VaR from Cumulative and Exceedance Distributions

Two Ways to Describe a Loss Distribution

Value at Risk can be found from cumulative loss probabilities or from exceedance probabilities. These describe the same distribution, but they answer different-looking questions: a cumulative distribution records the probability that loss is at most a threshold, while a strict exceedance probability records the probability that loss is greater than it. Keeping the inequality direction clear is especially important when the loss distribution has probability mass at a particular amount.

Definition: For a loss $X$, its cumulative distribution function (CDF) is $F_X(x)=P(X\leq x)$. Its strict exceedance function is $S_X(x)=P(X>x)$. The left limit of the CDF is $F_X(x-)=\lim_{y\uparrow x}F_X(y)=P(X<x)$. Consequently, $$ S_X(x)=1-F_X(x), \qquad P(X\geq x)=1-F_X(x-). $$

The distinction between $P(X>x)$ and $P(X\geq x)$ is precisely the probability at the threshold: $P(X=x)=F_X(x)-F_X(x-)$. If $X$ has no probability mass at $x$, then the two tail probabilities are equal. If there is mass at $x$, they differ. CDFs and exceedance functions are therefore interchangeable only when the inequality convention is also carried along.

VaR as a Cumulative or Exceedance Quantile

For confidence level $p$, where $0<p<1$, use the lower quantile definition of VaR:

Definition: The $p$-level Value at Risk of $X$ is $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\geq p\}. $$ Equivalently, since $F_X(x)=1-S_X(x)$, $$ \operatorname{VaR}_p(X)=\inf\{x:S_X(x)\leq 1-p\}. $$

The equivalence follows by rearranging the event probability at each fixed threshold: $F_X(x)\geq p$ if and only if $1-S_X(x)\geq p$, which is if and only if $S_X(x)\leq1-p$. The two definitions therefore select the same set of thresholds, and their infima are the same. This equivalence does not require a continuous distribution. It does require using the strict exceedance probability $P(X>x)$ in the second definition.

At a finite quantile $q=\operatorname{VaR}_p(X)$, the usual CDF boundary relations are $F_X(q-)\leq p\leq F_X(q)$. Thus $P(X>q)\leq1-p$, while $P(X\geq q)\geq1-p$. These inequalities help describe what happens at the quantile. They do not, by themselves, always prove that a proposed $x$ is the quantile: in particular, if $F_X(x-)=p$, the CDF may already have reached $p$ at some smaller loss.

Worked Example: Finding VaR from Both Tails

A hypothetical one-year portfolio loss $X$ can take only the values shown below. Find its 95% VaR, then calculate the strict and non-strict tail probabilities at that loss. Use the listed probability masses as the complete distribution.

Loss $x$ (thousands of dollars)	$P(X=x)$	$F_X(x)$	$P(X>x)$
0	0.40	0.40	0.60
20	0.30	0.70	0.30
90	0.18	0.88	0.12
210	0.08	0.96	0.04
500	0.04	1.00	0.00

Method. Accumulate the probability masses to obtain the CDF, then identify the first listed loss for which the CDF is at least $0.95$. The masses sum to $0.40+0.30+0.18+0.08+0.04=1.00$, as required.

At loss $90$, $F_X(90)=0.88<0.95$. At loss $210$, $F_X(210)=0.96\geq0.95$. Because the support is restricted to the listed losses, the first threshold meeting the quantile condition is $210$ thousand dollars. Therefore $\operatorname{VaR}_{0.95}(X)=\$210{,}000$.

The strict exceedance probability at $210$ is $P(X>210)=0.04$, which is below the $5\%$ tail allowance. The non-strict exceedance probability is $P(X\geq210)=0.08+0.04=0.12$. Equivalently, $F_X(210-)=0.88$, so $P(X\geq210)=1-0.88=0.12$. These checks agree with the table and with the CDF definition.

Common mistake avoided: Using $P(X\geq210)\leq0.05$ as the VaR condition would incorrectly reject this quantile. The quantile definition uses $P(X>q)\leq0.05$; losses equal to $q$ are included in $F_X(q)$.

