Comparing Tail Measures Across Confidence Levels
VaR and TVaR answer related but different questions about a loss distribution. VaR identifies a loss threshold at a selected confidence level; TVaR averages the losses in the upper tail beyond that threshold. Comparing both measures at several confidence levels shows how the indicated threshold and the average tail severity change as attention shifts to more extreme outcomes.
This comparison matters in risk management because a single VaR value does not describe the full severity of losses beyond its quantile. TVaR adds information about those losses, but its value also depends on the confidence level. Neither measure should be interpreted without specifying that level and the loss model.
The tail gap, \(\operatorname{TVaR}_p(X)-\operatorname{VaR}_p(X)\), describes how far the average loss in the upper tail lies above the VaR threshold. It is nonnegative, but it does not have to increase as \(p\) increases. VaR may jump between loss outcomes, while TVaR reflects an average over a tail whose probability shrinks as confidence rises.
Calculating a Discrete Upper-Tail Average
For a discrete loss, let \(x_p=\operatorname{VaR}_p(X)\). Write \(F_X(x_p-)=P(X<x_p)\) for the probability strictly below the VaR value. The worst \(1-p\) probability of losses contains all outcomes above \(x_p\), plus enough probability at \(x_p\) to make the total exactly \(1-p\). Thus the amount of probability at \(x_p\) included in the tail is \(F_X(x_p)-p\).
This formula is useful when the VaR value has positive probability. It does not mean that TVaR is always the ordinary conditional expectation \(E[X\mid X>x_p]\): that conditional expectation omits the fractional mass at the VaR boundary. For a continuous loss with no probability mass at the threshold, the upper-tail conditional mean and TVaR agree under the usual conditions.
Worked Example: Comparing Several Levels for One Loss Model
Worked Example: Comparing Several Levels for One Loss Model
A hypothetical insurer models one-year aggregate catastrophe losses, in thousands of dollars, as \(X=0,100,300,700,\) or \(1{,}500\), with respective probabilities \(0.50,0.25,0.15,0.08,\) and \(0.02\). Calculate VaR, TVaR, and the tail gap at confidence levels \(80\%,90\%,95\%,\) and \(99\%\). This is a one-year probability model; interest and mortality assumptions are not applicable.
Method. First form the cumulative probabilities, then locate each quantile. For TVaR, average exactly the upper-tail probability \(1-p\), taking only the needed fraction of an outcome's probability mass at the boundary.
| Loss (thousands) | Probability | Cumulative probability |
|---|---|---|
| 0 | 0.50 | 0.50 |
| 100 | 0.25 | 0.75 |
| 300 | 0.15 | 0.90 |
| 700 | 0.08 | 0.98 |
| 1,500 | 0.02 | 1.00 |
At \(p=0.80\), the first cumulative probability at least \(0.80\) is \(0.90\), so VaR is \(300\). The top \(20\%\) consists of all the \(700\) and \(1{,}500\) outcomes, with total probability \(0.10\), and \(0.10\) probability from the \(300\) outcome. Its tail-loss total is \(0.08(700)+0.02(1{,}500)+0.10(300)=56+30+30=116\). Therefore TVaR is \(116/0.20=580\), and the gap is \(580-300=280\) thousand dollars.
At \(p=0.90\), VaR is still \(300\), since the cumulative probability at \(300\) is exactly \(0.90\). The top \(10\%\) contains all losses of \(700\) and \(1{,}500\). Its tail-loss total is \(0.08(700)+0.02(1{,}500)=56+30=86\). TVaR is \(86/0.10=860\), and the gap is \(860-300=560\) thousand dollars.
At \(p=0.95\), VaR is \(700\), because the cumulative probability reaches \(0.95\) within the probability mass at \(700\). The top \(5\%\) contains all \(0.02\) probability at \(1{,}500\), plus \(0.03\) probability at \(700\). The tail-loss total is \(0.02(1{,}500)+0.03(700)=30+21=51\). TVaR is \(51/0.05=1{,}020\), and the gap is \(1{,}020-700=320\) thousand dollars.
At \(p=0.99\), VaR is \(1{,}500\). The top \(1\%\) is a fraction of the \(2\%\) probability mass at that same loss, so TVaR is also \(1{,}500\). The gap is zero.
Check. The VaR values are nondecreasing across the four levels, and each TVaR is at least its corresponding VaR. The gap grows from \(280\) to \(560\) thousand dollars, then falls to zero; it is not required to move in one direction. The common mistake avoided here is including the entire \(700\) mass at \(95\%\), even though the tail requires only \(0.03\) of its \(0.08\) probability.
What Changes as the Confidence Level Rises?
For any fixed loss distribution, \(\operatorname{VaR}_p(X)\) is nondecreasing in \(p\): a higher quantile cannot be below a lower quantile. TVaR is also nondecreasing in \(p\). Its definition averages the quantile function from \(p\) to \(1\); increasing \(p\) removes the lower part of that upper-quantile range, leaving an average that cannot be smaller.
For discrete losses, VaR is a step function of the confidence level. It can stay constant over an interval and then jump when the cumulative probability crosses an outcome. TVaR accounts for the probability mass at the boundary fractionally. Consequently it can change even while VaR stays fixed, and the tail gap can widen or narrow.
Worked Example: A Higher Confidence Level Can Change Both Measures
A hypothetical one-year equipment-loss model has outcomes \(20,80,200,\) and \(600\) thousand dollars with probabilities \(0.60,0.25,0.10,\) and \(0.05\), respectively. Calculate VaR and TVaR at \(80\%,90\%,\) and \(95\%\), and compare the tail gaps. The stated probabilities define the full loss distribution; interest and mortality assumptions are not applicable.
