VaR Under Shifts and Positive Scaling

Why Shifts and Scaling Matter

Risk calculations often begin with a loss variable and then express it in a different form. A fixed expense may be added to every outcome, or every loss may be converted using a common multiplier. Examples include adding a fixed claim-handling cost or expressing a loss in a different unit. It is useful to know whether such a change requires recalculating the full distribution or whether the original Value at Risk (VaR) can be transformed directly.

The key results are simple, but depend on using a consistent definition of VaR and interpreting the transformation correctly. A fixed addition moves every quantile by that amount. Multiplication by a positive constant multiplies every quantile by the same constant. These facts follow directly from how the cumulative distribution function changes.

Definition: For a loss random variable $X$, cumulative distribution function $F_X(x)=P(X\le x)$, and confidence level $0<\alpha<1$, define $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}$. This is the lower $\alpha$-quantile of the loss distribution. The results below use this quantile definition and assume the relevant quantiles are finite.

Adding a Fixed Amount

Let $c$ be a fixed real amount, and define the shifted loss $Z=X+c$. The shift changes every possible outcome by the same amount: if the original loss is $x$, the new loss is $x+c$. It does not change the order of outcomes or their probabilities.

Theorem: For a fixed real $c$, $\operatorname{VaR}_{\alpha}(X+c)=\operatorname{VaR}_{\alpha}(X)+c$. The result applies to positive, zero, or negative shifts, provided the loss amounts and $c$ use the same units.

To see why this holds, first relate the cumulative distribution functions. For any real $z$,

F_Z(z)=P(X+c\le z)=P(X\le z-c)=F_X(z-c).

The transformed loss reaches cumulative probability $\alpha$ at $z$ exactly when the original loss reaches that probability at $z-c$. Thus the set of $z$ values meeting the quantile condition is the original set shifted by $c$. Taking its infimum shifts the quantile by $c$ as well.

Worked Example: Adding a Fixed Claim Expense

A hypothetical portfolio has loss $X$, measured in thousands of dollars, with $X\sim\operatorname{Uniform}(0,100)$. A fixed expense of $15,000 is added to every portfolio outcome. Find the 90% VaR of the resulting loss $Z=X+15$, reporting the answer in dollars. Use the uniform distribution and its stated endpoints; the shift is exactly 15 thousand dollars.

Method. Find the original quantile, apply the shift property, and verify it using the transformed distribution.

For a uniform distribution on $[0,100]$, the cumulative distribution function on that interval is $F_X(x)=x/100$. The 90% quantile is therefore found by solving $x/100=0.90$:

\operatorname{VaR}_{0.90}(X)=0.90(100)=90 \quad\text{thousand dollars}.

Apply the fixed shift:

\operatorname{VaR}_{0.90}(Z)=90+15=105 \quad\text{thousand dollars}=\$105{,}000.

For a direct check, $Z$ is uniform on $[15,115]$, so $F_Z(z)=(z-15)/100$ on that interval. At $z=105$, $F_Z(105)=(105-15)/100=0.90$. The answer is checked. A common mistake is to add $15 rather than $15,000 to a result expressed in dollars; here both quantities were calculated in thousands before converting.

Multiplying by a Positive Constant

Now let $k>0$ and define $Y=kX$. Multiplying by a positive number preserves the ordering of outcomes: if one loss is larger than another before scaling, it remains larger afterward. A positive scale changes the magnitude of the outcomes but not which outcomes comprise the lowest $\alpha$ proportion of the distribution.

Theorem: If $k>0$, then $\operatorname{VaR}_{\alpha}(kX)=k\operatorname{VaR}_{\alpha}(X)$. If $k=0$, the transformed loss is identically zero and $\operatorname{VaR}_{\alpha}(0X)=0$.

For $k>0$, the cumulative distribution function of $Y$ is

F_Y(y)=P(kX\le y)=P(X\le y/k)=F_X(y/k).

Because division by $k$ preserves the inequality, $F_Y(y)\ge\alpha$ exactly when $F_X(y/k)\ge\alpha$. Every threshold $x$ satisfying the original quantile condition corresponds to the threshold $kx$ for the scaled loss. Since $k$ is positive, this correspondence preserves order and therefore preserves the infimum after scaling. This gives the stated VaR formula.

The zero case is separate: $0X=0$ with probability one, so its lower $\alpha$-quantile is zero for every $0<\alpha<1$. The positive-scaling formula is not a formula for negative multipliers. A negative multiplier reverses the ordering of outcomes, so the lower tail of the transformed loss comes from the upper tail of the original loss.

Worked Example: Scaling a Hypothetical Exponential Loss

A hypothetical loss $X$, measured in thousands of dollars, follows an exponential distribution with rate $0.05$ per thousand dollars. Thus $F_X(x)=1-e^{-0.05x}$ for $x\ge0$. A currency conversion and exposure adjustment together multiply every loss by $k=1.4$, giving $Y=1.4X$. Find the 75% VaR of $Y$, in dollars, to the nearest cent.

Method. Invert the original exponential cumulative distribution function, multiply the resulting quantile by the positive factor, then check against the transformed distribution.

First solve $F_X(x)=0.75$:

1-e^{-0.05x}=0.75 \quad\Longrightarrow\quad e^{-0.05x}=0.25 \quad\Longrightarrow\quad -0.05x=\ln(0.25).

Therefore,

\operatorname{VaR}_{0.75}(X) =\frac{-\ln(0.25)}{0.05} =\frac{\ln(4)}{0.05} =27.725887\ldots \quad\text{thousand dollars}.

