From an Exceedance Curve to TVaR
A loss exceedance-probability curve answers the question, “How likely is the loss to exceed this amount?” TVaR asks a different but related question: “How large is the loss on average in the worst specified fraction of outcomes?” The curve provides both pieces needed to answer it: the VaR threshold and the area representing expected loss in excess of that threshold.
This calculation is useful when a loss model is supplied graphically or as a table of exceedance probabilities rather than as a complete probability distribution. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. The key is to integrate the curve over loss amounts above the VaR threshold, then scale that area by the tail probability.
For a continuous loss distribution, if \(q=\operatorname{VaR}_p(X)\), the exceedance probability at \(q\) is \(1-p\). The area under the exceedance curve from \(q\) upward is the expected excess loss \(E[(X-q)^+]\), where \((z)^+=\max(z,0)\). Therefore,
The threshold \(q\) is a loss amount. The integral is also expressed in loss units, because it is probability multiplied by a range of loss amounts. Dividing the integral by the tail probability \(1-p\) gives the average excess above \(q\). Adding \(q\) converts that average excess into the average loss itself.
Why the Area Formula Holds
For any outcome with loss \(X\), the amount by which it exceeds \(q\) can be written as the length of the interval from \(q\) to \(X\), if \(X>q\), and as zero otherwise. Equivalently, it is the accumulated indicator of exceeding each level \(x\) above \(q\). Taking expectations turns each indicator into an exceedance probability:
For a continuous distribution, \(P(X>q)=1-p\), so conditioning on \(X>q\) gives
This establishes the formula for a continuous loss with a finite mean. The quantile-integral definition also gives the area formula when there is probability mass at the threshold, provided TVaR is interpreted using the fractional-quantile convention. In that case, simply calling TVaR \(E[X\mid X>q]\) can be wrong: the strict exceedance event may contain less than the required upper-tail probability. The quantile-integral definition avoids that ambiguity.
A Practical Calculation Sequence
A reliable calculation separates finding the threshold from measuring the area. First identify the confidence level and corresponding exceedance probability; then locate the loss amount at that probability. Only the area to the right of this threshold belongs in the excess-loss integral.
For confidence level \(p\), the upper-tail probability is \(1-p\).
Locate \(q\) where \(S_X(q)=1-p\), or use the specified quantile if the curve has a jump or the threshold is given directly.
For each straight-line segment, multiply its loss-width by the average of its endpoint exceedance probabilities. Add the segment areas.
Divide the total area by \(1-p\), then add \(q\). Keep the loss units unchanged throughout.
Worked Calculations
Worked Example: A Three-Segment Tail Curve
A hypothetical insurer models its one-year aggregate loss \(X\) in thousands of dollars. Assume the loss is nonnegative, has a continuous distribution with finite mean, and the listed exceedance curve is exact and piecewise linear above the threshold. The curve passes through \((100,0.05)\), \((200,0.03)\), \((400,0.01)\), and \((600,0)\), where each pair is (loss in thousands, exceedance probability). Find \(\operatorname{TVaR}_{0.95}(X)\).
Method and threshold. At \(p=0.95\), the tail probability is \(1-p=0.05\). The curve has exceedance probability \(0.05\) at \(q=100\), so \(\operatorname{VaR}_{0.95}(X)=100\) thousand dollars.
Segment areas. For a straight-line segment, the area is its width times the average of its endpoint probabilities:
From 100 to 200, the area is \(100(0.05+0.03)/2=100(0.04)=4\) thousand dollars. From 200 to 400, it is \(200(0.03+0.01)/2=200(0.02)=4\) thousand dollars. From 400 to 600, it is \(200(0.01+0)/2=200(0.005)=1\) thousand dollars. The total area is \(4+4+1=9\) thousand dollars.
Scale and add. The mean excess above the threshold is \(9/0.05=180\) thousand dollars. Thus \(\operatorname{TVaR}_{0.95}(X)=100+180=280\) thousand dollars, or $280,000.00.
Check. The answer exceeds the VaR threshold, as an upper-tail average should. The method avoids the common mistake of treating the area, 9 thousand dollars, as TVaR itself.
Worked Example: A Different Tail Probability and Loss Scale
A hypothetical company models a one-year loss \(X\) in thousands of dollars. Assume a nonnegative continuous loss with finite mean and a piecewise-linear exceedance curve above the stated threshold. The curve passes through \((50,0.10)\), \((100,0.06)\), \((150,0.04)\), \((250,0.01)\), and \((350,0)\). Calculate \(\operatorname{TVaR}_{0.90}(X)\).
Method and threshold. The upper-tail probability is \(1-0.90=0.10\). The exceedance probability is \(0.10\) at 50, so \(q=\operatorname{VaR}_{0.90}(X)=50\) thousand dollars.
Calculate each trapezoid. From 50 to 100, the area is \(50(0.10+0.06)/2=50(0.08)=4\). From 100 to 150, it is \(50(0.06+0.04)/2=50(0.05)=2.5\). From 150 to 250, it is \(100(0.04+0.01)/2=100(0.025)=2.5\). From 250 to 350, it is \(100(0.01+0)/2=100(0.005)=0.5\). The total excess-loss area is \(4+2.5+2.5+0.5=9.5\) thousand dollars.
