Why Select a Median Link Ratio?
A chain ladder factor at a given development age is selected from observed age-to-age link ratios. Those observations may not be alike in size or stability: one accident year can have a link ratio far from the rest. A median selection uses the middle value or values after sorting, so an unusually high or low observation generally has less influence on the selection than it would on a simple average.
The median is a selection method, not an automatic correction for poor data. It does not establish that the remaining observations are appropriate, and it does not show that every accident year has the same development pattern. The actuary still needs to check that the data are comparable and that the selected factor makes sense for the intended projection.
The ratio compares cumulative losses at successive development ages for the same accident year. It is dimensionless: a factor of 1.20 means the later cumulative amount is 120% of the earlier cumulative amount. Calculate and sort the ratios separately for each development interval. Do not combine 12-to-24-month ratios with 24-to-36-month ratios to find one median.
A conventional median selection requires usable observations. In particular, a zero denominator makes the ratio undefined. A negative cumulative amount or a change in the basis of the data can also make an ordinary link ratio difficult to interpret. The standard calculation below assumes positive denominators and comparable cumulative loss measures across the accident years being used.
Calculating the Median Selection
For an odd number of observations, sorting identifies a single middle ratio. For an even number, the selected median is halfway between the two middle ratios. The averaging in the even case applies only to those two central observations; it is not the simple average of the full set.
Worked Example: An Extreme 12-to-24-Month Ratio
A hypothetical insurer is reviewing five accident years of cumulative paid losses, in thousands of dollars. For the 12-to-24-month interval, the pairs of cumulative amounts are 1,000 to 1,200; 800 to 1,000; 1,250 to 1,600; 1,000 to 1,300; and 400 to 780. Calculate the median selection and compare it with the simple average. Then apply the median selection to a separate accident year with 2,400 thousand dollars of cumulative paid loss at 12 months.
Method: calculate each age-to-age ratio, sort the observations, and take the middle ratio. For each accident year, divide the 24-month cumulative amount by its 12-month cumulative amount:
The sorted ratios are 1.20, 1.25, 1.28, 1.30, and 1.95. There are five observations, so the third is the median: \(\hat f_{12}=1.28\).
For comparison, the simple average is \((1.20+1.25+1.28+1.30+1.95)/5=6.98/5=1.396\). Applying the median selection to 2,400 thousand dollars gives \(2{,}400\times1.28=3{,}072\) thousand dollars at 24 months. The calculation does not round any intermediate values.
Check: The selected factor exceeds 1, so the projected cumulative amount increases. Multiplying 2,400 by 1.28 gives 3,072, consistent with the stated factor.
Common mistake avoided: The 1.95 ratio is an observation, not a reason to discard the other ratios or replace the selected factor with the simple average. The median is 1.28 because it is the middle sorted value; the average is 1.396 because it incorporates the magnitude of every observation.
An even number of observations does not have one observed middle value. By definition, average the two central values after sorting. This convention gives a median between the two central observations and treats the lower and upper halves symmetrically.
Worked Example: Four Observations and Two Middle Values
For a hypothetical 24-to-36-month development interval, four accident years produce link ratios of 1.60, 1.10, 1.22, and 1.18. Find the median selection. A separate accident year has 5,000 thousand dollars of cumulative loss at 24 months; project it to 36 months using the selected factor.
Method: sort the observations, then average the two central ratios. In ascending order, the factors are 1.10, 1.18, 1.22, and 1.60. The two middle factors are 1.18 and 1.22, so:
The 36-month projected cumulative amount is \(5{,}000\times1.20=6{,}000\) thousand dollars. Thus the median selection is 1.20 and the projected cumulative loss is 6,000 thousand dollars.
Check: The factor is the midpoint of 1.18 and 1.22, and multiplying the 24-month amount by 1.20 increases it by 20% to 6,000.
Common mistake avoided: Selecting 1.18 alone because it is the lower middle value would be incorrect. With an even count, the median is the mean of the two middle values. The simple average of all four observations, 1.275, is not the median.
Why the Median Limits the Effect of an Extreme Ratio
The median depends on the ranks of the observations, not on how far the most extreme observation lies from the others. For example, in a sorted set of five observations, the third observation is the median. If the largest observation is replaced by an even larger value and remains largest, the third observation does not change. By contrast, the simple average changes whenever one observation changes.
