Why Two Averages Can Give Different Development Factors
For a given development interval, each accident period with observed cumulative losses at both ages provides an age-to-age link ratio. A simple average gives each observed ratio equal weight. A volume-weighted factor gives more influence to periods with larger cumulative losses at the start of the interval. The two methods can therefore produce different factors even when calculated from exactly the same triangle cells.
The distinction is not merely arithmetic. The simple average describes the mean of the observed row-level ratios, while the volume-weighted factor describes the change in the combined cumulative amounts. Neither calculation, by itself, establishes which factor should be selected for a projection. Here the focus is how to calculate and compare them.
For \(n\) accident periods with positive starting amounts and complete observations for the same development interval, the two calculations are
The ratio \(r_i\) is dimensionless, so either factor is dimensionless. If the cumulative losses are stated in dollars, thousands of dollars, or another common unit, the unit cancels in the volume-weighted ratio. The observations being averaged must refer to the same development interval; mixing, for example, one interval's 12-to-24-month ratios with another interval's 24-to-36-month ratios would not calculate a single age-to-age factor.
How Volume Weighting Works
A useful way to interpret the volume-weighted factor is to rewrite each next-age amount as its starting amount times its own link ratio. This gives the weighted-average identity below.
The weights \(w_i\) are the shares of the total starting cumulative amount. When the starting amounts are positive, each weight is positive and the weights sum to 1. Thus the volume-weighted factor is a weighted arithmetic mean of the individual link ratios, with each row's starting cumulative amount determining its weight.
This identity also proves the aggregate interpretation. The total at the next age is \(\sum_i C_{i,j+1}=\sum_i C_{i,j}r_i\). Dividing that total by the combined starting amount produces the growth factor for the combined amounts. In contrast, the simple average adds the ratios and divides by the number of rows, regardless of whether a row began with a small or large cumulative amount.
Use accident periods with observed cumulative amounts at both ages in the interval being analyzed.
For each eligible row, divide its next-age cumulative amount by its starting-age cumulative amount.
Average the ratios for the simple average; divide the sum of next-age amounts by the sum of starting-age amounts for the volume-weighted factor.
Confirm that the weighted calculation uses starting-age amounts as weights and that both answers concern the same age-to-age interval.
Worked Comparisons
Worked Example: Three Accident Periods With Unequal Ratios
A hypothetical cumulative loss triangle has the following amounts for a single development interval, all in thousands of dollars. Calculate the individual link ratios, the simple average, and the volume-weighted factor.
| Accident period | Starting age \(C_{i,j}\) | Next age \(C_{i,j+1}\) |
|---|---|---|
| A | 100 | 120 |
| B | 200 | 260 |
| C | 300 | 390 |
Step 1: Calculate each link ratio. For period A, \(r_A=120/100=1.20\). For period B, \(r_B=260/200=1.30\). For period C, \(r_C=390/300=1.30\).
Step 2: Calculate the simple average.
Step 3: Calculate the volume-weighted factor. The starting total is \(100+200+300=600\), and the next-age total is \(120+260+390=770\). Therefore,
Rounded to four decimal places, the simple average is \(1.2667\) and the volume-weighted factor is \(1.2833\). The larger ratio, 1.30, has more influence in the volume-weighted calculation because it occurs for periods with starting amounts of 200 and 300, rather than only for a small starting amount. As a check, the weighted expression gives \((100(1.20)+200(1.30)+300(1.30))/600=770/600\), the same result as the ratio of totals.
Method and mistake avoided. This example uses the arithmetic mean for the simple average and the ratio of aggregate amounts for the volume-weighted factor. It avoids treating the three rows as equally weighted in both calculations.
Worked Example: One Large Period and One Small Period
For a hypothetical development interval, one accident period has starting cumulative losses of 910 and a link ratio of 1.30. A second has starting cumulative losses of 90 and a link ratio of 1.02. Calculate both averages. Amounts are in thousands of dollars.
Step 1: Find the simple average. The two ratios receive equal weight:
Step 2: Recover the next-age amounts. The first next-age amount is \(910(1.30)=1{,}183\). The second is \(90(1.02)=91.8\).
