Why Grouped Loss Data Require an Assumption
A loss frequency table reports how many observations fall in each interval, not the individual losses within that interval. It therefore gives less information than ordered observations or a fully specified loss distribution. To estimate Tail Value at Risk (TVaR), an actuary first locates the confidence-level quantile, then determines how much of the upper tail lies in its class and how much lies in higher classes.
The calculation needs an assumption about losses within each class. A common working assumption is that losses are uniformly distributed within each finite class. This permits linear interpolation for the quantile and uses each class midpoint as its estimated mean. The result is an estimate, not a fact identified by the grouped frequencies alone.
Write the class intervals as \([a_j,b_j)\), with frequency \(f_j\), and let \(N=\sum_j f_j\). Define the cumulative frequency through class \(j\) as \(C_j=\sum_{k=1}^j f_k\). The target rank for level \(\alpha\) is \(\alpha N\). If it falls in class \(j\), then \(C_{j-1}<\alpha N\leq C_j\). Under the uniform-within-class assumption, the estimated quantile is
The fraction in this expression is the fraction of the class frequency that must be traversed to reach the target rank. This is interpolation across the class width; it is not a claim that the original losses are known individually.
Allocating the Tail and Calculating TVaR
The upper tail contains \(N(1-\alpha)\) observations in frequency terms. Under a continuous uniform model within each finite class, the part of the quantile class above the estimated VaR has estimated frequency \(f_j(b_j-\widehat{\operatorname{VaR}}_\alpha)/(b_j-a_j)\). Its average loss is the midpoint between the VaR estimate and the class upper endpoint. For a complete higher class \(k\), the estimated mean loss is \((a_k+b_k)/2\), and its full frequency contributes to the tail.
This formula assumes that the quantile class and all higher classes are finite and that losses are uniform within them. If the quantile lies exactly at a class boundary, the quantile-class contribution above the threshold is zero; the tail consists of the classes above that boundary. In grouped data with point masses or ties, the continuous uniform calculation does not represent those masses exactly. The appropriate treatment depends on the available information and the model being used.
A practical calculation keeps three quantities distinct: the target rank \(\alpha N\), the tail frequency \(N(1-\alpha)\), and the estimated loss total in that tail. Mixing the rank with the tail frequency is a frequent source of errors.
Worked Calculations Under Uniformity
Worked Example: Quantile Within a Class
A hypothetical portfolio has the following 200 losses, grouped in dollars. Estimate VaR and TVaR at level \(\alpha=0.8500\), assuming losses are uniform within each class.
| Loss interval | Frequency | Cumulative frequency |
|---|---|---|
| \([0,20)\) | 30 | 30 |
| \([20,40)\) | 50 | 80 |
| \([40,60)\) | 60 | 140 |
| \([60,100)\) | 40 | 180 |
| \([100,200)\) | 20 | 200 |
Method: locate the quantile class, interpolate within it, and average the allocated tail. The target rank is \(0.8500(200)=170\). Since \(140<170\leq180\), the quantile class is \([60,100)\). The rank is \(170-140=30\) observations into a class of frequency 40, so
The upper tail has \(200(1-0.8500)=30\) observations in frequency terms. Above \$90, the retained part of the quantile class is from \$90 to \$100. Its frequency is \(40(100-90)/(100-60)=10\), and its mean loss is \((90+100)/2=\$95.00\). The complete class \([100,200)\) contributes 20 observations with mean loss \((100+200)/2=\$150.00\). Therefore,
Check: the allocated tail frequencies sum to \(10+20=30\), as required. The estimated tail mean, \$131.67, exceeds the estimated VaR of \$90.00. The common mistake avoided here is to include the entire quantile class in the tail rather than only the portion above the quantile.
Worked Example: Quantile at a Class Boundary
A hypothetical insurer summarizes 250 losses as follows. Estimate VaR and TVaR at level \(\alpha=0.9200\), assuming uniform losses within finite classes.
| Loss interval | Frequency | Cumulative frequency |
|---|---|---|
| \([0,10)\) | 50 | 50 |
| \([10,30)\) | 70 | 120 |
| \([30,70)\) | 80 | 200 |
| \([70,150)\) | 30 | 230 |
| \([150,350)\) | 20 | 250 |
The target rank is \(0.9200(250)=230\). The cumulative frequency reaches 230 at the upper boundary of the \([70,150)\) class. Thus the estimated quantile is that boundary:
The tail frequency is \(250(1-0.9200)=20\). Since the threshold is at the boundary, none of the \([70,150)\) class lies above it under the continuous class model. The next class, \([150,350)\), supplies all 20 tail observations. Its uniform mean is \((150+350)/2=\$250.00\), giving
Check: the tail frequency is exactly the frequency of the class above the quantile boundary. The common mistake avoided is to interpolate into the \([150,350)\) class even though the target rank has already been reached at \$150.
