Translation Invariance and Positive Homogeneity

Why Shifts and Scales Matter

Risk calculations often need to be adjusted when a portfolio acquires a fixed expense or when every loss is multiplied by a common factor. A sound loss risk measure should respond predictably to these changes. In particular, adding a fixed amount to every possible loss should raise the measured risk by that amount, and multiplying every loss by a positive factor should multiply the measured risk by the same factor.

These properties can be checked directly for Value at Risk (VaR) and Tail Value at Risk (TVaR). The key is to track how the entire loss distribution—and therefore its quantiles—changes. That approach works for continuous distributions and for discrete distributions with probability atoms.

Definition: Let $X$ be a loss, and let $p$ be a confidence level with $0<p<1$. Define $\operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}$, using the left quantile convention. Define $\operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1\operatorname{VaR}_q(X)\,dq$, provided the upper-tail integral is finite. A loss risk measure $\rho$ is translation invariant if $\rho(X+k)=\rho(X)+k$ for every fixed amount $k$. It is positively homogeneous if $\rho(aX)=a\rho(X)$ for every $a\ge0$.

How Quantiles Change Under Shifts and Scaling

Write $Q_X(u)=\inf\{x:F_X(x)\ge u\}$ for the $u$-quantile of $X$. If a fixed amount $k$ is added, the distribution moves right by $k$: for any value $x$, $P(X+k\le x)=P(X\le x-k)$. Consequently, the $u$-quantile moves from $Q_X(u)$ to $Q_X(u)+k$.

For a positive multiplier $a$, multiplying all outcomes by $a$ preserves their ordering and multiplies their quantiles by $a$. If $a=0$, the result is the constant-zero loss, whose quantiles are all zero. These statements hold for atoms as well as for continuous distributions; they do not require a density.

Theorem: For any loss $X$, fixed amount $k$, and $a\ge0$, the quantiles satisfy \[ Q_{X+k}(u)=Q_X(u)+k \quad\text{and}\quad Q_{aX}(u)=aQ_X(u), \qquad 0<u<1. \] Therefore, wherever TVaR is finite, \[ \operatorname{VaR}_p(X+k)=\operatorname{VaR}_p(X)+k,\qquad \operatorname{TVaR}_p(X+k)=\operatorname{TVaR}_p(X)+k, \] and \[ \operatorname{VaR}_p(aX)=a\operatorname{VaR}_p(X),\qquad \operatorname{TVaR}_p(aX)=a\operatorname{TVaR}_p(X). \]

Why the Properties Hold

For VaR, the quantile transformation gives the result immediately by evaluating at $u=p$. For TVaR, insert the transformed quantiles into its quantile-integral definition. The constant shift contributes $k$ at every quantile level, while scaling multiplies every integrand by $a$.

\begin{aligned} \operatorname{TVaR}_p(X+k) &=\frac{1}{1-p}\int_p^1 Q_{X+k}(u)\,du\\ &=\frac{1}{1-p}\int_p^1 \bigl(Q_X(u)+k\bigr)\,du\\ &=\operatorname{TVaR}_p(X)+\frac{k(1-p)}{1-p}\\ &=\operatorname{TVaR}_p(X)+k. \end{aligned}

Similarly, for $a\ge0$,

\operatorname{TVaR}_p(aX) =\frac{1}{1-p}\int_p^1 aQ_X(u)\,du =a\operatorname{TVaR}_p(X).

The factor $1-p$ in the denominator is important: the added constant contributes $k(1-p)$ to the tail integral, so its contribution to the average is exactly $k$, not $k(1-p)$. These calculations establish the properties for the stated quantile-integral definition, including when a quantile level falls inside an atom.

Identify the transformation.
Write the new loss as $X+k$, $aX$, or $k+aX$, keeping the fixed shift and multiplier distinct.

Transform the risk measure.
Apply translation invariance for the shift and positive homogeneity for the nonnegative multiplier.

