VaR Versus TVaR in Heavy Tails

Why Compare VaR and TVaR?

Value at Risk (VaR) identifies a loss threshold at a chosen confidence level. Tail Value at Risk (TVaR) describes the average severity in the upper tail beyond that confidence level. Two loss models can therefore have the same VaR while implying quite different losses in unusually severe outcomes. This distinction is especially important for heavy-tailed losses, where a small number of outcomes can be far larger than the threshold.

Here, losses are nonnegative, and $\alpha$ is a confidence level with $0<\alpha<1$. Write $F_X(x)=P(X\le x)$ for the distribution function of loss $X$. Use the quantile convention $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}$. TVaR is defined using the average of quantiles in the upper tail. For a continuous loss distribution, it also equals the expected loss conditional on exceeding the VaR threshold.

Definition: For a nonnegative loss $X$ and $0<\alpha<1$, $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}, \qquad \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{p}(X)\,dp. $$ The integral definition applies whether or not the distribution is continuous. If $X$ has a continuous distribution at its $\alpha$-quantile, then $\operatorname{TVaR}_{\alpha}(X)=E[X\mid X>\operatorname{VaR}_{\alpha}(X)]$. With probability atoms, this conditional-expectation identity need not hold as written; the quantile-integral definition accounts for the appropriate fraction of mass at the threshold.

The upper-tail probability is $1-\alpha$, not $\alpha$. For example, at $\alpha=0.99$, VaR marks the 99th percentile and TVaR averages quantiles over the upper 1% of the probability distribution. VaR is a threshold; it is not itself an average of the worst losses.

Pareto Losses and Tail Severity

A Pareto Type I loss is a standard model for illustrating heavy tails. Let $X$ have scale $x_m>0$ and shape $r>0$, with distribution function $F_X(x)=0$ for $x<x_m$ and $F_X(x)=1-(x_m/x)^r$ for $x\ge x_m$. Equivalently, its survival function is $P(X>x)=(x_m/x)^r$ for $x\ge x_m$. This model has a polynomial, rather than exponentially decreasing, upper tail.

Result: For a Pareto Type I loss with scale $x_m$, shape $r$, and confidence level $0<\alpha<1$, $$ \operatorname{VaR}_{\alpha}(X)=\frac{x_m}{(1-\alpha)^{1/r}}. $$ If $r>1$, then $$ \operatorname{TVaR}_{\alpha}(X) =\frac{r}{r-1}\operatorname{VaR}_{\alpha}(X). $$ If $0<r\le 1$, then $\operatorname{TVaR}_{\alpha}(X)=\infty$, even though the VaR at every $\alpha<1$ is finite.

The factor $r/(r-1)$ makes the comparison direct: for a fixed Pareto model with finite mean, TVaR exceeds VaR by an amount that depends on the tail shape. As $r$ approaches 1 from above, the factor grows without bound. A smaller shape parameter means a heavier tail and, when the mean is finite, greater average severity above the VaR threshold.

Why the Pareto Formula Holds

The VaR formula follows by inverting the distribution function. For $q\ge x_m$, the quantile equation $F_X(q)=\alpha$ gives $1-(x_m/q)^r=\alpha$. Rearranging yields $q=x_m/(1-\alpha)^{1/r}$.

To obtain TVaR, substitute the quantile formula into the quantile integral. This calculation also identifies precisely when TVaR is finite.

Substitute the quantile.
For $p<1$, the Pareto quantile is $\operatorname{VaR}_{p}(X)=x_m(1-p)^{-1/r}$. Thus the TVaR integral is $x_m/(1-\alpha)$ times $\int_{\alpha}^{1}(1-p)^{-1/r}\,dp$.

Change variables.
Set $u=1-p$, so $dp=-du$. The integral becomes $\int_0^{1-\alpha}u^{-1/r}\,du$, interpreted as an improper integral at zero.

