Why Ties Require Care
A continuous severity model assigns probability zero to any exact loss amount. Consequently, a sample from an exactly continuous model has no ties with probability one. Yet actuarial loss data often contain repeated values because amounts are rounded to the nearest dollar, reported in broad bands, or recorded at a limited level of precision. The repeated values may therefore describe the recording process rather than a genuinely discrete loss distribution.
This distinction matters for goodness-of-fit testing. You can calculate a Kolmogorov–Smirnov (KS) or Anderson–Darling (AD) statistic from tied observations, but the usual continuous-model null distributions rely on continuous observations. Ties created by rounding change the sampling behavior of the statistic. A calculation may be arithmetically correct and still have an inappropriate critical value or p-value.
KS Discrepancies at Tied Observations
For ordered observations \(x_{(1)}\le \cdots \le x_{(n)}\) and a fully specified continuous null CDF \(F\), the one-sample KS statistic compares the empirical CDF with \(F\). The empirical CDF jumps by \(1/n\) at each observation. When several observations are equal, their jumps occur together, so the left and right sides of the whole jump—not intermediate ranks treated as distinct locations—determine the largest discrepancies.
This block formula follows directly from the empirical CDF’s jumps. Between consecutive observed values, the empirical CDF is constant while a continuous CDF \(F\) is nondecreasing. Thus the difference \(F_n(x)-F(x)\) decreases between jumps, and its extreme values occur immediately before or at a jump. At a tied value, the empirical CDF moves directly from \((r-1)/n\) to \(s/n\), which gives the two stated comparisons.
Worked Example: KS Calculation with Tied Continuous-Model Observations
A hypothetical severity sample has eight recorded losses. Assume the stated continuous model is exponential with rate \(\lambda=0.5\) per thousand dollars, so \(F(x)=1-e^{-0.5x}\) when \(x\) is measured in thousands of dollars. The observed losses are two copies of \(x_{0.10}\), one copy of \(x_{0.20}\), three copies of \(x_{0.45}\), one copy of \(x_{0.70}\), and one copy of \(x_{0.90}\), where \(x_p=-\ln(1-p)/0.5\). The corresponding loss amounts are approximately 0.210721, 0.446287, 1.195674, 2.407946, and 4.605170 thousand dollars. Find the KS statistic.
Method. Use tied rank blocks. By construction, the model CDF at each distinct loss is the indicated \(p\), so no rounded CDF approximation is needed. The rank blocks are 1–2, 3, 4–6, 7, and 8.
| \(F(t)\) | Ranks \(r\)–\(s\) | \(D^+\) candidate \(s/8-F(t)\) | \(D^-\) candidate \(F(t)-(r-1)/8\) |
|---|---|---|---|
| 0.10 | 1–2 | 0.25 − 0.10 = 0.150 | 0.10 − 0 = 0.100 |
| 0.20 | 3 | 0.375 − 0.20 = 0.175 | 0.20 − 0.25 = −0.050 |
| 0.45 | 4–6 | 0.75 − 0.45 = 0.300 | 0.45 − 0.375 = 0.075 |
| 0.70 | 7 | 0.875 − 0.70 = 0.175 | 0.70 − 0.75 = −0.050 |
| 0.90 | 8 | 1 − 0.90 = 0.100 | 0.90 − 0.875 = 0.025 |
The largest positive candidate is \(D^+=0.300\), and the largest negative candidate is \(D^-=0.100\). Therefore \(D=\max(0.300,0.100)=0.300\). This is the statistic computed from the tied observations. Because the ties here arise from recording at selected loss levels, the usual continuous-model KS calibration should not automatically be used to assign a p-value.
Rounding Creates a Discrete Observation Model
If a latent loss \(X\) is rounded to the nearest unit and the recorded value is \(j\), then—away from boundary conventions—the latent loss lies in the interval from \(j-0.5\) to \(j+0.5\). Under a continuous model with CDF \(F_X\), the recorded-value probability is the probability of that interval. For a bin \([a_j,b_j)\), this is \(F_X(b_j)-F_X(a_j)\). At the ends of a bounded range, or for a nonnegative loss, the first and last intervals must be adjusted to match the actual recording rule.
