Why Separate the Two Sources of Uncertainty?
An aggregate capital estimate depends both on the values assigned to model parameters and on the model used to combine risks. These are different questions. For example, an analyst might use a normal copula to represent dependence but be unsure whether the correlation is 0.20 or 0.35. That is parameter uncertainty. The analyst might instead be unsure whether a normal copula is an adequate representation of tail dependence at all. That is model uncertainty.
The distinction matters because the remedies differ. Resampling an estimated correlation can show how much a chosen model's result changes as that parameter varies. It cannot show whether the chosen dependence structure is appropriate. To examine model uncertainty, compare defensible alternative model specifications, such as different dependence structures or different aggregate-loss distributions. Any resulting range should be interpreted as conditional on the alternatives considered, not as a complete measure of all possible uncertainty.
Conditional Capital and Parameter Resampling
Suppose two losses \(X_1\) and \(X_2\) have fixed marginal distributions and are jointly normal, with standard deviations \(\sigma_1\) and \(\sigma_2\) and correlation \(\rho\). Then their sum \(S=X_1+X_2\) is normal. Its mean and variance are
For a continuous normal aggregate loss, the \(p\)-quantile, denoted \(\operatorname{VaR}_p(S)\), is the value \(s\) such that \(P(S\le s)=p\). Thus, conditional on a specified \(\rho\),
where \(z_p\) is the \(p\)-quantile of the standard normal distribution. This formula assumes the joint normal model, fixed marginal means and standard deviations, and the stated correlation. It is not a general formula for arbitrary marginal distributions or dependence structures.
Worked Example: Conditional Capital at a Resampled Correlation
A hypothetical insurer models two annual losses, in thousands of dollars, as jointly normal. Their means are 20 and 30, their standard deviations are 10 and 15, and one resampled correlation estimate is 0.20. Using \(z_{0.99}=2.3263\), calculate the conditional 99th-percentile aggregate loss.
Method. Use the normal-sum variance formula and then the normal quantile formula. The resampled correlation is treated as fixed for this conditional calculation; this is not yet a calculation of the full uncertainty across resamples.
The aggregate mean is
The aggregate variance, in squared thousands of dollars, is
Therefore, the aggregate standard deviation is \(\sqrt{385}=19.6214\) thousand dollars (rounded here for display). The conditional 99th percentile is
Rounded to two decimal places, the conditional 99th-percentile aggregate loss is 95.65 thousand dollars. The check is that the quantile exceeds the mean, as it should for a 99th percentile of a nondegenerate normal distribution.
Exam trap. Omitting the product of the standard deviations in the covariance term gives the incorrect variance \(100+225+2(0.20)=325.4\) and a quantile of about 91.95. The covariance is \(\rho\sigma_1\sigma_2\), not \(\rho\) alone. The correct variance is 385.
In practice, a parameter resampling procedure might produce many plausible values of \(\rho\), for example by resampling observations used to estimate dependence. Recalculate the conditional capital result for each resampled parameter value while holding the model specification and other inputs fixed. The resulting distribution of conditional estimates describes sensitivity to the estimated parameter, subject to the validity of the resampling procedure.
Worked Example: Summarizing Resampled Capital Estimates
For two hypothetical annual losses, in thousands of dollars, assume a fixed jointly normal model with means 40 and 25 and standard deviations 12 and 20. Three equally weighted illustrative resampled correlations are 0.10, 0.30, and 0.50. Using \(z_{0.99}=2.3263\), calculate the conditional 99th-percentile aggregate loss for each resample and summarize their variation.
Method. The aggregate mean is fixed at \(40+25=65\). For each correlation, calculate \(12^2+20^2+2\rho(12)(20)=544+480\rho\), then apply the normal quantile formula.
At \(\rho=0.10\), the variance is \(544+480(0.10)=592\), the standard deviation is \(\sqrt{592}=24.3311\), and the conditional quantile is \(65+2.3263(24.3311)=121.60\) thousand dollars.
