Tail Value-at-Risk for Continuous Losses

Why Average the Losses Beyond a Threshold?

A percentile threshold identifies a point in a loss distribution, but it does not describe how large losses tend to be after that point is crossed. Two portfolios can have the same value-at-risk at a chosen confidence level and very different average losses in the most adverse part of their distributions. Tail value-at-risk (TVaR) addresses this by averaging losses conditional on exceeding the value-at-risk threshold.

This tutorial treats continuous loss distributions with finite means. For such distributions, there is no probability mass at the percentile threshold, so the probability of exceeding it is exactly the upper-tail probability. This conditional-mean interpretation is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a continuous loss with cumulative distribution function $F_X$, survival function $S_X(x)=1-F_X(x)$, and finite mean. At confidence level $p$, where $0<p<1$, define the value-at-risk threshold $q_p$ by $F_X(q_p)=p$. The tail value-at-risk is $$ \operatorname{TVaR}_p(X)=E[X\mid X>q_p]. $$ For a continuous distribution, $P(X>q_p)=1-p$, so the conditional expectation averages the losses in the upper $1-p$ fraction of the distribution.

The subscript $p$ refers to the confidence level, not the proportion of observations retained in the tail. For example, at $p=0.95$, the threshold is the 95th percentile and the conditioning event has probability $0.05$. The losses being averaged are therefore the highest 5% of losses, not the highest 95%.

Conditional-Expectation Formula

For a continuous loss with density $f_X$, the conditional density given that $X>q_p$ is $f_X(x)/(1-p)$ for $x>q_p$. Integrating $x$ against that conditional density gives the calculation formula.

\operatorname{TVaR}_p(X) = \frac{1}{1-p}\int_{q_p}^{\infty} x f_X(x)\,dx, \qquad F_X(q_p)=p.

This formula follows directly from the definition of conditional expectation. The numerator is the expected loss contributed by outcomes in the tail, including zero contribution from outcomes outside it:

E[X\mid X>q_p] =\frac{E[X\,\mathbf{1}_{\{X>q_p\}}]}{P(X>q_p)} =\frac{\int_{q_p}^{\infty}x f_X(x)\,dx}{1-p}.

The denominator is $1-p$ because $F_X(q_p)=p$ and the distribution is continuous. Both parts of the calculation matter: first find the correct percentile threshold, then divide the expected tail loss by the probability of being in that tail. The formula requires a finite conditional mean in the tail; the finite-mean assumption on $X$ is sufficient when $X$ is a nonnegative loss.

Find the percentile threshold.
Solve $F_X(q_p)=p$, or equivalently $S_X(q_p)=1-p$, using the specified confidence level.

Calculate the tail probability.
For a continuous loss, verify that $P(X>q_p)=1-p$.

Calculate the tail first moment.
Evaluate $\int_{q_p}^{\infty}x f_X(x)\,dx$, or use an equivalent conditional-expectation calculation.

Normalize and check.
Divide by $1-p$. For an ordinary nonnegative loss distribution, the resulting tail mean should be at least $q_p$.

Worked Examples

Worked Example: Uniform Loss Above a High Percentile

A hypothetical equipment policy has loss $X\sim\operatorname{Uniform}(0,10{,}000)$, with amounts measured in dollars. Calculate $\operatorname{TVaR}_{0.80}(X)$. There is no interest or discounting, and no mortality assumption applies; the model is the stated continuous loss distribution.

Method. Find the 80th percentile and then take the mean conditional on exceeding it. For $0\leq x\leq10{,}000$, the distribution function is $F_X(x)=x/10{,}000$, and the density is $f_X(x)=1/10{,}000$.

Step 1: Find value-at-risk.

\frac{q_{0.80}}{10{,}000}=0.80 \quad\Longrightarrow\quad q_{0.80}=8{,}000.

Step 2: Confirm the tail probability. The probability above 8,000 is $(10{,}000-8{,}000)/10{,}000=0.20$, equal to $1-p$.

Step 3: Calculate the tail first moment and divide by the tail probability.

\operatorname{TVaR}_{0.80}(X) =\frac{\int_{8{,}000}^{10{,}000}x(1/10{,}000)\,dx}{0.20} =\frac{(10{,}000^2-8{,}000^2)/20{,}000}{0.20} =\frac{1{,}800}{0.20} =9{,}000.

The tail mean is $9,000.00, which exceeds the threshold of $8,000.00 and is the midpoint of the uniform interval from $8,000 to $10,000. This check is consistent with the conditional distribution in the tail.

Common mistake avoided: Dividing the tail first moment by $p=0.80$ would give $1{,}800/0.80=2{,}250$. That is not a conditional mean for the upper tail: the conditioning event has probability $1-p=0.20$, so the correct result is $1{,}800/0.20=9{,}000$.

Worked Example: Gamma Loss with a Specified Tail Threshold

A hypothetical repair portfolio has aggregate loss $X$ with a gamma distribution of shape $2$ and scale $\theta=\$1{,}000$. Its density is $f_X(x)=x e^{-x/1000}/1000^2$ for $x\geq0$. Calculate tail value-at-risk at the confidence level whose value-at-risk is $2,000. There is no interest or discounting and no mortality assumption; the loss model and its parameters are as stated.

Method. First calculate the confidence level from the survival probability at the stated threshold. Then integrate the tail first moment and divide by that tail probability. With $z=x/1000$, the gamma survival function for this shape and scale is $S_X(x)=e^{-z}(1+z)$.

Step 1: Determine the confidence level. At $q=2{,}000$, $z=2$, so the upper-tail probability is

P(X>2{,}000)=S_X(2{,}000)=e^{-2}(1+2)=3e^{-2}=0.4060\text{ (to four decimals)}.

