VaR and Three Coherence Properties

Introduction

Value at Risk (VaR) is a quantile of a loss distribution. Its interpretation is direct: at a selected confidence level, VaR is a loss threshold that the modeled loss does not exceed with at least that probability. Because VaR is defined from a distribution, its behavior under changes to the loss variable can be established by tracking how the distribution function changes.

This tutorial proves three properties: monotonicity, translation invariance, and positive homogeneity for nonnegative scaling. Each property describes a different change to losses: ordering one loss below another, adding a fixed amount, or multiplying losses by a nonnegative factor. The results apply to discrete, continuous, and mixed distributions.

Definition: Let $X$ be a real-valued loss with distribution function $F_X(x)=P(X\leq x)$. For a confidence level $p$, where $0<p<1$, define its VaR by the left quantile $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\geq p\}. $$ The inequality is inclusive: the quantile is the first loss threshold at which the cumulative probability reaches or exceeds $p$. The properties below concern real-valued losses with well-defined, finite quantiles.

Three Properties in Distributional Form

The properties are statements about random losses, not just about a particular list of observed outcomes. In the monotonicity property, $X\leq Y$ almost surely means that the inequality holds except possibly on an event of probability zero. This condition is stronger than saying that the two losses have similar distributions or that their expected values are ordered.

Theorem: For $0<p<1$, loss VaR has the following properties:

Monotonicity: If $X\leq Y$ almost surely, then $\operatorname{VaR}_p(X)\leq\operatorname{VaR}_p(Y)$.
Translation invariance: For any constant $b$, $\operatorname{VaR}_p(X+b)=\operatorname{VaR}_p(X)+b$.
Positive homogeneity: For any constant $c\geq0$, $\operatorname{VaR}_p(cX)=c\operatorname{VaR}_p(X)$.

Here $b$ is a deterministic amount added to the loss, while $c$ is a nonnegative scaling factor.

These are three of the properties used when assessing risk measures. They do not, by themselves, establish that VaR is coherent: coherence also requires subadditivity. The results here concern only the three properties stated above.

Why the Properties Hold

The distribution-function definition gives a short justification for each result. First suppose $X\leq Y$ almost surely. For any real $x$, the event $\{Y\leq x\}$ is contained, up to probability-zero outcomes, in $\{X\leq x\}$. Therefore,

F_Y(x)\leq F_X(x)\quad\text{for every }x.

The distribution of $X$ consequently reaches probability $p$ no later than the distribution of $Y$. More formally, if $r=\operatorname{VaR}_p(Y)$, then $F_Y(r)\geq p$, and hence $F_X(r)\geq F_Y(r)\geq p$. Thus $r$ is at least one threshold at which $F_X$ reaches $p$, so $\operatorname{VaR}_p(X)\leq r=\operatorname{VaR}_p(Y)$.

For translation, let $b$ be any real constant. For every $x$, the event $X+b\leq x$ is exactly the event $X\leq x-b$. Therefore,

F_{X+b}(x)=F_X(x-b).

The set of thresholds $x$ for which $F_{X+b}(x)\geq p$ is the set of thresholds $x$ for which $x-b$ is a qualifying threshold for $X$. Every qualifying threshold is shifted by $b$, giving $\operatorname{VaR}_p(X+b)=\operatorname{VaR}_p(X)+b$. The argument permits a negative $b$, such as a fixed reduction in loss.

For a positive scaling factor $c$, $cX\leq x$ is equivalent to $X\leq x/c$, so

F_{cX}(x)=F_X(x/c),\qquad c>0.

The qualifying quantile thresholds are therefore multiplied by $c$, which gives $\operatorname{VaR}_p(cX)=c\operatorname{VaR}_p(X)$. If $c=0$, then $cX$ is the constant zero loss and its VaR is zero, so the same formula holds. The restriction $c\geq0$ matters: multiplication by a negative number reverses the ordering of outcomes and is not the positive homogeneity property.

State the loss and confidence level.
Identify the random loss, the value of $p$, and the left-quantile definition being used.

Identify the transformation.
Check whether the problem orders two losses, adds a constant, or multiplies by a nonnegative factor.

Apply the matching property.
For a combined transformation $cX+b$, with $c\geq0$, use $\operatorname{VaR}_p(cX+b)=c\operatorname{VaR}_p(X)+b$.

Check the quantile against the distribution.
For a discrete loss, verify that the cumulative probability at the reported threshold is at least $p$, and that it is below $p$ immediately before that threshold.

