VaR and TVaR for Uniform and Exponential Losses

Why the Shape of the Loss Distribution Matters

VaR identifies a loss threshold at a chosen confidence level, while TVaR summarizes the losses beyond that threshold. The two measures depend on different features of a distribution: VaR is a quantile, and TVaR is an average over the upper tail. For uniform and exponential losses, both quantities have closed forms, but their formulas reveal very different tail behavior.

Throughout, let $p$ denote the confidence level, with $0<p<1$. For a loss $X$ with a continuous distribution, write $\operatorname{VaR}_p(X)$ for its $p$-quantile. In this tutorial, $\operatorname{TVaR}_p(X)$ is the average loss conditional on exceeding that quantile. Continuity ensures that there is no probability mass at the threshold, so this conditional-tail definition agrees with the quantile-integral definition.

Definition: For a continuous loss $X$, $\operatorname{VaR}_p(X)$ is the value $q_p$ satisfying $F_X(q_p)=p$. Its TVaR at confidence level $p$ is $\operatorname{TVaR}_p(X)=E[X\mid X>q_p]$, provided the conditional expectation exists. The event $X>q_p$ has probability $1-p$.

Uniform Losses

Suppose $X\sim \operatorname{Uniform}(a,b)$, where $a<b$. Its density is constant on $[a,b]$, and its cumulative distribution function increases linearly from zero at $a$ to one at $b$. The model is appropriate only when every loss in the interval is equally likely per unit of loss; it should not be used to imply that losses outside the specified interval are possible.

F_X(x)=\frac{x-a}{b-a}\quad (a\le x\le b), \qquad \operatorname{VaR}_p(X)=a+p(b-a).

The losses above the quantile are uniformly distributed between $q_p=a+p(b-a)$ and $b$. A uniform distribution on an interval has mean equal to the midpoint of that interval. Therefore,

\operatorname{TVaR}_p(X) =\frac{q_p+b}{2} =\frac{a+p(b-a)+b}{2}.

This formula is an average of the remaining upper interval, not the average of the original interval. As $p$ increases, the interval being averaged becomes shorter and closer to the upper endpoint $b$.

Worked Example: Uniform Loss With a Shifted Lower Bound

A hypothetical equipment policy has loss $X$, in thousands of dollars, with $X\sim\operatorname{Uniform}(12,72)$. Calculate VaR and TVaR at confidence level $p=0.80$. Give the answers in thousands of dollars to two decimal places.

Method and assumptions. The loss is continuous and uniform on the stated interval, so invert its linear cumulative distribution function to find VaR, then average the interval above that threshold to find TVaR.

The interval width is $72-12=60$ thousand dollars. Thus,

\operatorname{VaR}_{0.80}(X) =12+0.80(60) =12+48 =60

The upper-tail losses run from 60 to 72 thousand dollars. Their conditional distribution is uniform on that interval, so

\operatorname{TVaR}_{0.80}(X) =\frac{60+72}{2} =\frac{132}{2} =66

Therefore, the results are $\operatorname{VaR}_{0.80}(X)=\$60.00$ thousand and $\operatorname{TVaR}_{0.80}(X)=\$66.00$ thousand. As a check, the tail has width $72-60=12$ thousand dollars and its midpoint is 66 thousand dollars.

Common mistake avoided: The mean of the full uniform loss is $(12+72)/2=42$, but that is not TVaR. TVaR averages only losses above the 80th percentile.

Exponential Losses

Suppose $X$ has an exponential distribution with rate $\lambda>0$. Its support is nonnegative, its survival function is $S_X(x)=e^{-\lambda x}$ for $x\ge 0$, and its mean is $1/\lambda$. The rate must use the reciprocal of the same loss unit as $X$; for example, a rate per thousand dollars is used with losses measured in thousands of dollars.

F_X(x)=1-e^{-\lambda x}\quad (x\ge 0), \qquad \operatorname{VaR}_p(X) =-\frac{\ln(1-p)}{\lambda}.

