TVaR For Pareto Losses

Why the Pareto Tail Matters

A Pareto model is useful for representing losses with a heavy upper tail: very large losses are uncommon, but their probability decreases more slowly than it would under many lighter-tailed models. This distinction matters when assessing tail risk. At a chosen confidence level, VaR may be finite even when the average loss beyond that VaR is infinite.

The formulas below use the Type I Pareto distribution. Its scale parameter sets the minimum possible loss, while its shape parameter controls how quickly the tail probability declines. We use the quantile-integral definition of TVaR, which averages the upper quantiles beyond the chosen confidence level.

Definition: A loss $X$ has a Type I Pareto distribution with scale $\theta>0$ and shape $\alpha>0$ if $X\geq\theta$, its distribution function is $F_X(x)=1-(\theta/x)^\alpha$ for $x\geq\theta$, and $F_X(x)=0$ for $x<\theta$. Its survival function is $S_X(x)=(\theta/x)^\alpha$ for $x\geq\theta$.

For $0<p<1$, define $\operatorname{VaR}_p(X)$ as the $p$-quantile of $X$. For this continuous Pareto model, the quantile is unique. Define $\operatorname{TVaR}_p(X)$ as the average of the upper quantiles:

\operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_u(X)\,du.

This integral is allowed to diverge; in that case TVaR is infinite. Here $p$ is the confidence level and $1-p$ is the tail probability. Keeping those two quantities distinct is essential in the calculations.

VaR for a Pareto Loss

For $x\geq\theta$, the $p$-quantile $q_p$ satisfies $F_X(q_p)=p$. Substituting the Pareto distribution function and solving gives:

1-\left(\frac{\theta}{q_p}\right)^\alpha=p \quad\Longrightarrow\quad \left(\frac{\theta}{q_p}\right)^\alpha=1-p \quad\Longrightarrow\quad \operatorname{VaR}_p(X)=q_p=\theta(1-p)^{-1/\alpha}.

The quantile is finite for every $0<p<1$ and $\alpha>0$. As $p$ increases, $1-p$ decreases and the VaR increases. A small shape parameter makes the exponent $1/\alpha$ larger, producing a more rapidly increasing quantile as the confidence level approaches one.

Worked Example: VaR and TVaR with a Finite Tail Average

A hypothetical portfolio has annual loss $X$ with a Type I Pareto distribution, scale $\theta=\$1{,}000$, and shape $\alpha=2$. Calculate its 95% VaR and TVaR. Give monetary answers to the nearest cent, and verify the quantile.

Method. First find the 95% quantile using the Pareto formula, then substitute the quantile function into the upper-quantile integral. Since $\alpha=2>1$, the TVaR integral is finite.

Step 1: Calculate VaR. The tail probability is $1-p=1-0.95=0.05$, so $0.05^{-1/2}=\sqrt{20}\approx4.47213595$. Therefore $\operatorname{VaR}_{0.95}(X)=\$1{,}000\sqrt{20}\approx\$4{,}472.13595$, or \$4,472.14 to the nearest cent.

Step 2: Calculate TVaR. For $\alpha>1$, the formula derived below is $\operatorname{TVaR}_p(X)=\frac{\alpha}{\alpha-1}\operatorname{VaR}_p(X)$. Here the factor is $2/(2-1)=2$. Thus $\operatorname{TVaR}_{0.95}(X)=2(\$4{,}472.13595)=\$8{,}944.27190$, or \$8,944.27.

Check. At the calculated quantile, $S_X(q)=(\$1{,}000/\$4{,}472.13595)^2\approx0.05$, as required for a 95% quantile. The TVaR is twice the VaR because the shape parameter is two.

Common mistake avoided. Using $0.95$, rather than the tail probability $0.05$, inside the Pareto quantile formula would produce a value below the scale. The quantile equation uses $1-p$.

