Benchmarking Aggregate Capital Results

Why Benchmark Aggregate Capital?

An aggregate capital result is useful only if its basis is clear. A covariance formula, an exact convolution, a simulation, and a stress exercise can all produce defensible numbers while answering different questions. A benchmark compares their inputs and definitions before interpreting a difference as a model signal.

For a fair comparison, align the loss scope, valuation date, time horizon, gross or net basis, dependence assumptions, and risk measure. A one-year 99% Value at Risk (VaR) should not be compared as if it were the same quantity as a multi-year stress loss or a 99% Tail Value at Risk (TVaR). The methods may also differ in whether they represent parameter uncertainty or only variability under fixed parameters.

Definition: A benchmark of aggregate capital results is a structured comparison of estimates for a defined portfolio and risk measure. It checks both the calculation and the assumptions that determine what the calculation represents. A benchmark is not simply a ranking of methods from “best” to “worst.”

Covariance as a Fast Reference Calculation

For two losses $X_1$ and $X_2$ with finite variances, aggregate loss is $S=X_1+X_2$, and $\operatorname{Var}(S)=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+2\operatorname{Cov}(X_1,X_2)$. For $n$ losses, the corresponding expression sums all individual variances and twice each distinct pairwise covariance. If standard deviations and correlation are used, $\operatorname{Cov}(X_i,X_j)=\rho_{ij}\sigma_i\sigma_j$.

A covariance calculation gives the aggregate mean and variance, not a complete aggregate distribution. A common proxy assumes the aggregate loss is normal and estimates its $p$-quantile as $\mu_S+z_p\sigma_S$, where $z_p$ is the standard normal $p$-quantile. This is exact for a normally distributed aggregate and otherwise is an approximation; matching a mean and variance alone does not establish normality or accurate tail behavior.

Derivation: Expanding the square gives $(S-E[S])^2=\sum_i(X_i-E[X_i])^2+2\sum_{i<j}(X_i-E[X_i])(X_j-E[X_j])$. Taking expectations turns each squared term into a variance and each cross-product into a covariance. This yields the aggregate variance formula. The derivation requires finite second moments; it does not require independence.

Worked Example: Checking a Covariance-Based VaR

A hypothetical insurer models one-year losses $X_1$ and $X_2$ as jointly normal, with means $50 and $30 (in thousands), standard deviations $20 and $15 (in thousands), and correlation $0.40$. Calculate the normal 99% VaR of their sum, then compare it with the independent case. Assume all figures are on the same gross-loss basis and use $z_{0.99}=2.3263$.

Method and assumptions. Joint normality makes the aggregate normal. The covariance is $0.40(20)(15)=120$ (thousand dollars)$^2$, and the aggregate mean is $50+30=80$ thousand dollars.

Variance and standard deviation. $\operatorname{Var}(X_1+X_2)=20^2+15^2+2(120)=400+225+240=865$ (thousand dollars)$^2$. Thus $\sigma_S=\sqrt{865}=29.4109$ thousand dollars.

99% VaR. Using the displayed rounded inputs, $2.3263(29.4109)=68.4186$, so the normal proxy is $80+68.4186=148.4186$ thousand dollars, or $148,418.60 to the cent.

Independent comparison. If correlation is zero, the variance is $400+225=625$, the standard deviation is $25$ thousand dollars, and the 99% VaR is $80+2.3263(25)=138.1575$ thousand dollars, or $138,157.50 to the cent. The difference is $10,261.10, reflecting the covariance term under these assumptions.

Check and common mistake. The aggregate mean remains $80,000 in both cases; dependence changes the variance, not the sum of the means. Omitting the factor of two in the covariance term would incorrectly give variance $745$, rather than $865$, and understate this normal VaR.

Exact Convolution: A Distribution-Level Benchmark

For independent discrete losses, convolution combines their probability mass functions: for each possible total $s$, multiply the probabilities of component outcomes that sum to $s$, then add those products. Unlike a covariance-only calculation, convolution preserves the modeled shape of the aggregate distribution. Independence is essential to multiplying marginal probabilities this way; with dependent losses, the joint probabilities must be used instead.

