Why Variance and Standard Deviation Are Not Coherent

Why Test These Measures Against the Axioms?

Variance and standard deviation summarize spread, so they may seem like natural measures of risk. But a risk measure used to represent the capital required for a loss must also respond sensibly to changes in the loss itself. In particular, adding a certain amount to every possible loss should increase required capital by that amount, and a loss that is always larger should not receive a lower risk assessment. These requirements are captured by translation invariance and monotonicity.

Use a loss-based convention throughout: $X$ is a random loss, and $\rho(X)$ is the risk assigned to it. For variance and standard deviation, assume $X$ has a finite second moment. The comparisons below concern distributions under a specified probability model, not estimates from a finite sample.

Definitions: A loss risk measure $\rho$ is translation invariant if $\rho(X+c)=\rho(X)+c$ for every deterministic amount $c$. It is positively homogeneous if $\rho(\lambda X)=\lambda\rho(X)$ for every $\lambda\geq 0$. It is monotone if $X\leq Y$ almost surely implies $\rho(X)\leq\rho(Y)$. The almost-sure ordering means the inequality holds in every modeled state except possibly states with probability zero.

These are three of the four coherence properties; the fourth is subadditivity. This tutorial tests only the three named properties. Write $\operatorname{Var}(X)=E[(X-E[X])^2]$ and $\operatorname{SD}(X)=\sqrt{\operatorname{Var}(X)}$. Variance is measured in squared currency units, while standard deviation is measured in currency units. That distinction matters when interpreting either quantity as a capital amount.

What Translation and Scaling Do

The effects of adding a fixed amount or scaling a loss follow directly from the definitions of variance and standard deviation. Let $c$ be a deterministic amount and let $\lambda\geq 0$. The following identities hold whenever $X$ has a finite second moment.

\begin{aligned} \operatorname{Var}(X+c) &=E\left[\left((X+c)-E[X+c]\right)^2\right]\\ &=E\left[\left((X+c)-(E[X]+c)\right)^2\right]\\ &=E\left[(X-E[X])^2\right] =\operatorname{Var}(X),\\[4pt] \operatorname{Var}(\lambda X) &=E\left[\left(\lambda X-E[\lambda X]\right)^2\right]\\ &=E\left[\left(\lambda(X-E[X])\right)^2\right] =\lambda^2\operatorname{Var}(X). \end{aligned}

The first identity holds because adding $c$ shifts both every outcome and the mean by the same amount. It therefore leaves each deviation from the mean unchanged. The second holds because scaling multiplies every deviation by $\lambda$, and squaring those deviations multiplies their squares by $\lambda^2$. Taking square roots gives $\operatorname{SD}(X+c)=\operatorname{SD}(X)$ and, for $\lambda\geq0$, $\operatorname{SD}(\lambda X)=\lambda\operatorname{SD}(X)$.

Result: Variance is unchanged by a deterministic addition, so it fails translation invariance. It scales quadratically, so it fails positive homogeneity in general. Standard deviation is also unchanged by a deterministic addition, so it fails translation invariance; but it scales linearly for nonnegative $\lambda$, so it satisfies positive homogeneity.

Worked Examples

Worked Example: Adding a Certain Loss Amount

A hypothetical insurer models a one-period loss $X$ as $20 with probability $0.5$ and $60 with probability $0.5$. A contract change adds a certain $8 to the loss in either outcome. Calculate the variance and standard deviation before and after the change, then test translation invariance for each measure.

Method. Calculate the mean, then the probability-weighted squared deviations from that mean. The contract change produces $X+8$. A common mistake is to add $8 to the variance or standard deviation because the loss itself increased by $8.

Before the change. The mean loss is $E[X]=0.5(20)+0.5(60)=40$. The variance is $0.5(20-40)^2+0.5(60-40)^2=0.5(400)+0.5(400)=400$ dollars squared. Thus the standard deviation is $\sqrt{400}=20$ dollars.

After the change. The two losses are $28 and $68, each with probability $0.5$. Their mean is $0.5(28)+0.5(68)=48$. The variance is $0.5(28-48)^2+0.5(68-48)^2=0.5(400)+0.5(400)=400$ dollars squared, and the standard deviation is $\sqrt{400}=20$ dollars.

Test. Translation invariance would require the new variance to be $400+8=408$, but it is 400. It would require the new standard deviation to be $20+8=28$, but it is 20. Both tests fail. The calculation is checked by the unchanged deviations: each outcome and the mean rose by $8, so the deviations remain $-20$ and $20$.

Worked Example: Scaling the Exposure

A hypothetical specialty portfolio has loss $X$ equal to $10 or $30, each with probability $0.5$. An exposure expansion multiplies each loss by $1.5$. Find the variance and standard deviation before and after expansion, and test positive homogeneity.

Method. Calculate the two measures for each distribution and compare the results with $1.5$ times their original values. The common mistake is to assume that variance, like a dollar loss, scales linearly.

Before expansion. The mean is $E[X]=0.5(10)+0.5(30)=20$. Therefore, $\operatorname{Var}(X)=0.5(10-20)^2+0.5(30-20)^2=0.5(100)+0.5(100)=100$ dollars squared. The standard deviation is $\sqrt{100}=10$ dollars.

After expansion. The losses are $1.5(10)=15$ and $1.5(30)=45$, each with probability $0.5$. Their mean is $0.5(15)+0.5(45)=30$. The variance is $0.5(15-30)^2+0.5(45-30)^2=0.5(225)+0.5(225)=225$ dollars squared. The standard deviation is $\sqrt{225}=15$ dollars.