At \(80\%\). Cumulative probability is \(0.60\) at \(20\) and \(0.85\) at \(80\), so VaR is \(80\). The top \(20\%\) includes \(0.05\) probability at \(80\), all \(0.10\) at \(200\), and all \(0.05\) at \(600\). The tail-loss total is \(0.05(80)+0.10(200)+0.05(600)=4+20+30=54\). Thus TVaR is \(54/0.20=270\), and the gap is \(190\) thousand dollars.
At \(90\%\). Cumulative probability first reaches \(0.90\) at \(200\), so VaR is \(200\). The top \(10\%\) consists of \(0.05\) probability at \(200\) and \(0.05\) at \(600\). Its tail-loss total is \(0.05(200)+0.05(600)=10+30=40\). TVaR is \(40/0.10=400\), and the gap is \(200\) thousand dollars.
At \(95\%\). VaR is \(600\). The top \(5\%\) is exactly the probability mass at \(600\), so TVaR is \(600\) and the gap is zero.
Check. Both measures rise or stay level as confidence increases: VaR is \(80,200,600\), and TVaR is \(270,400,600\), all in thousands. The gap is \(190,200,0\); the zero at \(95\%\) occurs because the selected tail consists entirely of one loss amount.
Same VaR, Different Tail Severity
A useful comparison holds the confidence level fixed and asks whether two risks with the same VaR also have similar TVaR. They need not. If a small probability of extremely large losses lies above the VaR threshold, those outcomes may have a substantial effect on TVaR while leaving VaR unchanged.
Worked Example: Equal VaR Does Not Mean Equal TVaR
Two hypothetical one-year portfolios have the following losses, in thousands of dollars. Portfolio A loses \(0\) with probability \(0.90\), \(100\) with probability \(0.08\), and \(200\) with probability \(0.02\). Portfolio B has the same probabilities at \(0\) and \(100\), but its final outcome is \(1{,}000\) rather than \(200\). Compare VaR and TVaR at \(90\%\) and \(95\%\). Interest and mortality assumptions are not applicable.
At \(90\%\). For both portfolios, cumulative probability at zero is exactly \(0.90\), so VaR is zero. The upper \(10\%\) contains \(0.08\) probability at \(100\) and \(0.02\) at the final outcome. For A, TVaR is \((0.08(100)+0.02(200))/0.10=(8+4)/0.10=120\). For B, TVaR is \((0.08(100)+0.02(1{,}000))/0.10=(8+20)/0.10=280\).
At \(95\%\). For both portfolios, VaR is \(100\), because cumulative probability reaches \(0.98\) at that outcome. The upper \(5\%\) contains \(0.03\) probability at \(100\) and \(0.02\) at the final outcome. For A, TVaR is \((0.03(100)+0.02(200))/0.05=(3+4)/0.05=140\). For B, TVaR is \((0.03(100)+0.02(1{,}000))/0.05=(3+20)/0.05=460\).
Check. The portfolios have identical VaR at each level, but Portfolio B has much higher TVaR because its extreme outcome is more severe. The comparison shows why VaR alone cannot distinguish the severity of losses beyond its threshold.
Common Exam Trap: Using the Wrong Tail
A frequent error is to calculate \(E[X\mid X>\operatorname{VaR}_p(X)]\) for a discrete loss and call it TVaR. That expression conditions on exceeding VaR strictly and discards any part of the probability mass at the VaR value. In a discrete model, the upper-tail slice may require only a fraction of that mass.
For the first example at \(95\%\), VaR is \(700\). The top \(5\%\) comprises \(3\%\) probability at \(700\) and \(2\%\) at \(1{,}500\). The correct calculation is \((0.03(700)+0.02(1{,}500))/0.05=1{,}020\) thousand dollars. A strict conditional expectation \(E[X\mid X>700]\) would be \(1{,}500\), because it uses only the outcome above \(700\); that is not the required upper \(5\%\) average. The correction is to include exactly enough probability at the VaR boundary to make the tail probability \(1-p\).
Check Your Understanding
For each hypothetical one-year loss model, calculate the requested measures in the stated units. Assume no interest or mortality contingencies apply.
- A loss takes values \(0,50,250,\) and \(900\) thousand dollars with probabilities \(0.70,0.20,0.08,\) and \(0.02\). Find VaR and TVaR at \(90\%\). (Answer: VaR \(=50\), TVaR \(=380\) thousand dollars.)
- A loss takes values \(10,40,120,\) and \(500\) thousand dollars with probabilities \(0.40,0.35,0.20,\) and \(0.05\). Find VaR and TVaR at \(75\%\). (Answer: VaR \(=40\), TVaR \(=196\) thousand dollars.)
- A loss is \(0\) with probability \(0.80\), \(100\) with probability \(0.15\), and \(400\) with probability \(0.05\). Find VaR and TVaR at both \(90\%\) and \(95\%\). (Answer: At \(90\%\), VaR \(=100\), TVaR \(=250\); at \(95\%\), VaR \(=100\), TVaR \(=400\), all in thousands.)
- A loss is \(0\) with probability \(0.92\), \(300\) with probability \(0.06\), and \(1{,}200\) with probability \(0.02\). Find VaR, TVaR, and the tail gap at \(95\%\). (Answer: VaR \(=300\), TVaR \(=660\), gap \(=360\) thousand dollars.)
- Portfolio A has losses \(0,200,700\) thousand dollars with probabilities \(0.85,0.10,0.05\). Portfolio B has losses \(0,200,2{,}000\) thousand dollars with the same probabilities. Compare VaR and TVaR at \(90\%\). (Answer: Both have VaR \(=200\); TVaR is \(450\) for A and \(1{,}100\) for B, in thousands.)