Reading a Threshold When There Is Mass at It

For a candidate threshold $x$, the CDF value $F_X(x)$ includes any probability mass exactly at $x$, but $F_X(x-)$ does not. If $F_X(x-)<p\leq F_X(x)$, then $x$ is the $p$-level VaR: every lower threshold has cumulative probability below $p$, and $x$ reaches or exceeds $p$. If instead $F_X(x-)=p$, information at $x$ alone does not settle the question. The CDF may have reached $p$ at an earlier loss, so one must examine the distribution below $x$ or apply the infimum definition directly.

Worked Example: A Left Limit Does Not Settle the Boundary Level

In a hypothetical loss model, the complete probability distribution is $P(X=0)=0.60$, $P(X=70)=0.34$, $P(X=120)=0.035$, and $P(X=200)=0.025$, with losses in thousands of dollars. Let $x=120$. Determine the VaR levels for which $x$ is the quantile, and calculate both tail probabilities at $x$.

First accumulate the masses below and at $120$: $F_X(120-)=0.60+0.34=0.94$, and $F_X(120)=0.94+0.035=0.975$. Thus $P(X<120)=0.94$ and $P(X\leq120)=0.975$. The probability strictly above $120$ is $0.025$; the probability at least $120$ is $0.035+0.025=0.060$.

For any $p$ with $0.94<p\leq0.975$, the CDF is below $p$ at every loss below $120$, and reaches $p$ at $120$. Hence $\operatorname{VaR}_p(X)=120$ thousand dollars throughout that interval. For $p=0.94$, however, $F_X(70)=0.94$, so the infimum is $70$, not $120$. The left-limit and CDF values at $120$ therefore do not justify calling $120$ the 94% VaR.

Method. Calculate the cumulative probability strictly below the candidate, then compare $p$ with both the left limit and the CDF at the candidate. Common mistake avoided: Treating $F_X(120-)=0.94$ as sufficient to conclude $\operatorname{VaR}_{0.94}=120$ ignores the possibility that the CDF reached $0.94$ earlier. Checking $F_X(70)$ identifies the actual 94% quantile here.

What Exceedance Tables Can and Cannot Tell You

An exceedance table often gives $P(X>x)$ at selected thresholds. At any listed threshold $x$, a value $P(X>x)\leq1-p$ shows that $F_X(x)\geq p$, so $x$ is a qualifying threshold. It does not necessarily show that $x$ is the first qualifying threshold. To determine VaR from selected values alone, the information must also establish what happens at lower thresholds—for example, by specifying that the listed amounts are the only possible losses.

Worked Example: When the Listed Thresholds Are the Entire Support

A hypothetical portfolio can lose only $0$, $20$, $90$, $210$, or $500$ thousand dollars. Its strict exceedance probabilities at $20$, $90$, $210$, and $500$ thousand dollars are respectively $0.30$, $0.12$, $0.04$, and $0$. Find the 95% VaR.

Method. The support restriction means there are no unlisted possible losses between the reported thresholds. Convert the strict exceedance values to cumulative probabilities using $F_X(x)=1-P(X>x)$: the CDF values at $20$, $90$, $210$, and $500$ are $0.70$, $0.88$, $0.96$, and $1.00$, respectively. The 95% target is $p=0.95$.

At $90$, the CDF is $0.88<0.95$; at $210$, it is $0.96\geq0.95$. Since the stated support contains no possible loss between those amounts, the first threshold reaching the target is $210$ thousand dollars. Therefore $\operatorname{VaR}_{0.95}(X)=\$210{,}000$. The strict exceedance at that amount is $0.04\leq0.05$, which checks the result.

Common mistake avoided: Without the support restriction, the four exceedance values would not establish the first threshold meeting the 95% criterion. An unreported loss amount between $90$ and $210$ could already have a CDF of at least $0.95$. A table of selected tail probabilities is not a complete distribution unless additional information makes it one.

