From a Loss Model to Annual Trials
A frequency-severity model describes annual aggregate loss by combining the number of claims with the amount of each claim. Simulation turns that model into a sequence of possible annual outcomes. It is useful when a model is too complicated for a direct calculation, or when estimates are needed for several features of the loss distribution.
The aim is not to predict the exact losses of a particular future year. Rather, repeated simulated years provide estimates of quantities implied by the model. Those estimates have Monte Carlo error: a different set of random draws would generally produce slightly different results.
A standard compound model assumes that claim counts in different simulated years are independent and have the same frequency distribution; claim severities are independent draws from the same severity distribution; and the severity draws are independent of the count. These are modeling assumptions, not facts guaranteed by the simulation procedure. If the model specifies dependence, changing exposures, or different distributions in different periods, the simulation must represent those features instead. Unless stated otherwise, the examples below use one-year losses without discounting; interest and mortality are not part of these loss models.
How a Simulation Trial Is Constructed
A computer typically generates independent uniform random numbers on \((0,1)\) and transforms them into draws from the desired distributions. For a discrete claim count, a uniform draw is assigned to the interval associated with each possible count. For continuous or discrete severity, it is transformed according to the severity distribution. For example, if severity is exponential with mean \(\theta\), then \(Y=-\theta\ln(1-U)\) has that distribution when \(U\) is uniform on \((0,1)\).
A practical sequence is to generate one count for each simulated year, then generate exactly that many severities and add them. If the count is zero, record zero aggregate loss and do not generate severities for that year. Use fresh independent random numbers for each simulated year and each claim when the model requires independence.
State the frequency distribution, severity distribution, their parameters, and any dependence assumptions.
Generate \(N\) from the chosen frequency model for one trial year.
Generate \(N\) severities from the specified model and add them. If \(N=0\), the aggregate is zero.
Repeat for \(M\) years, then calculate the desired estimates from the \(M\) simulated aggregate losses.
Worked Example: Building Five Annual Loss Trials
Model and assumptions. For each hypothetical year, \(N\) is Poisson with parameter \(\lambda=1\). Conditional on the count, claim severities are independent of the count and of one another, with probabilities \(0.50\) for $500, \(0.30\) for $1,000, and \(0.20\) for $2,000. Trials are independent. There is no discounting and no mortality assumption is involved. Use the Poisson count probabilities \(P(N=0)=0.3679\), \(P(N=1)=0.3679\), and \(P(N=2)=0.1839\), with cumulative cutoffs \(0.3679\), \(0.7358\), and \(0.9197\). The remaining probability corresponds to three or more claims.
Method. For a uniform count draw, assign 0 claims below 0.3679, 1 claim from 0.3679 up to 0.7358, 2 claims from 0.7358 up to 0.9197, and 3 or more above 0.9197. For severity, assign $500 when \(U<0.50\), $1,000 when \(0.50\leq U<0.80\), and $2,000 when \(U\geq0.80\). The supplied uniform draws below are hypothetical inputs to the simulation.
| Trial | Count uniform | Count \(N\) | Severity uniforms, in claim order | Aggregate \(X\) |
|---|---|---|---|---|
| 1 | 0.10 | 0 | None | $0 |
| 2 | 0.50 | 1 | 0.20 | $500 |
| 3 | 0.80 | 2 | 0.60, 0.90 | $1,000 + $2,000 = $3,000 |
| 4 | 0.95 | 3 | 0.30, 0.55, 0.95 | $500 + $1,000 + $2,000 = $3,500 |
| 5 | 0.70 | 1 | 0.45 | $500 |
For example, the third count draw lies from 0.7358 to 0.9197, so it gives two claims. Its first severity draw lies from 0.50 to 0.80, giving $1,000; its second is at least 0.80, giving $2,000. The annual aggregate is therefore $3,000.
The five simulated aggregates sum to \(0+500+3{,}000+3{,}500+500=7{,}500\), so their sample mean is \(7{,}500/5=\)$1,500. The proportion strictly exceeding $2,000 is \(1/5=0.2000\). Sorting the aggregates gives $0, $500, $500, $3,000, $3,500. Using the nearest-rank convention, the empirical 80th percentile is the observation at rank \(\lceil5(0.80)\rceil=4\), or $3,000. These are summaries of five trials, not exact model quantities. The common mistake this method avoids is assigning a severity to a year with zero claims, which would incorrectly make its aggregate positive.
Turning Repeated Trials into Estimates
Suppose \(X_1,\ldots,X_M\) are the simulated annual aggregate losses. The simulation estimate of the model mean is the sample average. For a threshold \(t\), the estimate of the probability of exceeding \(t\) is the fraction of trials whose aggregate is greater than \(t\). To estimate a quantile, sort the simulated losses and select an order statistic under a stated convention.
Here \(I(X_m>t)\) equals 1 when the event is true and 0 otherwise. If \(p_t=P(X>t)\), the estimated standard error of \(\hat p_t\), under independent trials, is approximately \(\sqrt{\hat p_t(1-\hat p_t)/M}\). The standard error of the sample mean is estimated by \(s_X/\sqrt M\), where \(s_X\) is the sample standard deviation of the simulated aggregates. These quantify simulation variability; they do not account for errors in the assumed frequency or severity model.
For a reproducible empirical quantile convention, the nearest-rank estimate at probability \(q\) is the sorted observation at rank \(\lceil Mq\rceil\), with observations numbered from 1 to \(M\). This is a sample summary, especially coarse for small \(M\). Different quantile conventions can yield different answers in a finite sample, so state the convention being used.
