Why Compare Aggregation Methods?
An insurer may estimate risk capital for each of several business risks and then need a portfolio-level figure. A square-root formula is quick, moment matching summarizes the aggregate by its first two moments, and copula simulation models a joint distribution directly. These methods answer a similar question using different approximations and information, so their results need not agree.
This tutorial compares the three methods using the same hypothetical two-risk portfolio. The example is deliberately discrete: it makes the tail behavior visible and shows why a method that matches the mean and variance can still misstate a high quantile. The calculations are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
The Hypothetical Portfolio and the Three Approaches
Suppose risk A produces a loss of $100 thousand with probability 0.05 and otherwise no loss. Risk B produces a loss of $80 thousand with probability 0.05 and otherwise no loss. Their dependence is described by a Gaussian copula: the two standard normal variables underlying the copula are correlated, with the copula parameter calibrated so that the probability of both losses occurring is 0.02. Thus the marginal probabilities and the joint probability are specified explicitly.
Let \(I_A\) and \(I_B\) be the indicators of the two loss events. Then \(X_A=100I_A\), \(X_B=80I_B\), and \(S=X_A+X_B\), with all dollar amounts in thousands. The joint probability implies \(P(I_A=1,I_B=1)=0.02\). Each marginal event has probability 0.05, so the probability of A only is \(0.05-0.02=0.03\); likewise, B only has probability 0.03. The probability of neither is \(1-0.02-0.03-0.03=0.92\).
| Outcome | Probability | Aggregate loss (thousands) |
|---|---|---|
| Neither loss occurs | 0.92 | 0 |
| A only | 0.03 | 100 |
| B only | 0.03 | 80 |
| Both losses occur | 0.02 | 180 |
The standalone means are \(E[X_A]=0.05(100)=5\) and \(E[X_B]=0.05(80)=4\), so \(E[S]=9\). Each standalone 99% VaR is its nonzero loss: the probability of zero loss is only 0.95, below 0.99. The corresponding standalone risk capitals are therefore \(100-5=95\) and \(80-4=76\), a total of 171 thousand. This common baseline will be used to compare diversification benefits.
Method 1: Square-Root Aggregation
The square-root approach combines standalone capital estimates using correlations. In a two-risk version, if \(K_A\) and \(K_B\) are standalone risk capitals and \(\rho\) is the correlation used for aggregation, the portfolio capital estimate is \(K_S=\sqrt{K_A^2+K_B^2+2\rho K_AK_B}\). This is a capital approximation, not a general identity for VaR. Its interpretation depends on using a defensible correlation and comparable standalone capital measures.
Worked Example: Square-Root Capital for the Two Risks
Method. Use the loss correlation implied by the specified joint event probability, combine the standalone 99% risk capitals with the square-root formula, and add aggregate expected loss to express the result as an estimated gross VaR. All amounts are in thousands.
First calculate the variances. For a loss that is \(c\) with probability \(p\) and zero otherwise, its variance is \(c^2p(1-p)\). Thus \(\operatorname{Var}(X_A)=100^2(0.05)(0.95)=475\), and \(\operatorname{Var}(X_B)=80^2(0.05)(0.95)=304\).
The covariance is determined by the joint event: \(\operatorname{Cov}(X_A,X_B)=E[X_AX_B]-E[X_A]E[X_B]\). Here \(E[X_AX_B]=100(80)(0.02)=160\), while \(E[X_A]E[X_B]=5(4)=20\). Therefore the covariance is \(160-20=140\).
The loss correlation is \(\rho=140/\sqrt{475(304)}=140/380=0.368421\), since \(475(304)=144{,}400\) and \(\sqrt{144{,}400}=380\). The standalone risk capitals are \(K_A=95\) and \(K_B=76\). Hence \(K_S=\sqrt{95^2+76^2+2(0.368421)(95)(76)}\). The terms are \(95^2=9{,}025\), \(76^2=5{,}776\), and the cross term is \(5{,}320\), giving \(K_S=\sqrt{20{,}121}=141.85\) thousand, rounded to two decimals.