For a loss $S$, define $\operatorname{VaR}_p(S)=\inf\{s:F_S(s)\ge p\}$. In a discrete distribution, the cumulative probability can jump across $p$, so the VaR is the first loss amount at which the cumulative probability reaches or exceeds $p$. This convention matters when comparing a discrete result with a smooth normal approximation.

Worked Example: Comparing Convolution With a Normal Proxy

A hypothetical portfolio has independent one-year losses $A$ and $B$, measured in thousands of dollars. $A$ is 0 or 10, each with probability 0.50. $B$ is 0 or 20, each with probability 0.50. Find the exact 90% VaR by convolution and compare it with a normal proxy. The stated two-point distributions and independence are the complete model assumptions.

Convolution. The four outcome pairs are $(0,0),(10,0),(0,20),(10,20)$. Independence gives probability $0.50(0.50)=0.25$ to each pair. Therefore $S=A+B$ takes values 0, 10, 20, and 30, each with probability 0.25. Its cumulative probabilities at those amounts are 0.25, 0.50, 0.75, and 1.00. Hence $\operatorname{VaR}_{0.90}(S)=30$ thousand dollars.

Normal proxy. The component means are 5 and 10, so $E[S]=15$. Their variances are $0.5(0-5)^2+0.5(10-5)^2=25$ and $0.5(0-10)^2+0.5(20-10)^2=100$, in squared thousands of dollars. Independence gives aggregate variance 125 and standard deviation $\sqrt{125}=11.1803$ thousand dollars. Using $z_{0.90}=1.2816$, the proxy is $15+1.2816(11.1803)=29.3287$, or $29,328.70 to the cent.

The exact discrete VaR is $30,000, which is $671.30 higher than the normal proxy. This difference is not a calculation error: the proxy substitutes a normal tail shape for a distribution with only four possible outcomes. The common mistake is to treat a mean-and-variance approximation as though it were the exact convolution.

Simulation: Estimate and Sampling Variation

Monte Carlo aggregation draws component losses jointly from a specified model, sums each simulated set, and estimates a risk measure from the resulting aggregate sample. The dependence mechanism must be represented in the joint draws. For $N$ simulated aggregate losses ordered from smallest to largest, an empirical $p$-quantile can be defined as the observation at rank $\lceil Np\rceil$. That convention should be stated because software may use other interpolation rules.

A simulation result is an estimate, not an exact population quantile. Sampling variation is especially relevant in the tail, where relatively few simulations determine the estimate. At a fixed threshold, the estimated exceedance frequency $\hat q$ has approximate standard error $\sqrt{\hat q(1-\hat q)/N}$ when the draws are independent. This formula measures uncertainty in the exceedance frequency at that threshold; it is not itself a standard error for VaR.

Worked Example: Reading a Simulation Benchmark

A hypothetical portfolio has independent annual losses $C$ and $D$, in thousands of dollars. $C=0$ with probability 0.80 and $C=40$ with probability 0.20; $D=0$ with probability 0.90 and $D=100$ with probability 0.10. A simulation uses 5,000 independent draws from these stated distributions. Its aggregate outcome counts are 3,605 at 0, 895 at 40, 405 at 100, and 95 at 140. Find the exact and simulated 95% VaR, and quantify uncertainty in the simulated exceedance frequency above 40.

Exact benchmark. Independence gives probabilities $0.80(0.90)=0.72$ at 0, $0.20(0.90)=0.18$ at 40, $0.80(0.10)=0.08$ at 100, and $0.20(0.10)=0.02$ at 140. The cumulative probability is 0.72 at 0, 0.90 at 40, and 0.98 at 100. Thus exact $\operatorname{VaR}_{0.95}=100$ thousand dollars.

Simulation estimate. The supplied counts total $3,605+895+405+95=5,000$. The empirical cumulative count through 40 is $3,605+895=4,500$. The 95% rank is $\lceil 5,000(0.95)\rceil=4,750$. Ranks 4,501 through 4,905 have value 100, so the empirical VaR is also $100,000.

Exceedance check. There are $405+95=500$ simulated losses above 40, giving $\hat q=500/5,000=0.10$. Its approximate standard error is $\sqrt{0.10(0.90)/5,000}=0.00424$, or 0.424 percentage points. The modeled exceedance probability above 40 is $0.08+0.02=0.10$, consistent with the simulated frequency. Agreement here supports the implementation check; it does not prove that the model assumptions are correct. The common mistake is to call this exceedance-frequency standard error a VaR standard error.

