Choosing Tail Measures for Skewed and Heavy-Tailed Losses

Why the Choice of Tail Measure Matters

For a skewed loss distribution, a small set of severe outcomes can contribute a large share of total risk. A quantile measure records where a chosen tail begins; an average-tail measure also reflects how large losses can become beyond that point. Those are different questions, and they can produce different risk rankings even when the same confidence level is used.

This tutorial compares Value at Risk (VaR) and Tail Value at Risk (TVaR) for skewed, discrete, and heavy-tailed losses. The calculations are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. Throughout, $X$ denotes a loss, so larger values are worse, and $\alpha$ is a confidence level between 0 and 1.

Definition: For a loss $X$ with distribution function $F_X$, define its $\alpha$-quantile by $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}. $$ Define Tail Value at Risk by the quantile integral $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{p}(X)\,dp. $$ The integral definition applies to discrete and continuous distributions. A finite TVaR requires the relevant upper-tail mean to be finite.

VaR identifies a threshold: under the quantile convention above, the probability of a loss at or below $\operatorname{VaR}_{\alpha}(X)$ is at least $\alpha$. VaR alone does not describe the sizes of losses beyond that threshold. TVaR averages the losses across the upper $1-\alpha$ probability slice. It therefore uses more information about the far tail.

Calculating TVaR When the Distribution Has Atoms

A discrete loss can have positive probability at its VaR. In that case, the upper-tail slice may use only part of the probability mass at the threshold. If $q=\operatorname{VaR}_{\alpha}(X)$, then $$ \operatorname{TVaR}_{\alpha}(X)= \frac{E[X\mathbf{1}_{\{X>q\}}]+q\big(F_X(q)-\alpha\big)}{1-\alpha}. $$ Here $F_X(q)-\alpha$ is the amount of probability at $q$ needed to complete the upper $1-\alpha$ slice. This formula follows directly from the quantile integral: quantiles above $q$ contribute their full probability, while the integral starting at $\alpha$ includes only the remaining portion of the probability at $q$.

Worked Example: Taking the Correct Slice at a Discrete VaR

A hypothetical portfolio has loss distribution $P(X=0)=0.90$, $P(X=100)=0.09$, and $P(X=1{,}000)=0.01$. Calculate $\operatorname{VaR}_{0.95}(X)$ and $\operatorname{TVaR}_{0.95}(X)$. Report dollar amounts to the cent.

Method. First find the quantile from the cumulative distribution. Then average exactly the upper 5% of probability, including only the needed portion of any mass at the boundary.

The cumulative probabilities are $F_X(0)=0.90$, $F_X(100)=0.99$, and $F_X(1{,}000)=1.00$. Since $0.90<0.95\leq0.99$, $$ \operatorname{VaR}_{0.95}(X)=100. $$

The upper 5% slice starts at percentile 95%. The probability at loss 100 extends from percentile 90% to 99%; only the portion from 95% to 99%, or 4 percentage points, belongs in this slice. The 1% probability at loss 1,000 is also included. Thus $$ \operatorname{TVaR}_{0.95}(X) =\frac{0.04(100)+0.01(1{,}000)}{0.05} =\frac{4+10}{0.05} =280. $$ To the nearest cent, TVaR is $\$280.00$, while VaR is $\$100.00$.

Check. The slice probabilities total $0.04+0.01=0.05$. Its weighted loss is $14$, so dividing by $0.05$ gives $280$.

Common mistake avoided. Using the full 9% mass at loss 100 would average more than the upper 5% tail. The correct calculation uses 4 percentage points at 100, not the full 9% and not 5% of that 9% mass.

Why Equal VaR Can Hide Different Tail Severity

The following comparison isolates what VaR leaves out. Suppose two hypothetical portfolios each have a 90% probability of loss 0 and a 9% probability of loss 50. The remaining 1% is a severe outcome: loss 500 for portfolio A and loss 1,500 for portfolio B.

Worked Example: Same VaR, Different TVaR

For each portfolio, calculate VaR and TVaR at 95%. State which portfolio has the larger risk by each measure.

Method. Locate the 95th percentile first. Then assemble the upper 5% slice by adding probability from the highest losses downward.