A Reliable Calculation Procedure

When a problem supplies cumulative probabilities, exceedance probabilities, or a mixture of both, use a consistent sequence. State the confidence level, identify the exact inequality in each probability, and verify that the information is sufficient to locate the first qualifying threshold. The process is summarized below.

Set the tail target.
For a confidence level $p$, the corresponding upper-tail allowance is $1-p$.

Translate the supplied probabilities.
Use $F_X(x)=1-P(X>x)$ for strict exceedance. For a non-strict tail, use $P(X\geq x)=1-F_X(x-)$.

Locate the first qualifying loss.
Find the infimum of thresholds with $F_X(x)\geq p$, or equivalently $P(X>x)\leq1-p$. Check that the available distribution information covers lower thresholds.

Check the boundary.
At the proposed quantile, confirm that the strict exceedance probability is no greater than $1-p$, and distinguish it from the probability of a loss at least as large as the quantile.

Common Exam Traps

Switching strict and non-strict inequalities. $P(X>x)=1-F_X(x)$, but $P(X\geq x)=1-F_X(x-)$. They differ whenever $P(X=x)>0$.
Assuming one qualifying threshold is the quantile. VaR is the infimum of qualifying thresholds, not just any threshold at which the CDF reaches the target. Values at a few points may not identify that infimum.
Over-interpreting a left-limit equality. If $F_X(x-)=p$, $x$ need not be $\operatorname{VaR}_p$; the CDF might already equal $p$ at a smaller loss.
Ignoring the support or model assumptions. A list of thresholds cannot rule out a qualifying amount between them unless the problem specifies the possible losses or supplies enough distributional information.

Key takeaway: The CDF and strict exceedance function are related by $S_X(x)=1-F_X(x)$, so either can be used to find VaR. Keep strict and non-strict tails separate, and use the definition’s infimum condition—not a single threshold value alone—to establish the quantile.

Check Your Understanding

For each problem, identify the relevant tail convention and check that the supplied information is sufficient to determine the requested quantile.

A hypothetical loss, in thousands of dollars, has support $\{0,30,80,160\}$, with respective probabilities $0.50,0.28,0.17,0.05$. Find the 90% VaR and both tail probabilities at that loss. (Answer: $\operatorname{VaR}_{0.90}=80$ thousand dollars; $P(X>80)=0.05$, $P(X\geq80)=0.22$.)
For a hypothetical loss threshold $t=75$, suppose $F_X(y)\leq0.82$ for every $y<t$, $F_X(t-)=0.82$, and $F_X(t)=0.93$. For which $p$ with $0.82<p\leq0.93$ is $t$ the VaR, and what are the two tail probabilities at $t$? (Answer: $t$ is $\operatorname{VaR}_p$ for every $0.82<p\leq0.93$; $P(X>t)=0.07$, $P(X\geq t)=0.18$.)
A hypothetical loss can take only $0,25,60,140,$ or $400$ thousand dollars. Its strict exceedance probabilities at $25,60,140,$ and $400$ are $0.42,0.19,0.06,$ and $0$. Find its 90% VaR. (Answer: $\operatorname{VaR}_{0.90}=140$ thousand dollars, since $F_X(60)=0.81$ and $F_X(140)=0.94$.)
At a hypothetical threshold $x=300$, a report gives $F_X(x-)=0.96$ and $F_X(x)=0.99$. State the confidence levels for which $x$ is guaranteed to be the quantile from these facts, and calculate both tails at $x$. Does the information establish that $x$ is the 96% VaR? (Answer: $x$ is $\operatorname{VaR}_p$ for $0.96<p\leq0.99$; $P(X>x)=0.01$, $P(X\geq x)=0.04$; the 96% VaR is not established.)
A report gives $P(X>15)=0.35$, $P(X>55)=0.16$, $P(X>130)=0.07$, and $P(X>600)=0$, with amounts in thousands of dollars, but says nothing about losses between the listed thresholds. Can the 92% VaR be determined from these values alone? (Answer: No. Although the CDF at 130 is $0.93$, an unlisted threshold between 55 and 130 could be the first to reach $0.92$.)

What You'll Learn