Deriving TVaR and Its Finiteness Condition

The shape parameter determines whether the average upper quantile exists. Starting from the quantile formula, change variables to $t=1-u$. As $u$ runs from $p$ to 1, $t$ runs from $1-p$ down to 0; reversing the integration bounds gives:

\operatorname{TVaR}_p(X) =\frac{\theta}{1-p}\int_p^1(1-u)^{-1/\alpha}\,du =\frac{\theta}{1-p}\int_0^{1-p}t^{-1/\alpha}\,dt.

The integral at zero is finite exactly when the power of $t$ is greater than $-1$: $-1/\alpha>-1$, which is equivalent to $\alpha>1$. For that case, integrating and simplifying yields:

\int_0^{1-p}t^{-1/\alpha}\,dt =\frac{(1-p)^{1-1/\alpha}}{1-1/\alpha} =\frac{\alpha}{\alpha-1}(1-p)^{1-1/\alpha}, $$ $$ \operatorname{TVaR}_p(X) =\frac{\alpha}{\alpha-1}\theta(1-p)^{-1/\alpha} =\frac{\alpha}{\alpha-1}\operatorname{VaR}_p(X), \qquad \alpha>1.

If $\alpha=1$, the integral is proportional to $\int_0^{1-p}t^{-1}\,dt$, which diverges logarithmically. If $0<\alpha<1$, the integrand has exponent $-1/\alpha<-1$, so the integral also diverges at zero. Consequently, for every confidence level $0<p<1$, TVaR is infinite when $0<\alpha\leq1$, even though VaR remains finite.

The same finiteness condition can be understood from the conditional tail mean. For $q\geq\theta$, the Pareto survival function gives $S_X(x)/S_X(q)=(q/x)^\alpha$ for $x\geq q$. When $\alpha>1$, the tail integral is finite and $\operatorname{E}[X\mid X>q]=q+\int_q^\infty S_X(x)/S_X(q)\,dx=\alpha q/(\alpha-1)$. For this continuous distribution, averaging quantiles above $p$ has the same value as conditioning on $X>\operatorname{VaR}_p(X)$. This also explains the factor $\alpha/(\alpha-1)$; when $\alpha\leq1$, the tail integral does not converge.

Worked Example: A Finite TVaR with a Heavier Tail

A hypothetical insurer models a claim amount $X$ using a Type I Pareto distribution with scale $\theta=\$600$ and shape $\alpha=1.5$. Calculate VaR and TVaR at $p=0.80$, to the nearest cent, and check both the quantile and the relationship between the two measures.

Method. Use the Pareto quantile formula. Since $1.5>1$, apply the finite-TV aR formula, $\alpha/(\alpha-1)$ times VaR.

Step 1: Find the tail probability and VaR. The tail probability is $1-0.80=0.20$, so $0.20^{-1/1.5}=0.20^{-2/3}=5^{2/3}\approx2.924017738$. Therefore $\operatorname{VaR}_{0.80}(X)=\$600(2.924017738)\approx\$1{,}754.410643$, or \$1,754.41.

Step 2: Find TVaR. The multiplier is $\alpha/(\alpha-1)=1.5/0.5=3$. Thus $\operatorname{TVaR}_{0.80}(X)=3(\$1{,}754.410643)=\$5{,}263.231929$, or \$5,263.23.

Check. At the unrounded quantile, $S_X(q)=(600/1{,}754.410643)^{1.5}\approx0.20$. Also, the computed TVaR is three times VaR, matching the shape-based multiplier.

Common mistake avoided. The factor is $\alpha/(\alpha-1)$, not $\alpha-1$. It equals 3 here and must make TVaR greater than VaR.

Finite VaR Does Not Guarantee Finite TVaR

A heavy-tailed Pareto loss can have a well-defined quantile at any confidence level below one while lacking a finite mean. The same threshold, $\alpha>1$, governs whether the mean and the TVaR are finite. Thus, a finite reported VaR should not be taken as evidence that the average of more extreme losses is finite.

Worked Example: Finite VaR but Infinite TVaR

A hypothetical risk model assigns annual losses a Type I Pareto distribution with scale $\theta=\$250$ and shape $\alpha=0.5$. At $p=0.75$, calculate VaR and determine whether TVaR is finite. Give the VaR to the nearest cent.

