TVaR For Exponential Losses

Why the Exponential Model Gives a Simple Tail Formula

Tail Value at Risk (TVaR) describes the average loss in the upper tail beyond a selected confidence level. For an exponential loss, this average has a particularly direct interpretation: among losses that exceed the VaR threshold, the expected amount above that threshold is the same as the unconditional mean loss. The memoryless property of the exponential distribution is what makes this result possible.

This tutorial considers one nonnegative loss $X$ with an exponential distribution. It derives the VaR and TVaR formulas, then uses them in hypothetical risk calculations. The model is a single-period loss model; no discounting, interest, or mortality assumptions are involved.

Definition: Suppose $X$ has an exponential distribution with rate parameter $\lambda>0$. Its survival function and mean are $$ S_X(x)=P(X>x)=e^{-\lambda x},\quad x\geq 0,\qquad E[X]=\frac{1}{\lambda}. $$ The corresponding distribution function is $F_X(x)=1-e^{-\lambda x}$. The parameter $\lambda$ has units inverse to the units of loss; the mean $1/\lambda$ has the same units as $X$.

For a confidence level $p$ with $0<p<1$, $\operatorname{VaR}_p(X)$ is the $p$-quantile of the loss distribution. For a continuous distribution such as the exponential distribution, $\operatorname{TVaR}_p(X)$ is equivalently the average of the loss conditional on exceeding that quantile. It is also the average of the upper quantiles from level $p$ to level 1.

Definition: For a continuous loss distribution, $$ \operatorname{TVaR}_p(X)=E[X\mid X>\operatorname{VaR}_p(X)] =\frac{1}{1-p}\int_p^1\operatorname{VaR}_q(X)\,dq. $$ These expressions agree here because an exponential loss is continuous and has no probability mass at the VaR threshold.

Deriving VaR and TVaR for an Exponential Loss

To find the VaR, set the distribution function equal to the desired confidence level and solve for the loss threshold. To find TVaR, condition on exceeding that threshold. The key step is the exponential memoryless property: once the loss has exceeded a threshold, the distribution of the additional loss does not depend on the size of the threshold.

Solve for the quantile.
Set $F_X(v)=p$, where $v=\operatorname{VaR}_p(X)$. Thus $1-e^{-\lambda v}=p$, so $v=-\ln(1-p)/\lambda$.

Find the conditional exceedance distribution.
For $y\geq 0$, $P(X-v>y\mid X>v)=P(X>v+y)/P(X>v)$. The exponential survival function reduces this ratio to $e^{-\lambda y}$.

Take the conditional mean.
The excess $X-v$, conditional on $X>v$, is exponential with rate $\lambda$, so its mean is $1/\lambda$. Therefore the conditional mean of $X$ is $v+1/\lambda$.

\operatorname{VaR}_p(X)=\frac{-\ln(1-p)}{\lambda}, \qquad \operatorname{TVaR}_p(X)=\frac{1-\ln(1-p)}{\lambda} =\operatorname{VaR}_p(X)+\frac{1}{\lambda}.

The justification in Step 2 follows directly from the conditional probability definition:

P(X-v>y\mid X>v) =\frac{P(X>v+y)}{P(X>v)} =\frac{e^{-\lambda(v+y)}}{e^{-\lambda v}} =e^{-\lambda y}.

This is the survival function of an exponential random variable with rate $\lambda$, proving that the excess loss $X-v$, given $X>v$, has mean $1/\lambda$. Adding the threshold $v$ gives the conditional mean of the full loss, not just the excess. This establishes the TVaR formula.

If the model is specified by its mean $\theta=E[X]$ instead of its rate, then $\theta=1/\lambda$, and the same formulas become $\operatorname{VaR}_p(X)=\theta[-\ln(1-p)]$ and $\operatorname{TVaR}_p(X)=\theta[1-\ln(1-p)]$. Keep the parameterization and units explicit: a rate is not a mean.

Applying the Formula

Worked Example: A High-Confidence Cyber Loss Estimate

A hypothetical insurer models a one-year cyber loss $X$, measured in thousands of dollars, as exponential with mean $40$ thousand dollars. The rate is therefore $\lambda=1/40=0.025$ per thousand dollars. Calculate the 95% VaR and TVaR. Assume the model applies to the full one-year loss with no discounting.

Method. Use the mean-parameter form of the exponential formulas, first finding the quantile and then adding the mean to obtain TVaR.

VaR. Here $p=0.95$, so $1-p=0.05$. Thus $$ \operatorname{VaR}_{0.95}(X) =40[-\ln(0.05)] =40(2.995732) =119.8293 $$ thousand dollars.

TVaR. The conditional mean loss above this threshold is the threshold plus the mean: $$ \operatorname{TVaR}_{0.95}(X) =119.8293+40 =159.8293 $$ thousand dollars.

Rounded to two decimals in thousands of dollars, the 95% VaR is $119.83$ thousand dollars and the 95% TVaR is $159.83$ thousand dollars. Check the threshold: $P(X>119.8293)=e^{-0.025(119.8293)}=e^{-2.995732}=0.05$, as required. The method avoids treating the mean excess as the entire tail loss.

Worked Example: Converting a Rate to a Dollar TVaR

A hypothetical specialty insurer models an annual claim loss $X$ in dollars as exponential with rate $\lambda=0.00004$ per dollar. Find the 99% VaR and TVaR. Assume one annual loss, with no interest adjustment.

