VaR for Continuous Loss Models

From a Probability Level to a Loss Threshold

Value at Risk (VaR) is a quantile of a loss distribution. For a continuous loss model, the calculation often reduces to finding the loss threshold whose cumulative probability equals a specified level. The key is to use the cumulative distribution, not the exceedance probability, when identifying that threshold.

Write $F_X(x)=P(X\leq x)$ for the cumulative distribution function of loss $X$, and $S_X(x)=P(X>x)=1-F_X(x)$ for its survival, or exceedance, function. For a confidence level $\alpha$ between 0 and 1, VaR is the smallest threshold at which the cumulative probability reaches $\alpha$. If $F_X$ is continuous and strictly increasing at the relevant quantile, this threshold is found by solving $F_X(x)=\alpha$.

Definition: For $0<\alpha<1$, $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}$. If $F_X$ is continuous and strictly increasing around the quantile, then $\operatorname{VaR}_{\alpha}(X)=F_X^{-1}(\alpha)$, where $F_X^{-1}$ is the inverse cumulative distribution function. Consequently, $P(X\leq \operatorname{VaR}_{\alpha}(X))=\alpha$ and $P(X>\operatorname{VaR}_{\alpha}(X))=1-\alpha$.

The inverse-CDF shortcut requires care. A continuous distribution need not be strictly increasing everywhere, and a model with point masses need not satisfy the equality $F_X(\operatorname{VaR}_{\alpha})=\alpha$. This tutorial uses uniform and exponential models, for which the CDF is continuous and strictly increasing within the interior of the support. In those settings, direct inversion is valid.

Uniform Loss Models

Suppose $X$ is uniformly distributed between $a$ and $b$, where $a<b$. Every interval of the same length within this support has the same probability. The density is constant, and the CDF increases linearly from 0 at $a$ to 1 at $b$.

X\sim\operatorname{Uniform}(a,b),\qquad F_X(x)= \begin{cases} 0, & x<a,\\ \dfrac{x-a}{b-a}, & a\leq x\leq b,\\ 1, & x>b. \end{cases}

To find VaR at level $\alpha$, solve $(x-a)/(b-a)=\alpha$. The resulting threshold is a fraction $\alpha$ of the way from the lower endpoint to the upper endpoint. It is not generally $\alpha b$, unless the lower endpoint happens to be zero.

\operatorname{VaR}_{\alpha}(X)=a+\alpha(b-a),\qquad 0<\alpha<1.

Worked Example: Uniform Loss and a Separate Threshold Probability

A hypothetical one-year loss $X$, measured in dollars, is modeled as uniform on $\$200{,}000$ to $\$800{,}000$. Calculate the 95% VaR, then find the probability that the loss exceeds $\$740{,}000$. State how the threshold compares with the VaR.

Method. Use the uniform CDF to find the quantile, then evaluate the CDF at the separate loss threshold. The model has $a=\$200{,}000$, $b=\$800{,}000$, and $b-a=\$600{,}000$.

95% VaR. The threshold is $\$200{,}000+0.95(\$600{,}000)=\$200{,}000+\$570{,}000=\$770{,}000$. Thus, $\operatorname{VaR}_{0.95}(X)=\$770{,}000$.

Probability above $\$740{,}000$. Since this threshold is inside the support, $F_X(\$740{,}000)=(\$740{,}000-\$200{,}000)/\$600{,}000=\$540{,}000/\$600{,}000=0.90$. Therefore, $P(X>\$740{,}000)=1-0.90=0.10$.

Check and trap avoided. At $\$770{,}000$, the CDF is $(\$770{,}000-\$200{,}000)/\$600{,}000=0.95$, as required. The $\$740{,}000$ threshold is $\$30{,}000$ below the 95% VaR, and 10% of modeled losses exceed it. A common error is to calculate the quantile as $0.95b=\$760{,}000$, ignoring the nonzero lower endpoint; that amount gives CDF $(\$760{,}000-\$200{,}000)/\$600{,}000=0.9333$, not 0.95.