Apply the scale factor:

\operatorname{VaR}_{0.75}(Y) =1.4(27.725887\ldots) =38.816242\ldots \quad\text{thousand dollars}.

Converting to dollars and rounding only at the final step gives $\$38{,}816.24$. For a check, $Y$ is exponential with rate $0.05/1.4$ per thousand dollars, so its 75% quantile is $-\ln(0.25)/(0.05/1.4)=38.816242\ldots$ thousand dollars, as above.

Common mistake. Adding the multiplier to the original VaR would give $27.725887+1.4=29.125887$ thousand dollars. That is not the effect of multiplying a loss by 1.4; the correct calculation multiplies the entire quantile by 1.4.

Combining a Shift and a Positive Scale

A loss transformation may include both operations. If $Z=kX+c$ with $k>0$, first scale the loss and then add the fixed amount. Applying the two properties in sequence gives the quantile of the transformed loss. The order in this expression matters: the fixed amount is added after scaling, so it is not multiplied by $k$.

Result: For $k>0$ and fixed $c$, $\operatorname{VaR}_{\alpha}(kX+c)=k\operatorname{VaR}_{\alpha}(X)+c$.

This result can also be checked from the distribution. For any $z$, the inequality $kX+c\le z$ is equivalent to $X\le(z-c)/k$, because $k>0$. Hence $F_{kX+c}(z)=F_X((z-c)/k)$. Inverting at probability $\alpha$ gives the original quantile multiplied by $k$, then shifted by $c$.

Worked Example: Scaling and Then Adding a Fixed Charge

A hypothetical annual loss $X$, in thousands of dollars, is uniform on $[10,30]$. A new reporting measure is $Z=0.8X+7$: losses are reduced to 80% of their original amount and then a fixed 7-thousand-dollar charge is added. Find the 60% VaR, in thousands of dollars, and check it from the distribution of $Z$.

Method. Find the original uniform quantile, apply the combined transformation, and verify the cumulative probability at the result.

For a uniform distribution on $[10,30]$, the 60% quantile is

\operatorname{VaR}_{0.60}(X)=10+0.60(30-10)=10+12=22.

The combined shift-and-scale result gives

\operatorname{VaR}_{0.60}(Z)=0.8(22)+7=17.6+7=24.6 \quad\text{thousand dollars}.

For a direct check, $Z$ is uniform on $[0.8(10)+7,\;0.8(30)+7]=[15,31]$. Its 60% quantile is $15+0.60(31-15)=15+9.6=24.6$ thousand dollars. Thus the answer is checked.

A common mistake is to scale the fixed charge too, calculating $0.8(22+7)=23.2$. That would describe a different transformation, $0.8(X+7)$, rather than the stated $0.8X+7$. The order and placement of each part of the transformation must match the problem statement.

Using the Properties Carefully

These properties are useful when a loss model has already been analyzed and a new version of the loss differs only by a fixed shift or positive scale. They can avoid reconstructing and inverting a full cumulative distribution function. They also provide a quick check on a direct calculation: calculate VaR both by transforming the quantile and, where practical, by examining the transformed distribution.

Before applying either property, check the following:

Match units. A shift in dollars cannot be added directly to a quantile stated in thousands of dollars. Convert one quantity before combining them.
Identify the actual transformation. For $kX+c$, multiply the original quantile by $k$, then add $c$. Do not scale $c$ unless the stated transformation is $k(X+c)$.
Check the sign of the scale. The direct scaling property requires $k>0$; handle $k=0$ as a constant loss. A negative multiplier reverses outcome order.
Keep the confidence level fixed. The properties change the loss scale or location, not the probability level $\alpha$.

The formulas are quantile properties, not assumptions about a particular family of distributions. They hold whether the original loss is uniform, exponential, or another distribution, as long as VaR uses the stated lower-quantile definition and the relevant quantile is finite.

Key takeaway: A fixed shift moves VaR by the same fixed amount, while positive scaling multiplies VaR by the same positive factor. For $k>0$, the combined rule is $\operatorname{VaR}_{\alpha}(kX+c)=k\operatorname{VaR}_{\alpha}(X)+c$; unit consistency and transformation order are essential.

Check Your Understanding

For each problem, use the lower-quantile definition of VaR and keep the stated units consistent.

A hypothetical loss $X$, measured in thousands of dollars, is uniform on $[0,80]$. A fixed 12-thousand-dollar charge is added. Find the 90% VaR of the resulting loss, in dollars. (Answer: $\$84{,}000$)
A hypothetical loss is exponential with mean 25 thousand dollars. Every loss is multiplied by 0.6. Find the 80% VaR of the scaled loss, in dollars, to the nearest cent. (Answer: $\$24{,}141.57$)
A loss $X$ is uniform on $[4,24]$ thousand dollars and is transformed to $Z=2X+3$, with all amounts in thousands of dollars. Find the 35% VaR of $Z$. (Answer: 25 thousand dollars)
A loss $X$ is uniform on $[-20,40]$ thousand dollars. A fixed charge of 7 thousand dollars is subtracted from each outcome. Find the 75% VaR of the resulting loss. (Answer: 18 thousand dollars)
Let $X$ be uniform on $[0,10]$ thousand dollars and define $W=-X$. Find $\operatorname{VaR}_{0.90}(W)$, and explain why it is not obtained by multiplying $\operatorname{VaR}_{0.90}(X)$ by $-1$. (Answer: $-1$ thousand dollars; the negative multiplier reverses the ordering of outcomes, so the positive-scaling rule does not apply.)

What You'll Learn