Precision Planning for Compound Aggregate Loss

Why Plan Simulation Precision?

A simulated aggregate loss is random from one run to the next. Consequently, a reported mean or exceedance probability is also subject to Monte Carlo error. Planning a simulation means choosing the number of independent runs so that this error is suitably small for the decision being supported. The relevant number of runs depends not only on the target precision, but also on the variability of the simulated output.

For compound aggregate losses, that variability comes from two sources: the number of claims and the sizes of those claims. Treating the claim count as fixed, or using only the variance of an individual claim, can therefore lead to a materially inadequate simulation budget. This tutorial develops precision calculations for the aggregate mean and for the probability that aggregate loss exceeds a specified threshold.

Definition: In a compound frequency-severity model, $N$ is the number of claims in one period, $X_1,X_2,\ldots$ are independent and identically distributed claim severities, and the aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Assume $N$ is independent of the severities, and that the moments used below are finite. A simulation run generates one independent value of $S$.

Aggregate Moments for Precision Planning

Write $m_X=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$. Conditioning on the claim count gives the standard compound-model moment formulas. They quantify the dispersion of simulated aggregate losses, which is essential when planning precision for their mean.

E[S]=E[N]m_X,\qquad \operatorname{Var}(S)=E[N]\sigma_X^2+\operatorname{Var}(N)m_X^2.

The variance formula has an important interpretation. The first term reflects the variability of severities when the number of claims is given. The second reflects the additional variability caused by a random claim count. If $N\sim\operatorname{Poisson}(\lambda)$, then $E[N]=\operatorname{Var}(N)=\lambda$, so

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

For $n$ independent simulated aggregates $S_1,\ldots,S_n$, the estimated mean is $\bar S=n^{-1}\sum_{r=1}^n S_r$. Its Monte Carlo standard error (MCSE), when the model variance is known, is $\sqrt{\operatorname{Var}(S)/n}$. A normal-approximation confidence interval with confidence level approximately 95% has half-width $1.96\sqrt{\operatorname{Var}(S)/n}$. This approximation is most useful when the number of runs is large enough for the sampling distribution of $\bar S$ to be reasonably close to normal.

Planning formula: To target an approximate 95% confidence-interval half-width no greater than $h$ for the aggregate mean, plan

n\ \geq\ \frac{1.96^2\operatorname{Var}(S)}{h^2}.

Round the required count up to the next whole simulation. If $\operatorname{Var}(S)$ is not known, a pilot simulation can provide a planning estimate. The final MCSE should then use the sample standard deviation of the simulated aggregates divided by $\sqrt n$; the pilot-based planning calculation is not a guarantee that the realized interval will meet its target.

Worked Example: Planning Precision for a Compound-Poisson Mean

A hypothetical portfolio is modeled with $N\sim\operatorname{Poisson}(4)$ claims per year. Claim severities are independent exponential random variables with mean $\$1{,}000$, independent of $N$. How many independent simulated years are needed for an approximate 95% confidence interval for the mean annual aggregate loss to have half-width at most $\$100$? Check the achieved half-width using the selected count.

Method. Use the compound-Poisson aggregate variance and the normal-approximation sample-size formula. For an exponential severity with mean $\$1{,}000$, the variance is $\$1{,}000^2=\$1{,}000{,}000$, and $E[X^2]=\operatorname{Var}(X)+(E[X])^2=\$2{,}000{,}000$.

Thus $E[S]=4(\$1{,}000)=\$4{,}000$, and $\operatorname{Var}(S)=4(\$2{,}000{,}000)=\$8{,}000{,}000$. Substituting $h=\$100$,

n\geq \frac{1.96^2(8{,}000{,}000)}{100^2} =\frac{3.8416(8{,}000{,}000)}{10{,}000} =3{,}073.28.

The minimum whole-number count is therefore $n=3{,}074$. At this count, the MCSE is $\sqrt{8{,}000{,}000/3{,}074}\approx \$51.01$, and the approximate half-width is $1.96(\$51.01)\approx \$99.99$. This is just under the $\$100$ target. The result is consistent with the calculation: rounding the required count upward meets the target.

