resulting simulated aggregate losses can be used to estimate event probabilities, quantiles, and averages over the largest outcomes.

Simulation does not make a fitted model correct, nor does it remove uncertainty in its estimated parameters. It approximates quantities under the model and parameter values supplied. The central practical question is therefore twofold: what quantity is being estimated, and how much Monte Carlo error remains in that estimate? This simulation workflow is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: In a compound frequency–severity model, $N$ is the number of claims in the period, $X_1,X_2,\ldots$ are claim severities, and aggregate loss is $S=\sum_{j=1}^{N}X_j$. The sum is defined as zero when $N=0$. A standard model assumes that $N$ is independent of the severities and that the severities are independent and identically distributed. A simulation from fitted models treats the fitted frequency and severity distributions, and usually their parameter estimates, as fixed inputs.

From Fitted Models to Aggregate Outcomes

A simulation replicate represents one possible period of experience under the chosen model. First generate $N$ from the fitted frequency distribution. Then generate $N$ independent severities from the fitted severity distribution and add them. Repeat this process $M$ times to obtain $S_1,\ldots,S_M$. The replicate index matters: each $S_r$ is one complete period, not one claim.

Specify the fitted inputs.
Record the frequency family and parameter estimates, the severity family and parameter estimates, the time period, and any assumptions about independence or coverage adjustments.

Generate one claim count.
For replicate $r$, draw $N_r$ from the fitted frequency model. The draw is a nonnegative integer.

Generate and aggregate severities.
Draw $N_r$ values independently from the fitted severity model and calculate $S_r=\sum_{j=1}^{N_r}X_{rj}$. Set $S_r=0$ if $N_r=0$.

Repeat and summarize.
Use the $M$ aggregate values to calculate the estimate of interest, then report an appropriate measure of simulation precision.

For a discrete frequency model, one way to generate a count is to draw a uniform random number on $(0,1)$ and use the model’s cumulative probabilities to select the count. For a discrete severity model, use the same cumulative-probability principle to select a severity. For a continuous model, an inverse distribution function or another valid sampling method can be used. The method must reproduce the specified fitted distribution; rounding a continuous severity without a stated reason changes that model.

What the Simulation Estimates

Suppose the target is the probability that aggregate loss exceeds a specified amount $c$. Each replicate contributes an indicator: it is 1 if $S_r>c$, and 0 otherwise. The simulation estimate is the proportion of replicates exceeding $c$. The strict inequality is part of the definition; if the question instead asks for $P(S\ge c)$, equality must be counted as an exceedance.

\widehat{P}(S>c)=\frac{1}{M}\sum_{r=1}^{M} I(S_r>c)

A simulation quantile can be estimated by sorting the $M$ aggregate values and selecting an appropriate order statistic. For a simple empirical convention, the estimated $p$-quantile is the observation of rank $\lceil Mp\rceil$ in ascending order. Other quantile conventions interpolate between observations; when a convention is specified, use it consistently.

An upper-tail average can likewise be estimated from the largest simulated aggregate losses. For example, under a convention that averages the largest $k$ observations, sort the simulated losses and take the mean of those $k$ values. State the convention and the number of observations used, especially when the sample size does not divide evenly into the desired tail proportion. With a small simulation, a few large observations can strongly affect this average.

Monte Carlo Error for a Probability

The precision of a simulated probability estimate can be derived directly. Let $I_r=I(S_r>c)$. Under independent simulation replicates, each indicator has a Bernoulli distribution with probability $p=P(S>c)$, so its variance is $p(1-p)$. Since the estimate is the average of $M$ independent indicators, its variance is $p(1-p)/M$. Replacing the unknown $p$ by the estimate gives an estimated standard error.

\operatorname{SE}\bigl(\widehat{P}(S>c)\bigr) \approx \sqrt{\frac{\widehat{P}(S>c)\left(1-\widehat{P}(S>c)\right)}{M}}

This is Monte Carlo standard error: it measures variation from using a finite number of simulation replicates when the model is held fixed. It is not a measure of uncertainty in the fitted frequency or severity parameters. A rough 95% Monte Carlo interval is the estimate plus or minus $1.96$ estimated standard errors, though this normal approximation can be unreliable for very small $M$, or when the estimated probability is close to 0 or 1.

Worked Example: Estimating an Aggregate Exceedance Probability

A hypothetical fitted model has $N\sim\operatorname{Poisson}(2)$ claims per year, with every claim costing exactly $1,000. Treat the fitted parameter as fixed. Ten simulated claim counts are $0,1,2,3,1,0,2,1,4,2$. Estimate the probability that annual aggregate loss is at least $2,000, and calculate its Monte Carlo standard error.

Method. With a fixed $1,000 severity, aggregate loss is $1,000 times the claim count. The event $S\ge \$2{,}000$ is therefore exactly the event $N\ge2$. The simulated losses, in thousands of dollars, are $0,1,2,3,1,0,2,1,4,2$.

Five of the ten replicates have counts at least 2: the counts are 2, 3, 2, 4, and 2. Thus $\widehat{P}(S\ge \$2{,}000)=5/10=0.5$.

Substitute $0.5$ and $M=10$ in the standard-error formula: $\sqrt{0.5(1-0.5)/10}=\sqrt{0.025}=0.1581$, to four decimal places.

The estimated probability is 0.5000, with estimated Monte Carlo standard error 0.1581. This is a very imprecise estimate because only ten periods were simulated. Common mistake avoided: counting claims strictly greater than 2 would omit the three replicates with exactly two claims, even though the stated event is “at least” $2,000.

