Projecting Incremental Losses with Chain Ladder

Why Project Incremental Losses?

A loss triangle may record either incremental losses, the amount arising or paid during one development interval, or cumulative losses, the total accumulated through a development age. Chain ladder factors are applied to cumulative amounts. Therefore, when the input is incremental, the first task is to add the increments within each accident-year row. After projecting cumulative amounts, subtract adjacent cumulative values to recover the projected increments.

The distinction matters in practice: an estimate of cumulative loss at a future age is not itself the loss expected to emerge during that interval. That future increment is the difference between the projected cumulative total at the end of the interval and the cumulative total at its beginning. This tutorial is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Definition: Let \(I_{i,j}\) denote the incremental loss for accident year \(i\) in development interval \(j\), and let \(C_{i,j}\) denote the cumulative loss through development age \(j\). Then \(C_{i,1}=I_{i,1}\), and for \(j>1\), \(C_{i,j}=C_{i,j-1}+I_{i,j}\). Equivalently, \(I_{i,j}=C_{i,j}-C_{i,j-1}\). A selected age-to-age factor \(f_{j\to j+1}\) applies to cumulative losses at age \(j\) to project cumulative losses at age \(j+1\).

Projecting Cumulative and Incremental Amounts

For a row observed through development age \(j\), the chain ladder projection to the next age is \(C_{i,j+1}=C_{i,j}f_{j\to j+1}\). The projected increment for that interval is \(C_{i,j+1}-C_{i,j}\). The factor's starting age must match the age of the cumulative amount to which it is applied. For example, a factor from 24 months to ultimate is not a factor from 36 months to ultimate.

The standard chain ladder projection uses selected development factors as a common development pattern for the relevant rows. Its usual working assumptions are that the historical relationship between cumulative losses at successive ages is informative about future development, and that the underlying claim process and handling or payment practices are sufficiently stable for the selected factors to apply. A selected factor is not guaranteed to describe every accident year; unusual changes in coverage, claims handling, inflation, or mix can make a mechanical projection unsuitable without adjustment. The calculations below treat the supplied factors as appropriate and fixed.

If the selected factors are \(f_{j\to j+1},\ldots,f_{k-1\to k}\), then a cumulative amount at age \(j\) projects to age \(k\) by multiplying by each factor in order. If \(U\) denotes ultimate, a stated \(f_{k\to U}\) is applied to the cumulative amount at age \(k\), not to an amount from another age.

C_{i,k}=C_{i,j}\prod_{r=j}^{k-1} f_{r\to r+1}, \qquad I_{i,k}=C_{i,k}-C_{i,k-1}

These formulas require a consistent loss basis throughout the row: for example, paid losses are projected with paid-loss cumulative amounts and the corresponding paid development pattern. Do not add paid increments to incurred increments or switch bases midway through a projection.

Worked Example: Projecting an Incremental Paid Triangle

A hypothetical insurer has paid losses, in thousands of dollars, recorded at 12, 24, and 36 months. The incremental row for accident year A is \(100, 60, 30\). Selected factors are \(f_{12\to24}=1.60\), \(f_{24\to36}=1.20\), and \(f_{36\to U}=1.10\). Convert the row, project its ultimate, and identify each future incremental amount. Use the chain ladder method and avoid applying a factor from the wrong starting age.

Step 1: Convert observed increments to cumulative losses. At 12 months, \(C_{A,12}=100\). At 24 months, \(C_{A,24}=100+60=160\). At 36 months, \(C_{A,36}=160+30=190\). Thus the observed cumulative row is \(100,160,190\).

Step 2: Project from the latest observed age. The row is observed through 36 months, so the stated factor that applies directly is \(f_{36\to U}=1.10\). The projected ultimate is \(190(1.10)=209\) thousand dollars. The projected increment from 36 months to ultimate is \(209-190=19\) thousand dollars.