Step 3: Find the volume-weighted factor. The starting total is \(910+90=1{,}000\); the next-age total is \(1{,}183+91.8=1{,}274.8\). Thus,
The simple average is \(1.1600\), while the volume-weighted factor is \(1.2748\). This difference occurs because the higher ratio of 1.30 applies to 910 of the 1,000 starting units. Equivalently, the weights are \(910/1{,}000=0.91\) and \(90/1{,}000=0.09\), so \(0.91(1.30)+0.09(1.02)=1.2748\). The calculation checks against the ratio-of-totals result.
Method and mistake avoided. The volume-weighted method uses starting amounts as weights, not the number of rows and not the next-age amounts. The simple average remains the mean of the two ratios, regardless of their volumes.
Worked Example: Changing Volumes Without Changing the Ratios
Consider the same two link ratios, 1.30 and 1.02, but reverse which period has the larger starting amount. The 1.30 ratio now has a starting amount of 90, and the 1.02 ratio has a starting amount of 910. Calculate the two factors and compare them with the preceding example.
Step 1: Simple average. The ratios are unchanged, so
Step 2: Next-age amounts. The amounts are \(90(1.30)=117\) and \(910(1.02)=928.2\). The starting total is \(90+910=1{,}000\), and the next-age total is \(117+928.2=1{,}045.2\).
Step 3: Volume-weighted factor.
The simple average is still \(1.1600\), but the volume-weighted factor is now \(1.0452\), compared with \(1.2748\) when the larger volume had the higher ratio. The arithmetic verifies the point: changing volumes alone does not change the simple average if the ratios are fixed, but it can substantially change the volume-weighted factor.
Method and mistake avoided. This comparison isolates the effect of claim volume. It avoids concluding that the same set of ratios must produce the same volume-weighted factor when the row volumes change.
When the Averages Coincide and How to Check Them
The two averages coincide when the starting cumulative amounts are equal, because then the volume weights are all \(1/n\), just like the simple-average weights. They also coincide if all observed ratios are equal, even if the starting amounts differ: every weighted average of identical values equals that common value. In other cases, they may still happen to be equal, but equality should be verified rather than assumed.
A practical check is to calculate the volume-weighted factor in two equivalent ways: first as a ratio of summed next-age amounts to summed starting-age amounts, and then as a weighted average of the individual ratios using starting-age shares. Agreement catches many arithmetic or weighting errors. The simple average provides a separate check of the individual ratios and should not be confused with either form of the weighted calculation.
Common Exam Trap
A common mistake is to label an unweighted arithmetic mean as volume-weighted simply because it is being used in a loss triangle. For the amounts 910 and 90 with ratios 1.30 and 1.02, the incorrect calculation is
That is the simple average. The correction is to weight each ratio by its starting amount, or equivalently divide the total next-age amount by the total starting amount:
Another error is to weight by \(C_{i,j+1}\), the next-age amount. That does not reproduce the ratio of aggregate amounts. The identity \(\sum_i C_{i,j+1}=\sum_i C_{i,j}r_i\) shows why the weights belong to the starting-age amounts \(C_{i,j}\).
Check Your Understanding
For each problem, calculate the requested factor or comparison for one development interval. Round factors to four decimal places where needed.
- A hypothetical triangle has starting amounts 80 and 120, with corresponding next-age amounts 100 and 132. Calculate the two individual ratios, the simple average, and the volume-weighted factor. (Answer: ratios \(1.2500\) and \(1.1000\); simple average \(=1.1750\); volume-weighted \(=1.1600\))
- Two accident periods have starting amounts 400 and 100, and link ratios 1.15 and 1.35, respectively. Calculate both averages and state which ratio has greater influence on the volume-weighted factor. (Answer: simple average \(=1.2500\); volume-weighted \(=1.1900\); the 1.15 ratio has greater influence.)
- Two periods have equal starting amounts of 250 and link ratios 1.04 and 1.20. Calculate the simple-average and volume-weighted factors, and explain why they agree. (Answer: both factors \(=1.1200\); equal starting amounts give equal volume weights.)
- A hypothetical portfolio has starting amounts of 700 and 300 and ratios of 1.24 and 1.08. Find both factors. (Answer: simple average \(=1.1600\); volume-weighted \(=1.1920\))
- Two periods have ratios 1.12 and 1.40. In Scenario A their starting amounts are 950 and 50, respectively; in Scenario B the starting amounts are reversed. Calculate both volume-weighted factors and the simple average for each scenario. (Answer: simple average \(=1.2600\) in both scenarios; Scenario A volume-weighted \(=1.1340\); Scenario B volume-weighted \(=1.3860\))