Empirical VaR and TVaR Estimation

Why Empirical Tail Estimates Need a Convention

Aggregate-loss data may come from observed periods or from simulated outcomes under a specified model. When the data are ordered, VaR and TVaR can be estimated directly from those values without fitting a parametric distribution. The calculation is simple only after the finite-sample convention has been made explicit: different conventions can give different VaR estimates from the same observations.

This tutorial uses two conventions. First, the empirical-distribution convention treats each of \(n\) observations as having probability \(1/n\), defines VaR by the inverse empirical distribution function, and calculates TVaR by weighting exactly the required amount of upper-tail probability. Second, when an interpolated sample quantile is requested, the tutorial uses the Type 7 linear-interpolation convention. These are distinct methods; do not silently use one convention for VaR and another for TVaR. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let \(x_{(1)}\le x_{(2)}\le\cdots\le x_{(n)}\) be the observed aggregate losses in increasing order, including repeated observations. Under the empirical-distribution convention, each observation has probability \(1/n\). The empirical distribution function is \(F_n(x)=n^{-1}\sum_{j=1}^n \mathbf{1}(x_{(j)}\le x)\). At confidence level \(\alpha\), where \(0<\alpha<1\), empirical \(\operatorname{VaR}_\alpha\) is the smallest observed loss \(x\) for which \(F_n(x)\ge\alpha\).

Thus, using the order-statistic notation above, empirical \(\operatorname{VaR}_\alpha=x_{(\lceil n\alpha\rceil)}\). If multiple observations have the same value, they remain separate observations in the ordered list, even though they contribute probability to the same value of the empirical distribution. The confidence level is not the tail probability: the upper-tail probability used for TVaR is \(1-\alpha\).

Empirical TVaR and Fractional Tail Weights

For empirical TVaR, the relevant tail contains probability \(1-\alpha\), or \(n(1-\alpha)\) observations’ worth of probability. That quantity need not be a whole number. If it is fractional, include all observations strictly above the quantile and only the necessary fraction of the observations at the quantile. The resulting TVaR is a weighted tail average, with weights summing to the tail probability required.

m=n(1-\alpha),\qquad k=\lfloor m\rfloor,\qquad r=m-k

Here \(k\) is the number of complete observations in the upper tail and \(r\) is the fraction of one additional observation. The \(k\) largest observations receive full weight, and the next largest observation receives weight \(r\). When \(r=0\), no partial observation is needed. Dividing the weighted sum by \(m\) gives the empirical TVaR.

\widehat{\operatorname{TVaR}}_\alpha = \frac{ \displaystyle\sum_{j=0}^{k-1}x_{(n-j)}+r\,x_{(n-k)} }{m}

If \(k=0\), the sum is empty and the formula gives the largest observation: the requested tail is smaller than one observation’s empirical probability, so it takes only a fraction of that largest observation. The same result can be described using the empirical quantile function: integrate the quantile function from \(\alpha\) to \(1\), then divide by \(1-\alpha\).

Why fractional weighting is correct: The empirical quantile function is constant at \(x_{(j)}\) over a probability interval of width \(1/n\). The upper tail has width \(1-\alpha=m/n\). It therefore contains \(k\) complete intervals of width \(1/n\), plus a remaining interval of width \(r/n\). Its total loss integral is \(n^{-1}\big(\sum_{j=0}^{k-1}x_{(n-j)}+r x_{(n-k)}\big)\). Dividing this integral by its probability width \(m/n\) yields the stated weighted-average formula.

Worked Example: A Cutoff Within a Repeated Loss Value

A hypothetical portfolio has seven equally weighted simulated annual aggregate losses, recorded in thousands of dollars: \(1, 4, 4, 9, 12, 12, 30\). Estimate empirical VaR and TVaR at \(\alpha=0.80\). Report the estimates in thousands of dollars to two decimal places.

Method. Use the empirical-distribution convention for VaR and the fractional upper-tail formula for TVaR. There are \(n=7\) observations, so the empirical probability of each observation is \(1/7\).

