Comparing Capital Across Confidence Levels

Why Compare Confidence Levels?

Value-at-risk and tail value-at-risk summarize different parts of a loss distribution. A single reported value can hide how quickly the measure grows as the confidence level rises. Comparing several levels helps an actuary describe that growth, distinguish a light tail from a heavy one, and identify how much a capital result depends on assumptions about rare, severe losses.

Here $q$ denotes a confidence level between 0 and 1. For example, $q=0.99$ is a 99% confidence level. VaR at that level is a loss threshold that the distribution does not exceed with probability at least $q$, subject to the chosen quantile convention. TVaR summarizes losses in the upper tail beyond that level. Both are properties of a specified loss distribution; neither is meaningful without a clear model and confidence level.

Definition: For a loss $X$ with distribution function $F_X$, define $$ \operatorname{VaR}_q(X)=\inf\{x:F_X(x)\geq q\}. $$ For any distribution, define tail value-at-risk by the quantile integral $$ \operatorname{TVaR}_q(X)=\frac{1}{1-q}\int_q^1 \operatorname{VaR}_u(X)\,du. $$ If $X$ has a continuous distribution, then $\operatorname{TVaR}_q(X)=E[X\mid X>\operatorname{VaR}_q(X)]$. For a discrete distribution, the quantile-integral definition correctly includes only the required fraction of probability mass at a VaR threshold.

A capital comparison should also say what is meant by “capital.” The risk measure itself is one possible capital amount. In some applications, the additional amount above expected loss is reported instead. These are different quantities, so this tutorial writes $K_q^{\mathrm{VaR}}=\operatorname{VaR}_q(X)-E[X]$ and $K_q^{\mathrm{TVaR}}=\operatorname{TVaR}_q(X)-E[X]$ when discussing capital above expected loss. Subtracting the same expected loss at every level changes the vertical position of a curve, not the differences between its levels.

How to Read a Capital Curve

A capital curve plots a risk measure against confidence level $q$. Because higher quantiles cannot be lower than lower quantiles, VaR is nondecreasing as $q$ rises. TVaR is also nondecreasing: its defining integral averages the upper quantiles, and raising the lower endpoint removes the least severe part of that upper-tail average.

The size of the increase matters as much as the ranking. If VaR grows by similar amounts over equal changes in a transformed confidence scale, the curve may be comparatively regular. If it grows sharply near one, extreme quantiles are increasingly separated. TVaR is particularly informative about severity beyond the VaR threshold, but its value is still determined by the assumed tail distribution. A high confidence level does not make an inaccurate tail model reliable.

The comparisons below use hypothetical loss models and amounts in dollars. Results are rounded to the nearest cent only after the calculations.

Worked Example: An Exponential Capital Curve

A hypothetical portfolio has aggregate loss $X\sim\operatorname{Exponential}(\theta=200)$, where $\theta$ is the mean loss in dollars. Calculate VaR, TVaR, and capital above expected loss at 90%, 99%, and 99.9%. Compare the increases between confidence levels.

Method. For this exponential model, $S_X(x)=e^{-x/200}$ for $x\geq0$, and $E[X]=200$. Solving $S_X(x_q)=1-q$ gives $\operatorname{VaR}_q(X)=-200\ln(1-q)$. The exponential TVaR identity is justified below, and gives $\operatorname{TVaR}_q(X)=\operatorname{VaR}_q(X)+200$.

At 90%, $1-q=0.10$, so $\operatorname{VaR}_{0.90}(X)=-200\ln(0.10)=460.5170\ldots$, or $\$460.52$. Thus $\operatorname{TVaR}_{0.90}(X)=460.5170\ldots+200=\$660.52$. Capital above expected loss is $\$460.52-\$200.00=\$260.52$ on a VaR basis, and $\$660.52-\$200.00=\$460.52$ on a TVaR basis.

At 99%, $\operatorname{VaR}_{0.99}(X)=-200\ln(0.01)=921.0340\ldots=\$921.03$. Therefore $\operatorname{TVaR}_{0.99}(X)=921.0340\ldots+200=\$1,121.03$. The corresponding capitals above expected loss are $\$921.0340\ldots-\$200.00=\$721.03$ for VaR and $\$1,121.0340\ldots-\$200.00=\$921.03$ for TVaR.

At 99.9%, $\operatorname{VaR}_{0.999}(X)=-200\ln(0.001)=1,381.5511\ldots=\$1,381.55$. Hence $\operatorname{TVaR}_{0.999}(X)=1,381.5511\ldots+200=\$1,581.55$. Capitals above expected loss are $\$1,181.55$ for VaR and $\$1,381.55$ for TVaR.