Worked Example: Estimating a Quantile and an Upper-Tail Average

A hypothetical fitted model has $N\sim\operatorname{Poisson}(1)$ claims per period. Claim severities are independent, with each claim equal to $1,000 or $3,000 with probability 0.5 each, and are independent of $N$. Ten simulated replicates are: no claims; one claim of $1,000; two claims of $1,000 and $3,000; one claim of $3,000; two claims of $3,000 and $3,000; one claim of $1,000; no claims; two claims of $1,000 and $1,000; one claim of $3,000; and two claims of $1,000 and $3,000. Estimate the 80th percentile using the nearest-rank convention and estimate the mean of the largest 20% of the aggregate losses.

Method. Add the claim severities within each replicate. The resulting aggregate losses, in thousands of dollars and in replicate order, are $0,1,4,3,6,1,0,2,3,4$. In ascending order they are $0,0,1,1,2,3,3,4,4,6$.

For the nearest-rank 80th percentile, the rank is $\lceil10(0.80)\rceil=8$. The eighth ordered value is 4, so the estimated percentile is $4,000.

The largest 20% of ten observations are the two largest values, $4,000 and $6,000. Their mean is $(\$4{,}000+\$6{,}000)/2=\$5{,}000$. The estimated upper-tail average under the stated convention is therefore $5,000.

The calculations use all claims in each period before sorting the aggregate outcomes. Common mistake avoided: sorting individual severities instead would answer a claim-level question, not a question about the distribution of annual aggregate loss. These estimates are illustrative; ten replicates are too few to suggest fine precision.

Planning and Interpreting Simulation Size

For probability estimation, the standard error usually decreases in proportion to $1/\sqrt{M}$. Thus multiplying the simulation size by four approximately halves the Monte Carlo standard error. This improvement has a computational cost, and it does not address whether the fitted model is appropriate.

Worked Example: Assessing Precision for a Rare Aggregate Event

A hypothetical fitted model uses $N\sim\operatorname{Poisson}(1.5)$ claims per year and independent claim severities with an exponential distribution of mean $2,000. Parameters are treated as fixed. In a simulation of $M=40{,}000$ annual aggregate losses, 2,400 are greater than $10,000. Estimate the exceedance probability, its Monte Carlo standard error, and a rough 95% Monte Carlo interval.

Method. The estimated probability is $2{,}400/40{,}000=0.06$. The estimated standard error is $\sqrt{0.06(1-0.06)/40{,}000}=\sqrt{0.0564/40{,}000}=\sqrt{0.00000141}=0.0011874$, to four decimal places it is 0.0012.

Using the unrounded standard error, the 95% margin is $1.96(0.0011874)=0.0023273$. The approximate interval is $0.06\pm0.0023273$, or $[0.0577,0.0623]$ to four decimal places.

The estimate is 0.0600, and the approximate interval reflects simulation variation under the fixed fitted model. It does not include uncertainty in the estimated Poisson rate or severity mean, nor uncertainty about whether these distributions are suitable. Common mistake avoided: interpreting a narrow Monte Carlo interval as evidence that the fitted model itself is certain.

Common Simulation Traps

Mixing up claim and period units. A simulated claim count must correspond to the period being modeled. If a frequency estimate is annual, it cannot be used unchanged for a shorter period without an appropriate frequency adjustment.
Using the wrong event boundary. $S>c$ and $S\ge c$ can have different probabilities, particularly for discrete aggregate losses.
Simulating claims but not aggregating by replicate. The target aggregate distribution is built from one total per period, including zero totals when $N=0$.
Ignoring the fitted-model assumptions. The standard compound model assumes independent frequency and severities and independent severities. Dependence or changing claim conditions require a model that represents those features.
Confusing simulation precision with model precision. Increasing $M$ reduces Monte Carlo error, but it does not make estimated parameters or modeling choices more reliable.
Reporting excessive precision. A simulation estimate such as 0.0600 does not justify many extra decimal places when its Monte Carlo standard error is about 0.0012.

Key takeaway: Simulate complete periods from the fitted frequency–severity model, calculate the target from the resulting aggregate losses, and report Monte Carlo error separately from uncertainty in the fitted model. For probability estimates, the standard error is approximately $\sqrt{\widehat p(1-\widehat p)/M}$.

Check Your Understanding

Use the stated model and conventions in each problem. Answers are provided for checking.

A fitted annual model has $N\sim\operatorname{Poisson}(0.5)$ and fixed severity $4,000 per claim. In 20 simulated years, 6 aggregate losses exceed $4,000. Calculate the simulated exceedance probability and its Monte Carlo standard error. (Answer: 0.3000; standard error 0.1025.)
A fitted model has $N\sim\operatorname{Binomial}(2,0.5)$, with independent severities of $1,000 or $3,000, equally likely. Eight simulated aggregate losses, in thousands of dollars, are $0,1,3,2,4,3,6,1$. Estimate $P(S\ge \$4{,}000)$ and its Monte Carlo standard error. (Answer: 0.2500; standard error 0.1531.)
Ten simulated aggregate losses, in thousands of dollars, sorted from smallest to largest, are $0,1,1,2,2,3,3,4,5,8$. Using nearest rank, estimate the 80th percentile and the mean of the largest 20% of observations. (Answer: 4 thousand dollars; 6.5 thousand dollars.)
In 10,000 independent simulated periods, 400 aggregate losses exceed a specified threshold. Calculate the estimated probability, its Monte Carlo standard error, and the approximate 95% Monte Carlo interval. Round the interval endpoints to four decimals. (Answer: 0.0400; standard error 0.0020; interval [0.0362, 0.0438].)
A pilot simulation suggests an exceedance probability near 0.02. Approximately how many independent replicates are needed for an estimated standard error of 0.001, using that probability as the planning value? (Answer: 19,600 replicates.)

What You'll Learn