For TVaR, the key exponential property is memorylessness: conditional on exceeding a threshold $q$, the excess $X-q$ has the same exponential distribution as $X$. In particular, its conditional mean is $1/\lambda$. Since $q=\operatorname{VaR}_p(X)$, the average loss in the upper tail is the threshold plus the average excess.

Theorem: If $X$ is exponential with rate $\lambda>0$, then for $0<p<1$, $\operatorname{TVaR}_p(X)=\operatorname{VaR}_p(X)+1/\lambda$. The formula applies to an unshifted exponential loss with finite rate; it is not the formula for every distribution with an exponential-looking tail.

Why the Exponential TVaR Formula Holds

The result follows directly from conditioning on the tail threshold. For $q\ge 0$ and $y\ge 0$, the exponential survival function gives

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

The final expression is the survival function of an exponential random variable with rate $\lambda$. Therefore $X-q$, conditional on $X>q$, has mean $1/\lambda$. Taking $q=\operatorname{VaR}_p(X)$ gives

E[X\mid X>q] =q+E[X-q\mid X>q] =q+\frac{1}{\lambda}.

The same result can be expressed entirely in terms of the confidence level:

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

Worked Example: Exponential Loss at a High Confidence Level

A hypothetical portfolio has an exponential loss $X$, measured in thousands of dollars, with rate $\lambda=0.05$ per thousand dollars. Find VaR and TVaR at $p=0.90$, first in thousands of dollars and then in dollars. Report money to the nearest cent.

Method and assumptions. Use the exponential quantile formula for VaR and add the mean $1/\lambda$ to obtain TVaR. The loss and rate units are consistent: $\lambda$ is per thousand dollars.

First calculate the mean:

E[X]=\frac{1}{0.05}=20 \quad\text{thousand dollars}.

Because $1-p=0.10$ and $-\ln(0.10)=\ln(10)\approx2.302585$,

\operatorname{VaR}_{0.90}(X) =\frac{-\ln(0.10)}{0.05} =\frac{2.302585}{0.05} \approx46.05170 \quad\text{thousand dollars}.

Add the mean loss of 20 thousand dollars for TVaR:

\operatorname{TVaR}_{0.90}(X) =46.05170+20 =66.05170 \quad\text{thousand dollars}.

Thus VaR is $\$46,051.70$ and TVaR is $\$66,051.70$. As a check, TVaR exceeds VaR by exactly the exponential mean, $\$20,000.00$.

Common mistake avoided: The rate is per thousand dollars, so $1/\lambda=20$ is in thousands of dollars, not dollars. Mixing these units would make the tail excess appear 1,000 times too small or too large.

Comparing the Two Tail Shapes

A comparison is most informative when the models have the same mean. Consider a uniform loss on $[0,40]$ thousand dollars and an exponential loss with rate $\lambda=0.05$ per thousand dollars. Both have mean 20 thousand dollars: the uniform mean is $(0+40)/2=20$, and the exponential mean is $1/0.05=20$. Despite that match, the exponential model has no finite upper endpoint, while the uniform loss cannot exceed 40 thousand dollars.

Worked Example: Comparing Upper-Tail Averages at Two Confidence Levels

Using the two hypothetical loss models just specified, compare VaR and TVaR at confidence levels $p=0.90$ and $p=0.99$. Report all amounts in thousands of dollars to two decimal places.

Method and assumptions. Apply the uniform formulas to $X_U\sim\operatorname{Uniform}(0,40)$, and the exponential formulas to $X_E$ with rate $0.05$. Both models are continuous, so conditional tail means apply.

At $p=0.90$, for the uniform loss,

\operatorname{VaR}_{0.90}(X_U)=0+0.90(40)=36, \qquad \operatorname{TVaR}_{0.90}(X_U)=\frac{36+40}{2}=38.