Check the interpretation.
A fixed added loss changes the result dollar for dollar. A multiplier changes it in proportion to the multiplier.

Worked Examples

Worked Example: A Fixed Addition When the Tail Starts Inside an Atom

A hypothetical policy has loss $X$ equal to $0 with probability 0.6, $100 with probability 0.3, and $400 with probability 0.1. Find VaR and TVaR at $p=0.8$. Then find both measures after a fixed administrative loss of $50 is added to every outcome.

Method. Use the cumulative probabilities to find VaR, then average the upper 20% of quantiles for TVaR. The 80th percentile lies inside the probability mass at $100, so the tail average includes only the portion of that atom above the confidence level.

The cumulative probabilities are $F_X(0)=0.6$, $F_X(100)=0.9$, and $F_X(400)=1$. The first loss with cumulative probability at least 0.8 is $100, so $\operatorname{VaR}_{0.8}(X)=\$100$.

From quantile levels 0.8 to 0.9 the quantile is $100, and from 0.9 to 1 it is $400. Thus the tail integral is $0.1(100)+0.1(400)=10+40=50$. Dividing by the tail probability $1-0.8=0.2$ gives $\operatorname{TVaR}_{0.8}(X)=50/0.2=\$250$.

For $Y=X+50$, the possible losses are $50, $150, and $450 with the same probabilities. Therefore $\operatorname{VaR}_{0.8}(Y)=\$150$. Its upper-tail integral is $0.1(150)+0.1(450)=15+45=60$, so $\operatorname{TVaR}_{0.8}(Y)=60/0.2=\$300$.

Check. The shift raises VaR from $100 to $150 and TVaR from $250 to $300: both rise by exactly $50, as translation invariance requires. The fractional portion of the $100 atom in the tail is handled by integrating quantiles from 0.8, not by including the full atom.

Common mistake avoided: Including all 30% probability at $100 in the upper 20% tail would use too much probability mass and produce an incorrect TVaR.

Worked Example: Scaling and Shifting an Exponential Loss

Suppose a hypothetical repair loss $X$ has an exponential distribution with mean $100, so its rate is $0.01$ per dollar. A revised arrangement produces loss $Y=25+0.4X$. Find VaR and TVaR for $Y$ at $p=0.9$. Report money to the nearest cent.

Method. First calculate the two risk measures for the exponential loss, then apply the scale factor and fixed addition separately. For an exponential loss with mean $\theta$, the quantile is $Q_X(u)=-\theta\ln(1-u)$.

Here $\theta=100$. Therefore, $\operatorname{VaR}_{0.9}(X)=-100\ln(0.1)=100\ln(10)=\$230.2585$ before rounding.

For TVaR, integrate the quantile over the upper 10% of probability levels. With $t=1-u$, the result is $\operatorname{TVaR}_p(X)=\theta[1-\ln(1-p)]$. At $p=0.9$, this is $100[1-\ln(0.1)]=100[1+\ln(10)]=\$330.2585$.

Apply the transformation $Y=25+0.4X$. Its VaR is $25+0.4(230.2585)=25+92.1034=\$117.1034$, which rounds to $117.10. Its TVaR is $25+0.4(330.2585)=25+132.1034=\$157.1034$, which rounds to $157.10.

Check. The scale factor reduces the original measures, then the $25 fixed amount is added in full. The reported results preserve this order of operations.

Common mistake avoided: Calculating $0.4(\operatorname{VaR}_{0.9}(X)+25)$ would give $102.10, incorrectly scaling the fixed $25 addition. The correct transformation is $25+0.4\operatorname{VaR}_{0.9}(X)$.

Worked Example: Checking Both Properties for a Normal Loss

A hypothetical portfolio loss $X$ is normally distributed as $X\sim N(1000,200^2)$, with amounts in dollars. A stress scenario defines $Y=500+2X$. Find VaR and TVaR at $p=0.95$ for both $X$ and $Y$, using the standard normal values $z_{0.95}=1.644854$ and $\phi(z_{0.95})=0.103136$. Report money to the nearest cent.

