Sampling Uncertainty in VaR and TVaR

Why Tail-Risk Estimates Vary

A VaR or TVaR calculated from data is an estimate, not a fixed property known without error. A different sample from the same loss population would usually produce a different estimate. This variation is especially important at high confidence levels: a 99% empirical VaR is determined by observations near the largest 1% of losses, where a finite sample may contain very few data points.

This tutorial focuses on sampling uncertainty: how estimates can vary from sample to sample, and how order-statistic reasoning or resampling can help describe that variation. Neither method can recover information about losses beyond what the data and stated assumptions support.

Definition: For a loss $X$ with distribution function $F_X$, the lower-quantile convention for VaR at confidence level $p$ is $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}, \qquad 0<p<1. $$ For a distribution with finite mean, TVaR under the quantile convention is $$ \operatorname{TVaR}_p(X)=\operatorname{VaR}_p(X)+\frac{E[(X-\operatorname{VaR}_p(X))_+]}{1-p}, $$ where $z_+=\max(z,0)$. This definition also handles a probability mass at the VaR threshold by including only the appropriate fraction of that mass in the upper tail.

Given $n$ observed losses, sort them as $X_{(1)}\le \cdots \le X_{(n)}$. The empirical distribution assigns probability $1/n$ to each observation. Its empirical VaR at level $p$ is $X_{(k)}$, where $k=\lceil np\rceil$. The ceiling is important: it selects the first order statistic at which the empirical cumulative proportion reaches or exceeds $p$.

Order Statistics and Quantile Uncertainty

Suppose $X_1,\ldots,X_n$ are independent observations from the same continuous distribution, and let $q_p$ be its true $p$-quantile. The number of observations at or below $q_p$ has a binomial distribution with parameters $n$ and $p$. Therefore, the empirical VaR can fall above or below the true quantile depending on how many sample observations fall on either side of it.

Order-statistic result: If the distribution is continuous and $F_X(q_p)=p$, the number $B$ of sample observations below $q_p$ has $B\sim\operatorname{Bin}(n,p)$. The event $X_{(k)}\le q_p$ occurs exactly when at least $k$ observations are below $q_p$, so $$ P(X_{(k)}\le q_p)=P(B\ge k)=\sum_{j=k}^{n}\binom{n}{j}p^j(1-p)^{n-j}. $$ Continuity avoids ties at $q_p$. With atoms, the probability of equality must be handled explicitly.

This result gives a direct way to assess how variable a sample quantile’s rank can be without knowing the numerical shape of the loss distribution. It does not, by itself, give a dollar interval for the true VaR: that requires information about the distribution of losses around the quantile.

Set the quantile convention.
Use $k=\lceil np\rceil$ for the empirical lower-quantile VaR.

Count observations relative to the true quantile.
Under independent sampling from a continuous distribution, that count is binomial with success probability $p$.

Translate counts into rank uncertainty.
For example, $X_{(k)}\le q_p$ exactly when at least $k$ observations fall below $q_p$.

Worked Example: How Often a High Empirical VaR Falls Below the True Quantile

A hypothetical insurer has 10 independent annual loss observations from a continuous loss distribution. It estimates the 90% VaR using the empirical lower-quantile convention. Find the probability that the estimate is below the true 90% VaR.

Method. At $p=0.90$, the empirical order-statistic index is $$ k=\lceil np\rceil=\lceil 10(0.90)\rceil=\lceil 9\rceil=9. $$ The estimate is the ninth order statistic, or second-largest observation, not the sample maximum.

Let $B$ count observations below the true 90% quantile. Then $B\sim\operatorname{Bin}(10,0.90)$. With no ties, the empirical estimate is below the true quantile exactly when $B\ge9$: $$ P(X_{(9)}<q_{0.90})=P(B\ge9)=P(B=9)+P(B=10). $$

The two probabilities are $$ P(B=9)=\binom{10}{9}(0.90)^9(0.10)=10(0.387420489)(0.10)=0.387420489, $$ $$ P(B=10)=\binom{10}{10}(0.90)^{10}=0.3486784401. $$ Thus $$ P(X_{(9)}<q_{0.90})=0.387420489+0.3486784401=0.7360989291. $$ Rounded to four decimals, the probability is $0.7361$. The estimate exceeds the true quantile when $B\le8$, with probability $1-0.7360989291=0.2639010709$, or $0.2639$ to four decimals.