VaR. The order-statistic index is \(\lceil n\alpha\rceil=\lceil 7(0.80)\rceil=\lceil5.6\rceil=6\). The sixth ordered loss is \(x_{(6)}=12\), so \(\widehat{\operatorname{VaR}}_{0.80}=12\) thousand dollars. As a check, five of the seven observations are at or below 9, giving \(F_n(9)=5/7\approx0.7143\), while six are at or below 12, giving \(F_n(12)=6/7\approx0.8571\). The first cumulative probability is below 0.80, and the second is at least 0.80.

TVaR. The required tail contains \(m=7(1-0.80)=1.4\) observations’ worth of probability. Thus \(k=\lfloor1.4\rfloor=1\) and \(r=1.4-1=0.4\). Include the largest loss, 30, in full and 0.4 of the next largest loss, 12:

\widehat{\operatorname{TVaR}}_{0.80} = \frac{30+0.4(12)}{1.4} = \frac{34.8}{1.4} = 24.8571\ldots

Rounded to two decimal places, empirical TVaR is 24.86 thousand dollars. The weighted tail contains \(1+0.4=1.4\) observations, as required. The correction that matters here is retaining 0.4 of the loss at the cutoff; averaging only losses strictly above VaR would incorrectly return 30.

Linear Interpolation for a Sample Quantile

Some problems specify an interpolated sample quantile rather than the inverse empirical distribution quantile. A commonly used rule is Type 7. It assigns the position \(h=1+(n-1)\alpha\) to the requested quantile. If \(h=j+g\), where \(j=\lfloor h\rfloor\) and \(0\le g<1\), interpolate between adjacent order statistics as follows:

\widehat{Q}_{\alpha,\mathrm{Type\,7}} =(1-g)x_{(j)}+g x_{(j+1)}

At an integer position \(h\), the estimate is exactly the order statistic at that position; equivalently, \(g=0\). At the endpoints, the rule returns the minimum observation at \(\alpha=0\) and the maximum at \(\alpha=1\). This rule can produce a value that was not observed. It is not the inverse empirical distribution convention, which always selects an observed loss. Use Type 7 when the problem states that interpolation rule; otherwise, state the convention being used.

Worked Example: Comparing Empirical and Interpolated VaR

Six hypothetical annual aggregate losses, in thousands of dollars, are \(40, 50, 55, 70, 90, 130\). Find the VaR estimate at \(\alpha=0.75\) under both the inverse empirical distribution convention and the Type 7 convention. Then calculate empirical TVaR using the empirical-distribution convention. Report amounts in thousands of dollars to two decimal places.

Inverse empirical distribution VaR. Here \(n\alpha=6(0.75)=4.5\), so the order-statistic index is \(\lceil4.5\rceil=5\). Therefore, \(\widehat{\operatorname{VaR}}_{0.75}=x_{(5)}=90\).

Type 7 quantile. The interpolation position is \(h=1+(6-1)(0.75)=1+3.75=4.75\). Thus \(j=4\) and \(g=0.75\). Interpolate between \(x_{(4)}=70\) and \(x_{(5)}=90\):

\widehat{Q}_{0.75,\mathrm{Type\,7}} =(1-0.75)(70)+0.75(90) =17.5+67.5 =85

The two VaR estimates differ: the empirical-distribution estimate is 90, while the Type 7 estimate is 85. That difference is a consequence of the stated finite-sample convention, not an arithmetic error.

Empirical TVaR. The empirical upper tail has \(m=6(1-0.75)=1.5\) observations. It consists of the largest loss, 130, plus one-half of the next largest loss, 90:

\widehat{\operatorname{TVaR}}_{0.75} = \frac{130+0.5(90)}{1.5} = \frac{175}{1.5} = 116.6667\ldots

The empirical TVaR is 116.67 thousand dollars. This calculation is explicitly based on the empirical distribution; it does not combine the Type 7 VaR estimate of 85 with empirical tail weights.

Applying the Method Across Confidence Levels

For a fixed ordered sample, a higher confidence level usually selects a more extreme VaR and leaves a smaller TVaR tail to average. Checking the tail probability in observation units is particularly useful at high confidence levels, where the required tail may contain fewer than two observations. A small number of tail observations can make the empirical estimate sensitive to the largest recorded values.

