Translation Invariance and the Meaning of Cash Additions

Why a Sure Addition Matters

A risk measure assigns a number to a random loss. If a certain amount is added to every possible loss, the risk measure should reflect that change in level. This property is called translation invariance. It is especially useful for checking whether a proposed risk measure responds sensibly to a known liability or cash amount.

Throughout, $X$ denotes a loss: larger values are worse. A sure addition of $c$ means the new loss is $X+c$, where $c$ is a constant, not a random variable. Thus, adding a positive $c$ increases every loss by the same amount. A certain cash resource that offsets losses instead changes the loss to $X-c$.

Definition: A loss-based risk measure $\rho$ is translation invariant if, for every loss $X$ in its domain and every constant $c$, $$ \rho(X+c)=\rho(X)+c. $$ The convention matters: this definition treats $c$ as an addition to the loss. If a sure cash amount $c$ offsets a loss, then $\rho(X-c)=\rho(X)-c$.

This property concerns the level of a loss, not its uncertainty. The whole distribution is shifted, but its shape and the random differences between outcomes are unchanged. VaR and TVaR inherit the shift in level. Variance, which measures dispersion around the mean, does not.

Quantiles Shift by Exactly the Added Amount

Let $F_X(x)=P(X\leq x)$ be the cumulative distribution function of $X$. For confidence level $0<\alpha<1$, define the loss quantile, or VaR, by the generalized inverse $ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\} $. This definition applies to discrete as well as continuous losses; no assumption of a continuous distribution is needed.

F_{X+c}(x)=P(X+c\leq x)=F_X(x-c)

The equation says that the distribution function of $X+c$ is the distribution function of $X$ shifted to the right by $c$. Therefore its first point at which the cumulative probability reaches $\alpha$ is also shifted right by $c$.

\operatorname{VaR}_{\alpha}(X+c) =\inf\{x:F_X(x-c)\geq\alpha\} =\operatorname{VaR}_{\alpha}(X)+c

The generalized inverse is important at atoms, where the distribution has jumps. The shift result still holds there; it does not depend on finding a density or interpolating between possible loss values.

Translation Invariance for TVaR

For a loss with a finite upper-tail integral, define TVaR at confidence level $\alpha$ by the quantile integral $ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_u(X)\,du $. This definition is suitable for discrete distributions as well as continuous ones. In particular, at an atom it averages the appropriate portion of the quantile range rather than automatically treating TVaR as the conditional mean above VaR.

Result: For any constant $c$, VaR and quantile-integral TVaR are translation invariant, provided the quantities are defined and the TVaR upper-tail integral is finite: $$ \operatorname{VaR}_{\alpha}(X+c)=\operatorname{VaR}_{\alpha}(X)+c, \qquad \operatorname{TVaR}_{\alpha}(X+c)=\operatorname{TVaR}_{\alpha}(X)+c. $$

The TVaR result follows directly from the quantile shift. Substituting the shifted quantile into the defining integral gives

\begin{aligned} \operatorname{TVaR}_{\alpha}(X+c) &=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_u(X+c)\,du\\ &=\frac{1}{1-\alpha}\int_{\alpha}^{1}\bigl(\operatorname{VaR}_u(X)+c\bigr)\,du\\ &=\operatorname{TVaR}_{\alpha}(X) +\frac{c(1-\alpha)}{1-\alpha}\\ &=\operatorname{TVaR}_{\alpha}(X)+c. \end{aligned}

The constant contributes $c$ at every quantile level in the tail. The interval of integration has length $1-\alpha$, so dividing by that same length leaves exactly $c$. This is why TVaR shifts by the full addition, not by a fraction of it.

Worked Example: A Discrete Loss With an Atom at the VaR Boundary

A hypothetical equipment portfolio has annual loss $X$ equal to $0 with probability 0.70, $100 with probability 0.20, or $500 with probability 0.10. Calculate VaR and TVaR at $\alpha=0.80$. Then calculate both measures if a certain $40 liability is added to every outcome. Use the quantile-integral definition of TVaR.

Method and distribution check. The probabilities sum to $0.70+0.20+0.10=1.00$. The cumulative probabilities are 0.70 at $0, 0.90 at $100, and 1.00 at $500. Thus the quantile equals $0 up to probability level 0.70, $100 from 0.70 to 0.90, and $500 from 0.90 to 1.

VaR before the addition. The cumulative probability first reaches 0.80 at $100, so $ \operatorname{VaR}_{0.80}(X)=\$100 $.

TVaR before the addition. The upper 20% of quantile levels consists of 10 percentage points at $100 and 10 percentage points at $500. Therefore

\operatorname{TVaR}_{0.80}(X) =\frac{0.10(\$100)+0.10(\$500)}{0.20} =\frac{\$10+\$50}{0.20} =\$300.

After the addition. The three possible losses become $40, $140, and $540, with the same respective probabilities. The cumulative probabilities are still 0.70, 0.90, and 1.00. Hence $ \operatorname{VaR}_{0.80}(X+40)=\$140 $, and

\operatorname{TVaR}_{0.80}(X+40) =\frac{0.10(\$140)+0.10(\$540)}{0.20} =\frac{\$14+\$54}{0.20} =\$340.

Check and common mistake avoided. VaR increased by $40, from $100 to $140, and TVaR increased by $40, from $300 to $340. Using only the $500 outcome for TVaR would incorrectly discard the part of the tail between the 80th and 90th percentiles; the quantile integral includes that $100 portion.