Test. Positive homogeneity would require the new variance to be $1.5(100)=150$, not 225, so variance fails. For standard deviation, the required result is $1.5(10)=15$, which matches the calculation. Standard deviation satisfies this test. The result is also checked by the scaling identities: variance changes by $1.5^2=2.25$, while standard deviation changes by $1.5$.

Worked Example: A Larger Loss in Every State but Lower Dispersion

A hypothetical risk manager compares two portfolios under the same two equally likely economic states. Portfolio $X$ loses $0 in the first state and $100 in the second. Portfolio $Y$ loses $100 in the first state and $101 in the second. Test monotonicity for variance and standard deviation.

Method. First compare losses state by state; then calculate each measure from its own distribution. The trap is to infer the risk-measure ordering from the loss ordering without calculating what the measure captures.

Check the ordering. In the first state, $0\leq100$; in the second, $100\leq101$. Thus $X\leq Y$ in every modeled state. Monotonicity would require both $\operatorname{Var}(X)\leq\operatorname{Var}(Y)$ and $\operatorname{SD}(X)\leq\operatorname{SD}(Y)$.

Portfolio $X$. Its mean is $0.5(0)+0.5(100)=50$. Its variance is $0.5(0-50)^2+0.5(100-50)^2=0.5(2500)+0.5(2500)=2500$ dollars squared. Its standard deviation is $\sqrt{2500}=50$ dollars.

Portfolio $Y$. Its mean is $0.5(100)+0.5(101)=100.5$. Its variance is $0.5(100-100.5)^2+0.5(101-100.5)^2=0.5(0.25)+0.5(0.25)=0.25$ dollars squared. Its standard deviation is $\sqrt{0.25}=0.5$ dollars.

Conclusion and check. Although $X\leq Y$ state by state, $2500>0.25$ and $50>0.5$. Both variance and standard deviation therefore fail monotonicity. The outcomes of $Y$ are close together, which explains its low dispersion; neither measure records that its losses are higher in both states.

Summary of the Axiom Tests

Property	Variance	Standard deviation	Reason
Translation invariance	Fails	Fails	Adding a constant leaves dispersion unchanged, rather than adding that constant to the measure.
Positive homogeneity	Fails in general	Satisfies for $\lambda\geq0$	Variance scales by $\lambda^2$; standard deviation scales by $\lambda$.
Monotonicity	Fails	Fails	A state-by-state larger loss can have much less dispersion.

The failures have a practical interpretation. A deterministic increase in a portfolio's losses changes its mean but not its dispersion. A risk measure based only on dispersion therefore does not treat the increase as additional risk in the translation-invariant sense. Further, a portfolio can have consistently higher losses but lower dispersion, as in the third example. This is why neither variance nor standard deviation is a coherent loss risk measure.

Standard deviation does have one of the tested properties: it scales proportionally when exposure is multiplied by a nonnegative factor. That does not make it coherent, because coherence requires all the relevant axioms, not merely positive homogeneity. Variance also has a units problem when used directly as a capital amount: dollars squared cannot be interpreted as dollars of required capital. Standard deviation has dollar units, but correct units alone do not remedy its failures of translation invariance and monotonicity.

Common Exam Trap

A frequent error is to confuse “spread is unchanged by a constant shift” with translation invariance. They are not the same statement. Translation invariance for a loss risk measure requires $\rho(X+c)=\rho(X)+c$; unchanged variance or standard deviation gives $\rho(X+c)=\rho(X)$. For $c>0$, these equations disagree. Another error is to reject standard deviation's positive homogeneity because variance scales quadratically. Take the square root: for $\lambda\geq0$, $\sqrt{\lambda^2\operatorname{Var}(X)}=\lambda\operatorname{SD}(X)$.

Key takeaway: Variance fails translation invariance, positive homogeneity, and monotonicity. Standard deviation satisfies positive homogeneity for nonnegative scaling, but fails translation invariance and monotonicity. A measure that fails even one required coherence property is not coherent.

Check Your Understanding

For each hypothetical loss model, use the stated probabilities and report the requested property conclusion.

A loss is $5 or $25 with equal probability. Find its variance and standard deviation. (Answer: Mean $15; variance $100 dollars squared; standard deviation $10.)
A loss $X$ is $4 or $16 with equal probability. A certain $3 is added to every outcome. Calculate the new mean and variance, then state whether variance is translation invariant. (Answer: New mean $13; new variance $36 dollars squared, unchanged from the original; variance is not translation invariant.)
A loss $X$ is $8 or $20 with equal probability and is scaled by $2$. Find the original and scaled standard deviations and test positive homogeneity. (Answer: Original standard deviation $6; scaled standard deviation $12; standard deviation satisfies positive homogeneity.)
In two equally likely states, portfolio $A$ loses $20 and $80, while portfolio $B$ loses $80 and $82, respectively. Verify the state ordering and compare their standard deviations. (Answer: $A\leq B$ in both states; $\operatorname{SD}(A)=\$30$, $\operatorname{SD}(B)=\$1$; standard deviation violates monotonicity.)
Explain why the fact that standard deviation scales linearly does not make it coherent. (Answer: It satisfies positive homogeneity for nonnegative scaling but fails translation invariance and monotonicity, so it does not satisfy all coherence properties.)

What You'll Learn