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CAS Exam 9 · Aggregating Risks and Diversification · Tutorial 476 of 995

Square-Root Aggregation of Standalone Capital

Apply correlation-based square-root aggregation to standalone capital measures, check its assumptions, and interpret the resulting aggregate capital.

Advanced 10 min read

What You'll Learn

  • Define standalone capital and the correlation inputs required for square-root aggregation.
  • Derive the formula from the variance of a sum.
  • Aggregate capital across multiple risk categories with pairwise correlations.
  • Check whether a proposed correlation matrix is mathematically valid.
  • Identify when the formula is exact and when it is only an approximation.
  • Avoid common errors involving the factor of two, expected losses, and inconsistent bases.

Why Standalone Capital Needs an Aggregation Rule

An insurer may measure capital separately for several risks, such as property, liability, and market risk. Adding those standalone amounts assumes, implicitly, that every risk reaches its capital level at the same time. At the other extreme, treating the risks as independent may understate aggregate capital when losses move together. A correlation-based square-root formula provides a structured way to reflect their dependence.

The formula is useful only when its inputs are interpreted consistently. Standalone capital amounts are not automatically standard deviations, and a correlation matrix for one set of loss variables cannot be substituted without checking that it describes the same risks, time horizon, and loss basis.

Definition: Let \(K_i\) denote the standalone capital for risk \(i\), measured on a common basis, and let \(\rho_{ij}\) denote the correlation assigned to risks \(i\) and \(j\), with \(\rho_{ii}=1\). The correlation-based square-root aggregate is $$ K_{\mathrm{agg}}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{n}\rho_{ij}K_iK_j} =\sqrt{\sum_{i=1}^{n}K_i^2+2\sum_{i<j}\rho_{ij}K_iK_j}. $$

The second expression separates the standalone squares from the pairwise interaction terms. Each distinct pair appears once in the sum over \(i<j\), but its contribution is multiplied by 2. That factor is essential.

Why the Formula Holds

The formula follows directly from the variance of a sum when standalone capital is proportional to standard deviation using the same proportionality factor for every risk. Let \(L_i\) be the loss for risk \(i\), and center it by subtracting its mean. Write \(Z_i=L_i-E[L_i]\), so \(E[Z_i]=0\). If the risks are aggregated as \(Z=\sum_i Z_i\), then the variance of their sum is the sum of all variances and covariances.

1
Expand the variance.
\(\operatorname{Var}(Z)=\sum_i\operatorname{Var}(Z_i)+2\sum_{i<j}\operatorname{Cov}(Z_i,Z_j)\). The factor of 2 arises because the covariance of each pair appears in both orders in the full double sum.
2
Express covariance using correlation.
For each pair, \(\operatorname{Cov}(Z_i,Z_j)=\rho_{ij}\sigma_i\sigma_j\), where \(\sigma_i=\sqrt{\operatorname{Var}(Z_i)}\). Substitution gives \(\operatorname{Var}(Z)=\sum_i\sigma_i^2+2\sum_{i<j}\rho_{ij}\sigma_i\sigma_j\).
3
Scale standard deviations into capital.
If \(K_i=q\sigma_i\) for the same positive constant \(q\) across all risks, then \(q\sqrt{\operatorname{Var}(Z)}=\sqrt{\sum_i K_i^2+2\sum_{i<j}\rho_{ij}K_iK_j}\). The left side is the aggregate capital under the same scaling rule.

For example, if the centered losses are jointly normal and capital is the one-sided quantile of unexpected loss at a common confidence level, then \(q\) is the corresponding standard normal quantile. In that setting, the formula is exact for that capital measure. More generally, it may be used as a prescribed approximation, but it is not an identity for arbitrary standalone measures such as VaR or TVaR of differently shaped distributions. It is also not a formula for adding expected losses: expected loss and capital for unexpected loss must not be mixed.

Worked Aggregations

Worked Example: Three Risks Under a Common Normal Capital Measure

A hypothetical insurer measures three centered loss risks—property, liability, and market—using one-year 99% normal unexpected-loss capital. Assume the loss vector is jointly normal, the common quantile is \(q=2.326\), and the standalone capital amounts are \(K_1=30\), \(K_2=40\), and \(K_3=50\) million dollars. The loss correlations are \(\rho_{12}=0.25\), \(\rho_{13}=0.10\), and \(\rho_{23}=0.40\). Calculate aggregate capital. No interest-rate or mortality assumptions are relevant to this one-year loss calculation.

