How Heavy Tails Change Risk Measures

Why Tail Thickness Changes the Comparison

Two loss models can have similar average losses yet imply very different amounts of capital for unusually large losses. The difference can be modest at one confidence level and substantial at another. It also affects whether familiar summaries such as variance exist at all. Comparing value-at-risk, tail value-at-risk, and variance helps show what each measure reveals—and what it can miss.

The focus here is on continuous exponential and Pareto loss models. These provide a useful contrast: the exponential tail decreases at an exponential rate, while a Pareto tail decreases as a power of the loss. The calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For a loss $X$ and confidence level $q$, where $0<q<1$, define $\operatorname{VaR}_q(X)=\inf\{x:F_X(x)\ge q\}$. For a continuous loss distribution with finite conditional tail mean, define $\operatorname{TVaR}_q(X)=E[X\mid X>\operatorname{VaR}_q(X)]$. Variance is $\operatorname{Var}(X)=E[(X-E[X])^2]$, when the required moments are finite.

For continuous distributions, the probability of exceeding $\operatorname{VaR}_q(X)$ is $1-q$. TVaR therefore measures the average loss within that upper tail, not merely its threshold. VaR and TVaR are expressed in the same units as the loss, such as dollars. Variance is in squared units, such as dollars squared, so it is not directly comparable to a dollar capital amount.

Exponential and Pareto Tail Formulas

Suppose $X$ has an exponential distribution with rate $\lambda>0$, so $S_X(x)=e^{-\lambda x}$ for $x\ge 0$. Setting $S_X(x)=1-q$ gives its quantile. The exponential distribution also has the memoryless property: conditional on exceeding a threshold, the additional loss has the same exponential distribution.

\operatorname{VaR}_q(X)=\frac{-\ln(1-q)}{\lambda},\qquad \operatorname{TVaR}_q(X)=\operatorname{VaR}_q(X)+\frac{1}{\lambda},\qquad E[X]=\frac{1}{\lambda},\qquad \operatorname{Var}(X)=\frac{1}{\lambda^2}.

Now suppose $X$ has a Pareto Type I distribution with scale $x_m>0$ and shape $\alpha>0$, with survival function $S_X(x)=(x_m/x)^\alpha$ for $x\ge x_m$. Inverting the survival function gives VaR. If $\alpha>1$, the conditional mean beyond the quantile is finite and TVaR is a constant multiple of VaR. If $\alpha\le 1$, that conditional mean is infinite.

\operatorname{VaR}_q(X)=x_m(1-q)^{-1/\alpha},\qquad \operatorname{TVaR}_q(X)=\frac{\alpha}{\alpha-1}\operatorname{VaR}_q(X)\quad(\alpha>1).

The Pareto moment condition follows from integrating its density: $E[X^r]=\alpha x_m^r/(\alpha-r)$ when $r<\alpha$. Thus the mean is finite only when $\alpha>1$, and the variance is finite only when $\alpha>2$. When $\alpha>2$, subtracting the square of the mean from the second moment gives the variance formula below.

E[X]=\frac{\alpha x_m}{\alpha-1}\quad(\alpha>1),\qquad \operatorname{Var}(X)=\frac{\alpha x_m^2}{(\alpha-1)^2(\alpha-2)}\quad(\alpha>2).

These conditions matter separately. A Pareto model with $1<\alpha\le 2$ has finite mean and finite TVaR at each $q<1$, but infinite variance. A model with $\alpha\le1$ still has a finite VaR at any fixed $q<1$, but has infinite mean and infinite TVaR.

Worked Example: Comparing Equal-Mean Exponential and Pareto Losses

A hypothetical portfolio is modeled in two alternative ways. In the first model, $X$ is exponential with rate $\lambda=1/150$ per dollar. In the second, $X$ is Pareto Type I with $x_m=\$100$ and $\alpha=3$. Both models have mean loss $\$150$. Compare VaR, TVaR, and variance at $q=0.9900$. Round monetary amounts to the nearest cent.

Method and assumptions. Use the continuous-distribution formulas above. Both models have finite mean and variance; the Pareto shape satisfies $\alpha=3>2$.