Check and trap avoided. The two probabilities sum to $1$ with $P(B\le8)$. Calling the ninth order statistic the maximum would use the wrong rank and misstate the sampling behavior.

Bootstrap Uncertainty for Empirical VaR

The nonparametric bootstrap approximates sampling variation by treating the observed data as an empirical population. A bootstrap resample draws $n$ observations independently with replacement from the original $n$ losses. Repeating this process produces a collection of bootstrap VaR estimates. A percentile interval uses selected quantiles of that collection; it describes variability under resampling from the observed empirical distribution.

Resampling with replacement matters: observations can appear more than once or not at all in a bootstrap sample. In particular, a high empirical quantile is an order statistic within each resample, not automatically the resample maximum. The correct rank must be recalculated using that resample’s size and the stated quantile convention.

Worked Example: A Bootstrap Percentile Interval for the 80% VaR

A hypothetical portfolio has five observed annual losses, in thousands of dollars: $$ 5,\;5,\;7,\;11,\;20. $$ Use the empirical nonparametric bootstrap with resample size five and the lower-quantile convention to find the central 80% percentile interval for the bootstrap 80% VaR.

Original estimate. Here $n=5$, $p=0.80$, and $$ k=\lceil5(0.80)\rceil=4. $$ The original empirical VaR is the fourth ordered loss, $11$ thousand dollars.

Bootstrap rank. Every bootstrap resample also has size five, so its 80% VaR is the fourth order statistic of that resample. It is not the maximum. Let $C$ be the number of resampled values at or below a specified observed loss. The bootstrap VaR is at or below that loss exactly when $C\ge4$.

At $5$, two of the five original observations are at or below the threshold. A bootstrap draw is therefore at or below $5$ with probability $2/5=0.4$. Thus $C\sim\operatorname{Bin}(5,0.4)$, and $$ P(\text{bootstrap VaR}\le5)=P(C\ge4) =\binom54(0.4)^4(0.6)+(0.4)^5 =0.0768+0.01024=0.08704. $$

At $7$, three of the five observations are at or below the threshold, so the corresponding draw probability is $3/5=0.6$: $$ P(\text{bootstrap VaR}\le7)=\binom54(0.6)^4(0.4)+(0.6)^5 =0.2592+0.07776=0.33696. $$ At $11$, the draw probability is $4/5=0.8$: $$ P(\text{bootstrap VaR}\le11)=\binom54(0.8)^4(0.2)+(0.8)^5 =0.4096+0.32768=0.73728. $$ At $20$, every possible bootstrap draw is at or below the threshold, so the bootstrap VaR CDF is $1$.

The 10th percentile is the smallest value whose bootstrap CDF is at least $0.10$. Since the CDF is $0.08704$ at $5$ and $0.33696$ at $7$, the 10th percentile is $7$. The 90th percentile is $20$, since the CDF at $11$ is $0.73728$ and reaches $1$ at $20$. The central 80% percentile interval is therefore $[7,20]$ thousand dollars.

Check and trap avoided. The result uses the fourth order statistic in every resample. Using the maximum instead would give an interval for the bootstrap maximum, not for the bootstrap 80% VaR.

Why Empirical TVaR Can Be Especially Unstable

Empirical TVaR averages losses in the upper tail, so its estimate depends on the values of the largest observations as well as their ranks. At a confidence level $p$, the empirical tail contains roughly $n(1-p)$ observations’ worth of probability. If this quantity is small, one observed loss can have a large effect. With a discrete empirical distribution, the tail formula may assign fractional weight to an observation at the VaR threshold.

Empirical TVaR calculation: For observations $x_1,\ldots,x_n$, let $v$ be the empirical VaR. Then $$ \widehat{\operatorname{TVaR}}_p =v+\frac{\frac1n\sum_{i=1}^{n}(x_i-v)_+}{1-p}. $$ The excess-loss term accounts for the observations strictly above $v$; observations equal to $v$ contribute zero excess. This formula corresponds to the quantile-based TVaR convention and applies to finite empirical losses.

