Finding VaR by Inverting a Cumulative Distribution

Why Find VaR by Inverting the Cumulative Distribution?

For a loss $X$, Value at Risk at confidence level $\alpha$ is a loss threshold associated with cumulative probability $\alpha$. In practical calculations, the distribution may be given by a formula for its cumulative distribution function, or CDF. If that CDF is continuous and strictly increasing over the relevant range, finding VaR means solving the equation $F_X(x)=\alpha$ for $x$.

This is an inverse-probability calculation: the confidence level is the input, and the loss threshold is the answer. The upper-tail probability beyond the threshold is $1-\alpha$. Keeping those two probabilities distinct is essential, especially when a problem describes the threshold as being exceeded only rarely.

Definition: The cumulative distribution function of a loss $X$ is $F_X(x)=P(X\le x)$. For a continuous loss whose CDF is strictly increasing at the desired confidence level $\alpha$, the VaR threshold $x_\alpha$ is the unique solution to $F_X(x_\alpha)=\alpha$. The probability of exceeding this threshold is $P(X>x_\alpha)=1-\alpha$.

The restriction about strict increase matters: it makes the solution unique. For the continuous distributions considered here, the CDF increases throughout the interior of the loss range. The procedure is then to write down the model and its parameters, set its CDF equal to $\alpha$, solve for the loss, and check the result in the original CDF. A calculated threshold should not be accepted merely because the algebra looks plausible.

A Short Justification for the Inversion

Suppose $F_X$ is continuous and strictly increasing on the part of the loss range containing the desired quantile. Strict increase means that larger losses have larger cumulative probabilities, so each probability in that range corresponds to exactly one loss. Inverting the CDF therefore identifies the loss at which the probability of being at or below that loss is $\alpha$.

F_X(x_\alpha)=\alpha \quad\Longleftrightarrow\quad P(X\le x_\alpha)=\alpha.

For a continuous loss, the probability of taking any one exact value is zero. Thus $P(X=x_\alpha)=0$, and the probability of exceeding the threshold is the remaining probability:

P(X>x_\alpha)=1-P(X\le x_\alpha)=1-\alpha.

This establishes both the equation used to calculate VaR and a direct check on its interpretation. If an answer is supposed to be a 95% VaR, for example, its CDF value should be 0.95 and its probability of exceedance should be 0.05.

Uniform Losses

If a hypothetical loss is uniformly distributed between lower endpoint $a$ and upper endpoint $b$, where $a<b$, its CDF increases linearly from 0 to 1 across that interval. Within the interval, the cumulative probability at $x$ is the fraction of the interval lying at or below $x$. Equating that fraction to $\alpha$ gives a direct formula for the quantile.

F_X(x)=\frac{x-a}{b-a},\qquad a\le x\le b,

\frac{x_\alpha-a}{b-a}=\alpha \quad\Longrightarrow\quad x_\alpha=a+\alpha(b-a).

The formula applies when $0<\alpha<1$, and the resulting value must lie between $a$ and $b$. The lower endpoint is not generally zero: omitting it would incorrectly treat the distribution as uniform from zero to $b$.

Worked Example: A Uniform Loss with a Small Exceedance Probability

A hypothetical equipment warranty has annual loss $X$, measured in thousands of dollars, uniformly distributed from 10 to 70. Find the VaR at the confidence level corresponding to a 0.08 probability of exceeding the threshold. Check both probabilities at the answer.

Model and parameters. The loss is continuous and $X\sim\operatorname{Uniform}(10,70)$, with values measured in thousands of dollars. The stated exceedance probability is 0.08, so the cumulative confidence level is $ \alpha=1-0.08=0.92 $.

Method. For a uniform distribution, use $x_\alpha=a+\alpha(b-a)$. Here $a=10$, $b=70$, and $b-a=60$, all in thousands of dollars:

x_{0.92}=10+0.92(70-10) =10+0.92(60) =10+55.2 =65.2.

The VaR is 65.2 thousand dollars, or $65,200.00. No rounding beyond expressing the result in dollars and cents is needed.

Check. The CDF at 65.2 is $ (65.2-10)/(70-10)=55.2/60=0.92 $. The upper-tail probability is $1-0.92=0.08$, exactly as specified. The threshold is also within the support, since $10\le65.2\le70$.

Method and trap avoided. The method is to convert the exceedance probability to a cumulative confidence level before inverting the CDF. Using 0.08 directly as the confidence level would find the lower 8th percentile, not the threshold exceeded with probability 0.08.

Exponential Losses

An exponential loss with rate $\lambda>0$ has support $x\ge0$ and CDF $F_X(x)=1-e^{-\lambda x}$. Its mean is $1/\lambda$. If the mean is specified as $\theta$, then $\lambda=1/\theta$, so the same CDF can be written $1-e^{-x/\theta}$. These forms describe the same model; take care to distinguish the rate from the mean.

To find VaR, set the CDF equal to $\alpha$, isolate the exponential term, and take natural logarithms. For a confidence level strictly between 0 and 1, the result is nonnegative, as required by the support.

1-e^{-\lambda x_\alpha}=\alpha \quad\Longrightarrow\quad e^{-\lambda x_\alpha}=1-\alpha \quad\Longrightarrow\quad x_\alpha=-\frac{\ln(1-\alpha)}{\lambda}.

In mean parameterization, this is $x_\alpha=-\theta\ln(1-\alpha)$. The complement $1-\alpha$ appears inside the logarithm because the exponential term is the survival probability $P(X>x)$, not the CDF.