Method. Use the square-root formula because each standalone amount equals the same normal quantile \(q\) times that risk's standard deviation. The correlation matrix is \(\begin{pmatrix}1&0.25&0.10\\0.25&1&0.40\\0.10&0.40&1\end{pmatrix}\). Its 1-by-1 principal minors are 1. Its 2-by-2 principal minors are \(1-0.25^2=0.9375\), \(1-0.10^2=0.99\), and \(1-0.40^2=0.84\). Its determinant is \(1+2(0.25)(0.10)(0.40)-0.25^2-0.10^2-0.40^2=0.7875\). These are nonnegative, so the matrix is positive semidefinite.

Calculate the diagonal contribution. \(30^2+40^2+50^2=900+1{,}600+2{,}500=5{,}000\), in squared millions of dollars.

Calculate the pair contributions. The three terms, including the factor of 2, are \(2(0.25)(30)(40)=600\), \(2(0.10)(30)(50)=300\), and \(2(0.40)(40)(50)=1{,}600\). Their sum is \(2{,}500\).

Therefore \(K_{\mathrm{agg}}=\sqrt{5{,}000+2{,}500}=\sqrt{7{,}500}=86.6025\) million dollars. Rounded to three decimals in millions, aggregate capital is $86.603 million. As a check, the standalone standard deviations are \(30/2.326\), \(40/2.326\), and \(50/2.326\) million; aggregating their variance and multiplying the resulting standard deviation by 2.326 gives the same square-root expression.

Common mistake avoided. Omitting the factor of 2 gives \(5{,}000+300+150+800=6{,}250\), and a false result of \(\sqrt{6{,}250}=79.057\) million. Each pair's covariance contributes twice to the variance of a sum; the correct result is $86.603 million.

Worked Example: A Zero-Correlation Pair and Matrix Validity

A hypothetical insurer uses one-year 99.5% normal unexpected-loss capital for three centered risks: specialty property, professional liability, and cyber loss. The loss vector is jointly normal with common quantile \(q=2.5758\). Standalone capital, in millions of dollars, is \(K_1=20\), \(K_2=35\), and \(K_3=45\). The pairwise correlations are \(\rho_{12}=0\), \(\rho_{13}=0.50\), and \(\rho_{23}=0.25\). Find aggregate capital and verify that the correlation matrix is valid. Interest and mortality assumptions are not applicable.

The matrix is \(\begin{pmatrix}1&0&0.50\\0&1&0.25\\0.50&0.25&1\end{pmatrix}\). Its 1-by-1 principal minors are all 1, and its 2-by-2 principal minors are \(1-0^2=1\), \(1-0.50^2=0.75\), and \(1-0.25^2=0.9375\), all nonnegative. The determinant is \(1+2(0)(0.50)(0.25)-0^2-0.50^2-0.25^2=0.6875\), also nonnegative. Thus the matrix is positive semidefinite.

The diagonal contribution is \(20^2+35^2+45^2=400+1{,}225+2{,}025=3{,}650\). The pair terms are \(2(0)(20)(35)=0\), \(2(0.50)(20)(45)=900\), and \(2(0.25)(35)(45)=787.5\). The total under the square root is \(3{,}650+900+787.5=5{,}337.5\).

Thus \(K_{\mathrm{agg}}=\sqrt{5{,}337.5}=73.0582\) million dollars, or $73.058 million rounded to three decimals in millions. The check is that the zero correlation removes only the first pair's covariance term; it does not remove either risk's own squared capital or its interaction with the third risk.

Common mistake avoided. A zero pairwise correlation does not mean one risk can be dropped. Risk 1 still contributes \(20^2=400\), and its correlation with risk 3 contributes 900.