Stress Estimates Are Not Automatically Quantiles

A stress estimate applies a specified adverse condition to the portfolio and calculates the resulting loss. Unless a probability distribution or probability for that condition is supplied, the result is a conditional scenario loss, not a VaR or TVaR. Stress results are still valuable benchmarks: they can test whether modeled capital is informative under a coherent adverse scenario, but they cannot be compared numerically as though both figures had the same probability meaning.

Worked Example: Comparing a Normal Benchmark With a Stress Loss

Use the jointly normal hypothetical losses from the first example: means $50,000 and $30,000, standard deviations $20,000 and $15,000, and correlation 0.40. A separately specified one-year stress scenario produces losses of $100,000 and $70,000 for the two risks. Assume the scenario losses are simultaneous, on the same gross basis, and include no recoveries. Compare the stress total with the model’s 99% VaR.

Model benchmark. The covariance calculation gives a 99% normal VaR of $148,418.60, as computed above.

Stress calculation. The aggregate scenario loss is $100,000+70,000=$$170,000. The stress total exceeds the modeled 99% VaR by $170,000-148,418.60=$$21,581.40.

The difference is a useful comparison, but it does not mean the stress loss is the 99.9% VaR or any other quantile: no probability was assigned to the scenario. A common mistake is to label the $170,000 result “capital at 99%” solely because it is larger than a 99% model estimate.

A Practical Benchmarking Sequence

A disciplined comparison makes differences interpretable rather than merely visible. Use the following sequence before drawing a conclusion about aggregate capital:

Align the quantity.
Confirm that each result covers the same risks, horizon, valuation basis, gross or net treatment, and risk measure. Record any differences explicitly.

Recompute a transparent reference.
Check the mean, variances, covariance terms, and any normal quantile calculation. Verify units and whether correlations have been converted to covariances correctly.

Compare distribution-level results.
Where feasible, use exact convolution for a small discrete model or an independently checked simulation for a more complex joint model. Keep dependence assumptions visible.

Separate model output from scenario output.
Label probabilistic capital measures and conditional stress losses distinctly. A stress result becomes a quantile only if the scenario has a defensible probability interpretation within a specified model.

Explain the residual difference.
Attribute remaining gaps to identifiable features such as distribution shape, dependence, tail representation, finite simulation size, or different scope—not to a vague claim that one method is inherently conservative.

Benchmarking supports model validation and governance because it exposes mismatches that a single headline capital number can conceal. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. A sound conclusion states both the numerical result and what that result assumes; it does not mistake computational precision for certainty about the model.

Key takeaway: Covariance is a fast check on aggregate moments, convolution gives the exact distribution when its joint-probability assumptions are satisfied, simulation estimates results under a specified stochastic model, and stress testing reports losses under specified adverse conditions. Compare like with like, and label unlike outputs accurately.

Check Your Understanding

For each problem, keep the stated horizon, dependence assumptions, and risk measure in view.

Two losses have standard deviations 12 and 9 (thousands of dollars) and correlation 0.25. What is their covariance, and what covariance contribution enters the variance of their sum? (Answer: Covariance = 27 thousand-dollars squared; contribution = 54 thousand-dollars squared.)
Independent losses $M$ and $N$ each equal 0 or 8 with probability 0.50. Find the exact 75% VaR of their sum using the stated lower-quantile definition. (Answer: $8.)
A simulation has $N=2,000$ independent aggregate draws, of which 50 exceed a fixed threshold. Estimate the exceedance probability and its approximate standard error. (Answer: 0.0250; standard error $\sqrt{0.025(0.975)/2,000}=0.00349$, or 0.349 percentage points.)
A stress scenario produces simultaneous losses of $64,000 and $51,000. No scenario probability is assigned. What is the aggregate stress loss, and can it be called a 99% VaR? (Answer: $115,000; no, it is a conditional scenario loss, not a quantile without a probability model.)
A covariance model gives a one-year 99% VaR of $42 million, while a simulation gives a one-year 99% TVaR of $48 million for the same stated portfolio. Name the first comparability issue to resolve. (Answer: The risk measures differ—VaR versus TVaR—so the results are not directly comparable even though the horizon and portfolio match.)

What You'll Learn