Aggregation With Sparse Dependence Data

Why Sparse Dependence Data Require Explicit Assumptions

Aggregate loss estimates can depend materially on how risks move together. Yet an organization may have limited observations of joint losses: a new product has little history, a severe event has occurred only rarely, or business units have recorded losses on different bases. In those circumstances, a single estimated correlation can suggest more precision than the evidence supports.

A defensible response is to separate what is observed from what is assumed. Specify plausible dependence bounds, use comparable risks as proxies only with an explicit rationale, and calculate results under multiple feasible scenarios. The purpose is not to claim that one scenario is known to be correct. It is to make the consequences of limited evidence visible and reviewable.

Definition: A sparse-dependence estimate is an aggregate risk estimate made when direct evidence about joint risk behavior is limited. A dependence bound is a stated lower or upper value, or range of values, for an unknown dependence parameter. A proxy relationship is an observed dependence relationship for risks judged relevant to, but not identical to, the risks being modeled. A scenario is a specified set of dependence assumptions used to calculate a corresponding aggregate result.

These are modeling choices, not estimates automatically justified by the amount of data available. A range chosen by expert judgment is not a statistical confidence interval unless it has been constructed and interpreted using a suitable statistical method. Likewise, a proxy correlation is evidence about the proxy risks; it is not direct evidence about the target risks.

Bound the Unknown Relationship Before Aggregating

For two risks with standard deviations $\sigma_1$ and $\sigma_2$, and correlation $\rho$, the variance of their sum follows from the variance identity:

\operatorname{Var}(X_1+X_2) = \sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2.

If $\rho$ is unknown, an actuary can calculate the aggregate result over a defensible range rather than silently substitute zero or another convenient value. The mathematical bound $-1\leq \rho\leq 1$ is always necessary, but practical bounds may be narrower when supported by the nature of the risks, relevant observations, or documented expert judgment.

A useful scenario set often distinguishes a central assumption from less favorable and more favorable alternatives. The scenarios should be meaningful for the decision at hand, and their interpretation should be stated. For example, a “stress” scenario may represent a plausible adverse relationship, not the worst mathematically possible correlation. Do not assign scenario probabilities unless there is a defensible basis for doing so.

Worked Example: Bounding a Two-Risk Aggregate

A hypothetical insurer models two annual loss amounts as jointly normal. Their means are $12$ million and $9$ million dollars, and their standard deviations are $6$ million and $8$ million dollars. Sparse observations and underwriting judgment support examining correlations from $0$ to $0.50$. Calculate aggregate standard deviation and 99% VaR at correlations $0$, $0.25$, and $0.50$. Use $z_{0.99}=2.3263$, and report money in millions to two decimals.

Method and assumptions. Under the stated joint normal model, the aggregate mean is the sum of the means, $12+9=21$ million. The aggregate variance uses the two-risk variance identity. For a normal variable with mean $\mu$ and standard deviation $\sigma$, the 99% VaR is $\mu+2.3263\sigma$. The scenario endpoints are assumptions for analysis; the calculations do not establish that the true correlation lies in this range.

At $\rho=0$. The variance is $6^2+8^2+2(0)(6)(8)=36+64=100$, so the standard deviation is $\sqrt{100}=10$ million. VaR is $21+2.3263(10)=44.263$ million, or $\$44.26$ million.

At $\rho=0.25$. The variance is $36+64+2(0.25)(6)(8)=100+24=124$. The standard deviation is $\sqrt{124}=11.1355$ million. VaR is $21+2.3263(11.1355)\approx46.905$ million, or $\$46.91$ million.

At $\rho=0.50$. The variance is $36+64+2(0.50)(6)(8)=100+48=148$. The standard deviation is $\sqrt{148}=12.1655$ million. VaR is $21+2.3263(12.1655)\approx49.301$ million, or $\$49.30$ million.

Check and common mistake. The resulting VaRs increase as the assumed correlation increases, consistent with the positive covariance term. A common error is to report the standard deviation as the VaR, or to apply the normal quantile to the variance. The correct calculation uses the square root of the variance, then adds the quantile multiple to the aggregate mean.

Check That a Multi-Risk Scenario Is Feasible

Pairwise correlations cannot always be selected independently. For three standardized risks, a proposed correlation matrix must be positive semidefinite. In particular, its determinant and all principal minors must be nonnegative. A set of plausible individual correlations can therefore still imply an impossible joint model.

