Why Compare TVaR at a Common VaR?
Two loss models can agree at a selected percentile and still imply very different average losses beyond that percentile. This matters when a model is used to describe severe outcomes: matching a threshold does not establish that the models describe the same upper tail. Comparing their TVaR values at the same confidence level isolates an important difference in tail shape.
The comparison in this tutorial uses continuous loss distributions with finite expected loss in the tail being averaged. The losses are hypothetical. The methods are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
For a continuous distribution, the probability of exceeding \(q=VaR_p(X)\) is \(1-p\). A useful equivalent expression is the average of the upper quantiles:
This expression makes the comparison clear: VaR at level \(p\) uses one quantile, while TVaR averages all quantiles from \(p\) toward 1. Two models can therefore have the same \(VaR_p\) but different values of \(VaR_u\) for \(u>p\), and consequently different TVaR.
A Tail-Integral Justification
The following derivation gives a direct route from the loss distribution to TVaR. It assumes a continuous distribution, \(q=VaR_p(X)\), \(P(X>q)=1-p\), and a finite integral of the survival function above \(q\).
On the event \(X>q\), write \(X=q+(X-q)\). Therefore \(E[X\mid X>q]=q+E[(X-q)\mid X>q]\).
Since \((X-q)_+=(X-q)\) when \(X>q\) and is zero otherwise, \(E[(X-q)\mid X>q]=E[(X-q)_+]/P(X>q)\).
For a nonnegative excess, \(E[(X-q)_+]=\int_q^\infty S_X(x)\,dx\), where \(S_X(x)=P(X>x)\). Substitution gives the tail-integral form of TVaR.
The integral is the expected amount by which a loss exceeds \(q\), counted over all outcomes; division by \(1-p\) converts that amount to an average conditional excess. Adding \(q\) gives the average loss itself. If a distribution has an atom at \(q\), the simple conditional formula above need not equal the standard quantile-integral definition; the examples here avoid that complication.
Worked Comparisons
Worked Example: Exponential and Pareto Losses With the Same 95% VaR
A hypothetical portfolio team fits two continuous models to an annual loss \(X\), measured in thousands of dollars. Both models must have \(VaR_{0.95}=1{,}000\). Model E is exponential with scale parameter \(\theta\), so \(S_X(x)=e^{-x/\theta}\) for \(x\ge0\). Model P is Pareto Type I with minimum loss \(x_m=100\) and shape parameter \(\alpha>1\), so \(S_X(x)=(x_m/x)^\alpha\) for \(x\ge x_m\). Calibrate each model and calculate its TVaR at 95%.
Method. Set each model's 95th percentile equal to 1,000, then calculate its conditional tail mean. This avoids comparing tail averages at different thresholds.
Model E calibration. For an exponential loss, \(F_X(x)=1-e^{-x/\theta}\). Hence \(VaR_p=-\theta\ln(1-p)\). At \(p=0.95\), \(\ln(1/(1-p))=\ln20=2.995732\), so \(1{,}000=\theta\ln20\), giving \(\theta=1{,}000/2.995732=333.808\) thousand dollars.
Model E TVaR. Given \(X>q\), the exponential memoryless property gives an average excess of \(\theta\). Equivalently, the tail integral is \(\int_q^\infty e^{-x/\theta}dx=\theta e^{-q/\theta}\); dividing by \(P(X>q)=e^{-q/\theta}\) leaves \(\theta\). Thus \(TVaR_{0.95}=1{,}000+333.808=1{,}333.808\) thousand dollars.
Model P calibration. Its quantile is \(VaR_p=x_m(1-p)^{-1/\alpha}\). Therefore \(1{,}000=100(0.05)^{-1/\alpha}\), which implies \(10^\alpha=20\) and \(\alpha=\ln20/\ln10=1.301030\). This shape parameter exceeds 1, so the Pareto mean, including its tail mean, is finite.
Model P TVaR. For \(\alpha>1\), the conditional mean above \(q\ge x_m\) is \(\alpha q/(\alpha-1)\). Here \(\alpha-1=0.301030\), and \(\alpha/(\alpha-1)=\ln20/\ln2=4.321928\). Thus \(TVaR_{0.95}=4.321928(1{,}000)=4{,}321.928\) thousand dollars.
Check and interpretation. Both models have the same 95% VaR of 1,000 thousand dollars, but their TVaRs are 1,333.81 and 4,321.93 thousand dollars. The Pareto model assigns much more weight to very large losses beyond the matched percentile. The calculations are in thousands; rounding only at the final step, to two decimals, gives the stated TVaRs.
Common mistake avoided: Using the Pareto shape parameter from the exponential model, or using the same shape parameter without calibrating it, would not ensure a common VaR. Each distribution has its own quantile equation.
Worked Example: A Bounded Uniform Tail Versus an Exponential Tail
A hypothetical risk analyst compares two loss models, both in thousands of dollars, at the 90% confidence level. Model U is uniform on \([0,1{,}000]\). Model E is exponential with scale \(\theta\), calibrated so that both models have \(VaR_{0.90}=900\). Calculate and compare the two TVaRs.
Method. First verify the common quantile. Then average the losses strictly above it using the conditional expectation for each continuous distribution.
Model U. For a uniform distribution on \([0,b]\), \(F_X(x)=x/b\) on that interval. Thus \(VaR_{0.90}=0.90(1{,}000)=900\). Conditional on exceeding 900, the loss is uniform on \([900,1{,}000]\). Its mean is \((900+1{,}000)/2=950\) thousand dollars, so \(TVaR_{0.90}=950\) thousand dollars. The upper-tail average is below the maximum possible loss, as it must be.
