Positive Homogeneity Under Exposure Scaling

Why Proportional Scaling Matters

An insurer may compare a portfolio with the same risks at a larger exposure level: twice as many otherwise comparable policies, for example, or twice the insured amount on each policy. A useful risk measure should respond in a predictable way when the underlying loss is multiplied by a nonnegative constant. Positive homogeneity expresses one specific version of that expectation: if every loss is multiplied by $c$, the measured risk is multiplied by $c$ as well.

This is a property of a risk measure applied to a scaled loss random variable. It does not say that every real-world change in exposure produces an exact multiple of the original loss distribution. That depends on how the underlying policy terms, frequency, severity, and other model features change.

Definition: A loss-based risk measure $\rho$ is positively homogeneous if, for every loss $X$ in its domain and every constant $c\geq 0$, $$ \rho(cX)=c\rho(X). $$ Here $cX$ means that each possible loss outcome is multiplied by $c$. The case $c=0$ is included: all outcomes become zero, and the risk measure must assign zero to the zero loss for the property to hold.

VaR and TVaR Under Loss Scaling

Use the lower-quantile convention for Value at Risk. For a loss $X$ with cumulative distribution function $F_X(x)=P(X\leq x)$, and confidence level $0<\alpha<1$, define

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

This definition matters when there are atoms, such as a positive probability of a loss being exactly zero. In that situation, the quantile is the first loss value at which the cumulative probability reaches or exceeds $\alpha$; it is not necessarily the next positive loss value.

For this tutorial, define Tail Value at Risk by the quantile-integral convention:

\operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du.

This definition applies to discrete as well as continuous loss distributions and handles probability atoms through the quantile function. It averages the upper part of the loss distribution in quantile terms; it is not simply the conditional mean given that a loss exceeds VaR when the distribution has an atom at the threshold.

Theorem: For $c\geq0$, both VaR and quantile-integral TVaR are positively homogeneous: $$ \operatorname{VaR}_{\alpha}(cX)=c\operatorname{VaR}_{\alpha}(X), \qquad \operatorname{TVaR}_{\alpha}(cX)=c\operatorname{TVaR}_{\alpha}(X). $$

Why Quantiles Scale Proportionally

For $c>0$, the distribution of the scaled loss satisfies $F_{cX}(y)=F_X(y/c)$. Indeed, $cX\leq y$ is equivalent to $X\leq y/c$. Consequently, the set of values $y$ for which $F_{cX}(y)\geq\alpha$ is the set of $c$ times the values $x$ for which $F_X(x)\geq\alpha$. Since multiplication by positive $c$ preserves order, the infimum of the first set is $c$ times the infimum of the second. Thus:

\operatorname{VaR}_{\alpha}(cX) =\inf\{y:F_X(y/c)\geq\alpha\} =c\inf\{x:F_X(x)\geq\alpha\} =c\operatorname{VaR}_{\alpha}(X).

The same argument applies to every quantile level $u$ between $\alpha$ and 1. Substituting $\operatorname{VaR}_{u}(cX)=c\operatorname{VaR}_{u}(X)$ into the TVaR definition gives:

\operatorname{TVaR}_{\alpha}(cX) =\frac{1}{1-\alpha}\int_{\alpha}^{1}c\operatorname{VaR}_{u}(X)\,du =c\operatorname{TVaR}_{\alpha}(X).

For $c=0$, the loss $cX$ is zero with probability one, so every quantile and the quantile-integral TVaR equal zero. The scaling identities therefore hold at $c=0$ as well.

Worked Example: Scaling VaR and TVaR for a Discrete Loss

A hypothetical portfolio has annual loss $X$ equal to $0 with probability 0.20, $100 with probability 0.50, and $300 with probability 0.30. Find its VaR and TVaR at $\alpha=0.60$. Then find both measures when the entire loss is scaled by $c=2$.

Method. Build the cumulative distribution and identify the lower quantile. For TVaR, integrate the quantile function over the upper 40% of probability. This method avoids treating the probability at a quantile boundary as though it were automatically above that boundary.

The cumulative probabilities at losses $0, $100, and $300 are 0.20, 0.70, and 1.00, respectively. Since $F_X(0)=0.20<0.60$ and $F_X(100)=0.70\geq0.60$,

\operatorname{VaR}_{0.60}(X)=\$100.