Comparison and check. The VaR increase from 90% to 99% is $921.0340\ldots-460.5170\ldots=\$460.52$. The increase from 99% to 99.9% is $1,381.5511\ldots-921.0340\ldots=\$460.52$ as well. TVaR has the same increases because it is VaR plus a constant. Subtracting the expected loss also leaves those increments unchanged.

Why the Exponential TVaR Adds the Mean

For a continuous nonnegative loss with survival function $S_X$, the conditional-tail expectation can be written as $$ E[X\mid X>x_q]=x_q+\frac{\int_{x_q}^{\infty}S_X(t)\,dt}{S_X(x_q)}, $$ where $x_q=\operatorname{VaR}_q(X)$. This follows by writing the conditional excess $X-x_q$, given $X>x_q$, as a nonnegative random variable. Its expectation is the integral of its conditional survival probabilities; for $t\geq x_q$, that probability is $S_X(t)/S_X(x_q)$.

For the exponential model, $S_X(t)=e^{-t/\theta}$. Substitution gives $$ \frac{\int_{x_q}^{\infty}e^{-t/\theta}\,dt}{e^{-x_q/\theta}} = \frac{\theta e^{-x_q/\theta}}{e^{-x_q/\theta}} =\theta. $$ Thus $E[X\mid X>x_q]=x_q+\theta$, which proves the TVaR formula used in the example. The constant excess above VaR is a consequence of the exponential tail; it should not be assumed for other distributions.

Heavy Tails Can Bend the Curve More Sharply

For a Pareto loss with scale $x_m>0$ and shape $\alpha>0$, suppose $S_X(x)=(x_m/x)^\alpha$ for $x\geq x_m$. Solving $S_X(x_q)=1-q$ gives $\operatorname{VaR}_q(X)=x_m(1-q)^{-1/\alpha}$. When $\alpha>1$, the mean is finite and the continuous Pareto model has $\operatorname{TVaR}_q(X)=\frac{\alpha}{\alpha-1}\operatorname{VaR}_q(X)$. In this model, both measures increase proportionally to $(1-q)^{-1/\alpha}$; the tail multiplier for TVaR is greater than one.

Worked Example: Comparing Pareto Tail Levels

A hypothetical severity has Pareto survival function $S_X(x)=(50/x)^{2.5}$ for $x\geq50$, with losses measured in dollars. Calculate VaR and TVaR at 90%, 96%, and 99%, then compare the increases. State why the finite-mean condition matters.

Method. Here $x_m=50$ and $\alpha=2.5$. Since $\alpha>1$, TVaR is finite and equals $\frac{2.5}{2.5-1}\operatorname{VaR}_q(X)=\frac{5}{3}\operatorname{VaR}_q(X)$.

At 90%, $\operatorname{VaR}_{0.90}(X)=50(0.10)^{-0.4}=125.5943\ldots=\$125.59$. Then $\operatorname{TVaR}_{0.90}(X)=\frac{5}{3}(125.5943\ldots)=\$209.32$.

At 96%, $\operatorname{VaR}_{0.96}(X)=50(0.04)^{-0.4}=181.1949\ldots=\$181.19$. The TVaR is $\frac{5}{3}(181.1949\ldots)=301.9915\ldots=\$301.99$.

At 99%, $\operatorname{VaR}_{0.99}(X)=50(0.01)^{-0.4}=315.4787\ldots=\$315.48$. The TVaR is $\frac{5}{3}(315.4787\ldots)=525.7978\ldots=\$525.80$.

Comparison and check. VaR rises by $\$181.19-\$125.59=\$55.60$ from 90% to 96%, and by $\$315.48-\$181.19=\$134.29$ from 96% to 99%. The later increase is larger even though the confidence-level step is smaller. TVaR rises by $\$92.67$ and then $\$223.81$, respectively. The larger upper-level increases reflect the Pareto tail’s growth as $q$ approaches one. If $\alpha\leq1$, this unbounded Pareto distribution has infinite mean, so its TVaR is infinite; the finite-$\alpha>1$ formula cannot be used to report a finite TVaR.

Atoms at the Threshold: A Discrete Capital Curve

With discrete losses, VaR may equal a loss value that has positive probability. The quantile-integral definition of TVaR is essential in this case: the worst $1-q$ fraction of outcomes may include only part of the probability mass at the VaR threshold. Simply taking $E[X\mid X\geq\operatorname{VaR}_q(X)]$ can include too much mass and produce the wrong result.