Exponential model. The tail probability is $1-q=0.0100$, so

\operatorname{VaR}_{0.9900}(X) =-150\ln(0.0100) =150(4.6051702) =\$690.7755 \approx \$690.78.

The exponential mean is $1/\lambda=\$150$. Therefore,

\operatorname{TVaR}_{0.9900}(X) =\$690.7755+\$150 =\$840.7755 \approx \$840.78, \qquad \operatorname{Var}(X)=150^2=\$22{,}500.00\text{ dollars}^2.

Pareto model. Its quantile and TVaR are

\operatorname{VaR}_{0.9900}(X) =100(0.0100)^{-1/3} =100(4.6415888) =\$464.1589 \approx \$464.16, $$ $$ \operatorname{TVaR}_{0.9900}(X) =\frac{3}{2}(\$464.1589) =\$696.2383 \approx \$696.24.

As a check, the Pareto mean is $3(100)/(3-1)=\$150$, matching the exponential mean. Its variance is

\operatorname{Var}(X) =\frac{3(100)^2}{(3-1)^2(3-2)} =\frac{30{,}000}{4} =\$7{,}500.00\text{ dollars}^2.

At this particular confidence level, the exponential model has the larger VaR and TVaR, despite the Pareto model having the heavier tail. The comparison is not a contradiction: tail heaviness describes how the far tail behaves, not a guarantee that one model has larger quantiles at every finite confidence level. The common mistake is to assume “heavier tail” automatically means a larger risk measure at any selected $q$.

Why Pareto TVaR and Variance Behave Differently

The Pareto TVaR formula can be justified directly. For a continuous loss with threshold $t$, the tail-integral identity gives $E[X\mid X>t]=t+\int_t^\infty S_X(x)/S_X(t)\,dx$, provided this integral is finite. For Pareto $X$, the ratio of survival probabilities is $(t/x)^\alpha$. When $\alpha>1$, integrating gives $t/(\alpha-1)$. Hence the conditional mean is $t+t/(\alpha-1)=\alpha t/(\alpha-1)$. Taking $t=\operatorname{VaR}_q(X)$ yields the stated TVaR formula. When $\alpha\le1$, the integral diverges, so TVaR is infinite.

The same power-law tail explains the moment threshold. The integral defining $E[X^r]$ has a tail proportional to $\int_{x_m}^{\infty}x^{r-\alpha-1}\,dx$. It converges exactly when $r<\alpha$. Taking $r=1$ tests the mean, and taking $r=2$ tests the second moment needed for variance.

As $q$ approaches 1, Pareto VaR grows in proportion to $(1-q)^{-1/\alpha}$, whereas exponential VaR grows in proportion to $-\ln(1-q)$. The power-law increase eventually outpaces logarithmic growth. A smaller Pareto shape parameter produces faster growth and a larger TVaR multiplier $\alpha/(\alpha-1)$, as long as $\alpha>1$. This helps explain why conclusions at one confidence level should not be generalized to all levels.

Worked Example: Finite Pareto Variance and a High Quantile

A hypothetical equipment-loss model follows a Pareto Type I distribution with $x_m=\$120$ and $\alpha=2.5$. Calculate VaR and TVaR at $q=0.9900$, then calculate the variance. Round monetary amounts to the nearest cent.

Method and assumptions. This is a continuous Pareto model. Since $\alpha=2.5>2$, its mean and variance are finite, and since $\alpha>1$, its TVaR is finite.

The tail probability is $1-q=0.0100$. Thus,

\operatorname{VaR}_{0.9900}(X) =120(0.0100)^{-1/2.5} =120(100)^{0.4} =120(6.3095734) =\$757.1488 \approx \$757.15.

The TVaR multiplier is $\alpha/(\alpha-1)=2.5/1.5=5/3$. Therefore,

\operatorname{TVaR}_{0.9900}(X) =\frac{5}{3}(\$757.1488) =\$1{,}261.9147 \approx \$1{,}261.91.

For a check on variance, first calculate the mean and second moment:

E[X]=\frac{2.5(120)}{1.5}=\$200,\qquad E[X^2]=\frac{2.5(120)^2}{2.5-2} =\frac{36{,}000}{0.5} =\$72{,}000.