The quantile is $100 from probability level 0.20 up to 0.70, and $300 from level 0.70 up to 1. Therefore, the integral from 0.60 to 1 is:

\int_{0.60}^{1}\operatorname{VaR}_{u}(X)\,du =(0.70-0.60)(\$100)+(1-0.70)(\$300) =\$10+\$90=\$100.

Divide by $1-\alpha=0.40$:

\operatorname{TVaR}_{0.60}(X)=\frac{\$100}{0.40}=\$250.

After scaling by 2, the outcomes are $0, $200, and $600 with unchanged probabilities. Thus $\operatorname{VaR}_{0.60}(2X)=\$200$. The corresponding quantile integral is $(0.10)(\$200)+(0.30)(\$600)=\$20+\$180=\$200$, so $\operatorname{TVaR}_{0.60}(2X)=\$200/0.40=\$500$. The checks are $2(\$100)=\$200$ for VaR and $2(\$250)=\$500$ for TVaR, as required.

Common mistake avoided: The cumulative probability at $100 is 0.70, not 0.50. VaR is determined by cumulative probability, while the TVaR integral uses the portions of quantile levels above 0.60.

Variance Does Not Scale as a Degree-One Risk Measure

Variance is a measure of dispersion, but it responds to scaling differently from VaR and TVaR. For any random variable $X$ with finite variance and any constant $c$,

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

The reason is that scaling multiplies each deviation from the mean by $c$, and variance squares that deviation. Thus variance scales with the square of exposure, not proportionally with exposure. Unless $c=0$, $c=1$, or $\operatorname{Var}(X)=0$, it does not satisfy the positive-homogeneity identity $\operatorname{Var}(cX)=c\operatorname{Var}(X)$. Standard deviation, the square root of variance, does scale proportionally for $c\geq0$, but it is a different functional.

Worked Example: Testing Variance Under Scaling

A hypothetical annual loss is $0 with probability 0.25, $100 with probability 0.50, and $300 with probability 0.25. Calculate its variance in dollars squared, then calculate the variance after every loss is multiplied by 3.

Method. Calculate $E[X]$ and $E[X^2]$, then use $\operatorname{Var}(X)=E[X^2]-(E[X])^2$. This checks the degree of scaling directly rather than incorrectly applying a degree-one rule to variance.

The mean is:

E[X]=(0.25)(\$0)+(0.50)(\$100)+(0.25)(\$300)=\$0+\$50+\$75=\$125.

The second moment is:

E[X^2]=(0.25)(\$0^2)+(0.50)(\$100^2)+(0.25)(\$300^2) =\$0+\$5{,}000+\$22{,}500=\$27{,}500.

Therefore:

\operatorname{Var}(X)=\$27{,}500-(\$125)^2 =\$27{,}500-\$15{,}625 =11{,}875\text{ dollars}^2.

After multiplying losses by 3, the mean is $3(\$125)=\$375$, and the second moment is $9(\$27{,}500)=\$247{,}500$. Hence:

\operatorname{Var}(3X)=\$247{,}500-(\$375)^2 =\$247{,}500-\$140{,}625 =106{,}875\text{ dollars}^2.

Finally, $9(11{,}875)=106{,}875$, which verifies $\operatorname{Var}(3X)=3^2\operatorname{Var}(X)$. A degree-one calculation would give $3(11{,}875)=35{,}625$, which is incorrect.

Exposure Scaling Is Not Always Payment Scaling

The proportionality results above apply when the random variable being measured is scaled directly. A change to gross losses does not necessarily scale the insurer’s payment in the same way if contract terms remain fixed. A deductible or policy limit can change how much of the gross loss is retained or paid, so the net payment may not be $c$ times its original value.

For a hypothetical policy with per-loss deductible $d$ and payment limit $u$, payment for gross loss $X$ is $Y=\min((X-d)^+,u)$, where $(z)^+=\max(z,0)$. If gross losses are scaled but $d$ and $u$ are not, the new payment must be recalculated from the scaled gross loss. It cannot generally be found by multiplying the old payment by the exposure factor.

Worked Example: Fixed Contract Terms Break Payment Scaling

A hypothetical policy has gross loss $X$ equal to $0 with probability 0.40, $80 with probability 0.35, and $200 with probability 0.25. Its payment is $Y=\min((X-\$50)^+,\$100)$. Find $\operatorname{VaR}_{0.70}(Y)$. Then double gross losses while keeping the deductible and limit fixed, and find the new payment VaR.