The determinant gives a useful bound when two correlations are known and the third is uncertain. For the matrix below, the unknown correlation is $r$:

R= \begin{pmatrix} 1 & 0.5 & 0.2\\ 0.5 & 1 & r\\ 0.2 & r & 1 \end{pmatrix}, \qquad \det(R)=1+2(0.5)(0.2)r-(0.5)^2-(0.2)^2-r^2 =0.71+0.2r-r^2.

Because the diagonal entries are one, the one-by-one principal minors are positive. The two-by-two principal minors are $1-0.5^2=0.75$, $1-0.2^2=0.96$, and $1-r^2$. The last requires $-1\leq r\leq1$. Together with the determinant condition, these tests identify feasible values for this three-risk matrix.

Worked Example: Testing Three Dependence Scenarios

Three hypothetical annual losses have standard deviations $6$, $5$, and $3$ million dollars. Their correlations are $0.5$ between risks 1 and 2, $0.2$ between risks 1 and 3, and an uncertain value $r$ between risks 2 and 3. Test $r=0.1$, $0.4$, and $0.7$ for matrix feasibility. Then calculate the variance and standard deviation of the sum in each scenario.

Method. Use the determinant expression above and check the two-by-two minor $1-r^2$. Then apply the variance-of-a-sum identity, including all three covariance terms.

At $r=0.1$, the determinant is $0.71+0.2(0.1)-(0.1)^2=0.71+0.02-0.01=0.72$, and $1-r^2=0.99$. Both are nonnegative, so the matrix is feasible. The aggregate variance is $6^2+5^2+3^2+2[(0.5)(6)(5)+(0.2)(6)(3)+(0.1)(5)(3)]$ $=70+2(15+3.6+1.5)=110.2$ squared millions. The standard deviation is $\sqrt{110.2}=10.4976$ million.

At $r=0.4$, the determinant is $0.71+0.08-0.16=0.63$, and $1-r^2=0.84$, so the matrix is feasible. The aggregate variance is $70+2(15+3.6+6)=119.2$ squared millions, giving standard deviation $\sqrt{119.2}\approx10.9174$ million.

At $r=0.7$, the determinant is $0.71+0.14-0.49=0.36$, and $1-r^2=0.51$, so this matrix is also feasible. The aggregate variance is $70+2(15+3.6+10.5)=128.2$ squared millions, giving standard deviation $\sqrt{128.2}\approx11.3225$ million.

Check and common mistake. All three determinants and all three relevant two-by-two minors are positive, so all stated matrices are feasible. A sign error in the determinant’s cross term would incorrectly give $0.71-0.4r-r^2$. The correct cross term is $+2(0.5)(0.2)r=+0.2r$; using the incorrect expression could wrongly reject a valid scenario. The variance checks also show the expected values: $110.2$, $119.2$, and $128.2$ squared millions.

Use Proxies With a Clear Transfer Rationale

A proxy can help inform an assumption when direct target-risk data are scarce. Before transferring a proxy relationship, compare the risks’ drivers, exposure bases, time periods, loss definitions, and event conditions. A correlation between annual losses is not interchangeable with a correlation between event counts or between differently transformed variables. A proxy that shares only a broad label with the target risks may provide little useful evidence.

When the proxy is relevant but imperfect, it is often more defensible to use it to guide a scenario range than to copy its observed correlation exactly. Document why the target relationship might be higher or lower, and preserve scenarios that reflect that uncertainty. If proxy data are themselves limited or unstable, that is a reason to widen or qualify the assumptions, not to imply greater certainty.

Worked Example: Translating a Proxy Into Scenarios

A hypothetical insurer has little joint history for two target portfolios. A comparable pair of portfolios has an observed annual-loss correlation of $0.35$. The target portfolios have an additional shared exposure, so the actuary selects $0.20$, $0.45$, and $0.70$ as low, central, and stress assumptions. The target losses are jointly normal, with means $18$ and $10$ million dollars and standard deviations $7$ and $5$ million dollars. Calculate aggregate 99.5% VaR using $z_{0.995}=2.5758$, reporting millions to two decimals.

Method and assumptions. The proxy value motivates the central scenario but does not determine it: the increase to $0.45$ is an explicit judgment about the additional shared exposure. The three scenarios are not probability-weighted. The aggregate mean is $18+10=28$ million, and the variance is $7^2+5^2+2\rho(7)(5)=74+70\rho$.

At $\rho=0.20$, variance is $74+14=88$, and standard deviation is $\sqrt{88}=9.3808$. VaR is $28+2.5758(9.3808)\approx52.163$, or $\$52.16$ million.