Variance Does Not Measure a Shift in Level

For a loss with finite variance, the variance risk measure is $ \rho(X)=\operatorname{Var}(X) $. By definition, $ \operatorname{Var}(X)=E[(X-E[X])^2] $. Adding a constant changes both each outcome and the mean by the same amount, leaving every deviation from the mean unchanged:

\operatorname{Var}(X+c) =E\left[\bigl((X+c)-(E[X]+c)\bigr)^2\right] =E[(X-E[X])^2] =\operatorname{Var}(X).

Thus variance does not satisfy loss-based translation invariance: in general, it is not true that $ \operatorname{Var}(X+c)=\operatorname{Var}(X)+c $. This also reflects a difference in what is being measured. VaR and TVaR are expressed in units of loss, while variance is in squared units of loss.

Worked Example: Testing Variance Against a Sure Addition

A hypothetical warranty pool has annual loss $X=\$20$ with probability 0.75 and $X=\$100$ with probability 0.25. A certain $30 administrative liability is then added to every outcome. Calculate the mean and variance before and after the addition, and test translation invariance if the proposed risk measure is $\rho(X)=\operatorname{Var}(X)$.

Method. First calculate the mean, then sum the probability-weighted squared deviations from that mean. The probabilities sum to $0.75+0.25=1.00$.

Before the addition. The expected loss is $E[X]=0.75(\$20)+0.25(\$100)=\$15+\$25=\$40$. The deviations from the mean are $-\$20$ and $\$60$, so

\operatorname{Var}(X) =0.75(-\$20)^2+0.25(\$60)^2 =0.75(\$400)+0.25(\$3{,}600) =\$300+\$900 =1{,}200\text{ dollars}^2.

After the addition. The possible losses are $50 and $130. Their mean is $E[X+30]=0.75(\$50)+0.25(\$130)=\$37.50+\$32.50=\$70$. The deviations are again $-\$20$ and $\$60$, and therefore

\operatorname{Var}(X+30) =0.75(\$400)+0.25(\$3{,}600) =1{,}200\text{ dollars}^2.

The translation-invariance requirement would give $ \rho(X+30)=\rho(X)+\$30=1{,}200+\$30 $, which is not a meaningful equality of units: the variance is in dollars squared and does not change. The calculation confirms that variance fails this property. The common mistake is to add the sure liability directly to the variance; adding $30 to outcomes shifts the mean by $30 but does not alter their deviations from the mean.

Interpreting Cash Additions and Offsets

The sign of the change depends on whether the amount is added to the loss or used to offset it. If a new certain expense increases every loss by $c>0$, translation invariance gives $ \rho(X+c)=\rho(X)+c $ for a translation-invariant loss measure. If a certain reserve pays $c$ toward every loss, the net loss becomes $X-c$, so the risk measure falls by $c$, subject to its domain.

A common exam trap is to switch between loss and asset language without changing the sign. For a loss $X$, writing “$25 of certain cash is added” could mean either a $25 cash liability is added to loss, $X+25$, or a $25 resource offsets the loss, $X-25$. These are different scenarios:

\begin{aligned} \text{Added liability: }&\rho(X+\$25)=\rho(X)+\$25,\\ \text{Cash offset: }&\rho(X-\$25)=\rho(X)-\$25. \end{aligned}

For example, if a calculation starts with $\operatorname{VaR}_{0.95}(X)=\$800$, adding a certain $25 liability gives $\operatorname{VaR}_{0.95}(X+\$25)=\$825$. The incorrect calculation $\$800-\$25=\$775$ treats the liability as if it were cash available to reduce the loss.

Translation invariance is a test of how a risk measure treats a deterministic change in loss level; it is not a claim that the uncertain part of the loss has changed. When applying it, identify the loss convention, state whether the constant is added or deducted, and then apply the corresponding sign consistently. These distinctions are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, use the loss-based sign convention and the stated risk-measure definition.

A hypothetical liability has losses of $0, $80, and $200 with probabilities 0.60, 0.25, and 0.15. What are $\operatorname{VaR}_{0.80}(X)$ and $\operatorname{VaR}_{0.80}(X+12)$? (Answer: $80 and $92.)
A discrete loss is $0 with probability 0.50, $40 with probability 0.30, and $160 with probability 0.20. Calculate $\operatorname{TVaR}_{0.75}(X)$ using the quantile integral, then calculate it after adding a certain $20. (Answer: $136 and $156.)
Suppose $X$ is exponential with rate $\lambda=0.05$ per dollar, so $F_X(x)=1-e^{-0.05x}$ for $x\geq0$. At $\alpha=0.80$, calculate VaR and TVaR after adding a certain $10. Report dollars to the cent. (Answer: VaR $42.19; TVaR $62.19.)
A loss is $10 with probability 0.40 and $50 with probability 0.60. Calculate $\operatorname{Var}(X)$ and $\operatorname{Var}(X+25)$. Does variance satisfy translation invariance for this addition? (Answer: Both variances are 384 dollars squared; no.)
For any loss $X$ with finite variance, give the value of $\operatorname{Var}(X-c)-\operatorname{Var}(X)$ for a constant $c$, and state why this does not equal the translation-invariance change required of a loss-based risk measure. (Answer: The difference is 0; translation invariance would require a change of $-c$.)

What You'll Learn