Method. Use the rate-parameter formulas directly. First note that the mean is $1/\lambda=1/0.00004=\$25{,}000$.

VaR. At $p=0.99$, $1-p=0.01$, so $$ \operatorname{VaR}_{0.99}(X) =\frac{-\ln(0.01)}{0.00004} =\frac{4.605170}{0.00004} =\$115{,}129.25. $$

TVaR. Add the mean loss, $\$25{,}000$, to the VaR: $$ \operatorname{TVaR}_{0.99}(X) =\$115{,}129.25+\$25{,}000 =\$140{,}129.25. $$

Rounded to the nearest cent, the 99% VaR is $\$115{,}129.25$ and the 99% TVaR is $\$140{,}129.25$. As a check, the survival probability at the VaR is $e^{-0.00004(115{,}129.25)}$, which is approximately $e^{-4.605170}=0.01$. The method avoids confusing the rate $0.00004$ per dollar with the mean of $\$25{,}000$.

Worked Example: Finding the Confidence Level From a Target TVaR

A hypothetical portfolio has a one-period loss $X$ in thousands of dollars, modeled as exponential with mean $10$ thousand dollars, so $\lambda=0.1$ per thousand dollars. At what confidence level is TVaR equal to $50$ thousand dollars? Also find the VaR at that level.

Method. Rearrange the mean-parameter TVaR formula to solve for the upper-tail probability, then use the resulting confidence level to calculate VaR.

Set the target equal to the formula: $$ 50=10[1-\ln(1-p)]. $$ Dividing by 10 gives $5=1-\ln(1-p)$, so $-\ln(1-p)=4$. Exponentiating yields $1-p=e^{-4}=0.0183156$, and therefore $p=1-e^{-4}=0.9816844$.

The VaR at this confidence level is $$ \operatorname{VaR}_{p}(X)=10[-\ln(1-p)]=10(4)=40 $$ thousand dollars. Adding the mean gives $40+10=50$ thousand dollars, the specified TVaR. Rounded to four decimal places, the confidence level is $0.9817$, or $98.17\%$, and the VaR is $40.00$ thousand dollars. The method avoids attempting to infer the confidence level from the TVaR without accounting for the $1/\lambda$ excess.

Interpretation and Common Exam Trap

For an exponential loss, the gap between TVaR and VaR is always the unconditional mean loss: $$ \operatorname{TVaR}_p(X)-\operatorname{VaR}_p(X)=\frac{1}{\lambda}=E[X]. $$

This gap is independent of the confidence level $p$. A higher confidence level raises VaR, and therefore raises TVaR by the same amount, while leaving the difference between them unchanged. This is a distinctive consequence of the exponential model’s memoryless tail, not a general property of loss distributions.

Common mistake: reporting only the expected excess. In the second worked example, $1/\lambda=\$25{,}000$. Reporting $\operatorname{TVaR}_{0.99}(X)=\$25{,}000$ is wrong: this is the mean amount above the VaR threshold, not the mean full loss conditional on exceeding it.
Correction. Add the threshold back: $\operatorname{TVaR}_{0.99}(X)=\$115{,}129.25+\$25{,}000=\$140{,}129.25$.
Check the units. If $\lambda$ is measured per dollar, $1/\lambda$ is in dollars. If losses are in thousands of dollars, use a rate per thousand dollars or express the mean in thousands consistently.
Check the tail probability. The VaR at level $p$ must satisfy $P(X>\operatorname{VaR}_p(X))=1-p$ for this continuous model.

These calculations are useful when a risk measure is specified in terms of an exponential loss model and a confidence level, or when a target tail measure must be translated back to a confidence level. Their interpretation depends on the assumed exponential distribution; they should not be applied automatically to a loss model with a different tail shape. The formulas and interpretations here are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, treat the stated loss as a single-period exponential loss with no discounting. Round dollar answers to the nearest cent and probabilities to four decimal places.

A hypothetical equipment loss $X$, measured in thousands of dollars, is exponential with mean $8$ thousand dollars. Calculate $\operatorname{VaR}_{0.75}(X)$ and $\operatorname{TVaR}_{0.75}(X)$. (Answer: VaR $=11.09$ thousand dollars; TVaR $=19.09$ thousand dollars.)
A hypothetical liability loss in dollars is exponential with rate $0.0002$ per dollar. Calculate its 95% VaR and TVaR. (Answer: VaR $=\$14{,}978.66$; TVaR $=\$19{,}978.66$.)
A hypothetical loss is exponential with mean $\$12{,}000$. At the 90% confidence level, calculate VaR and TVaR, and state the expected excess above VaR conditional on exceeding it. (Answer: VaR $=\$27{,}631.02$; TVaR $=\$39{,}631.02$; expected excess $=\$12{,}000$.)
A hypothetical loss $X$ is exponential with rate $0.5$ per loss unit. The confidence level is $p=1-e^{-3}$. Calculate VaR and TVaR in loss units, and give $p$ to four decimal places. (Answer: VaR $=6$; TVaR $=8$; $p=0.9502$.)
A hypothetical portfolio loss is exponential with mean $\$30{,}000$. Find the confidence level at which TVaR is $\$120{,}000$, and calculate VaR at that level. (Answer: $p=0.9502$; VaR $=\$90{,}000$.)

What You'll Learn