Deriving Covariance From Shared Development-Factor Estimates

Why Shared Development Factors Create Covariance

In chain-ladder reserving, different accident years can use the same estimated development factor in their projections. Even when their future claim emergence is otherwise modeled separately, uncertainty in a shared estimate moves both projected ultimates in the same direction. This creates covariance that matters when calculating the prediction error of an aggregate reserve. The derivation and calculations here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

The focus is the covariance induced by shared development-factor estimates. This is distinct from the process variance for future claims within an accident year. For the calculations below, treat the latest observed cumulative amounts as fixed and the development-factor estimates as random. The resulting covariance is a first-order approximation to the parameter-estimation component of prediction error.

Definition: Let \(C_i\) be the latest observed cumulative amount for accident year \(i\), and let \(K_i\) be the set of future development ages needed to project it to ultimate. Its chain-ladder projected ultimate is \(\hat U_i=C_i\prod_{k\in K_i}\hat f_k\), where \(\hat f_k\) is the estimated development factor at age \(k\). A development-factor estimate is shared by accident years \(i\) and \(j\) when \(k\in K_i\cap K_j\).

Write \(\hat v_k=\operatorname{Var}(\hat f_k)\). If the estimates at different development ages are independent, only ages shared by both projections contribute to their covariance. A larger factor variance or larger projected ultimates generally means a larger covariance contribution.

Deriving the Shared-Factor Covariance

The product defining an ultimate estimate is nonlinear in the factors. A first-order Taylor approximation gives a convenient covariance formula. For a factor \(\hat f_k\) used in accident year \(i\), the sensitivity of the projected ultimate to that factor is \(\hat U_i/\hat f_k\): increasing the factor by a small amount changes the ultimate in proportion to the ultimate itself.

Linearize each ultimate.
Near the selected development factors, a small change in the factors gives \(d\hat U_i\approx\hat U_i\sum_{k\in K_i}d\hat f_k/\hat f_k\).

Take the covariance.
If factor estimates at different ages are independent, cross-age covariance terms vanish. A factor contributes only if it appears in both accident-year projections.

Sum the common contributions.
For each shared age, multiply the two sensitivities and the variance of that factor estimate.

\operatorname{Cov}(\hat U_i,\hat U_j) \approx \sum_{k\in K_i\cap K_j} \left(\frac{\hat U_i}{\hat f_k}\right) \left(\frac{\hat U_j}{\hat f_k}\right)\hat v_k = \hat U_i\hat U_j \sum_{k\in K_i\cap K_j}\frac{\hat v_k}{\hat f_k^2}.

This formula follows directly from the linearized changes. For each common factor \(k\), the two changes include terms \((\hat U_i/\hat f_k)d\hat f_k\) and \((\hat U_j/\hat f_k)d\hat f_k\). Their covariance is the product of the sensitivities times \(\operatorname{Var}(\hat f_k)=\hat v_k\). Independent factor estimates at other ages have zero covariance with this term, and a factor used in only one projection has no contribution to the pairwise covariance.

If factor estimates at different ages are not independent, the more general first-order expression includes cross-age terms:

\operatorname{Cov}(\hat U_i,\hat U_j) \approx \hat U_i\hat U_j \sum_{k\in K_i}\sum_{\ell\in K_j} \frac{\operatorname{Cov}(\hat f_k,\hat f_\ell)} {\hat f_k\hat f_\ell}.

The simpler shared-age formula is the special case in which estimates at different ages are uncorrelated. It is an approximation because it uses a first-order linearization; it is most useful when uncertainty in the factors is not so large that higher-order effects dominate.

Worked Examples

Worked Example: Two Accident Years Share One Factor

A hypothetical portfolio has latest cumulative amounts of \(80\) and \(120\) for two accident years. Each needs one common future development factor, with estimate \(\hat f=1.25\) and variance \(\hat v=0.0025\). Calculate the projected ultimates and their shared-factor covariance contribution.

Method. Project each latest amount and then multiply the two factor sensitivities by the variance of the common estimate.