Method. Calculate the quantile using the tail probability $1-p$, then test the shape parameter against the convergence condition. Do not use the finite-TV aR formula when its condition fails.

Step 1: Calculate VaR. Here $1-p=1-0.75=0.25$ and $1/\alpha=2$. Therefore $0.25^{-2}=1/(0.25^2)=1/0.0625=16$, so $\operatorname{VaR}_{0.75}(X)=\$250(16)=\mathbf{\$4{,}000.00}$.

Step 2: Determine whether TVaR is finite. Since $\alpha=0.5\leq1$, the upper-quantile integral diverges. Thus $\operatorname{TVaR}_{0.75}(X)=\infty$.

Check. The survival probability at \$4,000 is $(250/4{,}000)^{0.5}=(1/16)^{0.5}=0.25$, as required. The finite quantile and infinite TVaR are consistent: a finite threshold does not make the average of all losses beyond it finite.

Common mistake avoided. Replacing $0.25^{-2}$ by 4 gives an incorrect VaR of \$1,000. The correct reciprocal-square calculation is $1/(0.25^2)=16$, so the VaR is \$4,000.

Exam Traps and Practical Interpretation

For a Pareto TVaR calculation, proceed in a fixed order: identify the distribution parameterization, calculate $1-p$, find VaR, and then check whether $\alpha>1$ before applying the finite-TV aR formula. The scale $\theta$ is the lower endpoint of the loss distribution, not its mean or a location parameter that can be ignored.

Confusing confidence and tail probability. At $p=0.95$, the upper-tail probability is $0.05$, not $0.95$.
Using the wrong exponent. The quantile uses $(1-p)^{-1/\alpha}$; the exponent is the negative reciprocal of the shape.
Applying the finite formula without checking its condition. The factor $\alpha/(\alpha-1)$ applies only when $\alpha>1$. At or below one, TVaR is infinite.
Assuming finite VaR implies finite TVaR. The worked example with $\alpha=0.5$ shows why this inference fails.

For capital and risk comparisons, the shape parameter is therefore central: it affects both the size of upper quantiles and whether a finite average-tail measure exists. Results depend on the chosen Pareto model and parameterization; do not transfer these formulas to a different distribution without checking its survival function.

Key takeaway: For a Type I Pareto loss with scale $\theta$, shape $\alpha$, and confidence level $0<p<1$, $\operatorname{VaR}_p(X)=\theta(1-p)^{-1/\alpha}$. TVaR is $\frac{\alpha}{\alpha-1}\operatorname{VaR}_p(X)$ only when $\alpha>1$; for $0<\alpha\leq1$, it is infinite.

Check Your Understanding

For each problem, use the Type I Pareto parameterization defined above. Round finite monetary answers to the nearest cent.

A hypothetical portfolio has scale $\theta=\$400$ and shape $\alpha=2$. Find VaR and TVaR at $p=0.90$. (Answer: VaR = \$1,264.91; TVaR = \$2,529.82.)
A hypothetical claim model has scale $\theta=\$1{,}200$ and shape $\alpha=3$. Find VaR and TVaR at $p=0.96$. (Answer: VaR = \$3,508.82; TVaR = \$5,263.23.)
A hypothetical loss has scale $\theta=\$800$, shape $\alpha=2$, and confidence level $p=0.84$. Calculate both measures and state the multiplier of TVaR over VaR. (Answer: VaR = \$2,000.00; TVaR = \$4,000.00; multiplier = 2.)
A hypothetical loss has scale $\theta=\$500$ and shape $\alpha=1$. At $p=0.90$, calculate VaR and determine whether TVaR is finite. (Answer: VaR = \$5,000.00; TVaR = $\infty$.)
A hypothetical loss has scale $\theta=\$40$, shape $\alpha=0.5$, and confidence level $p=0.96$. Find VaR, then explain why the finite-TV aR formula cannot be used. (Answer: VaR = \$25,000.00; TVaR = $\infty$, because $\alpha\leq1$.)

What You'll Learn