Step 3: Check the projection. The ultimate exceeds the observed 36-month cumulative amount by \(209-190=19\), matching the projected future increment. The final results, in thousands of dollars, are an ultimate of \(209\) and future development of \(19\).

Method and trap avoided. This is a cumulative chain ladder projection followed by differencing. Applying the 24-to-36 factor to an incremental amount, or applying a 24-to-ultimate factor to the 36-month cumulative amount, would use the wrong base or age.

Why Differencing Gives the Increment

The conversion from projected cumulative losses back to increments follows directly from what cumulative loss means. At the end of an interval, cumulative loss includes all loss through that interval. At the beginning, it includes all loss through the preceding interval. Subtracting removes the amount already accumulated, leaving only the amount in the interval.

Start with the cumulative definition.
For an interval ending at age \(k\), \(C_{i,k}=C_{i,k-1}+I_{i,k}\).

Isolate the interval amount.
Subtract \(C_{i,k-1}\) from both sides to obtain \(I_{i,k}=C_{i,k}-C_{i,k-1}\).

Use projected cumulative values.
When \(C_{i,k}\) is projected by chain ladder, subtract the cumulative amount at the interval's beginning to get the projected increment. This gives the change during that interval, not the full projected cumulative total.

For example, if \(C_{i,24}=130\) and the projected \(C_{i,36}=143\), then the projected 24-to-36-month increment is \(143-130=13\). Reporting 143 as the incremental amount would count the earlier 130 again.

Worked Example: Projecting from an Observed 24-Month Amount

A hypothetical specialty liability portfolio has incurred losses, in thousands of dollars, with incremental values of \(80\) through 12 months and \(50\) from 12 to 24 months. Selected factors are \(f_{24\to36}=1.10\) and \(f_{36\to U}=1.05\). Find the projected 24-to-36-month increment, the 36-month cumulative amount, the 36-to-ultimate increment, and ultimate.

Step 1: Calculate the observed cumulative amount. \(C_{12}=80\), and \(C_{24}=80+50=130\) thousand dollars. The factor \(f_{24\to36}\) starts at the observed 24-month age, so it is appropriate to apply it to \(C_{24}\).

Step 2: Project to 36 months and difference. \(C_{36}=130(1.10)=143\) thousand dollars. The 24-to-36-month increment is \(143-130=13\) thousand dollars.

Step 3: Project from 36 months to ultimate. \(C_U=143(1.05)=150.15\) thousand dollars. The 36-month-to-ultimate increment is \(150.15-143=7.15\) thousand dollars.

Step 4: Reconcile future development. Total development after 24 months is \(13+7.15=20.15\) thousand dollars. Directly, \(C_U-C_{24}=150.15-130=20.15\), so the increments reconcile. The projected ultimate is \(150.15\) thousand dollars.

Method and trap avoided. The chain ladder method is applied to cumulative incurred losses, and each projected increment is found by differencing adjacent cumulative amounts. The observed 12-to-24 increment of 50 is not the base for the 24-to-36 factor; the cumulative base is 130.

Rows Observed at Different Development Ages

A triangle usually contains accident years with different latest observed ages. Project each row from its own latest cumulative age, using the factor that starts at that age. A row observed at 24 months may require several factors to reach ultimate. A row observed at 36 months should start with the 36-month factor to ultimate. It is not necessary to project every row through ages already observed, although doing so can be useful when the requested output includes intermediate projected increments.

Worked Example: Matching the Factor to Each Row's Age

A hypothetical workers compensation portfolio has cumulative incurred losses, in thousands of dollars. Accident year X is observed at 24 months with \(C_{X,24}=275\); accident year Y is observed at 36 months with \(C_{Y,36}=310\). Selected factors are \(f_{24\to36}=1.08\) and \(f_{36\to U}=1.04\). Project both rows to ultimate and show the future increments by interval.