Worked Example: Estimating a Mean and an Empirical Quantile
Model and assumptions. Consider a hypothetical compound model with \(N\sim\operatorname{Poisson}(1)\), and independent severities taking $500, $1,000, or $2,000 with probabilities 0.50, 0.30, and 0.20, respectively. Counts, severities, and different annual trials are independent as specified. There is no discounting and no mortality assumption is involved. Suppose ten simulation trials generated the following aggregates, shown in thousands of dollars: \(0,\ 0.5,\ 1,\ 1,\ 1.5,\ 2,\ 2.5,\ 0.5,\ 3,\ 1\). Each is a possible aggregate under the stated model; the list is a hypothetical simulation output.
Sample mean. The sum is \(0+0.5+1+1+1.5+2+2.5+0.5+3+1=13\) thousand dollars. Thus \(\bar X=13/10=1.3\) thousand dollars, or $1,300.
Estimated standard error of the mean. The sum of squared observations, in squared thousands of dollars, is \(0+0.25+1+1+2.25+4+6.25+0.25+9+1=25\). The sum of squared deviations from the mean can be calculated as \(25-10(1.3)^2=25-16.9=8.1\). The sample variance is \(8.1/(10-1)=0.9\) thousand-dollars squared, so \(s_X=\sqrt{0.9}=0.9487\) thousand dollars. Therefore, the estimated standard error is \(0.9487/\sqrt{10}=0.3000\) thousand dollars, or $300.00.
Empirical 90th percentile. In sorted order, the observations are \(0,\ 0.5,\ 0.5,\ 1,\ 1,\ 1,\ 1.5,\ 2,\ 2.5,\ 3\). The nearest-rank is \(\lceil10(0.90)\rceil=9\), so the estimate is the ninth observation: 2.5 thousand dollars, or $2,500. This method avoids the common error of using the unsorted tenth observation or silently changing quantile conventions.
Simulation Error and Threshold Estimates
A simulated estimate should be reported with its trial count and a clear event definition. “Exceeds $40,000” means \(X>40{,}000\); it does not include an aggregate exactly equal to $40,000. The estimate is more stable as the number of independent trials increases, but rare events can still be poorly represented unless the trial count is large enough.
The standard error formula for an exceedance estimate treats each trial as a Bernoulli outcome: exceed or do not exceed. It describes random variation in the estimated proportion under repeated simulations of the same model. It is not a guarantee that the underlying model is correct, nor does it measure uncertainty in fitted parameters.
Worked Example: Estimating a Rare-Event Probability
Model and assumptions. For this hypothetical calculation, each of 2,500 independent one-year trials is generated from \(N\sim\operatorname{Poisson}(0.8)\). Conditional on \(N\), severities are independent lognormal amounts with \(\ln(Y/1{,}000)\sim N(0,0.6^2)\), independent of the count. Thus the log-severity mean is 0 and its variance is \(0.6^2\) on the scale \(\ln(Y/1{,}000)\). There is no discounting and no mortality assumption is involved. Suppose 173 simulated annual aggregates strictly exceed $40,000.
Method. The estimated exceedance probability is the number of exceedances divided by the number of simulated years:
The estimated standard error is \(\sqrt{0.0692(1-0.0692)/2{,}500} =\sqrt{0.0692(0.9308)/2{,}500} =\sqrt{0.000025764544}\approx0.0051\), to four decimal places. The estimate is 6.92%, with an estimated simulation standard error of about 0.51 percentage points.
This calculation avoids a rounding trap: \(0.0692(0.9308)/2{,}500=0.000025764544\), which rounds to 0.00002576, not 0.00002577. Keeping the unrounded intermediate value gives the same final standard error, 0.0051 to four decimal places. The result estimates the probability under the stated model; it is not an observed industry rate.
Common Simulation Mistakes
- Using the wrong number of severity draws. Draw exactly \(N\) severities in a trial. A zero-count year has aggregate zero.
- Mixing up an event and its complement. For \(P(X>t)\), count aggregates strictly above \(t\); do not include ties. For \(P(X\geq t)\), ties are included.
- Applying frequency probabilities to severities. First map the count draw using the count distribution; then map each severity draw using the severity distribution.
- Treating simulation output as exact. A simulated mean, quantile, or probability changes with the random sample. Report the number of trials and, where useful, an estimated standard error.
- Assuming more trials repair a bad model. Increasing \(M\) reduces Monte Carlo error, but cannot correct unsuitable frequency, severity, or dependence assumptions.
Check Your Understanding
For each problem, use the stated simulation convention and keep the event definition precise.
- A hypothetical annual count is Poisson with \(\lambda=0.5\), and every claim has severity $800. Using the count cutoffs \(P(N=0)=0.6065\) and \(P(N\leq1)=0.9098\), map count uniforms 0.20, 0.65, and 0.90 to claim counts and calculate the three-year average aggregate. (Answer: Counts 0, 1, and 1; aggregates $0, $800, and $800; average $533.33.)
- Four simulated annual aggregates are $0, $400, $800, and $1,200. Calculate the sample mean and the simulated proportion strictly exceeding $800. (Answer: Mean $600; exceedance estimate \(1/4=0.2500\).)
- Eight simulated aggregates, sorted in thousands of dollars, are \(0,\ 0.4,\ 0.6,\ 0.8,\ 1.0,\ 1.2,\ 1.6,\ 2.0\). Using the nearest-rank convention, find the empirical 75th percentile and count the observations strictly above $1,000. (Answer: 75th percentile $1,200; \(3/8=0.3750\) strictly exceed $1,000.)
- In 1,000 independent simulated years, 50 annual aggregates exceed a specified threshold. Calculate the estimated probability and its estimated standard error, to four decimal places. (Answer: \(\hat p=0.0500\); standard error \(\sqrt{0.05(0.95)/1{,}000}=0.0069\).)