Empirical TVaR With Ties And Finite Samples

Why Finite-Sample Tails Need Care

When estimating tail risk from observed losses, the requested tail probability may correspond to only part of an observation. Ties can make the cutoff less obvious: several losses may equal the empirical VaR, even though only some of their combined empirical probability belongs in the upper tail. A sound calculation must represent the specified tail probability exactly.

The key is to treat the observed sample as an empirical distribution in which each of the $n$ observations has probability $1/n$. The upper tail therefore contains $n(1-p)$ observation-equivalents at confidence level $p$. If this number is not an integer, the boundary observation receives a fractional weight. If the boundary loss is tied, that fractional weight can be assigned to any of the tied observations at that loss value; the result is unchanged.

Definition: Let $X_1,\ldots,X_n$ be observed losses and let $X_{(1)}\leq\cdots\leq X_{(n)}$ be the observations ordered from smallest to largest, including repeated values. The empirical distribution assigns probability $1/n$ to each observation. At confidence level $p$, empirical TVaR is the average loss over the upper tail of probability $1-p$, with a fractional observation weight when needed to represent that probability exactly.

From Tail Probability to Fractional Weight

Let $m=n(1-p)$ be the number of observation-equivalents in the requested tail. Write $m=k+r$, where $k=\lfloor m\rfloor$ is an integer and $0\leq r<1$. Include the $k$ largest observations in full, then include a fraction $r$ of the next observation. When $r=0$, the tail consists of exactly $k$ observations and there is no fractional term.

\widehat{\operatorname{TVaR}}_p = \frac{ \displaystyle\sum_{j=n-k+1}^{n}X_{(j)} +rX_{(n-k)} }{m}, \qquad m=n(1-p)=k+r.

If $k=0$, the sum is empty: the formula uses only the fraction $r=m$ of the largest observation. The expression $X_{(n-k)}$ is the boundary observation when $r>0$. With tied boundary values, any one of those equal values gives the same result.

Why the Formula Works

For the empirical distribution, define the quantile $Q_n(u)$ to equal $X_{(j)}$ on the $j$th interval of probability $1/n$. Empirical TVaR is the average of those quantiles over the range from $p$ to 1. Each complete observation in the tail contributes one interval of width $1/n$, and a non-integer tail count leaves a partial interval.

\widehat{\operatorname{TVaR}}_p = \frac{1}{1-p}\int_p^1 Q_n(u)\,du = \frac{1}{m}\left( \sum_{j=n-k+1}^{n}X_{(j)} +rX_{(n-k)} \right).

For the second equality, measure interval widths in units of $1/n$: the tail length $1-p$ is $m/n$, so dividing the quantile integral by $1-p$ is equivalent to dividing its observation-weighted loss total by $m$. The $k$ full intervals contribute the $k$ largest losses; the remaining interval has width $r/n$ and contributes $r$ times the boundary loss. Ties do not alter the integral: equal-valued intervals contribute the same loss regardless of how the fractional interval is assigned among them.

Worked Example: A Fractional Tail Ending at a Tie

A hypothetical insurer reviews seven individual losses, measured in thousands of dollars: $2,5,8,10,10,10,25$. Estimate empirical TVaR at confidence level $p=0.8000$. Round the final dollar amount to the nearest cent. Method: order the observations, calculate the tail count, and apply a fractional boundary weight. This avoids treating all losses equal to the empirical VaR as either wholly in or wholly out of the tail.

The data are already ordered, so $n=7$. The specified tail probability is $1-p=0.2000$, giving $m=7(0.2000)=1.4$ observation-equivalents. Thus $k=1$ and $r=0.4$. The largest loss is $25$, and the boundary loss is $10$. There are three observations tied at $10$; any one of them can supply the fractional boundary weight.

The weighted tail loss total is $25+0.4(10)=25+4=29$ thousand dollars. Dividing by the tail count gives $29/1.4=20.7142857$ thousand dollars. Therefore, empirical TVaR is $20,714.29.

The check is that the selected weights total $1+0.4=1.4$ observations, representing $1.4/7=0.2000$ of the sample. The tied observations do not change the result because each has the same loss value.

Worked Example: A Non-Integer Tail Count Without Ties

A hypothetical portfolio has six observed losses, in thousands of dollars: $1,4,7,12,18,30$. Find empirical TVaR at $p=0.7500$. Method: use the same fractional-tail calculation, checking that the selected observation weights equal the required tail count. This avoids rounding the tail count to one or two observations.

