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Empirical Value-at-Risk and Tail Value-at-Risk

Calculate empirical VaR and TVaR consistently from ordered loss observations, including when the tail cuts through a group of tied losses.

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What You'll Learn

  • Define empirical VaR as a generalized-inverse quantile and identify its order-statistic index.
  • Distinguish the empirical quantile convention from interpolated sample-quantile conventions.
  • Calculate empirical TVaR by averaging exactly the required fraction of the observed tail.
  • Handle ties at the VaR threshold by including only the necessary fraction of their empirical mass.
  • Explain why averaging only losses strictly above VaR can give the wrong TVaR.
  • Check units, tail size, and the relationship between empirical VaR and TVaR.

Why the Empirical Tail Needs a Clear Convention

When losses are available as a finite list of observations, value-at-risk and tail value-at-risk can be estimated directly from the ordered data. The calculation looks simple, but its result depends on how the empirical quantile is defined and how much weight is assigned to observations tied at the tail threshold. Those choices matter especially when the confidence level does not leave an integer number of observations in the tail.

This tutorial treats the observed losses as an empirical distribution that places equal probability on each observation. Under this convention, VaR is an order statistic, while TVaR averages the upper tail using the quantile-integral definition. A fractional part of an observed loss may therefore enter the tail average; it represents a fraction of empirical probability mass, not a claim being physically divided.

Definition: Let \(x_1,\ldots,x_n\) be observed losses and let \(x_{(1)}\leq\cdots\leq x_{(n)}\) denote them in increasing order, retaining repeated values. The empirical distribution assigns probability \(1/n\) to each observation. For \(0<\alpha<1\), define empirical value-at-risk by the generalized-inverse convention $$ \widehat{\operatorname{VaR}}_\alpha =\inf\{x:F_n(x)\geq\alpha\} =x_{(\lceil n\alpha\rceil)}, $$ where \(F_n\) is the empirical distribution function. Define empirical tail value-at-risk by $$ \widehat{\operatorname{TVaR}}_\alpha =\frac{1}{1-\alpha}\int_\alpha^1 \widehat{\operatorname{VaR}}_u\,du. $$

The ceiling in the VaR formula is important. For example, if \(n\alpha=7.2\), the index is 8, not 7. This tutorial uses the empirical generalized-inverse convention throughout. Some software or texts use an interpolation rule between adjacent order statistics; that rule can produce a different VaR and should be identified rather than silently substituted.

Computing the Empirical Tail Average

The probability represented by the upper tail is \(1-\alpha\). In a sample with \(n\) equally weighted observations, that corresponds to \(m=n(1-\alpha)\) observation-units. The number \(m\) need not be an integer. Begin by finding the empirical VaR \(q=\widehat{\operatorname{VaR}}_\alpha\), then count observations strictly greater than \(q\).

Theorem: Let \(k\) be the number of observations strictly greater than \(q\), and let \(r=n(1-\alpha)-k\). Then $$ \widehat{\operatorname{TVaR}}_\alpha = \frac{\displaystyle\sum_{j:x_j>q}x_j+r q} {n(1-\alpha)}. $$ Here \(r\) is the number of observation-units at the threshold needed to complete the tail. It satisfies \(0\leq r\leq \#\{j:x_j=q\}\). If the data are recorded in dollars, the result is in dollars; if they are recorded in thousands of dollars, the result is in thousands.

This formula handles both fractional tail sizes and ties. Every observation strictly above the threshold is included in full. The observations equal to the threshold collectively supply just the remaining tail mass \(r\). Observations below the threshold contribute nothing.

Why the Fractional-Tail Formula Holds

The empirical quantile function is constant over intervals of width \(1/n\): it equals \(x_{(j)}\) for \((j-1)/n<u\leq j/n\). Therefore, integrating the quantile function from \(\alpha\) to 1 adds the loss values in the upper tail, each multiplied by the amount of quantile-probability interval included.