The formula follows by applying the population definition to the empirical distribution: its expectation of $(X-v)_+$ is the average of the $n$ observed positive excesses, each weighted by $1/n$. Dividing that average by $1-p$ scales the excess to the upper-tail probability, then adding $v$ gives TVaR.

Worked Example: Bootstrap Uncertainty When the Tail Has One Observation

Consider five hypothetical annual losses, in thousands of dollars: $2,4,6,10,30$. Calculate the empirical 80% TVaR and describe the bootstrap distribution of this estimate when resampling five observations with replacement.

Original empirical estimate. The 80% VaR rank is $k=\lceil5(0.80)\rceil=4$, so $v=10$. The positive excesses above $10$ are $0,0,0,0,20$, whose sum is $20$. Hence $$ \widehat{\operatorname{TVaR}}_{0.80} =10+\frac{(20/5)}{0.20} =10+\frac4{0.20} =10+20=30 $$ thousand dollars. Equivalently, the upper 20% tail has the weight of one observation, the largest loss.

Bootstrap distribution. Each bootstrap sample has five observations, so its empirical 80% TVaR is its maximum: the upper 20% tail has exactly one observation’s weight. The bootstrap maximum is at most a threshold $x$ only if all five draws are at most $x$. Therefore, for thresholds $2,4,6,10,30$, the cumulative probabilities are respectively $$ (1/5)^5=0.00032,\quad (2/5)^5=0.01024,\quad (3/5)^5=0.07776,\quad (4/5)^5=0.32768,\quad (5/5)^5=1. $$

The 10th percentile is $10$, because the CDF is $0.07776$ at $6$ and $0.32768$ at $10$. The 90th percentile is $30$, because the CDF remains $0.32768$ at $10$ and reaches $1$ at $30$. Thus the central 80% bootstrap percentile interval for TVaR is $[10,30]$ thousand dollars.

Check and trap avoided. The original TVaR is $30$, but resampling can omit the observed loss of $30$; the chance of omitting it is $(4/5)^5=0.32768$. This example shows how a tail estimate can vary sharply when the estimate is determined by one observation. It does not show that the true loss distribution has no values above $30$.

Interpreting Uncertainty and Avoiding Overconfidence

A bootstrap percentile interval is conditional on the observed sample. It captures one form of sampling variability under the assumption that the empirical distribution is a useful stand-in for the population. When the sample contains no very large losses, a nonparametric bootstrap cannot generate losses larger than the sample maximum. It can reveal that an estimate is sensitive to which observed tail values are selected, but it cannot quantify the possibility of unseen, more severe losses without additional assumptions or data.

Likewise, order-statistic probabilities describe rank uncertainty under a specified sampling model. They do not supply dollar-scale uncertainty by themselves. Translating a rank interval into loss amounts depends on the observed values or a model for the loss distribution. The separate question of whether a fitted tail model is appropriate requires diagnostic work beyond resampling the empirical sample.

Key takeaway: State the quantile convention, compute the order-statistic rank, and make the sampling assumptions explicit. Treat a bootstrap interval as a description of resampling variability—not as a guarantee about unobserved tail losses. These methods support integrated risk analysis aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Use the lower-quantile VaR convention. Unless a question says otherwise, assume independent observations and no ties where a population quantile is involved.

A sample of six losses, in thousands of dollars, is $3,4,4,8,9,15$. Find the empirical 75% VaR. (Answer: the fifth order statistic, $9$ thousand dollars.)
A sample is $1,2,3,4,5$. A bootstrap sample of size five is drawn with replacement. For the bootstrap 80% VaR, find the probability that the estimate is at most $3$. (Answer: $0.33696$.)
For $n=8$ observations, consider the empirical 87.5% VaR. Find the probability that it exceeds the true 87.5% quantile under a continuous population distribution. (Answer: $1-(0.875)^8-8(0.125)(0.875)^7=0.2637$, to four decimals.)
Five observed losses, in thousands of dollars, are $2,5,8,12,40$. Find the empirical 80% VaR and TVaR. (Answer: VaR $=12$ thousand dollars; TVaR $=40$ thousand dollars.)
For the sample $1,2,3,4,5$, resample five times with replacement and calculate the empirical 80% TVaR in each resample. What is the probability the bootstrap TVaR equals $5$? (Answer: $1-(4/5)^5=0.67232$.)

What You'll Learn