Worked Example: TVaR with Probability at VaR

A hypothetical policy portfolio has loss $X=0$ with probability 0.90, $X=\$1,000$ with probability 0.09, and $X=\$10,000$ with probability 0.01. Find VaR and TVaR at 90%, 95%, and 99%.

Method. The cumulative probabilities are $F_X(0)=0.90$, $F_X(1{,}000)=0.99$, and $F_X(10{,}000)=1$. Use the quantile definition for VaR and average exactly the worst $1-q$ fraction of probability for TVaR.

At 90%, $\operatorname{VaR}_{0.90}(X)=\$0$, because the cumulative probability at zero is already 0.90. The worst 10% comprises the 9% at $\$1,000$ and the 1% at $\$10,000$. Thus $\operatorname{TVaR}_{0.90}(X)=(0.09(1{,}000)+0.01(10{,}000))/0.10=\$1,900$.

At 95%, $\operatorname{VaR}_{0.95}(X)=\$1,000$. The worst 5% includes 4 percentage points of the 9% probability at $\$1,000$, and all 1 percentage point at $\$10,000$. Therefore $\operatorname{TVaR}_{0.95}(X)=(0.04(1{,}000)+0.01(10{,}000))/0.05=\$2,800$.

At 99%, $\operatorname{VaR}_{0.99}(X)=\$1,000$, since $F_X(1{,}000)=0.99$. The worst 1% consists entirely of the $\$10,000$ outcome, so $\operatorname{TVaR}_{0.99}(X)=\$10,000$.

Check. The calculated tail averages are nondecreasing: $\$1,900$, $\$2,800$, and $\$10,000$. At 90%, using the conditional mean given $X\geq \$0$ would average the entire distribution, not just its worst 10%; that is not the required tail average.

Common Mistakes When Comparing Levels

Confusing TVaR with capital above expected loss. In the exponential example at 99%, VaR is $\$921.03$, but the VaR-based capital above the $\$200$ expected loss is $\$921.03-\$200.00=\$721.03$. TVaR is $\$1,121.03$, not $\$921.03$, and TVaR-based capital above expected loss is $\$921.03$. Keep the risk measure and the subtraction separate.
Assuming equal confidence-level changes mean equal risk-measure changes. The exponential example has equal VaR increases across 90% to 99% and 99% to 99.9%; the Pareto example does not. This is a property of the distribution, not a general rule.
Using the continuous conditional-expectation shortcut for a discrete loss. At an atom, conditioning on $X\geq\operatorname{VaR}_q(X)$ can include more probability than the upper $1-q$ tail. Use the quantile integral or explicitly select the required fraction of the threshold mass.
Reporting tail results without checking the model’s moments. A Pareto VaR can be finite at every $q<1$ even when the mean and TVaR are infinite. Check the shape parameter before applying a finite-TV aR formula.

Key takeaway: Compare VaR and TVaR at clearly stated confidence levels, and label whether the reported capital is the risk measure itself or its excess over expected loss. The curve’s shape reflects tail behavior; high-confidence results can change substantially when tail assumptions change.

Check Your Understanding

For each hypothetical model, calculate before checking the answer. Amounts are in dollars; round monetary answers to the nearest cent.

A loss is exponential with mean $\$300$. Find VaR, TVaR, and VaR-based capital above expected loss at 95%. (Answer: VaR $=\$898.72$; TVaR $=\$1,198.72$; VaR-based capital above expected loss $=\$598.72$.)
A loss has Pareto survival function $S_X(x)=(80/x)^3$ for $x\geq80$. Find VaR and TVaR at 99%. (Answer: VaR $=80(0.01)^{-1/3}=\$371.33$; TVaR $=\frac{3}{2}(371.33\ldots)=\$556.99$.)
An exponential loss has mean $\$120$. By how much does VaR rise between 95% and 99.5%? (Answer: $120\ln(0.05/0.005)=120\ln(10)=\$276.31$.)
A discrete loss is $\$0$ with probability 0.80, $\$500$ with probability 0.15, and $\$4,000$ with probability 0.05. Find VaR and TVaR at 90%, then VaR and TVaR at 95%. (Answer: At 90%, VaR $=\$500$, TVaR $=\$2,250$; at 95%, VaR $=\$500$, TVaR $=\$4,000$.)
A Pareto loss has scale $\$100$, shape $\alpha=2$, and confidence level 96%. Find VaR, TVaR, and TVaR-based capital above expected loss. (Answer: VaR $=100(0.04)^{-1/2}=\$500$; TVaR $=2(500)=\$1,000$; since $E[X]=2(100)=\$200$, TVaR-based capital above expected loss $=\$800$.)

What You'll Learn