Thus $p=1-3e^{-2}=0.5940$ to four decimals, and the threshold is the $p$-quantile.

Step 2: Calculate the expected loss contributed by the tail. Substituting $t=x/1000$ gives $x=1000t$, $dx=1000\,dt$, and $x f_X(x)\,dx=1000t^2e^{-t}\,dt$. Therefore,

\int_{2{,}000}^{\infty}x f_X(x)\,dx =1{,}000\int_2^\infty t^2e^{-t}\,dt =1{,}000e^{-2}(2^2+2(2)+2) =10{,}000e^{-2}.

The integral uses $\int_z^\infty t^2e^{-t}\,dt=e^{-z}(z^2+2z+2)$, obtained by integration by parts twice. At $z=2$, the polynomial factor is $4+4+2=10$.

Step 3: Divide by the upper-tail probability.

\operatorname{TVaR}_{p}(X) =\frac{10{,}000e^{-2}}{3e^{-2}} =\frac{10{,}000}{3} =\$3{,}333.33.

The result exceeds the $2,000 threshold, as a tail conditional mean should for this nonnegative continuous loss. The exponential factors cancel, providing a useful arithmetic check.

Common mistake avoided: The confidence level is approximately 0.5940, not 0.80 or 0.95 simply because the threshold was supplied. It must be inferred from the model’s survival probability at $2,000. The denominator is the resulting tail probability, $3e^{-2}$.

Worked Example: Scaled Beta Loss at the Median

A hypothetical service contract has loss $X=1{,}000Z$ dollars, where $Z\sim\operatorname{Beta}(2,2)$ has density $f_Z(z)=6z(1-z)$, for $0<z<1$. Calculate $\operatorname{TVaR}_{0.50}(X)$. No interest or discounting or mortality assumption is relevant; the loss distribution is the scaled beta model specified here.

Method. Find the median, then calculate the first moment above that threshold. Scaling $Z$ by $1,000 scales both the threshold and the conditional mean by $1,000.

Step 1: Find the threshold. The beta distribution function is $F_Z(z)=3z^2-2z^3$. At $z=0.5$,

$$ F_Z(0.5)=3(0.5)^2-2(0.5)^3=0.75-0.25=0.50. $$

Therefore $q_{0.50}=1{,}000(0.5)=\$500$, and the upper-tail probability is 0.50.

Step 2: Calculate the tail first moment.

E[X\mathbf{1}_{\{X>500\}}] =1{,}000\int_{0.5}^{1}z\,6z(1-z)\,dz =6{,}000\left[\frac{z^3}{3}-\frac{z^4}{4}\right]_{0.5}^{1}.

At $z=1$, the bracketed expression is $1/3-1/4=1/12$. At $z=0.5$, it is $1/24-1/64=5/192$. Their difference is $1/12-5/192=16/192-5/192=11/192$. Hence the tail first moment is $6{,}000(11/192)=\$343.75$.

Step 3: Divide by the tail probability.

\operatorname{TVaR}_{0.50}(X) =\frac{343.75}{0.50} =\$687.50.

The result is above the median threshold of $500, which is a useful check. The calculation also confirms that the numerator is a probability-weighted tail loss, not already the conditional mean.

Common mistake avoided: The integral from 0.5 to 1 is the expected loss multiplied by the probability of being in that tail. It must be divided by 0.50 to obtain the mean conditional on being in the tail.

Interpretation and Limits

TVaR is expressed in the same units as loss. If $X$ is measured in dollars, then both the threshold and its tail conditional mean are dollar amounts. The percentile threshold marks where the upper tail begins; TVaR describes the average severity within that tail. It is not the probability of a loss, nor is it the probability-weighted tail first moment before normalization.

The continuous-distribution condition is important. With a continuous distribution, $P(X=q_p)=0$, so $P(X>q_p)=1-p$. If the loss has a positive probability at the threshold, the strict-exceedance event can have probability different from $1-p$; simply applying the formula here may then give the wrong tail measure. Discrete losses require treatment of the probability at the threshold, a distinct issue for discrete distributions.

A practical check for a nonnegative continuous loss is $\operatorname{TVaR}_p(X)\geq q_p$, since every loss included in the conditional average exceeds $q_p$. If a calculation produces a smaller amount, recheck the percentile, the tail probability, and the integration bounds. This check is not a replacement for the calculation, but it often exposes a swapped confidence level or an incorrect denominator.

Key takeaway: For a continuous loss, find $q_p$ from $F_X(q_p)=p$, calculate the expected loss over $X>q_p$, and divide by $1-p$. The upper-tail probability—not the confidence level—is the normalizing denominator.

Check Your Understanding

For each problem, identify the threshold and upper-tail probability before calculating the conditional mean.

A hypothetical loss is uniform on the interval from $200 to $1,200. Find its value-at-risk and TVaR at confidence level 0.90. (Answer: VaR = $1,100; TVaR = $1,150)
Let $X\sim\operatorname{Uniform}(0,1)$. Calculate $\operatorname{TVaR}_{0.60}(X)$. (Answer: 0.8000)
A nonnegative loss has a gamma distribution with shape 2 and scale $500. Its threshold is $1,000. Find the corresponding confidence level and TVaR. (Answer: $p=1-3e^{-2}=0.5940$ to four decimals; TVaR = $1,666.67)
Let $X=2{,}000Z$ dollars, where $Z\sim\operatorname{Beta}(2,2)$. Calculate $\operatorname{TVaR}_{0.50}(X)$. (Answer: $1,375.00)
A hypothetical loss is $X=300+2Y$ dollars, where $Y\sim\operatorname{Uniform}(0,1{,}000)$. Find $\operatorname{TVaR}_{0.95}(X)$. (Answer: $2,250.00)

What You'll Learn