Common mistake avoided: The severity variance alone is $\$1{,}000{,}000$, but it is not the aggregate variance per unit of frequency. For a Poisson count, the correct aggregate variance uses $E[X^2]=\$2{,}000{,}000$, giving $\$8{,}000{,}000$, not $\$4{,}000{,}000$. Omitting the frequency contribution would understate the planned simulation count.

Precision for an Exceedance Probability

For a threshold $c$, define $I_r=1\{S_r>c\}$ for simulation run $r$. Each indicator is Bernoulli with success probability $p=P(S>c)$. The simulated exceedance estimate is $\hat p=n^{-1}\sum_{r=1}^n I_r$. Since a Bernoulli variable has variance $p(1-p)$, the standard error of $\hat p$ is $\sqrt{p(1-p)/n}$. In practice, estimate it by $\sqrt{\hat p(1-\hat p)/n}$.

\operatorname{MCSE}(\hat p)\approx\sqrt{\frac{\hat p(1-\hat p)}{n}}, \qquad \text{approximate 95% half-width} =1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.

Solving the half-width requirement for $n$ gives the planning formula below, with $p$ replaced by a planning value such as a pilot estimate. It is an approximate formula, not a guarantee of coverage or a substitute for judgment when exceedances are very rare.

n\ \geq\ \frac{1.96^2p(1-p)}{h^2}.

Worked Example: Planning a Threshold-Exceedance Estimate

A hypothetical aggregate model has $N\sim\operatorname{Poisson}(2)$ claims per year and independent exponential severities with mean $\$1{,}000$. A pilot simulation estimates the probability of annual aggregate loss exceeding $\$6{,}000$ as $\hat p=0.0400$. Plan the number of independent annual simulations for an approximate 95% confidence-interval half-width no greater than $0.0100$.

Method. Treat each simulated year as a Bernoulli outcome, equal to 1 when its aggregate exceeds $\$6{,}000$. Use the pilot estimate as the planning value for $p$. The severity and frequency parameters describe the compound model; the tail probability used in this planning calculation is the pilot estimate, not an asserted exact probability.

Substitute $p=0.0400$ and $h=0.0100$:

n\geq \frac{1.96^2(0.0400)(0.9600)}{0.0100^2} =\frac{3.8416(0.0384)}{0.0001} =\frac{0.14751744}{0.0001} =1{,}475.1744.

Round up to $n=1{,}476$. The estimated MCSE at this count is $\sqrt{0.0384/1{,}476}\approx0.005101$, giving an approximate half-width $1.96(0.005101)\approx0.0100$. The corresponding expected number of exceedances, using the planning value, is $1{,}476(0.0400)=59.04$.

Common mistake avoided: The required count is not obtained by dividing the target half-width by $p$, or by using only $p$ as the indicator variance. The Bernoulli variance is $p(1-p)$, and the standard error decreases with the square root of $n$, which is why the half-width is squared when solving for $n$.

Why the Compound-Poisson Variance Has This Form

The moment formula is worth understanding because it identifies exactly what a precision calculation must include. Conditional on $N$, the aggregate is a sum of $N$ independent severities. Therefore $E[S\mid N]=Nm_X$ and $\operatorname{Var}(S\mid N)=N\sigma_X^2$. Apply the law of total expectation and the law of total variance:

Take the expectation.
$E[S]=E[E[S\mid N]]=E[Nm_X]=E[N]m_X.$

Separate the two sources of variance.
$\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])=E[N]\sigma_X^2+\operatorname{Var}(N)m_X^2.$

Use the Poisson property.
If $N\sim\operatorname{Poisson}(\lambda)$, then both $E[N]$ and $\operatorname{Var}(N)$ equal $\lambda$. Hence $\operatorname{Var}(S)=\lambda(\sigma_X^2+m_X^2)=\lambda E[X^2]$, because $E[X^2]=\operatorname{Var}(X)+(E[X])^2$.

This derivation also explains the common variance error: retaining only $E[N]\operatorname{Var}(X)$ ignores $\operatorname{Var}(N)(E[X])^2$. That term is not optional when the frequency is random. If frequency is fixed rather than Poisson, use the general formula with the actual $E[N]$ and $\operatorname{Var}(N)$; the Poisson simplification does not apply.