The projected ultimates are \(\hat U_1=80(1.25)=100\) and \(\hat U_2=120(1.25)=150\).

For both accident years, the sensitivity to the shared factor is the projected ultimate divided by \(1.25\): \(100/1.25=80\) and \(150/1.25=120\).

Therefore, \(\operatorname{Cov}(\hat U_1,\hat U_2)\approx80(120)(0.0025)=24\). Equivalently, \(100(150)(0.0025)/(1.25)^2=24\).

Check. The common estimate has positive variance and raises or lowers both projected ultimates together, so the covariance is positive. The two forms of the calculation agree. The common mistake avoided here is using the projected ultimates as sensitivities without dividing by the shared factor.

Worked Example: Only One of Two Future Ages Is Shared

For a hypothetical triangle, accident year 1 has a latest cumulative amount of \(50\) at age 1 and needs factors \(\hat f_1=2.00\) and \(\hat f_2=1.50\). Accident year 2 has a latest cumulative amount of \(90\) at age 2 and needs only \(\hat f_2=1.50\). The factor estimates are independent, with \(\operatorname{Var}(\hat f_1)=0.0400\) and \(\operatorname{Var}(\hat f_2)=0.0225\). Calculate the projected ultimates and their covariance from the shared factor.

Method. Find each projection’s future factor set, then include only the factor estimate in the intersection of those sets.

For accident year 1, \(K_1=\{1,2\}\), so \(\hat U_1=50(2.00)(1.50)=150\). For accident year 2, \(K_2=\{2\}\), so \(\hat U_2=90(1.50)=135\). The only shared age is age 2. Its estimate is \(1.50\), and its variance is \(0.0225\). Thus \[ \operatorname{Cov}(\hat U_1,\hat U_2) \approx \frac{150}{1.50}\frac{135}{1.50}(0.0225) =100(90)(0.0225) =202.5. \]

As a check, the equivalent projected-ultimate form gives \(150(135)(0.0225)/(1.50)^2=202.5\). The age-1 factor affects accident year 1’s uncertainty but not the covariance: accident year 2 does not use it. The common mistake avoided is including every factor from either projection instead of just the factors shared by both.

Worked Example: Summing Common and Pair-Specific Contributions

Three hypothetical accident years have projected ultimates \(100\), \(120\), and \(150\). All three use a shared factor \(\hat f=1.20\) with variance \(0.01\). In addition, pair \((1,2)\) shares a factor estimated as \(1.10\) with variance \(0.0121\); pair \((1,3)\) shares a factor estimated as \(1.30\) with variance \(0.0169\); and pair \((2,3)\) shares a factor estimated as \(1.10\) with variance \(0.0121\). No other factors are shared. Find the sum of the three pairwise covariance contributions and the corresponding amount added to aggregate MSEP.

Method. For each pair, add the contribution from the factor shared by all accident years and the contribution from that pair’s additional shared factor. Then double the sum when forming aggregate MSEP.

For pair \((1,2)\), the common factor contributes \(100(120)(0.01)/(1.20)^2=120/1.44=83.3333\). The additional factor contributes \(100(120)(0.0121)/(1.10)^2=145.2/1.21=120\). Thus the pair’s total covariance contribution is \(203.3333\).

For pair \((1,3)\), the common factor contributes \(100(150)(0.01)/(1.20)^2=150/1.44=104.1667\). The additional factor contributes \(100(150)(0.0169)/(1.30)^2=253.5/1.69=150\). Thus this pair contributes \(254.1667\).

For pair \((2,3)\), the common factor contributes \(120(150)(0.01)/(1.20)^2=180/1.44=125\). The additional factor contributes \(120(150)(0.0121)/(1.10)^2=217.8/1.21=180\). Thus this pair contributes \(305\).

The sum of pairwise covariance contributions is \(203.3333+254.1667+305=762.5\). For the aggregate ultimate, the variance identity includes twice each distinct pairwise covariance. Therefore the amount added to aggregate MSEP is \(2(762.5)=1{,}525\).