Here $n=6$, and the tail probability is $1-0.7500=0.2500$. Hence $m=6(0.2500)=1.5$, so $k=1$ and $r=0.5$. The largest loss is $30$; the next loss, which is the boundary loss, is $18$.

The weighted tail loss total is $30+0.5(18)=30+9=39$ thousand dollars. The empirical TVaR estimate is $39/1.5=26$ thousand dollars, or $26,000.00.

The selected weights total $1+0.5=1.5$, and $1.5/6=0.2500$, exactly the specified tail probability. Using just the largest observation would instead represent a tail probability of $1/6$, not $0.2500$.

Worked Example: Tied Losses at an Integer-Count Cutoff

A hypothetical risk team observes ten losses, in thousands of dollars: $3,5,7,9,12,12,12,20,20,35$. Calculate empirical TVaR at $p=0.6000$, then compare it with the average of observations strictly above the empirical VaR. Method: use the empirical upper-tail definition for the estimate and calculate the strict-exceedance average separately. This shows why the two calculations can differ when the cutoff is tied.

The tail probability is $1-0.6000=0.4000$, so $m=10(0.4000)=4$. Since the tail count is an integer, $k=4$ and $r=0$. The four largest observations are $12,20,20,35$; one of the three observations tied at $12$ is included.

Their total is $12+20+20+35=87$ thousand dollars. Therefore, empirical TVaR is $87/4=21.75$ thousand dollars, or $21,750.00. The weights total four observations, representing $4/10=0.4000$ of the sample.

The empirical 0.6000 quantile is $12$, the sixth ordered observation. Losses strictly above $12$ are $20,20,35$; their average is $(20+20+35)/3=25$ thousand dollars. That calculation excludes all three observations tied at the cutoff, so it uses only three of the four required tail observations. It is not the empirical TVaR at this confidence level.

Common Errors and Practical Checks

A tempting shortcut is to compute the empirical VaR and average only observations strictly greater than it. For a continuous distribution this can correspond to a familiar tail-mean description, but in a finite empirical distribution it can fail when the cutoff is tied or when the requested tail count is non-integer. The fractional-weight method instead follows the quantile-integral definition and preserves the specified tail probability.

Rounding the tail count: $m=n(1-p)$ is generally not an integer. Replacing it with the nearest integer changes the represented tail probability.
Including every tied cutoff observation: the tail needs a particular amount of probability, not every observation equal to the cutoff. Include only the required total weight.
Excluding every tied cutoff observation: strict exceedances alone can leave too little tail mass. The missing mass must be supplied at the cutoff value.
Dividing by the sample size: divide the weighted tail loss total by $m$, the number of observation-equivalents in the tail, not by $n$.

A useful check is to add the selected weights and verify that they total $n(1-p)$. Divide that number by $n$; it must equal the specified tail probability. Also check that the estimate lies between the smallest and largest losses included in the weighted tail.

Key takeaway: Sort all observations, calculate $m=n(1-p)$, include the $k=\lfloor m\rfloor$ largest observations in full, and give the next observation weight $r=m-k$. Ties at the boundary cause no ambiguity in the estimate because equal losses have equal values.

Check Your Understanding

For each problem, use the empirical upper-tail definition and retain fractional weights where needed.

A hypothetical sample of five losses, in thousands of dollars, is $2,4,9,14,20$. Find empirical TVaR at $p=0.7000$. (Answer: $18,000.00)
A hypothetical sample of nine losses, in thousands of dollars, is $1,3,5,8,10,10,16,16,31$. Find empirical TVaR at $p=0.8000$. (Answer: $24,333.33)
A hypothetical sample of eight losses, in thousands of dollars, is $0,4,6,11,11,11,19,27$. Find empirical TVaR at $p=0.6250$, and state how many observation-equivalents are in the tail. (Answer: $19,000.00; 3 observation-equivalents)
A hypothetical sample of twelve losses, in thousands of dollars, has ten observations at or below $20$, one loss of $30$, and one loss of $50$. Find empirical TVaR at $p=0.9000$. (Answer: $46,666.67)
A hypothetical sample of ten losses, in thousands of dollars, is $2,3,5,8,8,8,12,17,17,40$. Find empirical TVaR at $p=0.7500$. (Answer: $26,200.00)

What You'll Learn