Sample TVaR From Ordered Observations

Why Ordered Observations Matter

TVaR summarizes losses in the upper tail, so a calculation from observed data begins by identifying which observations belong to that tail. Sorting losses from smallest to largest makes the boundary visible. If the selected tail contains a whole number of observations, its average is direct. If it contains only part of an observation’s empirical probability, that boundary observation receives a fractional weight.

This distinction matters because an empirical distribution assigns probability $1/n$ to each of $n$ observations. A confidence level does not necessarily leave a tail containing an integer number of those equal probability masses. Averaging the largest observations without checking the tail size can therefore select too much of the sample.

Definition: Let $x_1,\ldots,x_n$ be observed losses and write their ordered values as $x_{(1)}\leq x_{(2)}\leq\cdots\leq x_{(n)}$. The empirical distribution assigns probability $1/n$ to each observation. Its empirical quantile function is $\widehat{Q}_n(p)=x_{(j)}$ for $(j-1)/n < p \leq j/n$. For confidence level $0<\alpha<1$, the empirical TVaR estimator is the average of this quantile function over the upper probability tail, $\widehat{\operatorname{TVaR}}_\alpha=\frac{1}{1-\alpha}\int_\alpha^1\widehat{Q}_n(p)\,dp$.

Deriving the Ordered-Observation Formula

The definition uses an integral, but the empirical quantile function is a step function. Each ordered loss $x_{(j)}$ contributes only over the portion of its probability interval that lies above $\alpha$. The length of that portion is

w_j=\max\left(0,\frac{j}{n}-\max\left(\alpha,\frac{j-1}{n}\right)\right).

Thus the estimator can be calculated directly from the ordered losses:

\widehat{\operatorname{TVaR}}_\alpha = \frac{1}{1-\alpha}\sum_{j=1}^{n}w_jx_{(j)}.

This formula follows by splitting the quantile integral at the points $j/n$. On the interval $((j-1)/n,j/n]$, the quantile is constant at $x_{(j)}$; only the overlap of this interval with $[\alpha,1]$ contributes. The overlap has length $w_j$, so its contribution to the integral is $w_jx_{(j)}$. Summing the contributions and dividing by the tail probability $1-\alpha$ gives the result.

A particularly useful form comes from measuring the tail size in observation units. Define $m=n(1-\alpha)$, $h=\lfloor m\rfloor$, and $\theta=m-h$. The tail includes the $h$ largest observations completely and a fraction $\theta$ of the next-largest observation:

\widehat{\operatorname{TVaR}}_\alpha = \frac{x_{(n)}+x_{(n-1)}+\cdots+x_{(n-h+1)}+\theta x_{(n-h)}}{m}.

When $h=0$, the sum of fully included observations is empty, and the formula uses only $\theta x_{(n)}$. When $m$ is an integer, $\theta=0$, so the boundary observation contributes nothing. In that case, the estimator is exactly the average of the $m$ largest observations. This is the empirical version of the upper-quantile integral used for TVaR.

Order the observations.
List the losses from smallest to largest, preserving their units and the sample size $n$.

Find the tail size.
For confidence level $\alpha$, calculate $m=n(1-\alpha)$. This is the number of observation-units in the empirical upper tail.

Select and weight the tail.
Include $\lfloor m\rfloor$ largest observations fully and, if needed, include the next-largest observation with weight $m-\lfloor m\rfloor$.

Divide by the tail size.
Divide the weighted loss total by $m$, not by the original sample size or by a rounded-up count.

Worked Calculations

Worked Example: An Integer-Sized Upper Tail

A hypothetical insurer records eight annual net losses, in thousands of dollars: $2, 4, 5, 7, 9, 11, 15, 27$. Calculate the empirical TVaR at confidence level $\alpha=0.75$, and identify the empirical VaR. No interest rate or mortality assumption applies; the empirical distribution assigns probability $1/8$ to each listed loss, with no parametric distribution assumed.

Method. First order the data. They are already in ascending order, so $n=8$ and the ordered observations are $x_{(1)}=2,\ldots,x_{(8)}=27$, all in thousands of dollars.

The upper-tail size is

$$ m=8(1-0.75)=8(0.25)=2. $$

The tail contains two whole observations. The two largest are $15$ and $27$, so the weighted loss total is $15+27=42$ thousand dollars. Therefore,

\widehat{\operatorname{TVaR}}_{0.75}=\frac{42}{2}=21 \quad\text{thousand dollars}=\$21{,}000.00.

Using the empirical inverse-quantile convention, $\widehat{\operatorname{VaR}}_{0.75}=x_{(\lceil 8(0.75)\rceil)}=x_{(6)}=11$ thousand dollars. The check is that the two observations strictly above this threshold are $15$ and $27$; their average is $21$. The common mistake to avoid is including $11$ in the tail as well as the two largest losses: that would select three observations, although the upper 25% of an eight-observation empirical distribution has size exactly two.

Worked Example: A Fractional Boundary Observation

A hypothetical specialty insurer observes ten annual claim losses, in thousands of dollars: $4, 6, 8, 10, 12, 16, 20, 24, 30, 40$. Calculate empirical TVaR at $\alpha=0.75$, and explain why simply averaging the three largest losses is not the estimator. No interest rate or mortality assumption applies; the empirical distribution places probability $1/10$ on each observation and does not assume a parametric loss distribution.