Combining Precision Requirements

A simulation may need to estimate more than one output. For example, a study may require both a precise aggregate mean and a precise exceedance probability. Calculate a required count for each output under its own assumptions and target, then use the largest count if the same independent runs will estimate both. This ensures that neither single-output requirement is weakened by the other.

Worked Example: Choosing One Count for Two Outputs

Use a hypothetical model with $N\sim\operatorname{Poisson}(4)$ and independent exponential severities of mean $\$1{,}000$, as in the first example. The required mean half-width is $\$200$. For a specified aggregate-loss threshold, a pilot estimate of the exceedance probability is $0.0100$; the required probability half-width is $0.0020$. Find a common simulation count for approximate 95% intervals for both outputs.

Mean requirement. The aggregate variance is $\$8{,}000{,}000$. Therefore

n_{\text{mean}}\geq\frac{3.8416(8{,}000{,}000)}{200^2} =\frac{30{,}732{,}800}{40{,}000}=768.32,

so $n_{\text{mean}}=769$.

Exceedance requirement. Using the pilot value $p=0.0100$,

n_{\text{exceed}}\geq \frac{3.8416(0.0100)(0.9900)}{0.0020^2} =\frac{0.03803184}{0.000004} =9{,}507.96,

so $n_{\text{exceed}}=9{,}508$. The common count must be the larger value: $n=9{,}508$. At this count, the mean half-width is approximately $1.96\sqrt{8{,}000{,}000/9{,}508}=\$56.85$, below $\$200$; the exceedance half-width using the planning value is approximately $1.96\sqrt{0.0099/9{,}508}=0.0020$. The probability requirement determines the simulation budget.

Common mistake avoided: Adding the two required counts is unnecessary when the same independent runs produce both outputs. The larger count meets both separate requirements, provided both outputs are calculated from every run.

Practical Checks and Limitations

Precision formulas are useful only when their inputs and assumptions match the simulation. Before committing a budget, check the following:

Use aggregate, not severity, variance for a mean. Include both frequency and severity variability through the appropriate compound-model formula.
Match the precision measure to the output. A dollar half-width for a mean and a probability half-width for an exceedance estimate have different units and different variance formulas.
Recognize plug-in uncertainty. If a pilot estimate is used for $\operatorname{Var}(S)$ or $p$, the resulting count is a planning estimate. A different pilot may give a different count, and the final standard error should be assessed from the production runs.
Do not interpret zero observed exceedances as zero uncertainty. When $\hat p=0$, the plug-in standard-error formula returns zero, but that does not establish that $p=0$. A pilot with no exceedances cannot, by itself, support a zero-width probability interval.
Check whether the normal approximation is credible. Very rare exceedances or highly variable aggregate losses may require special care; a nominal normal-approximation half-width should not be treated as an exact guarantee.

Careful planning links a stated accuracy target to the variability of the exact output being simulated. For compound aggregate loss, the core steps are to derive or estimate the aggregate moments, translate the target half-width into a run count, and verify that the resulting standard error is consistent with the target. This approach is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use an approximate 95% normal confidence interval where a half-width is requested. Round required simulation counts up to whole runs.

A hypothetical model has $N\sim\operatorname{Poisson}(5)$ and exponential claim severities with mean $\$400$. Find $E[S]$ and $\operatorname{Var}(S)$. (Answer: $E[S]=\$2{,}000$; $\operatorname{Var}(S)=\$1{,}600{,}000$.)
For simulated aggregate losses with known variance $\$500{,}000$, how many independent runs are needed for a mean half-width of $\$50$? (Answer: $769$ runs.)
A pilot estimates an aggregate exceedance probability as $0.0800$. Find the run count for a probability half-width of $0.0200$. (Answer: $707$ runs.)
A mean estimate has aggregate variance $\$2{,}000{,}000$ and needs a $\$100$ half-width. A separate exceedance estimate has pilot probability $0.0200$ and needs a $0.0050$ half-width. If both use the same runs, what count meets both requirements? (Answer: $3{,}012$ runs.)

What You'll Learn