Worked Example: Exponential Loss at the 95% Level

A hypothetical portfolio has a continuous exponential loss with mean $40,000. Find its 95% VaR in dollars, then verify the CDF and exceedance probability.

Model and parameters. The loss is exponential with mean $\theta=\$40{,}000$, so its rate is $ \lambda=1/40{,}000=0.000025 $ per dollar. The confidence level is $\alpha=0.95$.

Method. Use $x_\alpha=-\theta\ln(1-\alpha)$:

x_{0.95} =-40{,}000\ln(0.05) =40{,}000(2.99573227355399\ldots) =119{,}829.290942\ldots.

Rounding to the nearest cent gives a VaR of $119,829.29.

Check. Dividing the unrounded threshold by the mean gives $119{,}829.290942\ldots/40{,}000=2.99573227355399\ldots$. Therefore $e^{-x_{0.95}/40{,}000}=0.05$, and $F_X(x_{0.95})=1-0.05=0.95$. The exceedance probability is 0.05, as required.

Method and trap avoided. The method is to invert the CDF using the mean parameterization. Replacing $\ln(0.05)$ by $\ln(0.95)$ would use the confidence level in place of the survival probability and would not produce a 95% threshold.

Inverting Other Continuous CDFs

The same principle works when the CDF is not one of the two standard forms above. First identify the range on which the stated CDF formula applies. Then solve $F_X(x)=\alpha$ using that formula, and check that the solution lies in its range. If a CDF is given in pieces, using a formula outside its stated interval can produce a number that satisfies the algebra but is not a valid quantile.

Worked Example: A Continuous Loss with an Increasing Density

A hypothetical repair fund has loss $X$, in dollars, with CDF $F_X(x)=(x/120{,}000)^2$ for $0\le x\le120{,}000$, with CDF 0 below this range and 1 above it. Find the 81% VaR and verify that it has the stated confidence level.

Model and parameters. This is a continuous distribution on $[0,120{,}000]$, and the target probability is $\alpha=0.81$. On the stated interval, the CDF increases from 0 to 1, so the interior solution is unique.

Method. Set the given CDF equal to 0.81 and take the nonnegative square root because losses in the model are nonnegative:

\left(\frac{x_{0.81}}{120{,}000}\right)^2=0.81 \quad\Longrightarrow\quad \frac{x_{0.81}}{120{,}000}=\sqrt{0.81}=0.9 \quad\Longrightarrow\quad x_{0.81}=108{,}000.

The VaR is $108,000.00.

Check. The threshold is in the interval $[0,120{,}000]$, where the CDF formula applies. Substitution gives $F_X(108{,}000)=(108{,}000/120{,}000)^2=0.9^2=0.81$. Thus $P(X>108{,}000)=1-0.81=0.19$.

Method and trap avoided. The method is to solve the stated CDF equation and select the root within the model's support. Taking the negative square root would give an invalid loss outside the interval and would not represent the desired quantile.

Common Exam Traps and a Reliable Check

The most frequent error is confusing the confidence level with the probability of exceeding the VaR. At confidence level $\alpha$, the CDF is $\alpha$, while the upper-tail probability is $1-\alpha$. For the uniform example above, the wrong calculation $10+0.08(60)=14.8$ thousand uses the exceedance probability as if it were the cumulative probability. It gives $F_X(14.8)=(14.8-10)/60=0.08$, not 0.92. The correction is to use $1-0.08=0.92$, giving $10+0.92(60)=65.2$ thousand.

Other checks help catch different mistakes. Confirm that the answer is in the support of the loss; retain the stated units throughout; and substitute the unrounded answer into the CDF before rounding. For an exponential model, verify that the parameter used is the rate or mean as specified. A formula using the mean $\theta$ has the form $-\theta\ln(1-\alpha)$; a formula using the rate $\lambda$ has the form $-\ln(1-\alpha)/\lambda$.

Identify the target probability.
Use the stated confidence level as $\alpha$. If the problem gives an exceedance probability $q$, first set $\alpha=1-q$.

Write the model CDF.
State the distribution, its parameter values, units, and the range on which the CDF expression applies.

Invert the CDF.
Set $F_X(x)=\alpha$ and solve for $x$, taking account of the support and any applicable range restrictions.

Check the result.
Substitute the unrounded threshold into the CDF and confirm $F_X(x)=\alpha$; then confirm the exceedance probability is $1-\alpha$.

Finding VaR this way is useful whenever the distributional model is known and its CDF can be inverted directly. The final probability check is more than a formality: it confirms that the threshold answers the stated confidence question rather than its lower-tail or upper-tail counterpart. This calculation method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical loss model, find the requested VaR. Use the support and parameterization stated, and check whether the confidence level is cumulative or given as an exceedance probability.

A loss is uniform from $5,000 to $45,000. Find the VaR at confidence level 0.75. (Answer: $35,000.00)
A loss is exponential with mean $25,000. Find the VaR at confidence level 0.80, rounded to the nearest cent. (Answer: $40,235.95)
A loss, in thousands of dollars, is uniform from 12 to 92. Find the threshold that has probability 0.025 of being exceeded. (Answer: $90,000.00)
A continuous loss has CDF $F_X(x)=(x/50{,}000)^3$ for $0\le x\le50{,}000$. Find its 0.512 VaR. (Answer: $40,000.00)
A loss is exponential with rate $\lambda=0.00002$ per dollar. Find its 99% VaR, rounded to the nearest cent. (Answer: $230,258.51)

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