1
Set the tail probability.
The integral covers probability \(1-\alpha\), or \(n(1-\alpha)\) observation-units when each observation has empirical probability \(1/n\).
2
Include losses strictly above the threshold.
Each such loss contributes its full observation-unit to the tail. Their total is \(\sum_{j:x_j>q}x_j\), and there are \(k\) of them.
3
Complete the tail at the threshold.
The remaining mass is \(r=n(1-\alpha)-k\). On that portion, the empirical quantile equals \(q\), so its contribution to the integral, expressed in observation-units, is \(rq\).
4
Normalize by the tail probability.
Dividing the total tail loss by \(n(1-\alpha)\) gives the average quantile over \([\alpha,1]\), which is the stated empirical TVaR.

This is also why the shortcut \(E[X\mid X>q]\) is not generally the right empirical TVaR calculation. That conditional average discards observations equal to \(q\). For a discrete empirical distribution, the quantile-integral definition includes just enough threshold mass to make the upper-tail probability exactly \(1-\alpha\).

Worked Example: An Integer-Sized Tail With Tied Threshold Losses

A hypothetical insurer has ten observed claim severities, recorded in thousands of dollars: \(2,3,4,5,7,8,10,10,10,20\). Estimate VaR and TVaR at \(\alpha=0.80\), using the empirical generalized-inverse convention.

Method. Sort the losses, determine the order-statistic index, then apply the fractional-tail formula. The observations are already ordered, and \(n=10\).

VaR. The order-statistic index is \(\lceil n\alpha\rceil=\lceil10(0.80)\rceil=\lceil8\rceil=8\). Thus \(q=x_{(8)}=10\) thousand dollars.

Tail size. The tail contains \(n(1-\alpha)=10(0.20)=2\) observation-units. There is one observation strictly above \(q\), namely 20, so \(k=1\) and \(r=2-1=1\). The threshold value 10 occurs three times, so there is sufficient threshold mass to supply the remaining unit.

TVaR. The numerator is \(20+1(10)=30\) thousand dollars, and the denominator is 2. Therefore, \(\widehat{\operatorname{TVaR}}_{0.80}=30/2=15\) thousand dollars. The tail average is consistent with the two upper-tail units: one value of 20 and one unit at 10 have average 15.

Check and trap avoided. The result lies between the threshold 10 and the largest observation 20. Averaging only the losses strictly greater than VaR would give 20, not 15. That calculation incorrectly treats the strict exceedance set as the entire 20% tail and omits the required threshold mass.

Worked Example: A Fractional Number of Tail Observations

For a hypothetical specialty coverage portfolio, seven observed losses, in thousands of dollars, are \(1,3,4,7,9,14,20\). Estimate VaR and TVaR at \(\alpha=0.80\).

Method. Use the specified empirical quantile and allow the tail to contain a fractional observation-unit.

VaR. Here \(n\alpha=7(0.80)=5.6\), so the index is \(\lceil5.6\rceil=6\). Hence \(q=x_{(6)}=14\) thousand dollars.

Tail composition. The tail size is \(n(1-\alpha)=7(0.20)=1.4\) observation-units. One loss, 20, is strictly greater than 14, so \(k=1\). The remaining amount is \(r=1.4-1=0.4\) units at the threshold.

TVaR. The total tail loss is \(20+0.4(14)=20+5.6=25.6\) thousand dollars. Dividing by 1.4 gives \(25.6/1.4=18.285714\ldots\) thousand dollars. Rounding only at the final step to two decimal places gives \(\widehat{\operatorname{TVaR}}_{0.80}=18.29\) thousand dollars.

Check and trap avoided. The answer is above VaR and below the maximum observed loss. Taking only the largest observation would yield 20 thousand dollars, but would ignore the additional 0.4 tail units at 14. The fractional weight is required because the empirical tail probability is 0.20, not an integer multiple of \(1/7\).

