From Loss Triangle to Accident-Year Reserve
A chain ladder reserve calculation is not complete when an ultimate loss estimate has been projected. For each accident year, the actuary must compare that ultimate with the amount already paid and then sum the resulting unpaid liabilities. This tutorial works through that full process for both paid and incurred data. It is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.
A cumulative loss triangle records loss amounts by accident year and development age. Let \(C_{i,j}\) denote the cumulative amount for accident year \(i\) at development age \(j\). Paid triangles contain cumulative paid losses; incurred triangles contain cumulative incurred losses, commonly paid losses plus case outstanding. A blank cell means that the development age has not yet been observed at the valuation date.
The method assumes that the selected development pattern is applicable to the accident years being projected: at a given age, cumulative losses are expected to develop in proportion to their current amount. Its use also depends on the data and selections being appropriate—for example, there should not be an unaddressed change that makes the selected pattern unsuitable for the years being projected. A CDF can include a tail factor if development beyond the observed ages is modeled. In the examples below, the final modeled age is treated as ultimate, so no additional tail factor is applied.
Why the Projection Formula Works
Under the chain ladder assumption, the expected next cumulative amount is proportional to the current cumulative amount. The selected age-to-age factor \(f_j\) is that proportional multiplier. Applying the relationship successively gives the projection to ultimate:
Continuing the same multiplication through the last modeled transition gives the CDF. Thus an observed amount \(C_{i,j}\) is projected as follows:
where \(m\) is the final modeled age and \(\widehat{U}_i\) is the estimated ultimate loss for accident year \(i\). The formula follows from repeated application of the proportional projection, not from adding the factors.
When factors are estimated from several observed accident years, one common selection method is the volume-weighted age-to-age factor:
where the sums include accident years with observations at both ages. This ratio weights each observed transition in proportion to its cumulative amount at the starting age. Other selections may be appropriate after reviewing the data; the selected factors, not necessarily every raw observed ratio, are what the projection uses.
Worked Example: Paid and Incurred Reserves by Accident Year
Worked Example: Paid and Incurred Reserves by Accident Year
A hypothetical insurer reviews four accident years at a common valuation date. Loss amounts are in thousands of dollars. The final age shown is treated as ultimate, with no tail beyond it. Calculate paid and incurred chain ladder ultimates and unpaid liabilities by accident year, then total each reserve. Use the selected factors obtained below.
Paid triangle and factor selection. The cumulative paid triangle is:
| Accident year | Age 1 | Age 2 | Age 3 | Age 4 |
|---|---|---|---|---|
| AY 1 | 100 | 150 | 180 | 198 |
| AY 2 | 120 | 180 | 216 | — |
| AY 3 | 140 | 210 | — | — |
| AY 4 | 160 | — | — | — |
For age 1 to 2, the volume-weighted factor is \((150+180+210)/(100+120+140)=540/360=1.50\). For age 2 to 3, it is \((180+216)/(150+180)=396/330=1.20\). For age 3 to 4, only AY 1 has both ages, so the factor is \(198/180=1.10\). The selected paid factors are therefore 1.50, 1.20, and 1.10.
The CDFs are \(1.50\times1.20\times1.10=1.98\) at age 1, \(1.20\times1.10=1.32\) at age 2, \(1.10\) at age 3, and \(1.00\) at age 4. Apply each CDF to the latest observed paid amount:
- AY 1: latest paid is 198 at age 4. Ultimate \(=198\times1.00=198.00\); unpaid \(=198.00-198=0.00\).
- AY 2: latest paid is 216 at age 3. Ultimate \(=216\times1.10=237.60\); unpaid \(=237.60-216=21.60\).
- AY 3: latest paid is 210 at age 2. Ultimate \(=210\times1.32=277.20\); unpaid \(=277.20-210=67.20\).
- AY 4: latest paid is 160 at age 1. Ultimate \(=160\times1.98=316.80\); unpaid \(=316.80-160=156.80\).
The total paid chain ladder reserve is \(0.00+21.60+67.20+156.80=245.60\) thousand dollars.