Worked Example: Comparing Two Empirical Tail Levels

A hypothetical actuary orders ten equally weighted annual aggregate-loss outcomes, in thousands of dollars: \(3, 7, 7, 11, 16, 16, 24, 31, 50, 90\). Calculate empirical VaR and TVaR at \(\alpha=0.70\) and \(\alpha=0.85\), using the inverse empirical distribution convention. Report amounts in thousands of dollars to two decimal places.

At \(\alpha=0.70\). The VaR index is \(\lceil10(0.70)\rceil=7\), so VaR is \(x_{(7)}=24\). The tail contains \(m=10(0.30)=3\) observations. The three largest losses are 31, 50, and 90, so:

\widehat{\operatorname{TVaR}}_{0.70} = \frac{31+50+90}{3} = \frac{171}{3} = 57

At \(\alpha=0.85\). The VaR index is \(\lceil10(0.85)\rceil=\lceil8.5\rceil=9\), so VaR is \(x_{(9)}=50\). The tail contains \(m=10(0.15)=1.5\) observations: the maximum loss, 90, in full, and half of the next largest loss, 50. Therefore:

\widehat{\operatorname{TVaR}}_{0.85} = \frac{90+0.5(50)}{1.5} = \frac{115}{1.5} = 76.6667\ldots

The results are \(\widehat{\operatorname{VaR}}_{0.70}=24\), \(\widehat{\operatorname{TVaR}}_{0.70}=57.00\), \(\widehat{\operatorname{VaR}}_{0.85}=50\), and \(\widehat{\operatorname{TVaR}}_{0.85}=76.67\), all in thousands of dollars. The tail checks are three observations at 0.70 and one and a half observations at 0.85.

Common Mistakes and Practical Checks

Empirical estimates are transparent, but several small choices can change the result. State the confidence level, the quantile convention, the units, and whether each observation has equal weight. For each TVaR calculation, confirm that the weights sum to \(n(1-\alpha)\) observation units before dividing.

Using the wrong rank. Under the inverse empirical distribution convention, use \(\lceil n\alpha\rceil\), not \(n(1-\alpha)\), to locate VaR. The latter is the number of observations’ worth of upper-tail probability used for TVaR.
Dropping the cutoff mass. Averaging only losses strictly greater than VaR can omit part of the required tail. Include the needed fraction at the quantile, especially when values repeat.
Mixing quantile conventions. An interpolated Type 7 quantile can be between observed values; inverse empirical distribution VaR cannot. Do not pair an interpolated VaR with a TVaR calculation that assumes an empirical point mass at that interpolated value unless the problem defines that combined procedure.
Confusing confidence and tail levels. At confidence level \(\alpha\), TVaR averages the upper tail of probability \(1-\alpha\), not \(\alpha\).
Changing units unintentionally. If losses are given in thousands of dollars, calculations remain in thousands until the final conversion. Keep the reported unit explicit.

Key takeaway: Order the observations, state the finite-sample quantile convention, and calculate the upper-tail probability in observation units. For empirical TVaR, use full weights for observations above the cutoff and a fractional weight when the tail ends within an observation’s probability mass.

Check Your Understanding

Use the stated empirical-distribution convention unless a question explicitly requests Type 7 interpolation. Loss amounts are in thousands of dollars.

For the ordered sample \(5, 8, 12, 20, 35\), find empirical VaR and TVaR at \(\alpha=0.80\). (Answer: VaR \(=20\); TVaR \(=35\).)
For the ordered sample \(2, 4, 4, 10, 16, 25\), find empirical VaR and TVaR at \(\alpha=0.75\). (Answer: VaR \(=16\); TVaR \(=22\).)
For the ordered sample \(0, 3, 7, 7, 11, 18, 30, 50\), find empirical VaR and TVaR at \(\alpha=0.80\). (Answer: VaR \(=30\); TVaR \(=42.50\).)
For \(10, 14, 20, 29, 41\), find inverse empirical VaR and Type 7 quantile at \(\alpha=0.60\), and empirical TVaR at that level. (Answer: inverse empirical VaR \(=20\); Type 7 quantile \(=23.60\); empirical TVaR \(=35\).)
For the ordered sample \(1, 2, 2, 4, 8, 8, 13, 21, 40\), find empirical VaR and TVaR at \(\alpha=0.80\). (Answer: VaR \(=21\); TVaR \(=31.56\).)

What You'll Learn