Check convergence and evaluate.
The integral is finite exactly when $1/r<1$, or $r>1$. In that case it equals $(1-\alpha)^{1-1/r}/(1-1/r)$. Dividing by $1-\alpha$ and multiplying by $x_m$ gives $x_m r(1-\alpha)^{-1/r}/(r-1)$, which is $r/(r-1)$ times the VaR. For $r\le1$, the integral diverges, so TVaR is infinite.

This derivation explains an important contrast: a finite quantile only requires that the selected percentile be finite. TVaR averages quantiles all the way to probability 1, so an extremely severe upper tail can make that average infinite.

Worked Comparisons

Worked Example: A Finite-Mean Pareto Loss

A hypothetical insurer models the amount of a single large equipment loss by a Pareto Type I distribution with scale $x_m=\$250$ and shape $r=2$. Calculate VaR and TVaR at the 96% confidence level, and state how much larger TVaR is than VaR.

Method and assumptions. The loss has the continuous Pareto distribution specified above, with $x_m=250$, $r=2$, and $\alpha=0.96$. Since $r>1$, the finite-mean TVaR formula applies.

Step 1: Find VaR. The upper-tail probability is $1-\alpha=1-0.96=0.04$. Therefore, $$ \operatorname{VaR}_{0.96}(X) =\frac{250}{(0.04)^{1/2}} =\frac{250}{0.2} =\$1{,}250.00. $$

Step 2: Find TVaR. The multiplier is $r/(r-1)=2/(2-1)=2$. Hence, $$ \operatorname{TVaR}_{0.96}(X) =2(\$1{,}250.00) =\$2{,}500.00. $$

Step 3: Compare. TVaR exceeds VaR by $\$2{,}500.00-\$1{,}250.00=\$1{,}250.00$, and is twice the VaR. The result is consistent with the formula for shape $r=2$. A common mistake is to report the threshold, $\$1{,}250.00$, as the tail average; it is only the loss quantile, not the average of the upper 4%.

Worked Example: Equal VaR, Different Tail Averages

Two hypothetical portfolios each have a Pareto Type I modeled loss and a 99% VaR of $\$10{,}000$. Portfolio A has shape $r=1.2$; Portfolio B has shape $r=4$. Find each scale parameter and TVaR, then compare the tail averages.

Method and assumptions. For each portfolio, use the Pareto quantile formula at $\alpha=0.99$, then use $r/(r-1)$ for TVaR. Both shapes exceed 1, so both TVaRs are finite.

Step 1: Solve for Portfolio A's scale. Since $1-\alpha=0.01$, $$ 10{,}000=\frac{x_{m,A}}{(0.01)^{1/1.2}}, \qquad x_{m,A}=10{,}000(0.01)^{1/1.2} \approx \$215.44. $$ The calculation uses $(0.01)^{1/1.2}\approx0.0215443$.

Step 2: Calculate Portfolio A's TVaR. Its multiplier is $1.2/(1.2-1)=1.2/0.2=6$, so $$ \operatorname{TVaR}_{0.99}(X_A)=6(\$10{,}000)=\$60{,}000.00. $$

Step 3: Solve for Portfolio B's scale. $$ 10{,}000=\frac{x_{m,B}}{(0.01)^{1/4}}, \qquad x_{m,B}=10{,}000(0.01)^{1/4} \approx \$3{,}162.28, $$ because $(0.01)^{1/4}\approx0.316228$.

Step 4: Calculate Portfolio B's TVaR. Its multiplier is $4/(4-1)=4/3$, so $$ \operatorname{TVaR}_{0.99}(X_B) =\frac{4}{3}(\$10{,}000) \approx \$13{,}333.33. $$

Both portfolios have the same $\$10{,}000$ 99% VaR, but Portfolio A's TVaR is $\$60{,}000.00$, compared with Portfolio B's $\$13{,}333.33$. Thus the common VaR threshold does not capture the difference in modeled severity beyond that threshold. The shape parameter explains the contrast: Portfolio A's much smaller $r$ produces a heavier tail and a larger TVaR multiplier.

Worked Example: Finite VaR but Infinite TVaR

A hypothetical catastrophe loss is modeled as Pareto Type I with $x_m=\$5$, shape $r=0.9$, and confidence level $\alpha=0.99$. Calculate VaR and determine whether TVaR is finite.