Worked Example: Ties at the VaR Threshold in a Fractional Tail

A hypothetical reinsurer reviews twelve claim losses, in thousands of dollars: \(0,1,2,3,4,5,6,9,9,9,18,27\). Estimate VaR and TVaR at \(\alpha=0.80\), and compare TVaR with the average of losses strictly exceeding VaR.

VaR. The index is \(\lceil12(0.80)\rceil=\lceil9.6\rceil=10\). The tenth ordered loss is 9, so \(q=9\) thousand dollars.

Tail composition. The target tail size is \(12(0.20)=2.4\) observation-units. There are two losses strictly above 9: 18 and 27. Thus \(k=2\) and \(r=2.4-2=0.4\). Three observations equal the threshold, so using 0.4 units at 9 is permitted.

TVaR. The numerator is \(18+27+0.4(9)=45+3.6=48.6\) thousand dollars. The estimate is \(48.6/2.4=20.25\) thousand dollars.

Strict-exceedance comparison. The average of losses strictly greater than VaR is \((18+27)/2=22.50\) thousand dollars. It is not the quantile-integral TVaR because it averages only two observations, while the specified upper tail contains 2.4 observation-units. Including 0.4 units at the threshold reduces the tail average to 20.25 thousand dollars.

Check and trap avoided. The result exceeds VaR and is less than the strict-exceedance average, as expected when the threshold value is below both strict exceedances. Treating all three tied 9s as part of the tail would also be wrong: it would include more than the required 2.4 units.

Common Exam Traps and Practical Checks

An empirical estimate is reproducible only when the quantile convention, confidence level, and tail treatment are clear. Before finalizing a calculation, verify each of the following:

  • Use the ceiling index for this convention. For example, with \(n\alpha=5.6\), the empirical VaR is the sixth order statistic. Rounding down changes the stated quantile convention.
  • Retain duplicate observations. Ties are separate observations, each carrying empirical mass \(1/n\); do not collapse them into one value without preserving their total frequency.
  • Use the target tail size. Calculate \(n(1-\alpha)\) explicitly. Do not round it to the nearest whole number.
  • Handle threshold ties fractionally. Include every loss strictly above VaR, then only enough threshold mass to reach the specified tail probability.
  • Keep units consistent. If observations are in thousands of dollars, both the numerator and resulting risk measures are in thousands of dollars.
Key takeaway: Under the empirical-distribution convention, VaR is \(x_{(\lceil n\alpha\rceil)}\). TVaR is the average of exactly \(n(1-\alpha)\) observation-units in the upper tail, using fractional weight at the VaR threshold when needed. This treatment of ties is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical loss sample, use the empirical generalized-inverse quantile and quantile-integral TVaR conventions. Losses are in thousands of dollars; give both measures in the same units.

  1. Five ordered losses are \(2,5,7,11,20\). Find VaR and TVaR at \(\alpha=0.80\). (Answer: VaR \(=11\); TVaR \(=20\).)
  2. Six ordered losses are \(1,3,4,8,8,14\). Find VaR and TVaR at \(\alpha=0.75\), accounting for the threshold tie. (Answer: VaR \(=8\); TVaR \(=12\).)
  3. Eight ordered losses are \(2,4,6,6,10,13,13,19\). Find VaR and TVaR at \(\alpha=0.625\). (Answer: VaR \(=10\); TVaR \(=15\).)
  4. Nine ordered losses are \(1,2,5,7,7,7,10,16,22\). Find VaR and TVaR at \(\alpha=2/3\). (Answer: VaR \(=7\); TVaR \(=16\).)
  5. Twelve ordered losses are \(1,2,3,4,5,6,7,8,10,10,10,25\). Find VaR and TVaR at \(\alpha=0.90\), showing the fractional threshold weight. (Answer: VaR \(=10\); TVaR \(=22.50\).)

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