Adding the aggregate mean of 9 gives an estimated gross 99% VaR of \(141.85+9=150.85\) thousand. Relative to the standalone capital total of 171 thousand, estimated diversification benefit is \(171-141.85=29.15\) thousand.
Check and common mistake. The covariance and correlation are consistent because \(0.368421(380)=140\). A tempting but inconsistent calculation inserts the standalone gross VaRs, 100 and 80, directly into the square-root formula. That gives \(\sqrt{100^2+80^2+2(0.368421)(100)(80)}=149.31\) thousand, incorrectly treated as gross VaR. It mixes gross quantiles with a formula applied here to risk capitals and mishandles the aggregate mean. Use the capital inputs \(95\) and \(76\), then add the mean once.
Method 2: Moment Matching
Moment matching computes the aggregate mean and variance, then selects a distribution with those moments. For a normal approximation, the estimated \(p\)-quantile is \(E[S]+z_p\sqrt{\operatorname{Var}(S)}\), where \(z_p\) is the \(p\)-quantile of a standard normal distribution. Matching two moments does not establish that the true aggregate loss is normal; the resulting quantile is an approximation.
For two risks, the exact variance identity is \(\operatorname{Var}(X_A+X_B)=\operatorname{Var}(X_A)+\operatorname{Var}(X_B)+2\operatorname{Cov}(X_A,X_B)\). It follows by expanding \(E[(X_A+X_B-E[X_A]-E[X_B])^2]\): the squared terms give the two variances, and the two cross-products each have expectation \(\operatorname{Cov}(X_A,X_B)\). This identity is exact; the normal quantile calculation that may follow it is the approximation.
Worked Example: A Normal Moment-Matched 99% Quantile
Method. Match the aggregate mean and variance from the portfolio model, then use a normal 99% quantile. The standard normal 99th percentile is \(z_{0.99}=2.3263\). No assumption is made that the actual discrete portfolio loss is normal.
The aggregate mean is \(E[S]=5+4=9\) thousand. From the preceding covariance calculation, the aggregate variance is \(\operatorname{Var}(S)=475+304+2(140)=1{,}059\) thousand squared. Its standard deviation is \(\sqrt{1{,}059}=32.5423\) thousand.
The moment-matched normal estimate is therefore \(\operatorname{VaR}_{0.99}(S)\approx9+2.3263(32.5423)=9+75.70=84.70\) thousand, rounded to two decimals. The estimated aggregate risk capital is \(84.70-9=75.70\) thousand, and the implied diversification benefit is \(171-75.70=95.30\) thousand.
Check and common mistake. The aggregate variance includes \(2\operatorname{Cov}(X_A,X_B)=280\); omitting the covariance would be incorrect because the risks are dependent. The calculation includes it, but the resulting normal quantile is still a model approximation. In the exact discrete distribution, losses of 180 thousand occur with probability 0.02, so the normal estimate of 84.70 thousand is not the exact 99% VaR.
Method 3: Gaussian-Copula Simulation
A Gaussian copula simulation first generates dependent uniform random variables with the desired copula, then maps each uniform through its marginal loss distribution. For this example, generate independent standard normal variables \(Z_1\) and \(\varepsilon\), set \(Z_2=\rho_C Z_1+\sqrt{1-\rho_C^2}\varepsilon\), and transform to \(U_A=\Phi(Z_1)\) and \(U_B=\Phi(Z_2)\). Here \(\Phi\) is the standard normal cumulative distribution function, and \(\rho_C\) is the latent Gaussian correlation, calibrated so that the joint probability of both uniforms exceeding 0.95 is 0.02. This latent correlation is not the same quantity as the 0.368421 correlation between the dollar losses.
Map the uniforms to losses using \(X_A=100\) if \(U_A>0.95\), and zero otherwise; use \(X_B=80\) if \(U_B>0.95\), and zero otherwise. Repeating the draw-and-map process produces simulated aggregate losses. The empirical quantile is obtained by sorting those losses and selecting the required order statistic. Monte Carlo estimates vary from run to run; a reported result should be interpreted as a simulation estimate, not as a universal exact value.
