Why Loss Data Are Arranged in Triangles
Losses from an accident period do not all become known or get paid at once. As time passes, an insurer can record additional payments and revise its estimate of what remains to be paid. A loss development triangle organizes observations by accident period and development age, so a reader can see both which period a value belongs to and how far that period has developed.
At a single valuation date, older accident periods have had more time to develop than newer ones. The newest periods therefore have fewer observed development ages. This unequal maturity creates the triangular pattern: the older rows extend farther to the right, while the newer rows stop sooner.
In this tutorial, annual accident periods and annual development ages are used for clarity. A row labelled 2020 refers to losses arising from accidents in 2020. A column labelled development age 2 refers to the second annual development observation for each accident period. The exact date conventions used to define an age must be stated in an actual data set; do not infer them from a label alone.
A triangle may contain paid losses or incurred losses. In a paid triangle, the cell records payments made by the valuation time. In an incurred triangle, the cell records an incurred estimate, commonly defined as cumulative paid amounts plus case reserves at that time. The two triangles describe different measures, even when their row and column labels match. Incurred estimates can change as reserves are revised; they are not necessarily increasing from one development age to the next.
Reading the Rows, Columns, and Boundary
Read the row label first to identify the accident period, then the column label to identify the development age, and finally the cell value to identify the recorded loss measure. For a triangle with \(n\) annual accident periods and a common valuation date, label the rows from \(i=1\) for the oldest to \(i=n\) for the newest. Label the development ages \(j=1,\ldots,n\). Under a regular annual layout, the cell \((i,j)\) is observed when
This condition is a compact way to describe the triangular shape. The oldest row, \(i=1\), can have \(n\) observations, since \(j\leq n\). The newest row, \(i=n\), can have only the first observation, since \(j\leq 1\). The condition describes which cells could have been observed by the valuation date; it does not guarantee that every expected observation was successfully recorded.
The observed boundary also forms a diagonal. In a table with older periods at the top and later development ages to the right, the most recent available observations run from the upper right toward the lower left. These boundary cells correspond to the same valuation date, though they refer to different accident periods and different development ages. This diagonal is a useful visual check: if a purported common-date triangle has an irregular edge, determine whether the data use uneven valuation dates, omit observations, or follow a different layout.
Worked Example: Locating Observations and Unavailable Cells
Worked Example: Locating Observations and Unavailable Cells
A hypothetical insurer prepares a paid loss triangle at the end of 2022. It includes accident periods 2019 through 2022 and development ages 1 through 4. The recorded values, in thousands of dollars, are shown below. Identify the value for accident period 2021 at development age 2, determine which cells are structurally unavailable, and count the observed cells.
| Accident period | Age 1 | Age 2 | Age 3 | Age 4 |
|---|---|---|---|---|
| 2019 | 100 | 150 | 180 | 200 |
| 2020 | 110 | 160 | 190 | — |
| 2021 | 120 | 175 | — | — |
| 2022 | 130 | — | — | — |
Step 1: Read the requested row and column. Accident period 2021 is the third row from the top. Development age 2 is the second development column. Their intersection is \(175\), or $175,000 because the table is in thousands of dollars.
Step 2: Identify the structurally unavailable cells. The dashes are 2020 at age 4; 2021 at ages 3 and 4; and 2022 at ages 2, 3, and 4. Each represents an age not yet reached by the valuation date, not a recorded zero.
Step 3: Count the observed cells. The rows contain 4, 3, 2, and 1 observed cells. Thus the total is \(4+3+2+1=10\). There are \(4\times4=16\) positions in the rectangular display, so \(16-10=6\) positions are structurally unavailable, matching the six dashes.
Check. The requested cell is within the observed region, the number of observed cells plus unavailable cells is \(10+6=16\), and the newest row has only one observed age. The common mistake this method avoids is treating the six dashes as zero-dollar observations.
Paid and Incurred Triangles Are Different Views
Paid and incurred triangles can use identical accident-period and development-age labels, but their values should not be interchanged. Paid amounts track what has been paid by the relevant valuation. Incurred amounts include an estimate of unpaid obligations as well as paid amounts, according to the stated incurred basis. The same position in two triangles can therefore have different values without either table being inconsistent.
Worked Example: Comparing Paid and Incurred Cells
For a hypothetical insurance portfolio valued at the end of 2022, the following paid and incurred amounts are reported in thousands of dollars. At development age 2 for accident period 2021, identify both measures, compare them, and explain whether a decrease in an incurred value from one age to the next is possible.
| Accident period | Paid: Age 1 | Paid: Age 2 | Paid: Age 3 | Incurred: Age 1 | Incurred: Age 2 | Incurred: Age 3 |
|---|---|---|---|---|---|---|
| 2020 | 80 | 120 | 145 | 100 | 142 | 138 |
| 2021 | 90 | 130 | — | 115 | 150 | — |
| 2022 | 105 | — | — | 125 | — | — |
Step 1: Confirm the requested position. The row for accident period 2021 and the column for age 2 are observed in both tables. The paid amount is $130,000; the incurred amount is $150,000.