Consequently,

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =72{,}000-200^2 =\$32{,}000.00\text{ dollars}^2.

The result is internally consistent: the shape parameter exceeds 2, so the variance is finite. A common exam error is rounding the VaR before multiplying it to get TVaR. Keeping the intermediate VaR unrounded gives TVaR of $\$1{,}261.91$, not $\$1{,}261.92$.

Worked Example: Finite TVaR but Infinite Variance

A hypothetical specialty-property loss has a Pareto Type I distribution with $x_m=\$60$ and $\alpha=1.8$. Find VaR and TVaR at $q=0.9500$, and state whether the mean and variance exist. Round monetary amounts to the nearest cent.

Method and assumptions. The model is continuous and has $\alpha>1$, so the mean and TVaR are finite. Because $\alpha<2$, the second moment and variance are infinite.

The tail probability is $1-q=0.0500$, giving

\operatorname{VaR}_{0.9500}(X) =60(0.0500)^{-1/1.8} =60(20)^{5/9} =\$316.9171 \approx \$316.92.

The TVaR multiplier is $1.8/(1.8-1)=1.8/0.8=2.25$, so

\operatorname{TVaR}_{0.9500}(X) =2.25(\$316.9171) =\$713.0635 \approx \$713.06.

The mean exists and equals $1.8(60)/0.8=\$135$. The variance does not exist as a finite number because its second moment requires $\alpha>2$, while here $\alpha=1.8$. Thus VaR and TVaR are finite at this confidence level even though variance is infinite. Do not infer that TVaR must be infinite merely because variance is infinite; the moment thresholds differ.

Interpreting the Measures in Risk Decisions

VaR gives a loss threshold associated with a confidence level, but it does not describe the size of losses beyond that threshold. TVaR incorporates the severity of that upper tail, making it particularly informative when tail losses matter. Variance summarizes squared deviations around the mean across the whole distribution. It can be useful for dispersion analysis, but it is neither a quantile nor a tail-only measure.

For heavy-tailed models, moment conditions should be checked before a variance-based calculation is attempted. If a fitted Pareto shape estimate is at or below 2, reporting a finite variance is not justified under the model. If it is at or below 1, even the mean and the finite-level TVaR fail to exist. A finite VaR alone does not resolve either issue.

The scale also matters in comparisons. VaR and TVaR are monetary amounts; variance is measured in squared monetary units. Taking the square root produces standard deviation in monetary units, but it still describes overall dispersion rather than the average of the upper tail. No single measure answers every capital question, so a comparison should state the model, confidence level, and intended interpretation.

Key takeaway: Tail thickness affects how rapidly extreme-loss measures grow, while the Pareto shape parameter also determines whether moments exist. Calculate VaR, TVaR, and variance using their separate definitions and conditions; do not treat them as interchangeable measures of capital.

Check Your Understanding

For each hypothetical model, identify the relevant tail or moment conditions before calculating. Round monetary answers to the nearest cent unless the answer is stated as a formula.

A loss is exponential with mean $\$80$. Calculate VaR and TVaR at $q=0.9500$, and give the variance. (Answer: VaR $=\$239.66$; TVaR $=\$319.66$; variance $=\$6{,}400.00$ dollars squared.)
A loss is Pareto Type I with $x_m=\$50$ and $\alpha=4$. Calculate VaR and TVaR at $q=0.9900$, and give the variance. (Answer: VaR $=\$158.11$; TVaR $=\$210.82$; variance $=\$555.56$ dollars squared.)
A loss is Pareto Type I with $x_m=\$100$ and $\alpha=2$. Calculate VaR and TVaR at $q=0.9600$, and determine whether its mean and variance are finite. (Answer: VaR $=\$500.00$; TVaR $=\$1{,}000.00$; mean is finite at $\$200.00$, but variance is infinite.)
A loss is Pareto Type I with $x_m=\$40$ and $\alpha=0.8$. Calculate VaR at $q=0.9000$, and determine whether the mean, TVaR, and variance are finite. (Answer: VaR $=\$711.31$; mean, TVaR, and variance are infinite.)

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