Once those probabilities are calculated, the recorded data can be treated as observations from a discrete distribution. A discrete-model KS discrepancy compares cumulative observed proportions with the model’s cumulative probabilities at category boundaries. It must account for both sides of each jump: just before category \(j\), compare the cumulative proportions and probabilities for earlier categories; at category \(j\), compare the cumulative proportions and probabilities through that category.
Worked Example: Discrete KS Statistic After Rounding
A hypothetical latent loss \(X\) is uniform on \([0,4]\) thousand dollars and is recorded to the nearest whole thousand dollars, producing values 0, 1, 2, 3, or 4. In a sample of 20, the recorded counts are 2, 3, 8, 5, and 2, respectively. Calculate the model probabilities and the discrete-model KS statistic.
Step 1: Find the recorded-value probabilities. The rounding interval for 0 has length 0.5, each interval for 1, 2, or 3 has length 1, and the interval for 4 has length 0.5. The total uniform range has length 4. Hence the probabilities are \(0.5/4=0.125\), \(1/4=0.250\), \(0.250\), \(0.250\), and \(0.125\).
Step 2: Compare cumulative probabilities. The model cumulative probabilities at recorded values 0 through 4 are 0.125, 0.375, 0.625, 0.875, and 1.000. The observed cumulative counts are 2, 5, 13, 18, and 20; dividing each by 20 gives 0.100, 0.250, 0.650, 0.900, and 1.000. The differences (observed minus model) at the category values are −0.025, −0.125, 0.025, 0.025, and 0.
Step 3: Check the left limits as well. Immediately before each category, the observed cumulative proportions are 0, 0.100, 0.250, 0.650, and 0.900. The model left-limit cumulative probabilities are 0, 0.125, 0.375, 0.625, and 0.875. Their differences are 0, −0.025, −0.125, 0.025, and 0.025. The largest absolute difference on either side is therefore \(D=0.125\).
This statistic describes the discrepancy for the recorded discrete model. The common mistake is to treat the recorded values as if they were unrounded observations from a continuous distribution and then apply continuous-model KS calibration. The rounding intervals, not just the displayed values, determine the model probabilities.
Anderson–Darling with Tied or Rounded Data
For ordered observations from a continuous model, the usual one-sample AD statistic uses the model CDF values at the ordered observations. With \(Z_i=F(x_{(i)})\), its standard form is
This formula can still be evaluated when some observed values are tied, provided all the CDF values used are strictly between 0 and 1. But ties make the ordinary continuous-model reference distribution inappropriate: the formula’s numerical value does not itself restore the assumptions behind its calibration. With rounded observations, a principled approach is to calculate the discrete probabilities induced by the rounding rule and calibrate an appropriate statistic under that recorded-data model. Calibration can be performed by generating or enumerating samples from the specified discrete probabilities; it must reflect the same rounding and any parameter-fitting steps used in the analysis.
Worked Example: Computing AD with Tied CDF Values
A hypothetical severity model is exponential with rate \(\lambda=1\) per unit. Four recorded losses consist of two copies of \(x_{0.20}=-\ln(0.80)\approx0.223144\) and two copies of \(x_{0.60}=-\ln(0.40)\approx0.916291\). Treat these as tied recorded levels and compute the continuous-form AD statistic, then state its limitation.
The ordered model CDF values are \(Z_1=Z_2=0.20\) and \(Z_3=Z_4=0.60\), with \(n=4\). The weighted logarithm sum has four terms involving \(\ln(0.20)+\ln(0.40)=\ln(0.08)\), with combined weight \(1+3=4\), and twelve terms involving \(\ln(0.60)+\ln(0.80)=\ln(0.48)\), with combined weight \(5+7=12\). Thus
Therefore \(A^2=-4-(-18.910545/4)\approx0.727636\), or 0.7276 to four decimal places. The calculation is complete, but the ordinary continuous-model AD calibration is not justified merely because this formula produced a number. If the ties result from rounding, the model for the recorded categories and a matching calibration are required.