Step 2: Compare on the same basis. The incurred amount exceeds the paid amount by \(150-130=20\) thousand dollars, or $20,000. This difference is consistent with the incurred measure including an estimate of unpaid obligations, under the definition specified for this example.
Step 3: Interpret the 2020 incurred values over time. The incurred amount changes from 142 at age 2 to 138 at age 3, a decrease of \(142-138=4\) thousand dollars. This is possible because the estimated outstanding obligations can be revised downward. The table gives no basis to conclude that a paid amount decreased; it shows the incurred estimate changing.
Check. Both requested age-2 cells are observed, and their difference is $20,000. The age-2 to age-3 incurred decrease is $4,000. The common mistake this method avoids is assuming every measure in every triangle must rise with development.
Why the Last Observations Form a Diagonal
The triangular boundary follows from a simple observation rule. At a common valuation date, an older accident period has had more development time than a newer one. In the regular annual layout, moving down one row to a newer accident period leaves one fewer development age available. The endpoint of each row therefore shifts one column to the left.
For example, with four accident periods, the row lengths are \(4,3,2,1\). Their final observed cells are at \((1,4),(2,3),(3,2),(4,1)\), where the first coordinate is the row index and the second is the development-age index. The row index increases by one as the age index decreases by one, so these endpoints form the upper-right-to-lower-left diagonal.
This gives a quick audit of the layout. If there are four equally spaced annual accident periods and the oldest has four available ages at the stated valuation date, then the next rows should have three, two, and one available ages under the same regular convention. If not, do not force the data into the expected shape. Check the valuation dates, period definitions, missing values, and whether the table is meant to display all available ages.
Worked Example: Checking a Common-Date Boundary
A hypothetical analyst presents a four-row paid triangle at one common valuation date. The observed row lengths are 4, 3, 2, and 1, from oldest accident period to newest. The final observed cell values, in thousands of dollars, are 240, 205, 160, and 125, respectively. Identify the development age of each final observation and state which observations form the valuation-date diagonal.
Step 1: Map row lengths to final ages. A row with four observed cells ends at age 4, a row with three ends at age 3, a row with two ends at age 2, and a row with one ends at age 1.
Step 2: Locate the endpoints. The four final observations are therefore the oldest period at age 4 with value 240, the second-oldest at age 3 with value 205, the third-oldest at age 2 with value 160, and the newest at age 1 with value 125.
Step 3: Check the boundary pattern. Their indexed locations are \((1,4),(2,3),(3,2),(4,1)\). Each step down one row moves one column left, so these endpoints form the diagonal boundary. The observations are all recorded at the common valuation date, but they are not observations of the same accident period or the same development age.
Check. The row lengths sum to \(4+3+2+1=10\) observed cells. The common mistake this method avoids is reading the endpoint values as one row of comparable development ages; each endpoint belongs to a different age.
Common Reading Errors
- Reading a column as an accident period. Rows identify accident periods; columns identify development ages. Always use the labels rather than relying on the visual position alone.
- Treating an unavailable cell as zero. A future age is not yet observed. A zero should be used only when the data actually record a zero under the stated measure.
- Confusing missing data with future data. A dash beyond the boundary can be structural. A blank where the triangle indicates an observation should exist may signal an incomplete record, a data problem, or a different convention.
- Assuming paid and incurred values mean the same thing. They are different measures. Check the triangle heading and the definition of the recorded amount before comparing values.
- Assuming incurred values must increase. Incurred estimates may be revised downward. The direction of change is not enough to determine whether a cell has been read correctly.
- Assuming every displayed diagonal has the same interpretation. The boundary diagonal is useful because it marks the latest available age in each row. Other diagonals may group cells differently; identify the row and column before interpreting one.
Check Your Understanding
For each problem, identify the relevant rows, columns, and observed boundary before calculating or interpreting a value.
- A three-period triangle has accident periods 2022, 2023, and 2024, in that order, with three development-age columns. At one valuation date, the rows contain 3, 2, and 1 observed cells. How many cells are structurally unavailable, and how many are observed? (Answer: 3 unavailable and 6 observed.)
- In a four-period triangle, the oldest row is the first row and the newest is the fourth. Which development age is the final observed age in the third row under the regular common-date layout? (Answer: Age 2.)
- A paid triangle shows $72,000 in the 2023 row at age 2, while the corresponding cell is a dash in the 2024 row. The table is valued at the end of 2024. What does each entry tell you? (Answer: The 2023 age-2 cell is an observed $72,000; the 2024 age-2 dash is structurally unavailable, not a zero.)
- At a stated common valuation date, four row lengths are 4, 3, 2, and 1. Give the indexed locations of the final observed cells, where indices count rows from oldest to newest and columns from age 1 upward. (Answer: \((1,4),(2,3),(3,2),(4,1)\).)
- For one accident period, incurred amounts at ages 1, 2, and 3 are $110,000, $155,000, and $149,000. Is the age-3 value necessarily an error because it is below the age-2 value? State the change. (Answer: No; incurred estimates can decrease, and the change is a decrease of $6,000.)