This rank-based feature has a second useful justification: the median minimizes the sum of absolute deviations from a proposed central value. To see why, sort observations as \(x_{(1)}\leq x_{(2)}\leq\cdots\leq x_{(n)}\), and consider the total distance \(D(a)=\sum_{r=1}^{n}|x_{(r)}-a|\) from a proposed center \(a\). As \(a\) moves to the right, each observation below \(a\) contributes a slope of \(+1\) to \(D(a)\), while each observation above \(a\) contributes a slope of \(-1\). The total distance is minimized where at least half the observations are on either side. That is precisely the median condition. With an even number of observations, every point between the two central values minimizes this total distance; their arithmetic mean is the conventional median selection.
This property explains the median’s resistance to an extreme ratio, but it does not mean that the median always produces the best actuarial selection. If the extreme value reflects a genuine change in development rather than an isolated unusual observation, the median can suppress information relevant to the projection. Nor does a median resolve differences in data quality, exposure, or development conditions among accident years.
Selecting Medians Across Multiple Development Ages
Calculate one median for each development interval using only the link ratios observed at that interval. If a projection continues across several intervals, the selected factors are chained by multiplication. This preserves the interpretation of each factor as the multiplier for its own interval.
Worked Example: Two Median Factors and a Cumulative Projection
A hypothetical portfolio has the following observed link ratios: for 12 to 24 months, 1.15, 1.18, 1.20, 1.25, and 1.30; for 24 to 36 months, 1.08, 1.10, 1.12, and 1.15. Select a median factor for each interval, then project a separate accident year’s 12-month cumulative loss of 5,000 thousand dollars to 36 months using both selections.
Method: sort and find the median separately for each development interval. The 12-to-24-month observations are already sorted. There are five, so the third is the median: \(\hat f_{12}=1.20\).
The 24-to-36-month observations are also sorted. There are four, so average the second and third values:
The 12-to-36-month cumulative multiplier is \(1.20\times1.11=1.332\). Applying the two factors in sequence gives \(5{,}000\times1.20=6{,}000\) thousand dollars at 24 months, followed by \(6{,}000\times1.11=6{,}660\) thousand dollars at 36 months.
Check: The chained multiplier is also \(6{,}660/5{,}000=1.332\), which agrees with the product of the selected age-to-age factors.
Common mistake avoided: Do not pool all nine ratios into one median. The 12-to-24 and 24-to-36 observations describe different intervals, so each interval needs its own selection before the factors are chained.
Common Calculation Traps
- Taking the median before sorting. Identify the middle position only after arranging ratios from smallest to largest.
- Using the wrong middle value for an even count. Average the two central observations; do not choose just the lower or upper one.
- Confusing a median with a simple average. The median is determined by position. The arithmetic mean uses the magnitudes of all observations.
- Mixing development ages. Select separate medians for separate intervals, then multiply selected factors when projecting through multiple intervals.
- Using an undefined or incomparable ratio. A zero denominator prevents calculation, and inconsistent cumulative loss measures undermine comparability. Do not treat a numerical median as meaningful without usable data.
Check Your Understanding
For each problem, identify the relevant observations, calculate the requested median selection, and keep development intervals separate where required.
- A hypothetical 12-to-24-month review gives five ratios: 1.12, 1.09, 1.15, 1.11, and 1.80. What is the median selection? (Answer: 1.12)
- Four ratios for a hypothetical 24-to-36-month interval are 1.04, 1.12, 1.08, and 1.40. Find the median, and state the projected 36-month amount if a separate accident year has 2,500 thousand dollars at 24 months. (Answer: median 1.10; projected amount 2,750 thousand dollars)
- Six hypothetical ratios for one development interval are 0.98, 1.02, 1.05, 1.06, 1.08, and 1.50. Find the median and explain whether replacing 1.50 with 3.00 changes it. (Answer: median 1.055; replacing 1.50 with 3.00 does not change it)
- A hypothetical portfolio has 12-to-24-month ratios of 1.10, 1.20, and 1.30, and 24-to-36-month ratios of 1.05, 1.15, 1.25, and 1.35. Select a median for each interval and find the 36-month projection from 4,000 thousand dollars at 12 months. (Answer: selections 1.20 and 1.20; projected amount 5,760 thousand dollars)
- Three hypothetical 12-to-24-month ratios are 1.08, 1.12, and 1.40. A proposed calculation reports a median of 1.20 by averaging all three ratios. Identify the error and give the correct median. (Answer: 1.20 is the simple average, not the median; the correct median is 1.12)