Worked Applications

For a discrete loss, arranging the outcomes from smallest to largest and accumulating their probabilities is a direct way to find the left quantile. The worked examples use this calculation to check the property-based results against the underlying distributions.

Worked Example: Comparing Ordered Losses

A hypothetical two-part portfolio has three equally defined scenarios with probabilities $0.4000$, $0.3500$, and $0.2500$. In those scenarios, loss $X$ is respectively $\$0$, $\$5$, and $\$9$, while loss $Y$ is respectively $\$1$, $\$7$, and $\$12$. Find both VaRs at confidence level $p=0.7500$, and check monotonicity.

Method. Compare the losses scenario by scenario, then accumulate each loss distribution up to the stated confidence level. This avoids treating a comparison of average losses as if it were the almost-sure ordering required by monotonicity.

In each scenario, $X\leq Y$: $0\leq1$, $5\leq7$, and $9\leq12$. Thus $X\leq Y$ almost surely. For $X$, the cumulative probabilities at the ordered losses are $0.4000$ at $\$0$, $0.4000+0.3500=0.7500$ at $\$5$, and $1.0000$ at $\$9$. The first cumulative probability that reaches $0.7500$ occurs at $\$5$, so $\operatorname{VaR}_{0.7500}(X)=\$5$.

For $Y$, the cumulative probabilities are $0.4000$ at $\$1$, $0.4000+0.3500=0.7500$ at $\$7$, and $1.0000$ at $\$12$. Hence $\operatorname{VaR}_{0.7500}(Y)=\$7$.

The result satisfies $\$5\leq\$7$, as monotonicity requires. The cumulative probability at each reported quantile is exactly $0.7500$, confirming the inclusive quantile convention.

Worked Example: Adding a Fixed Loss Amount

A hypothetical contract has loss $X$ equal to $\$0$, $\$100$, or $\$500$, with probabilities $0.2000$, $0.5000$, and $0.3000$, respectively. A fixed administrative charge of $\$40$ is added in every outcome, making the revised loss $Z=X+\$40$. Find the $0.7000$-VaR of $X$ and $Z$, both directly and by translation invariance.

Method. Find the quantile from the cumulative distribution, then shift every outcome by the fixed charge. This checks that the deterministic addition affects the threshold by the same amount, not by a probability-weighted amount.

For $X$, the cumulative probability is $0.2000$ at $\$0$, then $0.2000+0.5000=0.7000$ at $\$100$, and $1.0000$ at $\$500$. Therefore, $\operatorname{VaR}_{0.7000}(X)=\$100$.

Adding $\$40$ gives $Z$ the outcomes $\$40$, $\$140$, and $\$540$, with the same probabilities. Its cumulative probability at $\$40$ is $0.2000$, and at $\$140$ it is $0.2000+0.5000=0.7000$. Thus the direct calculation gives $\operatorname{VaR}_{0.7000}(Z)=\$140$.

Translation invariance gives the same result: $\operatorname{VaR}_{0.7000}(Z)=\$100+\$40=\$140$. Both calculations agree, and the quantile is reached exactly at the second outcome.

Worked Example: Scaling a Loss

A hypothetical equipment loss $X$ is $\$0$, $\$1{,}200$, or $\$4{,}000$, with probabilities $0.2500$, $0.5000$, and $0.2500$, respectively. A coverage adjustment scales each loss by $c=1.5$, so the adjusted loss is $W=1.5X$. Find the $0.7500$-VaR before and after the adjustment.

Method. Accumulate the original probabilities to locate the quantile, then multiply each outcome by the positive scaling factor. This avoids scaling the confidence level or applying the factor to probabilities.

For $X$, the cumulative probability at $\$0$ is $0.2500$. At $\$1{,}200$, it is $0.2500+0.5000=0.7500$. Therefore, $\operatorname{VaR}_{0.7500}(X)=\$1{,}200$.

The adjusted outcomes are $1.5(\$0)=\$0$, $1.5(\$1{,}200)=\$1{,}800$, and $1.5(\$4{,}000)=\$6{,}000$. Their probabilities are unchanged. The cumulative probability at $\$1{,}800$ is still $0.2500+0.5000=0.7500$, so direct calculation gives $\operatorname{VaR}_{0.7500}(W)=\$1{,}800$.

Positive homogeneity gives $1.5(\$1{,}200)=\$1{,}800$, matching the direct result. The factor is nonnegative, as required, and the probability at the adjusted quantile remains $0.7500$.

