Worked Precision-Planning Case for Aggregate Risk

Introduction: One Simulation, Several Precision Targets

A simulation study of aggregate loss may need to estimate several quantities at once: the expected loss, the chance that loss exceeds a specified threshold, and a high quantile such as 99% Value-at-Risk. Each quantity has its own sampling variability, so a run count adequate for one result may be too small for another. A useful design starts with an explicit precision target for each quantity, calculates the corresponding run count, and then chooses a total that meets all targets.

The hypothetical case below uses independent simulated aggregate losses. Pilot estimates are used to plan the production run. This is a planning calculation, not a claim that the pilot estimates are known population values. The method is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X_1,\ldots,X_n$ be independent, identically distributed simulated aggregate losses from the specified model. The Monte Carlo estimate of $E[X]$ is $\bar X=n^{-1}\sum_{j=1}^n X_j$. For a threshold $c$, the estimate of $P(X>c)$ is $\hat p=n^{-1}\sum_{j=1}^n I(X_j>c)$, where $I(\cdot)$ is 1 when its condition holds and 0 otherwise. The empirical $\alpha$-quantile estimates $\operatorname{VaR}_{\alpha}(X)=F_X^{-1}(\alpha)$.

A two-sided normal-approximation interval with confidence level 95% uses $z_{0.975}=1.96$. Its approximate half-width is the standard error multiplied by 1.96. These intervals describe Monte Carlo error conditional on the model and its parameters. They do not include uncertainty from estimating the model or its parameters, nor do they address whether the model is appropriate.

Worked Case: Plan the Mean Estimate

Suppose a hypothetical insurer simulates aggregate annual losses in dollars. An independent pilot study suggests $\operatorname{Var}(X)=40{,}000{,}000$ dollars squared. The design requirement is a 95% confidence half-width no greater than $55 for the mean. A proposed production run has 56,320 independent simulations.

Worked Example: Mean Precision at the Proposed Run Count

Method. For independent observations with finite variance, the standard error of $\bar X$ is estimated using $\sqrt{\operatorname{Var}(X)/n}$. Using the pilot variance as a planning value, the required run count is the smallest integer that makes the normal-approximation half-width no greater than the target.

Step 1: Calculate the minimum run count for a $55 half-width.

n \geq \frac{1.96^2(40{,}000{,}000)}{55^2} = \frac{153{,}664{,}000}{3{,}025} \approx 50{,}797.02.

Because the run count must be an integer and must meet or exceed the calculated value, the minimum is $50{,}798$ simulations.

Step 2: Check the proposed 56,320-run design. Its estimated standard error is

\sqrt{\frac{40{,}000{,}000}{56{,}320}} = \sqrt{710.2273} \approx 26.6501\text{ dollars}.

The 95% half-width is $1.96(26.6501)\approx \$52.23$, which is below the target of $55. The proposed run count meets the mean-precision requirement.

Step 3: Illustrate reporting. If the simulated mean were $12,480.00, the approximate interval would be $$12{,}480.00\pm \$52.23$, or $[\$12{,}427.77,\$12{,}532.23]$. The endpoints are rounded to the nearest cent; the interval remains an approximation based on the planning variance.

Check. The proposed run count exceeds the calculated minimum by $56{,}320-50{,}798=5{,}522$ runs, and the computed half-width is less than $55.

Plan the Exceedance Probability

For each simulation, the event $X>c$ is a Bernoulli outcome: its indicator equals 1 with probability $p=P(X>c)$ and 0 otherwise. A Bernoulli variable has variance $p(1-p)$. Consequently, the standard error of the sample proportion is approximately $\sqrt{p(1-p)/n}$. For planning, use a pilot estimate or a conservative value for $p$; the resulting normal-approximation run count is

n \geq \frac{z_{0.975}^2p(1-p)}{h^2},

where $h$ is the required absolute half-width. This formula targets absolute error in probability units, not relative error. It is an approximation, particularly when the expected number of exceedances is small. In that setting, a normal interval may perform poorly and a suitable binomial interval or a more conservative design should be considered.