For the exponential loss, $-\ln(0.10)=2.302585$, and its mean is 20:

\operatorname{VaR}_{0.90}(X_E) =\frac{2.302585}{0.05} =46.05170, \qquad \operatorname{TVaR}_{0.90}(X_E) =46.05170+20 =66.05170.

At $p=0.99$, $-\ln(0.01)=4.605170$. The uniform results are

\operatorname{VaR}_{0.99}(X_U)=0+0.99(40)=39.6, \qquad \operatorname{TVaR}_{0.99}(X_U)=\frac{39.6+40}{2}=39.8.

The exponential results are

\operatorname{VaR}_{0.99}(X_E) =\frac{4.605170}{0.05} =92.10340, \qquad \operatorname{TVaR}_{0.99}(X_E) =92.10340+20 =112.10340.

The comparison, rounded to two decimal places, is:

Confidence level	Uniform VaR	Uniform TVaR	Exponential VaR	Exponential TVaR
90%	36.00	38.00	46.05	66.05
99%	39.60	39.80	92.10	112.10

All amounts are in thousands of dollars. The uniform VaR and TVaR approach the finite endpoint of 40 as confidence increases. The exponential VaR and TVaR continue to increase without a finite upper bound. The calculations also check internally: in the exponential model, TVaR exceeds VaR by 20 at both confidence levels; in the uniform model, TVaR is the midpoint between VaR and 40.

Interpretation and Common Traps

These formulas show why neither the mean nor VaR alone describes the severity of upper-tail losses. Two models can have the same mean and still produce substantially different high-confidence TVaR values. The uniform model imposes a hard maximum, whereas the exponential model assigns positive probability to losses above any finite threshold. The selected model and its parameters therefore matter when interpreting a tail average.

Using the wrong tail interval. For a uniform loss, average from $q_p$ to the upper endpoint $b$, not across the full interval $[a,b]$.
Dropping the exponential tail excess. Exponential TVaR is VaR plus $1/\lambda$, not VaR alone. The added mean is the expected amount by which a loss exceeds the threshold, conditional on exceeding it.
Confusing rate and mean. An exponential rate $\lambda$ corresponds to mean $1/\lambda$. A larger rate means a smaller mean and smaller VaR at a fixed confidence level.
Mixing loss units. If loss is in thousands of dollars, both the exponential rate and reported results must use compatible units.

Key takeaway: For $X\sim\operatorname{Uniform}(a,b)$, VaR is $a+p(b-a)$ and TVaR is the midpoint between that quantile and $b$. For an exponential loss with rate $\lambda$, VaR is $-\ln(1-p)/\lambda$ and TVaR is VaR plus $1/\lambda$.

Check Your Understanding

For each continuous model, calculate the requested quantities using the stated confidence level and units.

A hypothetical loss is uniform on $[5,25]$ dollars. Find VaR and TVaR at $p=0.75$. (Answer: VaR $=\$20.00$; TVaR $=\$22.50$.)
A hypothetical claim loss is uniform on $[0,120]$ thousand dollars. Find VaR and TVaR at $p=0.95$. (Answer: VaR $=114.00$ thousand dollars; TVaR $=117.00$ thousand dollars.)
A hypothetical loss is exponential with mean 8 thousand dollars. Find VaR and TVaR at $p=0.75$. (Answer: VaR $=11.0904$ thousand dollars; TVaR $=19.0904$ thousand dollars.)
A hypothetical loss is exponential with rate $0.04$ per thousand dollars. Find VaR and TVaR at $p=0.90$. (Answer: VaR $=57.5646$ thousand dollars; TVaR $=82.5646$ thousand dollars.)
Compare a uniform loss on $[0,50]$ thousand dollars with an exponential loss of mean 25 thousand dollars, both at $p=0.80$. Give VaR and TVaR for each model. (Answer: Uniform VaR $=40.00$ and TVaR $=45.00$ thousand dollars; exponential VaR $=40.24$ and TVaR $=65.24$ thousand dollars.)

What You'll Learn