Incurred triangle and factor selection. The cumulative incurred triangle is:
| Accident year | Age 1 | Age 2 | Age 3 | Age 4 |
|---|---|---|---|---|
| AY 1 | 220 | 264 | 277.20 | 291.06 |
| AY 2 | 240 | 288 | 302.40 | — |
| AY 3 | 250 | 300 | — | — |
| AY 4 | 260 | — | — | — |
For age 1 to 2, the volume-weighted factor is \((264+288+300)/(220+240+250)=852/710=1.20\). For age 2 to 3, it is \((277.20+302.40)/(264+288)=579.60/552=1.05\). For age 3 to 4, it is \(291.06/277.20=1.05\). The incurred CDFs are therefore \(1.20\times1.05\times1.05=1.323\) at age 1, \(1.05\times1.05=1.1025\) at age 2, \(1.05\) at age 3, and \(1.00\) at age 4.
An incurred ultimate is not itself the unpaid liability. Subtract paid losses to date from the projected incurred ultimate. The latest paid amounts by accident year are 198, 216, 210, and 160, respectively, from the paid triangle.
- AY 1: ultimate \(=291.06\times1.00=291.06\). Unpaid liability \(=291.06-198=93.06\).
- AY 2: ultimate \(=302.40\times1.05=317.52\). Unpaid liability \(=317.52-216=101.52\).
- AY 3: ultimate \(=300\times1.1025=330.75\). Unpaid liability \(=330.75-210=120.75\).
- AY 4: ultimate \(=260\times1.323=343.98\). Unpaid liability \(=343.98-160=183.98\).
The total incurred chain ladder reserve is \(93.06+101.52+120.75+183.98=499.31\) thousand dollars. As a check, the incurred ultimates sum to \(291.06+317.52+330.75+343.98=1{,}283.31\), and paid-to-date losses sum to \(198+216+210+160=784\). Their difference is \(1{,}283.31-784=499.31\), matching the accident-year reserve total.
Method and mistake avoided. The example applies selected age-to-age factors to each year’s latest cumulative amount, then subtracts paid-to-date losses from each ultimate. The key trap is to subtract incurred-to-date from the incurred ultimate and call the result the total unpaid liability. That difference estimates further development beyond current incurred amounts, not all unpaid losses; the case outstanding already included in incurred losses must also be accounted for.
Two More Reserve Calculations
Worked Example: Projecting a Paid Amount Through Two Remaining Ages
A hypothetical equipment warranty portfolio has 480 thousand dollars of cumulative paid losses for a recent accident year at age 2. The selected age 2 to 3 factor is 1.125, and the selected age 3 to ultimate factor is 1.04. Assume the latest paid amount is the only amount paid to date for this accident year. Find the projected ultimate and unpaid liability.
First project to age 3: \(480\times1.125=540\) thousand dollars. Then project from age 3 to ultimate: \(540\times1.04=561.60\) thousand dollars. Equivalently, the age 2 CDF is \(1.125\times1.04=1.17\), and \(480\times1.17=561.60\). The unpaid liability is \(561.60-480=81.60\) thousand dollars.
The check is that projected development beyond the observed paid amount is \(540-480=60\) through age 3 plus \(561.60-540=21.60\) thereafter, totaling \(81.60\). The method avoids stopping after the first factor: age 3 is not ultimate because a second selected transition remains.
Worked Example: Separating Case Outstanding and IBNR
At a common valuation date, a hypothetical insurer has an AY 1 incurred amount of 520 thousand dollars at age 2 and paid losses of 410 thousand dollars. The selected CDF from age 2 to ultimate is 1.20. For AY 2, incurred losses are 300 thousand dollars at age 1, paid losses are 180 thousand dollars, and the selected age 1 CDF is 1.50. Calculate each ultimate, total unpaid liability, and—within this estimate—the case outstanding and additional development beyond incurred.