At \(\rho=0.30\), the variance is \(544+480(0.30)=688\), the standard deviation is \(\sqrt{688}=26.2298\), and the conditional quantile is \(65+2.3263(26.2298)=126.02\) thousand dollars.
At \(\rho=0.50\), the variance is \(544+480(0.50)=784\), the standard deviation is \(\sqrt{784}=28\), and the conditional quantile is \(65+2.3263(28)=130.14\) thousand dollars.
The equally weighted average of the three conditional quantiles is \((121.60+126.02+130.14)/3=125.92\) thousand dollars. Their range is \(130.14-121.60=8.54\) thousand dollars. These summaries describe variation in the conditional capital estimate across the specified resamples. They are not automatically a 99th percentile or other risk measure of a predictive aggregate-loss distribution.
Exam trap. A positive change in correlation raises this model's variance because both standard deviations are positive, but the spread across only three resamples is not a confidence interval unless the resampling design and interval construction justify that interpretation. Here, the three equally weighted values are illustrative inputs, not a claim about a sampling distribution.
Why Averaging Conditional Results Is Not Enough
Parameter resampling can be used in more than one way. One approach reports the distribution of conditional capital estimates across parameter draws. Another constructs a predictive mixture: first draw a parameter value, then draw a loss from the model conditional on that value. The predictive mixture has its own aggregate-loss distribution and should be evaluated directly if the intended quantity is its VaR.
A useful identity for understanding the mixture is the law of total variance. Let \(\theta\) be an uncertain parameter and \(S\) the aggregate loss conditional on it. Write \(m(\theta)=E[S\mid\theta]\). Decompose the deviation from the unconditional mean:
Squaring and taking expectations gives the sum of two variance terms plus a cross term. The cross term is zero because \(E[S-m(\theta)\mid\theta]=0\); consequently its expectation after multiplication by any function of \(\theta\) is also zero. Therefore,
The first term is average variability within the conditional models; the second is variability in their conditional means. This identity concerns variance, not VaR. In general, the VaR of a mixture is not the average of the conditional VaRs, because quantiles do not combine linearly.
Worked Example: Adding Parameter Variation to Predictive Variance
A hypothetical aggregation analysis has two equally likely parameter states. Conditional on the first state, aggregate loss has mean 100 and variance 400; conditional on the second, it has mean 120 and variance 400. All amounts are in thousands of dollars. Calculate the unconditional mean and variance of the predictive mixture.
Method. Apply the law of total expectation and the law of total variance, treating the parameter state as a random variable with probability one-half in each state.
The unconditional mean is
The expected conditional variance is \(\tfrac12(400)+\tfrac12(400)=400\). The conditional means, 100 and 120, differ from their average 110 by \(-10\) and \(10\), so their variance is \(\tfrac12(10^2)+\tfrac12(10^2)=100\). Thus,
The predictive standard deviation is \(\sqrt{500}=22.36\) thousand dollars, rounded to two decimal places. The result exceeds the conditional variance of 400 because parameter variation adds 100 of variance. This example calculates predictive moments only; those moments alone do not determine a predictive VaR without a distributional assumption.
Common mistake. Reporting variance 400 as the predictive variance ignores variation between the two conditional means. The correction is to add the between-state variance, giving 500.
Quantifying Model Uncertainty
Model uncertainty is assessed by comparing alternative, clearly specified models rather than by varying a parameter inside only one model. Alternatives may differ in marginal distribution, dependence structure, treatment of common events, or aggregation approximation. To make the comparison informative, hold the risk measure and confidence level fixed, and identify which assumptions change.
The next example compares two hypothetical aggregate-loss distributions with the same mean and variance but different tail shapes. This isolates the consequence of distributional form in the comparison. The distributions are illustrative alternatives; matching two moments does not establish that either is a suitable model for a particular portfolio.