Worked Example: Reading the Simulated 99% Quantile
Method. Consider an illustrative 100,000-draw run from the calibrated copula model. Suppose the mapped outcomes are 92,000 draws with neither loss, 3,000 with A only, 3,000 with B only, and 2,000 with both. Use the empirical left-quantile convention: the 99% quantile is the observation at sorted position \(\lceil0.99(100{,}000)\rceil=99{,}000\).
The counts sum to \(92{,}000+3{,}000+3{,}000+2{,}000=100{,}000\). The simulated event proportions are 5,000/100,000 = 0.05 for A, 5,000/100,000 = 0.05 for B, and 2,000/100,000 = 0.02 for both, as specified by the marginals and joint calibration.
The sorted losses are 0 in positions 1 through 92,000; 80 thousand in positions 92,001 through 95,000; 100 thousand in positions 95,001 through 98,000; and 180 thousand in positions 98,001 through 100,000. Thus position 99,000 has loss 180 thousand, so the empirical 99% VaR is 180 thousand. The simulated mean is \((3{,}000(100)+3{,}000(80)+2{,}000(180))/100{,}000=9\) thousand. Simulated risk capital is therefore \(180-9=171\) thousand, and the estimated diversification benefit is \(171-171=0\) thousand.
Check and common mistake. The non-both outcomes total 98,000 draws, so the empirical cumulative proportion at 100 thousand is 0.98, below 0.99. The quantile must be 180 thousand. Using the 98,000th observation, or selecting the 99th observation after expressing the simulation as percentages, would give the wrong order statistic.
Compare the Results on the Same Basis
All three results below are expressed as 99% VaR and risk capital relative to the same 9-thousand expected aggregate loss. The standalone risk capital total is 171 thousand. The square-root result is a capital approximation, the moment-matched result is a normal approximation, and the simulation result is an empirical quantile from the stated illustrative run.
| Method | Estimated aggregate 99% VaR (thousands) | Aggregate risk capital (thousands) | Diversification benefit (thousands) |
|---|---|---|---|
| Square-root aggregation | 150.85 | 141.85 | 29.15 |
| Normal moment matching | 84.70 | 75.70 | 95.30 |
| Copula simulation | 180.00 | 171.00 | 0.00 |
These differences are not arithmetic contradictions. The square-root method compresses each risk into a capital amount and a correlation. Moment matching preserves the aggregate mean and variance but smooths the portfolio into a normal distribution. Copula simulation retains the modeled joint event structure and its discrete outcomes. In this example, a 2% probability of a joint 180-thousand loss is enough to place that outcome above the 99th percentile, even though the first two methods produce lower estimates.
A useful comparison should therefore ask what each method preserves and omits, rather than choosing the smallest number. Check that standalone and aggregate estimates use the same confidence level, time horizon and gross-or-net basis; verify the dependence input; and inspect whether the aggregate tail is represented adequately by the approximation. A large divergence is a prompt to investigate model assumptions, not a reason to average the results mechanically.
Check Your Understanding
For each problem, identify the requested aggregation method and keep units and capital conventions consistent.
- Two standalone risk capitals are 60 and 80 thousand, with aggregation correlation 0.25. Using the square-root formula, calculate aggregate risk capital. (Answer: 111.36 thousand)
- Two independent losses are each zero or a fixed amount: risk A is 50 thousand with probability 0.10, and risk B is 30 thousand with probability 0.20. Use normal moment matching and \(z_{0.99}=2.3263\) to estimate aggregate 99% VaR. (Answer: 55.69 thousand)
- Risk A pays 100 thousand with probability 0.04; risk B pays 200 thousand with probability 0.06. Their joint payment probability is 0.025. What is the exact aggregate 99% VaR? (Answer: 300 thousand)
- In a 20,000-draw copula simulation, losses are 0 in 18,300 draws, 60 thousand in 500 draws, 90 thousand in 700 draws, and 150 thousand in 500 draws. Find the empirical 99% VaR using the left-quantile order statistic. (Answer: 150 thousand)