Model E calibration. For the exponential model, \(VaR_{0.90}=\theta\ln10\). Setting this equal to 900 gives \(\theta=900/\ln10=390.865\) thousand dollars.
Model E TVaR. The conditional mean above the threshold is \(q+\theta\), so \(TVaR_{0.90}=900+390.865=1{,}290.865\) thousand dollars. This exceeds the uniform model's TVaR even though the two VaRs agree.
Check and interpretation. The uniform tail is bounded above by 1,000, while the exponential tail has no finite upper bound. That difference in upper-tail shape explains why the exponential model has the larger TVaR. Rounding only at the final step to two decimals gives 950.00 and 1,290.87 thousand dollars.
Common mistake avoided: The uniform model's TVaR is not its upper bound of 1,000. TVaR averages the losses in the tail; for a continuous uniform distribution over \([900,1{,}000]\), that average is 950.
Worked Example: Normal and Exponential Models Matched at the 99th Percentile
A hypothetical severity study considers two continuous annual loss models in thousands of dollars. Model N is normal with mean \(\mu=100\) and standard deviation \(\sigma=100\). Model E is exponential with scale \(\theta\), chosen to match Model N's 99% VaR. Calculate both 99% VaRs and TVaRs, using \(z_{0.99}=2.32635\) and standard normal density \(\phi(2.32635)=0.026652\).
Method. Calculate the normal quantile first, then calibrate the exponential scale to that amount. For the normal tail, use the standard normal conditional-mean formula; for the exponential, use the mean excess.
Model N VaR. With \(X\sim N(\mu,\sigma^2)\), the 99% quantile is \(\mu+z_{0.99}\sigma\). Therefore \(VaR_{0.99}=100+(2.32635)(100)=332.635\) thousand dollars.
Model N TVaR. For a standard normal variable \(Z\), the identity \(E[Z\mid Z>z]=\phi(z)/(1-p)\) holds when \(z\) is its \(p\)-quantile. Rescaling and shifting gives \(TVaR_p(X)=\mu+\sigma\phi(z_p)/(1-p)\). Here the conditional standardized tail mean is \(0.026652/0.01=2.6652\). Thus \(TVaR_{0.99}=100+(100)(2.6652)=366.520\) thousand dollars.
Model E calibration and TVaR. For an exponential loss, \(VaR_{0.99}=\theta\ln100\). Hence \(\theta=332.635/\ln100=72.231\) thousand dollars. Its TVaR is \(VaR_{0.99}+\theta=332.635+72.231=404.866\) thousand dollars.
Check and interpretation. Both models have \(VaR_{0.99}=332.635\) thousand dollars, while the exponential TVaR is larger: 404.87 versus 366.52 thousand dollars. The comparison concerns the particular 99% tail, not every possible percentile or all features of the distributions. Rounding only at the final step to two decimals gives the reported TVaRs.
Common mistake avoided: Adding \(z_{0.99}\sigma\) to the normal mean gives the VaR, not the TVaR. The TVaR calculation uses the conditional tail mean, \(\mu+\sigma\phi(z)/(1-p)\).
How to Read a Common-VaR Comparison
The examples illustrate a disciplined comparison. Fix the confidence level \(p\) and the common threshold \(q\); calibrate each distribution so that \(VaR_p=q\); then compute each TVaR using a formula appropriate to that distribution. This separates the effect of tail shape from the effect of using different thresholds.
| Question | What it tells you |
|---|---|
| Do the models have the same \(VaR_p\)? | Whether they place the selected percentile at the same loss threshold. |
| Do they have the same \(TVaR_p\)? | Whether their average losses beyond that threshold also agree. |
| Is one TVaR larger? | Which model gives the larger average loss in this specified tail, not necessarily which is larger at every loss level. |
A single matched quantile does not prove that one distribution has a heavier tail everywhere. It establishes only that the distributions meet at the selected percentile. Their behavior farther into the tail depends on their complete model forms and parameters. Accordingly, report the confidence level, common VaR, distribution assumptions, parameter values, and TVaRs together.
Common comparison trap: A calculation such as \(E[X\mid X>VaR_{0.99}]\) for one model compared with \(E[Y\mid Y>VaR_{0.95}]\) for another does not compare the same percentile tail. Even if the numerical thresholds happen to be equal, the exceedance probabilities differ. State both the confidence level and threshold before comparing results.
Check Your Understanding
For each hypothetical continuous loss model, calibrate or identify the common percentile before calculating or interpreting the tail average. Amounts are in thousands of dollars unless stated otherwise.
- An exponential loss has scale \(\theta\) and \(VaR_{0.95}=600\). Find \(\theta\) and \(TVaR_{0.95}\). (Answer: \(\theta=200.28\); \(TVaR_{0.95}=800.28\).)
- A uniform loss is distributed on \([0,800]\). Find its 87.5% VaR and TVaR. (Answer: \(VaR_{0.875}=700\); \(TVaR_{0.875}=750\).)
- A Pareto Type I loss has \(x_m=200\) and \(\alpha=2\). Find its 90% VaR and TVaR. (Answer: \(VaR_{0.90}=200\sqrt{10}=632.46\); \(TVaR_{0.90}=400\sqrt{10}=1{,}264.91\).)
- A normal loss has \(\mu=50\), \(\sigma=40\), \(z_{0.95}=1.64485\), and \(\phi(z_{0.95})=0.10314\). An exponential loss is calibrated to have the same 95% VaR. Calculate both TVaRs, using \(\ln20=2.99573\). (Answer: common \(VaR_{0.95}=115.79\); normal \(TVaR_{0.95}=132.51\); exponential \(TVaR_{0.95}=154.45\).)