Worked Example: Comparing Alternative Aggregate-Loss Models
Two hypothetical models specify aggregate loss \(S\), in thousands of dollars. Model A assumes \(S\sim N(100,400)\), where the second normal parameter is the variance. Model B assigns probability 0.98 to 97.1429 and probability 0.02 to 240. Compare their 99th-percentile losses and verify that the models have the same mean and variance.
Method. Calculate each model's mean and variance directly, then use the definition of a quantile. For Model A, use \(z_{0.99}=2.3263\). For Model B, use its discrete cumulative distribution.
For Model A, the mean is 100 and the variance is 400. Its 99th percentile is
or 146.53 thousand dollars rounded to two decimal places.
For Model B, its mean is
The lower outcome is approximately \(100-2.8571\), and the upper outcome is \(100+140\). The variance is therefore approximately
For Model B, the cumulative probability at 97.1429 is 0.98, which is less than 0.99. The cumulative probability at 240 is 1.00, so its 99th percentile is 240 thousand dollars. The two models have equal means and variances, but their 99th-percentile results differ by \(240-146.53=93.47\) thousand dollars. The comparison demonstrates that matching the first two moments does not remove model uncertainty in the tail.
Common mistake. Substituting the shared mean and variance into the normal VaR formula for both models would incorrectly assign Model B a 99th percentile of 146.53. That formula applies to Model A's normal distribution, not to Model B's discrete distribution.
Practical Interpretation and Exam Traps
A clear analysis labels each result by what was held fixed. “Conditional 99th percentile under the normal model at correlation 0.30” is more informative than “99th-percentile capital.” When resampling, report the parameter being resampled, what remains fixed, and how the resulting estimates are summarized. When comparing models, state the changed assumptions and avoid implying that the alternatives exhaust all plausible structures.
Several distinctions are especially important:
- Parameter range is not model range. A range across correlation resamples describes variation in that parameter under a chosen model. It does not test the adequacy of the dependence model.
- Conditional VaR is not predictive-mixture VaR. The average of conditional VaRs is generally not the VaR of the mixture distribution. Construct the mixture distribution, or simulate it, when that is the target quantity.
- Moment agreement does not imply tail agreement. Models with the same mean and variance can have materially different high quantiles, as the discrete-versus-normal comparison shows.
- Do not overstate a limited comparison. A model range covers the alternatives actually examined. It is not a probability interval unless probabilities and a valid construction for the alternatives have been specified.
Separating these sources of uncertainty supports better decisions about what additional analysis is needed. More data may reduce uncertainty in an estimated correlation; they do not, by themselves, settle whether the selected model captures relevant tail behavior. Conversely, comparing alternative models does not quantify the sampling uncertainty of the parameters inside each one.
Check Your Understanding
For each problem, identify whether the calculation concerns parameter uncertainty or model uncertainty, and check the stated assumptions before using a formula.
- Two jointly normal annual losses, in thousands of dollars, have means 12 and 18, standard deviations 5 and 12, and correlation 0.25. Calculate the conditional 95th-percentile aggregate loss using \(z_{0.95}=1.6449\). (Answer: 53.21 thousand dollars.)
- Two jointly normal losses have means 10 and 15, standard deviations 6 and 8, and resampled correlations 0 and 0.40. Calculate each conditional 99th-percentile aggregate loss using \(z_{0.99}=2.3263\), then their equally weighted average. (Answer: 48.26 and 52.37 thousand dollars; average 50.32 thousand dollars.)
- An aggregate loss has mean 80 and variance 225 under two alternative models. Model A is normal; Model B has probability 0.94 at \(80-\sqrt{225(0.06/0.94)}\) and probability 0.06 at \(80+\sqrt{225(0.94/0.06)}\). Find each model's 95th percentile using \(z_{0.95}=1.6449\) for Model A. (Answer: Model A, 104.67; Model B, approximately 139.37, thousand dollars.)
- A parameter has two equally likely states. The conditional aggregate means are 50 and 70, and the conditional variance is 900 in each state. Use the law of total variance to find the predictive mean and variance. (Answer: Mean 60 thousand dollars; variance 1,000 \((\text{thousand dollars})^2\).)