For both portfolios, the cumulative probability at loss 0 is 0.90 and at loss 50 is 0.99. Therefore, $$ \operatorname{VaR}_{0.95}(A)=\operatorname{VaR}_{0.95}(B)=50. $$

For portfolio A, the upper 5% comprises 1% at loss 500 and 4% at loss 50: $$ \operatorname{TVaR}_{0.95}(A) =\frac{0.01(500)+0.04(50)}{0.05} =\frac{5+2}{0.05} =140. $$ For portfolio B, it comprises 1% at loss 1,500 and 4% at loss 50: $$ \operatorname{TVaR}_{0.95}(B) =\frac{0.01(1{,}500)+0.04(50)}{0.05} =\frac{15+2}{0.05} =340. $$

Check. Each tail slice has probability 5%. Both VaR values are 50, but portfolio B's larger extreme loss raises its TVaR from 140 to 340.

Common mistake avoided. Equal VaR does not imply equal tail risk. VaR sees the shared threshold but not the difference between the two losses beyond it.

Risk Rankings Can Reverse

A second comparison shows that the choice of measure can reverse the order of two portfolios. VaR ranks only the threshold at a specified confidence level. TVaR ranks the average loss in the entire selected tail. A portfolio can therefore have a higher threshold but a less severe average tail.

Worked Example: A Reversal Between VaR and TVaR

Portfolio A has the distribution from the first example: losses 0, 100, and 1,000 with probabilities 0.90, 0.09, and 0.01. Portfolio B has losses 0, 120, and 140 with probabilities 0.94, 0.05, and 0.01. Compare their 95% measures.

Method. For each distribution, find the first loss at which cumulative probability reaches 0.95. Then average the upper 5% slice, including only the needed threshold mass.

For A, cumulative probability reaches 0.99 at loss 100, so $\operatorname{VaR}_{0.95}(A)=100$. Its TVaR is $$ \operatorname{TVaR}_{0.95}(A) =\frac{0.04(100)+0.01(1{,}000)}{0.05} =280. $$ For B, cumulative probability is 0.94 at loss 0 and 0.99 at loss 120. Hence $\operatorname{VaR}_{0.95}(B)=120$. The upper 5% contains 4 percentage points at loss 120 and 1 percentage point at loss 140: $$ \operatorname{TVaR}_{0.95}(B) =\frac{0.04(120)+0.01(140)}{0.05} =\frac{4.8+1.4}{0.05} =124. $$

Check. VaR ranks B higher because $120>100$. TVaR ranks A higher because $280>124$. The tail probabilities used for each portfolio sum to 0.05.

Common mistake avoided. Do not assume that the distribution with the higher VaR must have the higher TVaR. The loss amounts beyond the VaR threshold determine the average tail severity.

Heavy Tails and a Pareto Illustration

A useful model for illustrating heavy-tailed loss severity is the Pareto distribution. For scale $x_m>0$ and shape $k>0$, let $X$ have survival function $P(X>x)=(x_m/x)^k$ for $x\geq x_m$. The smaller the shape parameter, the more slowly this tail decays. The mean is finite only when $k>1$, which is required for finite TVaR in the calculation below.

Result: For a Pareto loss with $k>1$, $$ \operatorname{VaR}_{\alpha}(X)=\frac{x_m}{(1-\alpha)^{1/k}}, \qquad \operatorname{TVaR}_{\alpha}(X)=\frac{k}{k-1}\operatorname{VaR}_{\alpha}(X). $$

For this continuous distribution, there is no probability mass at the threshold, so TVaR equals the conditional mean $E[X\mid X>q]$, where $q=\operatorname{VaR}_{\alpha}(X)$. The quantile formula follows by solving $P(X>q)=1-\alpha$. To justify the conditional-mean formula for TVaR, use the tail-integral identity for a nonnegative loss: $$ E[X\mid X>q]=q+\frac{1}{P(X>q)}\int_q^\infty P(X>x)\,dx. $$ Substituting the Pareto survival function gives an integral of $x_m^k x^{-k}$. For $k>1$, it equals $x_m^k q^{1-k}/(k-1)=qP(X>q)/(k-1)$. Dividing by $P(X>q)$ and adding $q$ yields $kq/(k-1)$, as claimed. When there is an atom at $q$, the conditional-mean shortcut need not equal quantile-integral TVaR; use the boundary-mass formula instead.