Exponential Loss Models

An exponential model is often used for a nonnegative loss with a decreasing density. Use the scale parameter $\theta>0$, so that $\theta$ is the mean loss. The model has support $x\geq0$, density $f_X(x)=\theta^{-1}e^{-x/\theta}$, CDF $F_X(x)=1-e^{-x/\theta}$, and exceedance probability $S_X(x)=e^{-x/\theta}$. These formulas do not describe a loss model with a different support or distributional shape.

For a VaR level $\alpha$, set the CDF equal to $\alpha$, then isolate $x$. Equivalently, the probability of exceeding the quantile is $1-\alpha$, which can be used with the survival function.

1-e^{-x/\theta}=\alpha \quad\Longrightarrow\quad e^{-x/\theta}=1-\alpha \quad\Longrightarrow\quad \operatorname{VaR}_{\alpha}(X)=-\theta\ln(1-\alpha).

The logarithm is applied to the tail probability $1-\alpha$, not to $\alpha$. As the confidence level increases, the tail probability decreases, its logarithm becomes more negative, and the VaR threshold increases.

Worked Example: Exponential VaR at a High Confidence Level

For a hypothetical annual loss $X$, assume an exponential distribution with mean $\theta=\$3{,}000{,}000$. Calculate the 98% VaR in dollars, rounded to the nearest cent, and verify the result using the exceedance probability.

Method. The exponential parameter is the scale $\theta$, not the rate $1/\theta$. At $\alpha=0.98$, the tail probability is $1-\alpha=0.02$. Therefore, $\operatorname{VaR}_{0.98}(X)=-\$3{,}000{,}000\ln(0.02)=\$3{,}000{,}000\ln(50)$.

Using $\ln(50)=3.912023005428\ldots$, the unrounded result is $\$3{,}000{,}000(3.912023005428\ldots)=\$11{,}736{,}069.016284\ldots$. Rounded to the nearest cent, the 98% VaR is $\$11{,}736{,}069.02$.

Check. Before rounding, the quantile $q$ satisfies $q/\theta=\ln(50)$, so $e^{-q/\theta}=e^{-\ln(50)}=1/50=0.02$. Thus $F_X(q)=1-0.02=0.98$, confirming the quantile. The rounded dollar amount is the reported value; the exact probability identity uses the unrounded quantile.

Trap avoided. Substituting $0.98$ inside the logarithm would give $-\theta\ln(0.98)$, a small value corresponding to a 2% quantile, not a 98% quantile. The correct formula uses $\ln(1-\alpha)$.

Why Inversion Works

The quantile formula is not a separate rule to memorize for each distribution. It follows by translating the definition of VaR into an equation and solving it. For a continuous, strictly increasing CDF, the set of losses with cumulative probability at least $\alpha$ begins at the unique value $q$ satisfying $F_X(q)=\alpha$. Hence $q=F_X^{-1}(\alpha)$. The uniform and exponential formulas result from inserting their respective CDFs and rearranging.

Set the cumulative probability.
For the desired level $\alpha$, write $F_X(q)=\alpha$. The value $q$ is the candidate VaR threshold.

Solve for the threshold.
For a uniform model, solve $(q-a)/(b-a)=\alpha$. For an exponential model, solve $1-e^{-q/\theta}=\alpha$.

Check the probability.
Substitute $q$ back into the CDF and confirm $F_X(q)=\alpha$. Equivalently, check $S_X(q)=1-\alpha$.

This argument depends on the CDF increasing through $\alpha$, so that the solution is the smallest threshold reaching the requested cumulative probability. If a CDF has a flat interval or a jump, the generalized-quantile definition remains the safe definition; blindly solving an equality may fail or give the wrong threshold.

Worked Example: Finding a Percentile and a VaR from an Exponential Model

A hypothetical portfolio loss $X$ follows an exponential distribution with mean $\theta=\$400{,}000$. First find the percentile level associated with a loss threshold of $\$1{,}200{,}000$. Then calculate the 95% VaR, rounded to the nearest cent, and compare the two thresholds.

Method. To find the percentile of a stated threshold, evaluate the CDF there. To find the VaR at a stated percentile, invert the CDF. For the threshold, $x/\theta=\$1{,}200{,}000/\$400{,}000=3$, so $F_X(x)=1-e^{-3}$.