Method. For a normal loss with mean $\mu$ and standard deviation $\sigma$, $\operatorname{VaR}_p=\mu+\sigma z_p$, where $z_p$ is the $p$-quantile of the standard normal distribution. The quantile-integral formula gives $\operatorname{TVaR}_p=\mu+\sigma\phi(z_p)/(1-p)$.

For the TVaR formula, write each normal quantile as $\mu+\sigma z_q$, then substitute $q=\Phi(z)$ in the integral. The tail contribution from the standardized quantiles is $\int_{z_p}^{\infty}z\phi(z)\,dz=\phi(z_p)$, since the derivative of $\phi(z)$ is $-z\phi(z)$.

For $X$, VaR is $1000+200(1.644854)=1000+328.9708=\$1,328.97$. TVaR is $1000+200(0.103136/0.05)$. Since $0.103136/0.05=2.06272$, this is $1000+412.544=\$1,412.54$.

For $Y=500+2X$, apply the fixed shift and multiplier: its VaR is $500+2(1,328.9708)=\$3,157.94$, and its TVaR is $500+2(1,412.544)=\$3,325.09$. Equivalently, $Y$ has mean $500+2(1000)=2500$ and standard deviation $2(200)=400$, which gives the same results from the normal formulas.

Check. Both transformed values equal $500 plus twice the corresponding value for $X$, confirming translation invariance and positive homogeneity together.

Common mistake avoided: The second parameter in $N(1000,200^2)$ is the variance, not the standard deviation. The standard deviation used in the formulas is $200.

Conditions and Common Traps

The multiplier must be nonnegative for the stated positive-homogeneity result. A negative multiplier reverses the ordering of outcomes, so the quantiles of $aX$ cannot generally be obtained by simply multiplying the quantiles of $X$ by $a$. Also distinguish a fixed amount added to every outcome from a random additional loss: only the former is the constant shift in translation invariance.

Do not scale the shift by accident. For $k+aX$, the risk measure is $k+a\rho(X)$, not $a(k+\rho(X))$.
Use the same confidence level. The shift and scale rules compare VaR or TVaR at the same $p$; changing $p$ is a separate change to the measure.
Respect the loss convention. These formulas treat larger values as larger losses. Recasting losses as gains can change signs and interpretations.
Keep the tail definition consistent. The TVaR result here uses the average of quantiles from $p$ to 1, including the appropriate fractional contribution at an atom.

These properties are useful for quick calculations and consistency checks: if every possible loss receives the same fixed surcharge, a reported VaR or TVaR should increase by that surcharge. If all losses are doubled, each measure should also double. This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the loss-based definitions and the stated quantile-integral convention. Answers are rounded only where indicated.

A hypothetical loss is $0 with probability 0.7, $60 with probability 0.2, and $300 with probability 0.1. Find VaR and TVaR at $p=0.8$. (Answer: VaR = $60; TVaR = $180.)
A loss $X$ is $20 or $80, each with probability 0.5. For $Y=X+15$, find VaR and TVaR at $p=0.75$. (Answer: VaR = $95; TVaR = $95.)
A hypothetical loss $X$ is uniform on the interval from $0 to $500. For $Y=10+0.6X$, find VaR and TVaR at $p=0.9$. (Answer: VaR = $280; TVaR = $295.)
A hypothetical exponential loss has mean $40. For $Y=8+1.25X$, find VaR and TVaR at $p=0.95$, to the nearest cent. (Answer: VaR = $157.79; TVaR = $207.79.)
A hypothetical normal loss is $X\sim N(500,100^2)$. At $p=0.9$, use $z_{0.9}=1.281552$ and $\phi(z_{0.9})=0.175498$ to find VaR and TVaR for $Y=3X$, to the nearest cent. (Answer: VaR = $1,884.47; TVaR = $2,026.49.)

What You'll Learn