Randomized Probability Transforms for Discrete Models
A useful way to connect a discrete recorded-data model to a uniform reference is the randomized probability transform. If the recorded value is \(Z=z\), let \(F_Z(z^-)=P(Z<z)\) and \(p_z=P(Z=z)\). Draw an independent \(V\) uniformly on \([0,1]\), and define \(U=F_Z(z^-)+Vp_z\). Under the stated discrete model, \(U\) is uniform on \([0,1]\): each atom spreads its probability mass uniformly across the corresponding interval in the unit interval.
Worked Example: Randomized Transform for a Rounded Exponential Loss
Let the latent loss have an exponential distribution with rate \(\lambda=1\) per unit and be recorded to the nearest whole unit, with nonnegative values. The recorded category 0 corresponds to \(0\le X<0.5\); category 1 corresponds to \(0.5\le X<1.5\). Find the probability of recorded value 1 and the randomized transform for an observed value of 1 when \(V=0.4000\).
The probability of recorded value 0 is \(1-e^{-0.5}\approx0.393469\). The probability of recorded value 1 is \(e^{-0.5}-e^{-1.5}\approx0.606531-0.223130=0.383401\). Therefore \(F_Z(1^-)=0.393469\). The randomized transform is \(U=0.393469+0.4000(0.383401)=0.393469+0.153360=0.546829\), or 0.5468 to four decimal places.
The transformed value lies within the probability interval assigned to category 1, from approximately 0.393469 to 0.776870. A different independent draw of \(V\) would give a different transformed value for the same recorded loss, which is why the randomization must be included in any calibration.
Common Mistakes and Practical Choices
- Breaking a tie into imaginary intermediate jumps. For a block occupying ranks \(r\) through \(s\), use the empirical CDF immediately before the block and after the entire block. The empirical CDF does not take the intermediate values at that same observed amount.
- Using continuous-model critical values automatically. Ties caused by rounding alter the null behavior of KS and AD statistics. A computed statistic alone does not justify the usual continuous reference distribution.
- Evaluating the latent CDF only at the displayed rounded value. The probability of a recorded category is the probability of its full rounding interval, \(F_X(b)-F_X(a)\), not generally the density or CDF at the displayed number.
- Ignoring endpoint bins. For nonnegative losses, the first rounding interval begins at zero; it does not extend to negative losses. Match all bins to the actual recording rule.
- Treating randomization as a cure for model misspecification. A randomized transform is uniform only when the discrete category probabilities are correct. It cannot fix an incorrect severity model or an incorrectly specified rounding mechanism.
For actuarial loss data, the practical sequence is to identify whether repetitions are genuine atoms or arise from limited precision, state the recording rule, and then select the model for the observations actually available. Continuous KS and AD calculations remain useful for understanding discrepancies, but inference must respect how the data were produced. This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify the observation model before choosing a calculation.
- A discrete null model assigns probabilities 0.10, 0.30, 0.40, and 0.20 to four ordered categories. A sample of 12 has counts 1, 5, 4, and 2. Calculate the two-sided discrete KS discrepancy using both category values and left limits. (Answer: \(D=0.1000\).)
- A latent uniform loss on \([0,6]\) is rounded to the nearest integer, producing categories 0 through 6. State the probabilities of recorded values 0, 1, and 6. (Answer: \(1/12,\ 1/6,\ 1/12\), respectively.)
- A latent exponential loss has rate \(\lambda=0.25\) per unit and is rounded to the nearest unit. Find the probability that the recorded value is 1, whose rounding interval is \([0.5,1.5)\). (Answer: \(e^{-0.125}-e^{-0.375}\approx0.1952\).)
- For \(n=4\), suppose the ordered continuous-model CDF values are 0.10, 0.10, 0.50, and 0.50. Evaluate the continuous-form AD statistic to four decimal places and state whether ties permit automatic use of its usual continuous calibration. (Answer: \(A^2\approx1.3913\); no, the tied-data calibration must be justified.)
- A discrete category has probability 0.20 and cumulative probability 0.30 strictly before it. For randomized-transform value \(V=0.75\), calculate \(U\). (Answer: \(U=0.30+0.75(0.20)=0.4500\).)