Since $e^{-3}=0.04978706837\ldots$, the CDF is $1-0.04978706837\ldots=0.9502129316\ldots$, or 0.9502 to four decimal places. Thus, a loss of $\$1{,}200{,}000$ is at approximately the 95.02nd percentile.

For the 95% VaR, the tail probability is $0.05$, so $q=-\$400{,}000\ln(0.05)=\$400{,}000\ln(20)$. Using $\ln(20)=2.995732273554\ldots$, $q=\$1{,}198{,}292.909421\ldots$, which rounds to $\$1{,}198{,}292.91$.

Check and comparison. The threshold $\$1{,}200{,}000$ is slightly above the 95% VaR because its CDF, 0.9502129, is slightly above 0.95. At the unrounded 95% VaR, $e^{-q/\theta}=e^{-\ln(20)}=0.05$, so the CDF is exactly 0.95. Rounding the VaR to cents does not change the stated percentile calculation.

Trap avoided. Do not round the percentile 0.9502129 to 0.95 and then conclude that the $\$1{,}200{,}000$ threshold is exactly the 95% quantile. The unrounded CDF shows that it is above, not equal to, the 95% VaR.

Common Tail-Probability Error

A confidence level $\alpha$ is the probability at or below the VaR threshold. The corresponding probability above it is $1-\alpha$. Confusing these probabilities reverses the quantile: it can produce a low threshold when a high percentile was requested.

Exam trap: For an exponential model, the correct equation is $e^{-q/\theta}=1-\alpha$, so $q=-\theta\ln(1-\alpha)$. Using $e^{-q/\theta}=\alpha$ instead gives the quantile at level $1-\alpha$, not at level $\alpha$.

For example, with $\theta=\$300{,}000$ and $\alpha=0.99$, the incorrect tail substitution gives $-\$300{,}000\ln(0.99)=\$3{,}015.10$. That is the 1% quantile because its CDF is $1-0.99=0.01$. The correct 99% VaR is $-\$300{,}000\ln(0.01)=\$300{,}000\ln(100)=\$1{,}381{,}551.06$, rounded to the nearest cent. Its exceedance probability is $e^{-q/\theta}=0.01$, as required.

The same discipline applies to uniform models: use $\alpha$ in the CDF equation, but use $1-\alpha$ if calculating the probability beyond the resulting threshold. A brief check by substitution often catches a reversed tail before it affects a reported capital amount. These inversion techniques are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each hypothetical model, calculate the requested quantity and use the CDF or exceedance probability to check your result. Dollar answers are rounded to the nearest cent where needed.

A loss is uniform on $\$100{,}000$ to $\$500{,}000$. Calculate its 90% VaR and the probability that it exceeds $\$420{,}000$. (Answer: VaR $=\$460{,}000$; exceedance probability $=0.2000$.)
A loss is exponential with mean $\theta=\$150{,}000$. Calculate its 95% VaR and verify the exceedance probability at the unrounded quantile. (Answer: VaR $=\$449{,}359.84$; exceedance probability $=0.0500$.)
A loss is uniform on $\$250{,}000$ to $\$1{,}250{,}000$. Find the percentile level of a $\$1{,}000{,}000$ threshold, then calculate the 99% VaR. (Answer: percentile $=0.7500$; 99% VaR $=\$1{,}240{,}000$.)
A loss is exponential with mean $\theta=\$800{,}000$. Calculate the 90% VaR and the exceedance probability at a separate threshold of $\$1{,}600{,}000$. (Answer: VaR $=\$1{,}842{,}068.07$; exceedance probability $=e^{-2}=0.1353$.)
Compare the 95% VaR for two hypothetical models with the same mean loss of $\$500{,}000$: a uniform loss on $\$0$ to $\$1{,}000{,}000$, and an exponential loss with $\theta=\$500{,}000$. Calculate both VaRs and state which is larger. (Answer: uniform VaR $=\$950{,}000$; exponential VaR $=\$1{,}497{,}866.14$; the exponential VaR is larger.)

What You'll Learn