Why Tied Losses Need to Be Grouped
An empirical distribution summarizes the loss values actually recorded in a sample. When a loss amount occurs more than once, each record contributes to the sample, but all records with that same amount contribute to a single jump location in the empirical distribution function. The jump’s size reflects the number of tied records, not just the fact that the amount appears.
This distinction matters whenever the sample contains repeated exact amounts, as can happen when payments are rounded to whole dollars or when several claims have the same benefit amount. A correct calculation keeps both parts of the information: the distinct observed loss values determine where jumps occur, and the frequency at each value determines how large each jump is.
The observations are assumed to be exact and equally weighted: each recorded loss contributes one observation. The formula does not, without modification, represent grouped intervals, censored records, or observations with unequal weights. Those data structures require additional treatment. Here, a tie means that two or more exact records have the same loss amount.
From Records to Jump Sizes
The calculation has two counts that should not be confused. The sample size \(n\) is the total number of records, counting every repetition. The number \(k\) is the number of distinct observed loss amounts. Thus \(k\) can be smaller than \(n\). For each distinct amount, its count \(m_j\) gives the jump size \(m_j/n\); adding the jump sizes in increasing loss order gives the cumulative distribution values.
Set \(n\) equal to the number of exact loss observations, including repeats.
Sort the different observed values from smallest to largest. A repeated value appears once in this list.
For each distinct value \(x_j\), count its frequency \(m_j\) in the original sample.
The jump is \(m_j/n\); the cumulative value is the current jump plus all earlier jumps.
Worked Example: Three Distinct Loss Amounts With Different Frequencies
A hypothetical insurer has 12 exact claim losses, in dollars: 40, 70, 70, 100, 100, 100, 150, 180, 180, 240, 240, 300. Find the jump size at each distinct loss and the empirical distribution value at each such amount.
Method. Count all 12 records, then group equal amounts. The distinct amounts are 40, 70, 100, 150, 180, 240, and 300, so there are seven jump locations. Their frequencies are \(1,2,3,1,2,2,1\), which add to \(12\).
The jump sizes are the frequencies divided by \(n=12\). For example, the jump at 100 is \(3/12\), since three records equal 100. The cumulative value there includes the records at 40, 70, and 100: \((1+2+3)/12=6/12\).
| Loss \(x_j\) | Frequency \(m_j\) | Jump \(m_j/12\) | Cumulative count | \(\widehat F_{12}(x_j)\) |
|---|---|---|---|---|
| 40 | 1 | 1/12 | 1 | 1/12 |
| 70 | 2 | 2/12 | 3 | 3/12 |
| 100 | 3 | 3/12 | 6 | 6/12 |
| 150 | 1 | 1/12 | 7 | 7/12 |
| 180 | 2 | 2/12 | 9 | 9/12 |
| 240 | 2 | 2/12 | 11 | 11/12 |
| 300 | 1 | 1/12 | 12 | 12/12\) |
For instance, at 180 the cumulative count is \(1+2+3+1+2=9\), so \(\widehat F_{12}(180)=9/12=0.7500\). At the largest observed amount, the cumulative count is all 12 records, giving \(12/12=1\). As a check, the seven jump numerators sum to \(12\), so the jumps sum to 1.
The common mistake is to make a separate jump location for each individual record at 100. There are three records there, but only one distinct loss amount of 100; its single jump is \(3/12\), not three separate jumps at different locations.
Why a Tie Produces One Combined Jump
The indicator-sum definition explains the jump directly. Consider two values \(a\) and \(b\) with no observed loss strictly between them, and suppose \(m\) observations equal \(b\). When \(x\) moves from just below \(b\) to \(b\), each of those \(m\) observations changes its indicator from 0 to 1. Every other observation’s indicator stays unchanged over that step. The sum of indicators therefore increases by \(m\), and division by \(n\) produces a change of \(m/n\).
Here, \(\widehat F_n(b^-)\) denotes the value immediately to the left of \(b\). The difference between the two indicator sums counts exactly the observations equal to \(b\). This also shows why splitting a tie into several jump locations would be wrong: those observations do not have different loss amounts at which to enter the empirical distribution.
The cumulative value at a loss amount is a different quantity from the jump at that amount. The jump counts records equal to the amount; the cumulative value counts records at that amount or at smaller observed amounts. Keeping these separate prevents a frequent arithmetic error: reporting the current frequency divided by \(n\) as if it were the cumulative value.
Worked Example: Converting a Frequency Table Into the Empirical Distribution
A hypothetical portfolio has 10 exact losses with the following frequency table: 25 occurs twice, 50 occurs once, 75 occurs four times, and 120 occurs three times. Calculate each jump and the cumulative value at each distinct amount. Also check that the table accounts for the full sample.