Worked Example: Combining Scaling and Translation

A hypothetical loss $X$ is $\$40$, $\$100$, or $\$300$, with probabilities $0.2000$, $0.5000$, and $0.3000$. A revised loss is $W=2.5X+\$50$. Find the $0.7000$-VaR of $W$ using both a direct distribution calculation and the two properties.

Method. First locate the quantile of $X$; then apply the positive scale and fixed addition. A direct check confirms the order of the transformed outcomes and the quantile at the atom.

For $X$, the cumulative probabilities are $0.2000$ at $\$40$, $0.2000+0.5000=0.7000$ at $\$100$, and $1.0000$ at $\$300$. Thus $\operatorname{VaR}_{0.7000}(X)=\$100$.

Transforming the outcomes gives $2.5(\$40)+\$50=\$150$, $2.5(\$100)+\$50=\$300$, and $2.5(\$300)+\$50=\$800$. The probabilities remain $0.2000$, $0.5000$, and $0.3000$. The cumulative probability at $\$300$ is $0.2000+0.5000=0.7000$, so the direct VaR is $\$300$.

Applying the properties in sequence gives $2.5\operatorname{VaR}_{0.7000}(X)+\$50=2.5(\$100)+\$50=\$300$. The two methods agree; the scale is positive, and the fixed amount is added after scaling.

Common Quantile Traps

A frequent error is to define a discrete quantile as the first outcome whose cumulative probability is strictly greater than $p$. That rule can move VaR to the next loss even when the cumulative probability already equals $p$. For the loss in the translation example, cumulative probability at $\$100$ is $0.7000$. At confidence level $0.7000$, the left-quantile definition gives $\$100$, not $\$500$.

The incorrect calculation is to require $F_X(x)>0.7000$, observe $F_X(\$100)=0.7000$, and select $\$500$ as the first loss where the cumulative probability is strictly greater. The correction is to use the stated definition $F_X(x)\geq p$: since $F_X(\$100)=0.7000$, the VaR is $\$100$. Translation then gives $\$140$, not $\$540$.

Another important distinction is between an ordering assumption and a comparison of distribution summaries. Monotonicity follows when $X\leq Y$ almost surely, as in the scenario-by-scenario comparison above. Merely knowing that one loss has a smaller mean, or that its outcomes are smaller in some scenarios, does not establish this condition. Check the actual premise before applying the theorem.

Key takeaway: With the left-quantile definition, VaR respects almost-sure ordering, shifts by a fixed amount added to losses, and scales by any nonnegative factor. For discrete losses, verify the inclusive probability threshold at the reported quantile.

Check Your Understanding

For each discrete loss, use the left-quantile convention and the stated confidence level. Check the answer against the relevant property or cumulative probability.

A hypothetical loss $A$ is $\$2$ with probability $0.2500$ and $\$9$ with probability $0.7500$. Loss $B$, on the same scenarios, is $\$3$ and $\$10$, respectively. Find both $0.7500$-VaRs and state whether the values agree with monotonicity. (Answer: $\operatorname{VaR}_{0.7500}(A)=\$9$, $\operatorname{VaR}_{0.7500}(B)=\$10$; yes.)
A loss $X$ is $\$0$, $\$50$, or $\$200$, with probabilities $0.3000$, $0.5000$, and $0.2000$. A fixed $\$25$ is added. Find the original and revised $0.8000$-VaRs. (Answer: $\$50$ and $\$75$.)
A loss $X$ is $\$10$, $\$30$, or $\$80$, with probabilities $0.2000$, $0.4000$, and $0.4000$. Every loss is scaled by $0.4$. Find the original and scaled $0.6000$-VaRs. (Answer: $\$30$ and $\$12$.)
A loss $X$ is $\$0$, $\$40$, or $\$100$, with probabilities $0.2000$, $0.5000$, and $0.3000$. A revised loss is $Z=3X+\$50$. Find $\operatorname{VaR}_{0.7000}(Z)$ using the properties, and identify the loss outcome at the quantile. (Answer: $\$170$, corresponding to $X=\$40$.)
A loss $X$ has cumulative probabilities $0.4500$ at $\$20$, $0.8500$ at $\$60$, and $1.0000$ at $\$150$. A fixed $\$10$ is added, then the result is multiplied by $2$. Find the $0.8500$-VaR after both transformations. (Answer: $\$140$.)

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