Worked Example: Comparing Pareto Tails at a Common VaR

Compare two hypothetical Pareto loss models at confidence level 75%. Model A has $k=2$ and $x_m=500$. Model B has $k=1.25$ and $x_m=1{,}000(0.25)^{0.8}$, approximately 329.88. Calculate VaR and TVaR for each. Report dollar amounts to the nearest cent.

Method. Apply the Pareto formulas, noting that both shapes exceed 1, so each model has finite mean and finite TVaR.

For A, the tail probability is $1-0.75=0.25$, so $$ \operatorname{VaR}_{0.75}(A) =\frac{500}{(0.25)^{1/2}} =\frac{500}{0.5} =1{,}000. $$ Its TVaR is $$ \operatorname{TVaR}_{0.75}(A) =\frac{2}{2-1}(1{,}000) =2{,}000. $$

For B, the scale was chosen so its VaR is also 1,000: $$ \operatorname{VaR}_{0.75}(B) =\frac{1{,}000(0.25)^{0.8}}{(0.25)^{1/1.25}} =\frac{1{,}000(0.25)^{0.8}}{(0.25)^{0.8}} =1{,}000. $$ Its TVaR is $$ \operatorname{TVaR}_{0.75}(B) =\frac{1.25}{1.25-1}(1{,}000) =5(1{,}000) =5{,}000. $$

Check. The VaRs match at $\$1{,}000.00$. The smaller shape parameter in B produces a heavier tail and a larger TVaR, $\$5{,}000.00$ versus $\$2{,}000.00$. Both results are reported to two decimal places.

Common mistake avoided. A common error is to treat the common VaR as if it determined tail risk. The Pareto calculation shows that the shape parameter still matters: it changes the average severity beyond the same quantile.

Interpreting the Measures and Avoiding Traps

For losses with finite upper-tail mean, the quantile integral also shows that $\operatorname{TVaR}_{\alpha}(X)\geq\operatorname{VaR}_{\alpha}(X)$: quantiles at levels $p\geq\alpha$ are at least as large as the $\alpha$-quantile, and TVaR averages those quantiles. This inequality does not mean TVaR is always the preferred measure. It means the measures summarize different features of the same loss distribution.

When comparing risks, make the confidence level, time horizon, loss definition, and currency basis consistent. Then consider what the decision requires. If the question concerns a threshold that is exceeded with a specified probability, VaR directly reports that threshold. If the question concerns the average severity in the selected upper tail, TVaR provides that information. Neither measure alone fully describes the distribution, and a single confidence level does not summarize every feature of a heavy tail.

Quantile trap: Do not interpret VaR as the maximum loss or as the expected loss beyond the threshold.
Boundary trap: With discrete losses, include only enough probability at the VaR to complete the upper-tail slice.
Conditional-mean trap: $E[X\mid X>\operatorname{VaR}_{\alpha}(X)]$ is a shortcut for continuous cases without a boundary atom, not a universal definition of TVaR.
Ranking trap: A higher VaR does not guarantee a higher TVaR; inspect the losses beyond the quantile.

Key takeaway: VaR measures a tail threshold; TVaR measures the average across the probability slice beyond that threshold. For skewed or heavy-tailed losses, outcomes far beyond VaR can dominate the comparison, so equal VaR does not imply equal tail risk.

Check Your Understanding

For each problem, calculate the requested measures using the quantile definition and the appropriate upper-tail slice.

A loss is 0 with probability 0.80, 40 with probability 0.15, and 200 with probability 0.05. Find $\operatorname{VaR}_{0.90}$ and $\operatorname{TVaR}_{0.90}$. (Answer: VaR $=40$; TVaR $=120$.)
A loss is 0 with probability 0.94, 80 with probability 0.05, and 400 with probability 0.01. Find both measures at 95%. (Answer: VaR $=80$; TVaR $=144$.)
Portfolio C has losses 0, 60, and 300 with probabilities 0.90, 0.09, and 0.01. Portfolio D has losses 0, 60, and 900 with the same probabilities. Compare their 95% VaR and TVaR. (Answer: Both VaRs are $60$; TVaRs are $108$ for C and $228$ for D, so D has the larger TVaR.)
A Pareto loss has $x_m=200$ and $k=3$. Find VaR and TVaR at 87.5%. (Answer: VaR $=400$; TVaR $=600$.)

What You'll Learn