Worked Example: Inferring a Pairwise Correlation

For a hypothetical two-risk portfolio, centered losses are jointly normal and use one-year 99% unexpected-loss capital with \(q=2.326\). Standalone capital is $60 million for equipment breakdown and $80 million for cyber risk. The proposed aggregate capital is $110 million. Find the implied correlation and assess whether it is feasible. Interest and mortality assumptions do not apply.

For two risks, \(K_{\mathrm{agg}}^2=K_1^2+K_2^2+2\rho_{12}K_1K_2\). Substituting the stated amounts gives \(110^2=60^2+80^2+2\rho_{12}(60)(80)\), or \(12{,}100=3{,}600+6{,}400+9{,}600\rho_{12}\). Hence \(2{,}100=9{,}600\rho_{12}\), so \(\rho_{12}=2{,}100/9{,}600=0.21875\).

This value is feasible because a two-risk correlation must lie between \(-1\) and 1. Substitution checks the proposed aggregate: \(3{,}600+6{,}400+9{,}600(0.21875)=10{,}000+2{,}100=12{,}100\), whose square root is $110 million. The implied correlation is 0.2188 to four decimal places. The calculation is valid under the stated common normal scaling; with other standalone measures, it would be an implied parameter of the approximation, not necessarily an observed loss correlation.

Conditions, Interpretation, and Exam Traps

The formula needs inputs that describe the same risk aggregation problem. In particular:

  • Common scope and basis: the risks should use a consistent horizon, currency, gross-or-net treatment, and definition of loss. Any adjustment, such as reinsurance, must be applied consistently before aggregation.
  • Compatible capital measures: exact aggregation requires each \(K_i\) to be the same positive multiple of its risk's standard deviation. When standalone measures do not have this property, the square-root expression is an approximation or a specified aggregation rule, not a general theorem about those measures.
  • Valid dependence inputs: pairwise correlations must be consistent with a positive semidefinite correlation matrix. Individually plausible correlations can still form an invalid matrix when considered together.
  • Correct interpretation of dependence: \(\rho_{ij}\) is the linear correlation between the losses on the chosen basis. It is not automatically a latent copula parameter, a rank correlation, or the correlation between standalone capital numbers.

The pairwise terms explain how dependence changes the result. Positive correlation raises the quadratic expression relative to zero correlation for positive \(K_i\) and \(K_j\); negative correlation lowers it. But the square root must still be real, and a valid correlation matrix ensures the variance expression cannot be negative. Correlation is not a guarantee that extreme outcomes behave like ordinary outcomes: the square-root rule summarizes dependence through second moments and may not capture tail dependence adequately.

A frequent exam trap is to apply correlations to capital values without asking whether the capital measure supports that interpretation. If standalone capital includes expected loss, plugging it directly into an unexpected-loss aggregation can misstate the result. Another is to mix periods—for example, one-year capital for one risk and a multi-year amount for another. First identify what each \(K_i\) measures; then confirm the basis and the role of the supplied correlations before calculating.

Key takeaway: Square-root aggregation combines standalone capital amounts as \(\sqrt{\sum_i K_i^2+2\sum_{i<j}\rho_{ij}K_iK_j}\). It is exact when capital is a common multiple of standard deviation and the correlation matrix describes the corresponding losses; otherwise, treat it as an approximation or prescribed rule and state that limitation.

Check Your Understanding

Use the square-root formula. Unless a question says otherwise, treat the stated correlations as correlations of compatible loss measures and report the final aggregate in the same units as standalone capital.

  1. Two risks have standalone capitals of 25 and 40, with correlation 0.30. Calculate aggregate capital. (Answer: 53.1507)
  2. Three risks have standalone capitals 10, 20, and 30. Their correlations are \(\rho_{12}=0\), \(\rho_{13}=0.20\), and \(\rho_{23}=0.40\). Calculate aggregate capital. (Answer: 44.7214)
  3. Two risks have standalone capitals 50 and 70, with correlation \(-0.25\). Calculate aggregate capital. (Answer: 75.1665)
  4. Three risks each have standalone capital 40, and every pair has correlation 0.50. Calculate aggregate capital. (Answer: 97.9796)
  5. Two risks have standalone capitals 30 and 50, and their aggregate capital is 65. Solve for the implied correlation and state whether it is feasible as a two-risk correlation. (Answer: 0.2750; feasible)
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