Worked Example: Comparing Within-Class Assumptions
A hypothetical risk register contains 100 losses, grouped as shown below. Estimate TVaR at \(\alpha=0.8000\) first under uniformity within each class and then using each class midpoint as the representative loss for every observation in that class.
| Loss interval | Frequency | Cumulative frequency |
|---|---|---|
| \([0,25)\) | 40 | 40 |
| \([25,50)\) | 30 | 70 |
| \([50,100)\) | 20 | 90 |
| \([100,200)\) | 10 | 100 |
Uniform-within-class estimate. The target rank is \(0.8000(100)=80\), which is 10 observations into the \([50,100)\) class. The estimated VaR is
The tail frequency is 20. The portion from \$75 to \$100 has frequency \(20(100-75)/(100-50)=10\) and mean loss \((75+100)/2=\$87.50\). The complete \([100,200)\) class has frequency 10 and mean loss \$150.00. Hence
Class-midpoint approximation. The class representatives are \$12.50, \$37.50, \$75.00, and \$150.00, with frequencies 40, 30, 20, and 10. The upper 20 observations comprise all 10 observations represented by \$150 and 10 of the 20 represented by \$75. Using fractional allocation at the quantile,
Check: both calculations use 20 tail observations, but their TVaR estimates differ by \$6.25. The difference arises because the uniform model spreads losses through each interval, whereas the midpoint method places every loss at one representative value. Neither estimate is determined by the grouped frequencies alone.
Assumptions, Limitations, and Exam Traps
The uniform-within-class method is transparent and convenient, but it should be reported as an assumption. Wider classes give more scope for the within-class shape to affect the result. In particular, a high quantile or a class with a large frequency may make the estimate sensitive to how losses near the class endpoints are represented.
An open-ended highest class, such as \([100,\infty)\), has no finite midpoint. Grouped frequencies alone then do not specify the mean of that class or a finite TVaR contribution from it. A tail model or additional information is needed; silently assigning a midpoint is not justified. Likewise, a midpoint approximation is a separate estimation convention, not a consequence of uniformity.
A common exam trap is to use the number of observations below VaR as the denominator for TVaR. The denominator is the upper-tail frequency, \(N(1-\alpha)\), not \(\alpha N\). Another is to count every loss in the quantile class as an upper-tail loss. Under the continuous uniform assumption, only the part of that class above the estimated quantile belongs in the tail.
Common Mistake: Using the Wrong Tail Count
In the first worked example, the correct tail frequency is \(200(1-0.8500)=30\), and the estimated tail loss total is \(10(95)+20(150)=3{,}950\). Dividing by the target rank of 170 would give the incorrect calculation
This is not a tail average: 170 is the approximate number of observations at or below the quantile, not the number in the tail. The correction is to divide by 30:
The corrected result is above the \$90.00 VaR, as an upper-tail average should be in this continuous-loss setting.
Check Your Understanding
For each problem, use uniformity within finite classes unless another method is specified. Money is in dollars.
- A table has classes \([0,20)\), \([20,50)\), \([50,80)\), and \([80,160)\), with frequencies 20, 40, 30, and 10. Estimate VaR and TVaR at level \(\alpha=0.9000\). (Answer: VaR \(=\$80.00\); TVaR \(=\$120.00\).)
- A table has classes \([0,10)\), \([10,30)\), \([30,50)\), and \([50,90)\), with frequencies 24, 36, 36, and 24. Estimate VaR and TVaR at level \(\alpha=0.7500\). (Answer: VaR \(=\$46.67\); TVaR \(=\$65.67\).)
- A table has classes \([0,10)\), \([10,30)\), \([30,60)\), and \([60,100)\), with frequencies 20, 20, 24, and 16. Estimate VaR and TVaR at level \(\alpha=0.7500\). (Answer: VaR \(=\$55.00\); TVaR \(=\$75.50\).)
- A table has classes \([0,40)\), \([40,100)\), and \([100,300)\), with frequencies 50, 30, and 20. Estimate VaR and TVaR at level \(\alpha=0.9000\). (Answer: VaR \(=\$200.00\); TVaR \(=\$250.00\).)
- A grouped table has frequencies 40 in \([0,10)\), 35 in \([10,30)\), and 25 in \([30,\infty)\). At level \(\alpha=0.9000\), can the grouped table and the uniform-within-class assumption for finite classes determine a finite TVaR? State why. (Answer: No; the quantile is in the open-ended highest class, whose upper-tail mean is not determined by the grouped frequencies or a finite-class uniform assumption.)