Obtain TVaR. The average excess is \(9.5/0.10=95\) thousand dollars. Adding the threshold gives \(50+95=145\) thousand dollars, or $145,000.00.
Check. The threshold is the point where exceedance probability equals 10%, not 90%. Using 0.90 as the tail probability would incorrectly produce \(50+9.5/0.90\), because the confidence level and the upper-tail probability are complements.
Worked Example: A High-Confidence Threshold Within the Curve
For a hypothetical portfolio, let \(X\) be a nonnegative one-year loss in thousands of dollars. Assume a continuous distribution with finite mean. Its exact piecewise-linear exceedance curve above the threshold passes through \((150,0.025)\), \((200,0.018)\), \((300,0.006)\), and \((500,0)\). Find \(\operatorname{TVaR}_{0.975}(X)\).
Method and threshold. The tail probability is \(1-0.975=0.025\). The curve reaches that probability at \(q=150\), so the requested VaR is 150 thousand dollars.
Integrate the upper tail. From 150 to 200, the area is \(50(0.025+0.018)/2=50(0.0215)=1.075\). From 200 to 300, it is \(100(0.018+0.006)/2=100(0.012)=1.2\). From 300 to 500, it is \(200(0.006+0)/2=200(0.003)=0.6\). The total area is \(1.075+1.2+0.6=2.875\) thousand dollars.
Scale and add. The average excess is \(2.875/0.025=115\) thousand dollars. Therefore, TVaR is \(150+115=265\) thousand dollars, or $265,000.00.
Check. The 2.5% tail is relatively small, so dividing by 0.025 makes the average excess much larger than the unscaled area. This does not mean the calculation is unstable by definition; it means that the area must be correctly measured and divided by the specified tail probability.
Reading the Curve Without Losing the Units
The curve’s horizontal axis is measured in loss amounts and its vertical axis in probabilities. A trapezoid’s area is therefore a loss amount multiplied by a probability, which has the same units as a loss. Dividing by a probability leaves loss units, as required for the excess component of TVaR.
| Quantity | What it represents | Units in a curve measured in thousands |
|---|---|---|
| \(q=\operatorname{VaR}_p(X)\) | Loss threshold at the selected confidence level | Thousands of dollars |
| \(\int_q^\infty S_X(x)\,dx\) | Expected loss in excess of \(q\) | Thousands of dollars |
| \(\frac{1}{1-p}\int_q^\infty S_X(x)\,dx\) | Average excess in the upper tail | Thousands of dollars |
| \(\operatorname{TVaR}_p(X)\) | Threshold plus average excess | Thousands of dollars |
If a curve is supplied only as plotted points, the calculation requires an interpolation assumption. When the problem specifies straight-line interpolation, use trapezoids. If it does not specify how the curve behaves between observations, a numerical area is not uniquely determined by the points alone; state the interpolation assumption before calculating. Likewise, the upper-tail curve must be integrated far enough to capture its remaining area. Stopping while the exceedance probability is still positive omits expected excess loss.
Common Exam Traps
- Using \(p\) instead of \(1-p\). A 95% VaR has an upper-tail probability of 5%, not 95%. The threshold is found at the 5% exceedance probability.
- Reporting the area as TVaR. The area is expected excess loss, not the average upper-tail loss. Divide by \(1-p\), then add the threshold.
- Adding the threshold before scaling the area. The formula is \(q+\text{area}/(1-p)\), not \((q+\text{area})/(1-p)\). Only the excess area is normalized by the tail probability.
- Integrating below the threshold. The required area begins at \(q\). Area under the curve below \(q\) does not measure excess loss above VaR.
- Mixing dollar units. If the horizontal axis is in thousands of dollars, the segment widths, area, threshold, and final answer must remain in thousands until any final conversion.
- Assuming strict conditional expectation when there is an atom. If there is probability mass at VaR, use the quantile-integral or fractional-quantile definition rather than automatically replacing TVaR with \(E[X\mid X>q]\).
Check Your Understanding
For each question, assume the stated curve is exact and piecewise linear above the threshold, and round only if needed at the final step.
- A hypothetical loss is measured in thousands of dollars. At confidence level 80%, the curve passes through \((80,0.20)\), \((120,0.12)\), \((200,0.04)\), and \((240,0)\). Calculate TVaR. (Answer: 148 thousand dollars.)
- A hypothetical loss \(X\) has a continuous distribution and is measured in thousands. For confidence level 99%, the curve passes through \((400,0.01)\), \((600,0.006)\), \((900,0.002)\), and \((1100,0)\). Calculate \(\operatorname{TVaR}_{0.99}(X)\). (Answer: 700 thousand dollars.)
- A hypothetical portfolio has a 95% VaR of $2 million. Its exceedance curve is piecewise linear from \((2,0.05)\) to \((3,0.03)\), then \((5,0.01)\), then \((7,0)\), where losses on the horizontal axis are in millions of dollars. Calculate TVaR in millions. (Answer: 3.8 million dollars.)
- A hypothetical loss is measured in thousands of dollars. The curve has points \((100,0.05)\), \((160,0.035)\), \((260,0.015)\), and \((360,0)\). First identify the confidence level corresponding to the threshold of 100, then calculate TVaR at that level. (Answer: 216 thousand dollars at the 95% confidence level.)