Method. Apply the deductible and limit to each gross-loss outcome first; then form the payment distribution and calculate the lower quantile. This avoids assuming that scaled gross losses imply proportionally scaled payments.

For the original gross loss of $0, the payment is $0. For $80, the payment is $\min(\$80-\$50,\$100)=\$30$. For $200, it is $\min(\$200-\$50,\$100)=\$100$. Thus payments $0, $30, and $100 have probabilities 0.40, 0.35, and 0.25. The cumulative probability at $0 is 0.40, and at $30 it is $0.40+0.35=0.75$. Since $0.40<0.70\leq0.75$,

\operatorname{VaR}_{0.70}(Y)=\$30.

After doubling gross losses, the outcomes are $0, $160, and $400. With the same $50 deductible and $100 limit, the payments are $0, $\min(\$160-\$50,\$100)=\$100$, and $\min(\$400-\$50,\$100)=\$100$. The payment distribution is now $0 with probability 0.40 and $100 with probability 0.60. Its cumulative probability at $0 is 0.40, below 0.70, while at $100 it is 1.00. Therefore, the new payment VaR is $100.

If the payment itself had scaled by 2, its VaR would have been $2(\$30)=\$60$. The actual result is $100, not $60, because the fixed deductible and limit changed the payment mapping. The probability check also confirms that the quantile is not $0: the cumulative probability at $0 is only 0.40.

Exam Checks and Practical Interpretation

When asked whether a measure is positively homogeneous, first identify the random variable that is being scaled. If it is the loss $X$ itself, apply the definition $\rho(cX)=c\rho(X)$ and check the functional’s scaling rule. If the question changes exposure through insurance terms, construct the new payment distribution before evaluating the measure.

VaR: scales by $c$ for nonnegative $c$, using the lower-quantile definition.
Quantile-integral TVaR: scales by $c$, including when the loss distribution has atoms.
Variance: scales by $c^2$, so it is not generally positively homogeneous of degree one.
Payments with fixed terms: need not scale proportionally when gross loss exposure changes.

A frequent exam trap is to read a probability at a loss point as though it were already the cumulative probability. For example, if a loss is $0 with probability 0.70 and $200 with probability 0.30, then $F_X(0)=0.70$, so the lower $\operatorname{VaR}_{0.70}$ is $0. Doubling the loss still gives probability 0.70 at zero, so the scaled VaR remains $0, not $400. Always compute the cumulative distribution at the candidate quantile and verify that it reaches the confidence level there, but remains below it at all smaller values.

Key takeaway: Positive homogeneity is a proportionality test on the risk measure applied to the scaled loss. VaR and quantile-integral TVaR pass this test; variance scales quadratically. Fixed deductibles and limits can prevent gross exposure scaling from producing proportional payment scaling.

Check Your Understanding

For each problem, identify whether the calculation scales the loss itself or changes a payment with fixed terms. Use the lower-quantile convention where a VaR is requested.

A hypothetical loss is $20 with probability 0.20, $80 with probability 0.50, and $200 with probability 0.30. Find $\operatorname{VaR}_{0.60}(X)$ and the VaR after scaling all losses by 1.5. (Answer: $80 and $120.)
A loss is uniformly distributed between $0 and $100. Find $\operatorname{VaR}_{0.90}$ and $\operatorname{TVaR}_{0.90}$, then find both measures after scaling the loss by 4. (Answer: Original VaR $90 and TVaR $95; scaled VaR $360 and TVaR $380.)
A loss is $0 with probability 0.60, $50 with probability 0.20, and $150 with probability 0.20. Find $\operatorname{VaR}_{0.60}(X)$ and $\operatorname{VaR}_{0.60}(2X)$. (Answer: $0 and $0.)
A loss is $0 with probability 0.50 and $40 with probability 0.50. Find its variance and the variance after scaling all losses by 0.5. (Answer: $400 dollars squared and $100 dollars squared.)
A hypothetical policy pays $Y=\min((X-\$40)^+,\$150)$. Gross loss $X$ is $0 with probability 0.30, $100 with probability 0.40, and $300 with probability 0.30. Find $\operatorname{VaR}_{0.65}(Y)$, then double gross losses but keep the policy terms fixed and find the new payment VaR. (Answer: $60 and $150.)

What You'll Learn