Worked Example: Precision for a One-Percent Exceedance Probability

Method. The pilot estimate for the hypothetical event $X>\$50{,}000$ is $p=0.01$. The design requires a 95% half-width no greater than 0.001, equivalent to 0.1 percentage point. Use the Bernoulli standard error and round the run count upward.

Step 1: Calculate the required count.

n \geq \frac{1.96^2(0.01)(0.99)}{0.001^2} = \frac{0.03803184}{0.000001} = 38{,}031.84.

Thus the minimum integer run count is $38{,}032$. The multiplication in the numerator is $3.8416(0.0099)=0.03803184$.

Step 2: Check the common 56,320-run design. If the pilot probability is used as the planning value, its half-width is

1.96\sqrt{\frac{0.01(0.99)}{56{,}320}} \approx 0.000822.

This is less than 0.001, so the proposed run count meets the target. For illustration, if 563 of the 56,320 production simulations exceed $50,000, then $\hat p=563/56{,}320\approx0.0099964$, which is 0.0100 to four decimal places. Using that observed proportion, the approximate standard error is $\sqrt{0.0099964(1-0.0099964)/56{,}320}\approx0.0004192$. The corresponding interval is approximately $0.0099964\pm1.96(0.0004192)$, or $[0.0092,0.0108]$ to four decimal places.

Check. The count 563 is consistent with the reported proportion: $563/56{,}320$ is approximately 0.0100. The interval half-width is about 0.0008, below the planned maximum of 0.001.

Plan Precision for a High Quantile

For a continuous loss distribution with a positive density near its $\alpha$-quantile $q_\alpha$, the large-sample standard error of the empirical quantile is approximately

\operatorname{SE}(\widehat{\operatorname{VaR}}_{\alpha}) \approx \frac{\sqrt{\alpha(1-\alpha)/n}}{f_X(q_\alpha)}.

This approximation requires independent observations, a well-defined quantile, and a density that is positive and sufficiently smooth around that quantile. It can be unreliable with atoms, a very sparse tail, or a poorly estimated density. The density has units of inverse dollars when losses are measured in dollars, so the resulting standard error is in dollars. Planning also treats the pilot estimate of the density as fixed; uncertainty in that estimate is not included in this simple calculation.

Justification: Near a continuous quantile $q_\alpha$, a small change $\Delta q$ changes the cumulative probability by approximately $F_X(q_\alpha+\Delta q)-F_X(q_\alpha)\approx f_X(q_\alpha)\Delta q$. The empirical cumulative probability at $q_\alpha$ has approximate standard error $\sqrt{\alpha(1-\alpha)/n}$, because it is an average of Bernoulli indicators with success probability $\alpha$. Dividing this probability-scale error by the local rate of change $f_X(q_\alpha)$ gives the stated quantile standard error.

Worked Example: Planning a 99% VaR Estimate

Method. A hypothetical pilot gives $\alpha=0.99$ and an estimated density $f_X(q_{0.99})=0.000002$ per dollar near the 99% quantile. The desired 95% half-width is at most $500. Use the asymptotic quantile standard error, with the pilot density as a planning estimate.

Step 1: Solve for the run count. Requiring $1.96\operatorname{SE}\leq500$ gives

n \geq \frac{1.96^2(0.99)(0.01)} {500^2(0.000002)^2} = \frac{0.03803184}{0.000001} = 38{,}031.84.

The minimum integer count is $38{,}032$. The denominator is $500^2(0.000002)^2=250{,}000(0.000000000004)=0.000001$.

Step 2: Assess the proposed 56,320-run design. Its approximate 95% half-width is

\frac{1.96\sqrt{0.99(0.01)/56{,}320}}{0.000002} \approx \$410.88.