At $\rho=0.45$, variance is $74+31.5=105.5$, and standard deviation is $\sqrt{105.5}\approx10.2713$. VaR is $28+2.5758(10.2713)\approx54.457$, or $\$54.46$ million.

At $\rho=0.70$, variance is $74+49=123$, and standard deviation is $\sqrt{123}\approx11.0905$. VaR is $28+2.5758(11.0905)\approx56.567$, or $\$56.57$ million.

Check and common mistake. Each VaR uses the same aggregate mean and quantile; only the assumed dependence changes. It would be a mistake to treat the proxy correlation $0.35$ as if it were a measured target correlation, or to describe the three VaRs as a confidence interval. They are conditional model results under stated assumptions.

Why the Variance Formula Holds—and What It Does Not Say

The covariance term in aggregation follows directly from the definition of variance. For two risks with finite second moments, expand the squared deviation of their sum:

\begin{aligned} \operatorname{Var}(X_1+X_2) &=E\left[\{(X_1-E[X_1])+(X_2-E[X_2])\}^2\right]\\ &=\operatorname{Var}(X_1)+\operatorname{Var}(X_2) +2\operatorname{Cov}(X_1,X_2)\\ &=\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2. \end{aligned}

The last equality uses $\operatorname{Cov}(X_1,X_2)=\rho\sigma_1\sigma_2$. This identity does not require normality; finite variances are sufficient. Normality is needed in the worked examples only to convert aggregate mean and standard deviation into the stated normal-model VaR. In non-normal models, matching the first two moments does not in general determine a tail quantile.

For more than two risks, the same expansion includes every pairwise covariance. A correlation matrix should be checked for feasibility before it is used in a covariance-based aggregation calculation. Even a feasible matrix, however, does not establish that its correlations are empirically supported or that it captures all relevant tail behavior.

Communicating Sparse-Data Results

A useful report distinguishes the evidence, the judgment, and the calculation. It should identify the source and limitations of any proxy, explain the rationale for lower, central, and stress assumptions, confirm matrix feasibility where relevant, and show how aggregate results change across scenarios. State the loss basis, time horizon, distributional assumptions, and risk measure so a reviewer can reproduce the result.

Avoid presenting a scenario range as a statistical interval, calling the central case “best estimate” without support, or assuming independence merely because joint observations are unavailable. Independence is a dependence assumption too. If the data do not distinguish among scenarios, report that limitation and use the resulting range to inform decisions rather than disguising uncertainty as a precise point estimate.

Key takeaway: With sparse dependence data, make assumptions visible: establish defensible bounds, use proxies with a transfer rationale, verify that proposed correlation matrices are feasible, and report aggregate results conditionally under multiple scenarios.

Check Your Understanding

For each problem, state the method and retain sufficient precision in intermediate calculations.

Two loss amounts have standard deviations of $3$ and $4$ million dollars and assumed correlation $0.25$. Calculate the aggregate variance and standard deviation. (Answer: $31$ squared million dollars; $\sqrt{31}\approx5.5678$ million dollars.)
Two risks have standard deviations $2$ and $5$ million dollars. A defensible correlation range is $[-0.2,0.6]$. Calculate the corresponding lower and upper aggregate variances and identify which endpoint gives each. (Answer: lower $25$, at $\rho=-0.2$; upper $41$, at $\rho=0.6$, both squared million dollars.)
A three-risk correlation matrix has $\rho_{12}=0.5$, $\rho_{13}=0.2$, and $\rho_{23}=-0.8$. Using $\det(R)=0.71+0.2r-r^2$, determine whether it is feasible. (Answer: No; the determinant is $0.71-0.16-0.64=-0.09$, so the matrix is not positive semidefinite.)
Two risks have standard deviations $4$ and $3$ million dollars. A relevant proxy suggests correlation $0.4$, which is adopted as the stated target assumption. Calculate aggregate variance and explain why this does not verify the target correlation empirically. (Answer: $16+9+2(0.4)(4)(3)=34.6$ squared million dollars; the value is conditional on transferring the proxy assumption, not a direct target observation.)
Two annual losses are jointly normal with means $5$ and $8$ million dollars, standard deviations $2$ and $3$ million dollars, and assumed correlation $0.5$. Using $z_{0.95}=1.6449$, calculate aggregate 95% VaR. (Answer: variance $=4+9+2(0.5)(2)(3)=19$; VaR $=13+1.6449\sqrt{19}\approx\$20.17$ million.)

What You'll Learn