Method. The sample size is \(n=2+1+4+3=10\). In increasing order, the distinct values are 25, 50, 75, and 120. Divide each frequency by 10 for the jump, then add frequencies from the top of the table down for the cumulative count.
| Loss | Frequency | Jump calculation | Jump | Cumulative value calculation | Cumulative value |
|---|---|---|---|---|---|
| 25 | 2 | 2/10 | 0.2000 | 2/10 | 0.2000 |
| 50 | 1 | 1/10 | 0.1000 | (2+1)/10 | 0.3000 |
| 75 | 4 | 4/10 | 0.4000 | (2+1+4)/10 | 0.7000 |
| 120 | 3 | 3/10 | 0.3000 | (2+1+4+3)/10 | 1.0000 |
At 75, the jump is \(4/10=0.4000\), while the cumulative value is \(7/10=0.7000\). These numbers answer different questions: one uses only the four records tied at 75, and the other uses all seven records at 75 or below.
To check, the frequencies total \(2+1+4+3=10\), matching the stated sample size. The jump sizes total \(0.2000+0.1000+0.4000+0.3000=1.0000\), and the final cumulative value is 1. Both checks are satisfied. The common mistake is to use \(4/10\) as the cumulative value at 75; the correction is to include the earlier frequencies as well.
Adding a Record at an Existing Loss Amount
When another exact observation is added, both the denominator and the relevant frequency may change. If the added loss ties an existing value, that value’s jump numerator increases by one, while the sample size also increases by one. Consequently, the new jump is not simply the old jump plus \(1/n\); it must be recalculated using the new denominator. Cumulative counts at that loss and at larger observed losses also include the added record.
Worked Example: Updating a Jump When a Loss Is Tied
A hypothetical sample of eight exact losses is 60, 90, 90, 120, 150, 150, 150, 210. A ninth record is then added, also equal to 150. Find the new jump at 150, the new cumulative value at 150, and the new jump sizes at all distinct loss values.
Method. Before the new record, the frequencies are 1, 2, 1, 3, and 1 at losses 60, 90, 120, 150, and 210. After adding a record at 150, the frequency at 150 becomes \(3+1=4\), and the sample size becomes \(8+1=9\). The updated frequencies are therefore 1, 2, 1, 4, and 1.
The new jump at 150 is \(4/9=0.4444\), rounded to four decimal places. The cumulative count through 150 is \(1+2+1+4=8\), so the cumulative value is \(8/9=0.8889\), rounded to four decimal places.
| Loss | Updated frequency | Updated jump | Cumulative count | Cumulative value |
|---|---|---|---|---|
| 60 | 1 | 1/9 = 0.1111 | 1 | 1/9 = 0.1111 |
| 90 | 2 | 2/9 = 0.2222 | 3 | 3/9 = 0.3333 |
| 120 | 1 | 1/9 = 0.1111 | 4 | 4/9 = 0.4444 |
| 150 | 4 | 4/9 = 0.4444 | 8 | 8/9 = 0.8889 |
| 210 | 1 | 1/9 = 0.1111 | 9 | 9/9 = 1.0000 |
The updated frequencies sum to \(1+2+1+4+1=9\), and the jump fractions sum to \(9/9=1\). The largest observed loss has cumulative value 1, as required. The common mistake is to retain the old denominator of eight, which would produce a jump of \(4/8\) at 150. The correct denominator is nine because the new record is part of the sample.
Common Errors and Checks
Use the following checks whenever you construct empirical jumps from exact loss records:
- Count observations, not distinct amounts, for \(n\). A sample with repeated values still includes every repeated record in its total size.
- Use one row per distinct loss amount. Combine all records tied at that value before calculating the jump.
- Keep jump and cumulative values distinct. A jump uses the frequency at one amount; a cumulative value uses the frequencies at that amount and all smaller observed amounts.
- Check the total frequency. The sum of all grouped frequencies must equal the number of original records.
- Check the total jump size. Since the grouped frequencies sum to \(n\), their proportions must sum to \(n/n=1\). The cumulative value at the largest observed loss must also equal 1.
These checks can identify errors even when a table looks plausible. For example, if tied observations were accidentally omitted, the grouped frequencies would sum to less than \(n\). If a tie was counted at more than one location, the grouped frequencies could exceed \(n\), and the jump sizes would sum to more than 1.
Check Your Understanding
For each set of exact, equally weighted loss records, group the ties before finding the requested jump or cumulative value. Round decimal probabilities to four places.
- A sample has losses 20, 40, 40, 80, 80, 80. Find the jump at 40 and the cumulative value at 40. (Answer: Jump \(2/6=0.3333\); cumulative value \(3/6=0.5000\).)
- Nine losses comprise two records of 10, three of 30, one of 50, and three of 70. Find the jump at 30 and the cumulative value at 30. (Answer: Jump \(3/9=0.3333\); cumulative value \(5/9=0.5556\).)
- Eight losses are 5, 5, 5, 10, 15, 15, 20, 20. Find the jump at 15 and the cumulative value at 15. (Answer: Jump \(2/8=0.2500\); cumulative value \(6/8=0.7500\).)
- Eleven losses comprise two records of 25, one of 40, five of 60, and three of 90. Find the jump at 60 and the cumulative value at 60. (Answer: Jump \(5/11=0.4545\); cumulative value \(8/11=0.7273\).)
- Nine losses comprise one record of 30, two of 60, two of 90, one of 120, and three of 150. Find the jump at 90 and the cumulative value at 90. (Answer: Jump \(2/9=0.2222\); cumulative value \(5/9=0.5556\).)