Method. The observations are ordered and $n=10$. The empirical upper-tail size is

$$ m=10(1-0.75)=2.5. $$

There are $h=\lfloor2.5\rfloor=2$ fully included observations and a fractional weight $\theta=2.5-2=0.5$ on the next-largest observation. The two largest losses are $40$ and $30$; the next-largest is $24$. The weighted total is

40+30+0.5(24)=40+30+12=82 \quad\text{thousand dollars}.

Divide by the tail size $2.5$:

\widehat{\operatorname{TVaR}}_{0.75}=\frac{82}{2.5}=32.8 \quad\text{thousand dollars}=\$32{,}800.00.

The empirical VaR is $x_{(\lceil10(0.75)\rceil)}=x_{(8)}=24$ thousand dollars. The upper quantile integral includes half of the probability interval associated with $24$, then the full intervals associated with $30$ and $40$, which verifies the fractional weighting. Simply averaging the three largest observations gives $(24+30+40)/3=94/3=31.3333\ldots$ thousand dollars, not the empirical TVaR estimator: it uses three full observation-units instead of the required 2.5.

Worked Example: A Tail Smaller Than Two Observations

A hypothetical reinsurer reviews twelve annual recoveries, in thousands of dollars: $1, 2, 3, 4, 5, 6, 7, 9, 12, 18, 25, 50$. Find empirical TVaR at $\alpha=0.90$, and state the empirical VaR. No interest rate or mortality assumption applies; each observed recovery has empirical probability $1/12$, and no parametric distribution is assumed.

Method. The losses are ordered and $n=12$. The tail size is

$$ m=12(1-0.90)=12(0.10)=1.2. $$

Here $h=\lfloor1.2\rfloor=1$ and $\theta=1.2-1=0.2$. Include the largest loss, $50$, fully and the next-largest, $25$, with weight $0.2$. The weighted total is $50+0.2(25)=50+5=55$ thousand dollars. Thus,

\widehat{\operatorname{TVaR}}_{0.90}=\frac{55}{1.2} =45.8333\ldots\quad\text{thousand dollars}.

Rounded only at the final step to two decimal places, the estimate is $\$45{,}833.33$. The empirical VaR is $x_{(\lceil12(0.90)\rceil)}=x_{(11)}=25$ thousand dollars. The result exceeds the largest observation, $50$ thousand dollars? It does not: $45.8333$ thousand dollars is below $50$ thousand dollars, as an average of the weighted tail values should be. The common mistake is to average $50$ and $25$ equally, obtaining $37.5$; the second-largest observation contributes only $0.2$ of an observation-unit, not one full unit.

Interpretation and Common Traps

The empirical estimator describes the upper tail of the empirical distribution: it is an average of observed losses, weighted according to how much of each observation’s probability mass lies above the selected confidence level. If $n(1-\alpha)$ is an integer, this reduces to the familiar average of the largest $n(1-\alpha)$ observations. Otherwise, the boundary observation must be fractionally weighted to preserve the requested tail probability.

The empirical VaR identifies a threshold, while empirical TVaR averages the upper quantiles. They are not interchangeable. In a finite empirical sample, the quantile at the confidence level can itself contribute only part of its probability interval to the TVaR calculation. A strict conditional average of losses above the threshold can therefore use a different set of weights; the upper-quantile integral definition gives the weighting used here.

Rounding the tail count too early. Calculate $m=n(1-\alpha)$ exactly before deciding how many observations are fully included. A noninteger $m$ calls for fractional weighting.
Using the wrong observations. TVaR concerns the upper tail of losses. Sort in ascending order, then work downward from the largest observation.
Dividing by the wrong count. Divide the weighted total by $m$. Dividing by $n$ gives a contribution to the sample mean, while dividing by $\lceil m\rceil$ changes the tail definition.
Mixing units. If losses are recorded in thousands of dollars, the calculated average remains in thousands until converted to dollars.

The arithmetic requires only an ordered sample and a chosen confidence level. Interpreting the estimate as informative about future losses additionally requires that the observations reasonably represent the loss process of interest. The calculation itself does not establish that representativeness or measure sampling uncertainty.

Key takeaway: For $n$ ordered observations and confidence level $\alpha$, calculate $m=n(1-\alpha)$. Average the largest $m$ observation-units: include whole losses fully and use a fractional weight on the boundary loss whenever $m$ is not an integer.

Check Your Understanding

For each problem, use the empirical upper-quantile integral convention and keep the stated loss units.

A sample of six ordered annual losses, in thousands of dollars, is $3, 5, 8, 9, 14, 21$. Calculate empirical TVaR at $\alpha=2/3$. (Answer: $17$ thousand dollars.)
Eight ordered losses, in dollars, are $2, 4, 7, 9, 12, 18, 25, 36$. Calculate empirical TVaR at $\alpha=0.875$. (Answer: $\$36.00$.)
Ten ordered losses, in thousands of dollars, are $1, 3, 4, 6, 7, 9, 13, 17, 22, 31$. Calculate empirical TVaR at $\alpha=0.80$. (Answer: $26.5$ thousand dollars.)
Eight ordered losses, in thousands of dollars, are $2, 4, 6, 8, 10, 14, 18, 30$. Calculate empirical TVaR at $\alpha=0.70$, showing the fractional boundary weight. (Answer: $22.33$ thousand dollars, rounded to two decimal places.)
Twelve ordered losses, in thousands of dollars, are $2, 3, 5, 7, 9, 12, 15, 19, 24, 32, 50, 80$. Calculate empirical TVaR at $\alpha=0.85$, showing the weighted tail total and tail size. (Answer: $66.67$ thousand dollars, rounded to two decimal places.)

What You'll Learn