Check. The common-factor contributions sum to \(83.3333+104.1667+125=312.5\); the pair-specific contributions sum to \(120+150+180=450\). Together they give \(312.5+450=762.5\). The common mistake avoided is omitting the pair-specific factors or reporting the pairwise sum as the aggregate MSEP addition without doubling it. The covariance sum is \(762.50\), and the aggregate MSEP addition is \(1{,}525.00\).

From Pairwise Covariance to Aggregate MSEP

Let the aggregate reserve estimate be the sum of the accident-year reserve estimates. Subtracting fixed observed amounts from projected ultimates does not change their covariance. For any collection of prediction errors, the variance of their sum is the sum of their individual variances plus twice the sum of their distinct pairwise covariances. Hence, if individual MSEPs are already calculated in a consistent framework, the shared-factor contribution to aggregate MSEP is added as follows:

\operatorname{MSEP}_{\mathrm{aggregate}} = \sum_i \operatorname{MSEP}_i + 2\sum_{i<j}\operatorname{Cov}(\hat U_i,\hat U_j),

where the covariance terms here represent the shared-factor component under the stated approximation. If a broader prediction-error model already includes this parameter-estimation covariance, do not add it a second time. Nor does this calculation replace the future-claim process variance: it addresses the additional dependence created by common factor estimates.

The sum is over unordered pairs \(i<j\), so each pair is counted once before multiplication by two. When a pair shares several independent development factors, calculate and add each common factor’s contribution. When a factor is shared by many accident years, it contributes to every pair among those years.

Common Exam Traps

Using a factor that is not shared. A factor used only by one accident year contributes to that year’s variance, not to its covariance with another year.
Forgetting the squared factor in the denominator. The shared-age term is \(\hat U_i\hat U_j\hat v_k/\hat f_k^2\), because both sensitivities contain a division by \(\hat f_k\).
Leaving out one of several shared factors. Calculate each shared-age contribution; do not stop after the factor common to every accident year if a pair shares an additional factor.
Confusing a pairwise covariance sum with the aggregate MSEP addition. The variance of a sum contains \(2\operatorname{Cov}\) for each distinct pair.
Mixing variance components. Shared-factor covariance is not a substitute for process variance, and should not be counted again if already included in the MSEP components being aggregated.

Key takeaway: Under a first-order approximation with independent factor estimates, multiply the two projected ultimates by the sum of \(\hat v_k/\hat f_k^2\) over development ages shared by the pair. Sum each distinct pair once, then multiply that covariance sum by two for its contribution to aggregate MSEP.

Check Your Understanding

For each problem, use the first-order shared-factor approximation. Unless stated otherwise, factor estimates at different ages are independent.

Two accident years have projected ultimates \(64\) and \(120\). Both use the same estimated factor \(1.60\), whose variance is \(0.0064\). Find their covariance contribution. (Answer: \(19.20\))
Accident year A has latest cumulative amount \(40\) at age 1 and uses factors \(1.50\) and \(1.20\). Accident year B has latest cumulative amount \(75\) at age 2 and uses only the factor \(1.20\). The age-2 factor estimate has variance \(0.0036\). Find the projected ultimates and their covariance contribution. (Answer: projected ultimates \(72\) and \(90\); covariance \(16.20\))
Three projected ultimates are \(80\), \(100\), and \(125\). All share a factor \(1.25\) with variance \(0.0100\). Only pair \((1,3)\) also shares a factor \(1.10\) with variance \(0.00484\). Find the sum of the three pairwise covariance contributions and the resulting aggregate MSEP addition. (Answer: pairwise sum \(235.20\); aggregate addition \(470.40\))
Two accident years both use two factors, with estimates \(1.20\) and \(1.50\). Their projected ultimates are \(180\) and \(240\). The factor variances are \(0.0036\) and \(0.0100\), and the covariance between the two factor estimates is \(0.0010\). Use the general first-order formula to find the covariance between the ultimates. (Answer: \(348.00\))

What You'll Learn