Accident year X, observed at 24 months. First, \(C_{X,36}=275(1.08)=297\) thousand dollars, so the 24-to-36-month increment is \(297-275=22\). Next, \(C_{X,U}=297(1.04)=308.88\) thousand dollars, so the 36-month-to-ultimate increment is \(308.88-297=11.88\). Total future development after 24 months is \(22+11.88=33.88\), which also equals \(308.88-275\).

Accident year Y, observed at 36 months. Its latest cumulative amount is already at the starting age of \(f_{36\to U}\). Therefore, \(C_{Y,U}=310(1.04)=322.40\) thousand dollars, and its future increment is \(322.40-310=12.40\) thousand dollars.

Check and method. Both rows use the chain ladder pattern, but they start at different ages. For X, the projected increments sum to \(33.88\), the difference between ultimate and the observed 24-month cumulative amount. For Y, \(12.40\) equals ultimate less the observed 36-month cumulative amount. Applying \(f_{24\to36}\) to Y's 36-month amount would be an age mismatch.

Common Mistakes and Practical Checks

A particularly tempting error is to take a factor with a familiar label and apply it to whichever cumulative amount is available. The factor is defined by its starting development age, so the age attached to the base amount matters just as much as the factor value.

For example, suppose a row is observed at 36 months with cumulative loss \(160\), and the only stated factor is \(f_{24\to U}=1.20\). The calculation \(160(1.20)=192\) is not supported: \(160\) is a 36-month amount, whereas the factor starts at 24 months. If the row is actually observed at 36 months, a \(f_{36\to U}\) factor is needed to project from the observed amount. Alternatively, the 24-to-ultimate factor can be applied to a 24-month cumulative amount, but that produces a projection from 24 months, not future development from the observed 36-month amount.

Other useful checks are:

Check the cumulative conversion. Each cumulative value should equal the sum of the row's increments through that age.
Check the factor age. Confirm that the factor's starting age matches the cumulative amount being multiplied.
Check each projected increment. Subtract the cumulative amount at the start of the interval from the projected cumulative amount at its end.
Reconcile totals. The sum of projected future increments should equal ultimate minus the latest observed cumulative amount, when all future intervals are included.
Keep the basis consistent. Use one loss basis and its corresponding development factors throughout a row.

These checks help prevent double counting and age mismatches. They also make results easier to interpret: the projected cumulative triangle describes totals through each age, while the projected incremental triangle describes the amount assigned to each individual development interval.

Key takeaway: Convert increments to cumulative losses before applying chain ladder factors. Project from the latest observed age using a factor with the same starting age, then obtain each future increment by subtracting successive cumulative amounts.

Check Your Understanding

For each problem, work in the units stated and verify projected increments against the change in cumulative loss.

A hypothetical property portfolio has incremental paid losses of \(90\) and \(45\), in thousands of dollars, at 12 and 24 months. What are the cumulative amounts at those ages? (Answer: \(90\) and \(135\) thousand dollars.)
A hypothetical accident year has \(C_{12}=200\) thousand dollars and \(f_{12\to24}=1.30\). Find the projected 24-month cumulative amount and the 12-to-24-month increment. (Answer: cumulative \(260\) thousand dollars; increment \(60\) thousand dollars.)
A hypothetical liability row has \(C_{24}=240\) thousand dollars, \(f_{24\to36}=1.15\), and \(f_{36\to U}=1.08\). Find the projected 24-to-36 increment, the 36-to-ultimate increment, and ultimate. (Answer: increments \(36\) and \(22.08\) thousand dollars; ultimate \(298.08\) thousand dollars.)
A hypothetical incurred row is observed at 36 months with cumulative loss \(425\) thousand dollars. The selected factor is \(f_{36\to U}=1.06\). Find ultimate and future development, and state why a 24-to-36 factor is not needed for this projection. (Answer: ultimate \(450.50\) thousand dollars; future development \(25.50\) thousand dollars; the row is already observed at 36 months.)

What You'll Learn