This is below $500. If the simulated empirical 99% VaR were $48,120.00, the approximate interval would be $[\$47{,}709.12,\$48{,}530.88]$. The result is rounded to the nearest cent; the interval is an asymptotic Monte Carlo interval, not a guarantee that the model's true VaR lies within those endpoints.

Check. The half-width scales inversely with the density. The calculation uses density per dollar and reports a dollar half-width, as required.

Combine the Requirements and Report Carefully

When one simulation run produces all three measures, each target yields a minimum run count. The production count must be at least the largest of those requirements. For the hypothetical case, the mean target requires $50{,}798$ runs, the exceedance target requires $38{,}032$, and the VaR target requires $38{,}032$. Therefore, $56{,}320$ runs satisfy all three planning criteria. The mean is the binding requirement in this case.

Specify the outputs and tolerances.
Write down the mean, event probability, or quantile to be estimated and express each desired error in the appropriate units: dollars for a mean or VaR, probability units for an exceedance estimate.

Choose planning inputs.
Use pilot estimates of the variance, event probability, and density at the quantile. State the confidence level and the assumptions behind each approximation.

Calculate each run count separately.
Apply the standard error formula for each output, then round every calculated minimum upward to an integer.

Select the production count and verify it.
Take the maximum required count, substitute it back into each standard-error formula, and check that every half-width is within its target.

The intervals above are individual approximate 95% intervals. Three separate 95% intervals do not generally provide 95% simultaneous coverage for all three quantities. If a joint coverage requirement is specified, the confidence allocation should be adjusted accordingly rather than treating the individual intervals as a joint guarantee. Also, a production simulation should use the same aggregate-loss model and parameter values assumed in the planning exercise; a change in model or dependence structure can invalidate the run-count calculations.

Common Exam Traps

Using the wrong scale for probability precision. A half-width of 0.001 is 0.1 percentage point, not 0.001 percent. Convert a stated percentage-point target to probability units before using the formula.
Rounding down a run count. A calculated requirement of 38,031.84 means at least 38,032 runs. Rounding to the nearest integer could fail the stated target.
Forgetting the density in a quantile calculation. The probability-scale quantile error must be divided by $f_X(q_\alpha)$. Omitting it produces the wrong units and an invalid VaR standard error.
Confusing Monte Carlo error with total uncertainty. These calculations quantify simulation error conditional on a model. They do not account for parameter estimation or model misspecification.
Assuming every normal approximation is reliable. Sparse exceedances, heavy tails, atoms, or a small density near the target quantile can make the approximation inaccurate. Check whether its conditions are plausible.

Key takeaway: Plan each aggregate-risk output using its own variance or local distributional information. Round each required run count upward, use the maximum when one simulation supports several outputs, and report the uncertainty as Monte Carlo uncertainty under clearly stated assumptions.

Check Your Understanding

For each problem, use a two-sided 95% normal approximation with $z=1.96$, unless stated otherwise. Round run counts upward and report probabilities to four decimal places where relevant.

A pilot estimates $\operatorname{Var}(X)=9{,}000{,}000$ dollars squared. What run count is required for a mean half-width no greater than $30? (Answer: 38,416 simulations.)
A pilot exceedance probability is 0.04. Find the approximate run count needed for an absolute half-width of 0.002. (Answer: 36,880 simulations.)
For a 95% VaR estimate, a pilot density estimate is 0.00001 per dollar. What run count gives an approximate half-width of $200? (Answer: 45,619 simulations.)
One design requires a mean half-width of $75 with pilot variance $25{,}000{,}000$ dollars squared. A second requires an exceedance-probability half-width of 0.0015 with pilot probability 0.02. What single run count meets both targets? (Answer: 33,465 simulations; the probability requirement is larger than the mean requirement of 17,074.)
At a planned count of 10,000, an event is observed in 40 simulations. What is the estimated probability, and what approximate 95% half-width follows from the plug-in Bernoulli standard error? (Answer: $\hat p=0.0040$; half-width $\approx0.0012$, using $1.96\sqrt{0.004(0.996)/10{,}000}$.)

What You'll Learn