For AY 1, ultimate is \(520\times1.20=624\). Current case outstanding is incurred minus paid: \(520-410=110\). Development beyond incurred to ultimate is \(624-520=104\). Total unpaid liability is \(624-410=214\), which checks as \(110+104=214\).
For AY 2, ultimate is \(300\times1.50=450\). Case outstanding is \(300-180=120\), and development beyond incurred is \(450-300=150\). Total unpaid liability is \(450-180=270\), which checks as \(120+150=270\).
The two accident-year liabilities total \(214+270=484\) thousand dollars. The method avoids confusing estimated development beyond incurred with the full unpaid liability: the total includes both existing case outstanding and further estimated development.
Reading and Checking the Reserve Result
A paid and an incurred chain ladder calculation can produce different estimates because their triangles develop differently and because incurred losses include case estimates that may change. A difference between the estimates is not, by itself, an arithmetic error. The calculations should be reviewed on a consistent basis: use the same valuation date, keep paid amounts and incurred amounts distinct, and apply the CDF corresponding to each accident year’s actual maturity.
For each accident year, locate the rightmost nonblank cumulative amount in the triangle being analyzed.
Multiply only the factors from that observed age forward. A more mature year uses fewer remaining factors than an immature year.
For a paid triangle, subtract paid-to-date from projected ultimate. For an incurred triangle, also subtract paid-to-date—not incurred-to-date—to obtain total unpaid liability.
Sum the accident-year liabilities and independently check that total ultimate minus total paid equals the same reserve.
The final check is particularly useful for an incurred estimate. If \(\widehat{U}_i\) is the incurred chain ladder ultimate and \(P_i\) is paid-to-date, the estimated unpaid liability is \(\widehat{U}_i-P_i\). Summing across accident years gives:
A reserve total should also be interpreted in light of the selected factors and assumptions. It is an estimate produced by the chosen development pattern, not a guaranteed final payment amount. In particular, the calculation above makes no adjustment for a tail beyond the final modeled age; if a tail is needed, its factor must be included in the relevant CDFs before projecting ultimates.
Check Your Understanding
For each problem, report amounts in the stated units and identify the calculation that produces the unpaid liability.
- A hypothetical paid triangle has 400 thousand dollars paid at age 2. The selected factors are 1.25 from age 2 to 3 and 1.08 from age 3 to ultimate. Find the ultimate and unpaid liability. (Answer: ultimate 540 thousand dollars; unpaid liability 140 thousand dollars.)
- A paid accident year has 250 thousand dollars paid at age 1. Its selected factors are 1.40 from age 1 to 2, 1.10 from age 2 to 3, and 1.02 from age 3 to ultimate. Calculate the age 1 CDF, ultimate, and unpaid liability. (Answer: CDF 1.5708; ultimate 392.70 thousand dollars; unpaid liability 142.70 thousand dollars.)
- Two paid accident years have latest paid amounts of 500 thousand dollars at age 3 and 300 thousand dollars at age 2. The selected factors are 1.04 from age 3 to ultimate, 1.10 from age 2 to 3, and 1.04 from age 3 to ultimate. Calculate both unpaid liabilities and their total. (Answer: AY 1 liability 20 thousand dollars; AY 2 liability 43.2 thousand dollars; total 63.2 thousand dollars.)
- A hypothetical incurred triangle has 460 thousand dollars incurred at the latest age, with a selected CDF of 1.05. Paid-to-date is 390 thousand dollars. Calculate the incurred ultimate, the estimated total unpaid liability, and development beyond incurred. (Answer: ultimate 483 thousand dollars; total unpaid liability 93 thousand dollars; development beyond incurred 23 thousand dollars.)
- For one accident year, incurred-to-date is 600 thousand dollars, paid-to-date is 470 thousand dollars, and the incurred CDF to ultimate is 1.10. Find the case outstanding, development beyond incurred, ultimate, and total unpaid liability. Check the liability by both subtraction and addition. (Answer: case outstanding 130 thousand dollars; development beyond incurred 60 thousand dollars; ultimate 660 thousand dollars; total unpaid liability 190 thousand dollars, equal to 130+60.)