Method and assumptions. The loss follows the stated continuous Pareto model. Use the Pareto quantile formula for VaR. Since $r=0.9\le1$, the quantile-integral test, rather than the finite-mean TVaR formula, determines TVaR.

Step 1: Calculate VaR. Here $1-\alpha=0.01$ and $1/r=1/0.9=1.111111\ldots$. Therefore, $$ \operatorname{VaR}_{0.99}(X) =\frac{5}{(0.01)^{1/0.9}} \approx \frac{5}{0.00599484} \approx \$834.05. $$

Step 2: Check TVaR. Because $r=0.9\le1$, the integral of the upper-tail quantiles diverges. Consequently, $\operatorname{TVaR}_{0.99}(X)=\infty$, even though the 99% VaR is finite.

The calculation illustrates why a high but finite quantile does not guarantee a finite tail average. A common error is to apply $r/(r-1)$ mechanically here; that factor is not a finite TVaR formula when $r\le1$.

Interpreting the Difference and Avoiding Traps

For a finite-mean Pareto loss, the ratio $\operatorname{TVaR}_{\alpha}/\operatorname{VaR}_{\alpha}=r/(r-1)$ does not depend on $\alpha$. The dollar amounts do depend on the confidence level, because raising $\alpha$ raises the quantile. For example, at a fixed Pareto shape, a higher confidence level selects a higher VaR and averages a smaller, more extreme portion of the distribution.

Using $\alpha$ instead of $1-\alpha$. At 99% confidence the upper-tail probability is 1%, so the Pareto quantile uses $0.01$, not $0.99$.
Calling VaR a tail average. VaR is a percentile threshold. TVaR summarizes the quantiles above that percentile.
Assuming every finite VaR gives finite TVaR. A Pareto model with $r\le1$ has finite VaR at any $\alpha<1$, but infinite TVaR.
Using conditional tail expectation at an atom without adjustment. The identity $E[X\mid X>\operatorname{VaR}_{\alpha}(X)]$ applies directly here because the Pareto distribution is continuous. In a discrete or mixed model, use the quantile-integral definition or the appropriate fractional tail mass.

A heavy-tailed model can make TVaR far greater than VaR, or infinite, so these measures answer different questions. VaR asks for a threshold that losses exceed with probability at most $1-\alpha$, subject to the quantile convention. TVaR asks how severe the upper tail is on average. Comparing them at the same confidence level is a useful way to see how much information a threshold alone leaves out. This comparison is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each Pareto model, use the stated scale, shape, and confidence level. Round dollar amounts to the nearest cent only at the final step.

A hypothetical repair loss has Pareto Type I scale $x_m=\$400$, shape $r=2$, and confidence level $\alpha=0.75$. Find VaR, TVaR, and the difference between them. (Answer: VaR = $800.00; TVaR = $1,600.00; difference = $800.00.)
A modeled loss has scale $x_m=\$1{,}200$, shape $r=3$, and confidence level $\alpha=0.992$. Find VaR and TVaR. (Answer: VaR = $6,000.00; TVaR = $9,000.00.)
A Pareto loss has shape $r=1.5$, confidence level $\alpha=0.99$, and a 99% VaR of $20,000. Find its scale parameter and TVaR. (Answer: $x_m=20{,}000(0.01)^{2/3}\approx\$928.32$; TVaR = $60,000.00.)
A hypothetical severity model has scale $x_m=\$2$, shape $r=0.8$, and confidence level $\alpha=0.95$. Find the VaR and determine whether TVaR is finite. (Answer: VaR $=2/(0.05)^{1.25}\approx\$84.59$; TVaR = $\infty$.)
Two Pareto losses both have a 99% VaR of $5,000. The first has shape $r=1.5$, and the second has shape $r=2.5$. Find both TVaRs and identify which model has the heavier tail. (Answer: First TVaR = $15,000.00; second TVaR $\approx\$8,333.33$; the first model has the heavier tail.)

What You'll Learn