Building Two-Outcome Counterexamples to Axiom Claims

Why Two-State Examples Are Useful

A proposed axiom can sound plausible in general terms but fail for a particular risk measure. A two-state model is often enough to expose the failure: it keeps the arithmetic manageable while making the role of probabilities, loss ordering, and joint outcomes visible. The aim is not to find a complicated model. It is to design the smallest example that meets the axiom’s conditions and contradicts its conclusion.

For loss variables, monotonicity means that if $X\leq Y$ in every state, then $\rho(X)\leq\rho(Y)$. Subadditivity means that for two losses $X$ and $Y$ defined on the same probability space, $\rho(X+Y)\leq\rho(X)+\rho(Y)$. A counterexample must satisfy the condition on the left and show that the claimed inequality on the right fails.

Definition: A two-state loss model has outcomes in states $A$ and $B$, with probabilities $p$ and $1-p$, where $0<p<1$. A loss $X$ is specified by its state-contingent values $x_A$ and $x_B$. To establish $X\leq Y$ state by state, verify both $x_A\leq y_A$ and $x_B\leq y_B$.

The states must be aligned when comparing losses or forming an aggregate. For example, if $X$ and $Y$ both depend on the same two-state event, then the value of $X+Y$ in state $A$ is $x_A+y_A$, and its value in state $B$ is $x_B+y_B$. The marginal distributions alone do not identify these joint outcomes; the pairing of outcomes matters.

A Construction Method

Begin with the axiom you want to test. For monotonicity, choose two loss vectors in which every outcome of $X$ is no larger than the corresponding outcome of $Y$. For subadditivity, specify both losses in each shared state, then calculate the risk of the aggregate as well as the risks separately. Choose probabilities that make the calculations simple, but do not assume that equal probabilities are necessary.

Write the axiom as an inequality.
For monotonicity, the required condition is $X\leq Y$ in every state and the claimed conclusion is $\rho(X)\leq\rho(Y)$. For subadditivity, compare $\rho(X+Y)$ with $\rho(X)+\rho(Y)$.

Choose state probabilities and losses.
Use two states and record each loss in each state. For monotonicity, check the loss ordering before calculating the risk measure. For subadditivity, make the joint state pairing explicit.

Calculate both sides completely.
Use the proposed formula for each loss and for the aggregate. A counterexample is valid only if the required condition holds and the resulting inequality is reversed.

Explain the mechanism.
Identify which probabilities or outcome changes make the risk measure move in the wrong direction. This distinguishes a genuine counterexample from an arithmetic coincidence.

A Useful Two-Outcome Calculation

Let a loss take values $a$ and $b$ with probabilities $p$ and $1-p$. Its mean is $pa+(1-p)b$. Its variance can be calculated directly from the definition, or by the equivalent expression below.

\begin{aligned} \operatorname{Var}(X) &=p\bigl(a-E[X]\bigr)^2+(1-p)\bigl(b-E[X]\bigr)^2\\ &=p(1-p)(b-a)^2. \end{aligned}

To verify the second line, substitute $E[X]=pa+(1-p)b$. Then $a-E[X]=-(1-p)(b-a)$ and $b-E[X]=p(b-a)$. Squaring and weighting gives $p(1-p)^2(b-a)^2+(1-p)p^2(b-a)^2=p(1-p)(b-a)^2$. This identity shows how both the probability split and the distance between outcomes determine variance. It also makes two-state counterexamples quick to check.

Worked Example: Monotonicity Failure for Variance

A proposed risk score is $\rho(X)=\operatorname{Var}(X)$, with losses measured in dollars squared. In state $A$, with probability $0.20$, let $X=0$ and $Y=4$. In state $B$, with probability $0.80$, let $X=10$ and $Y=10$. Test monotonicity.

Method. First check the required ordering in each state. In state $A$, $0\leq4$; in state $B$, $10\leq10$. Therefore $X\leq Y$ state by state, so monotonicity requires $\rho(X)\leq\rho(Y)$.

For $X$, the outcomes are $0$ and $10$, with probabilities $0.20$ and $0.80$. The two-outcome variance formula gives $\operatorname{Var}(X)=0.20(0.80)(10-0)^2=0.16(100)=16$.

For $Y$, the outcomes are $4$ and $10$, with the same probabilities. Thus $\operatorname{Var}(Y)=0.20(0.80)(10-4)^2=0.16(36)=5.76$. The proposed risk scores are $\rho(X)=16$ and $\rho(Y)=5.76$, so $\rho(X)>\rho(Y)$, contrary to the required inequality.

Check. The losses are ordered in every state, but the variance is larger for $X$. Replacing the loss of $0$ by $4$ in state $A$ narrows the gap between the two possible outcomes, reducing variance. The example disproves monotonicity for variance; it does not imply that every risk measure fails monotonicity.

Common mistake avoided. Comparing only the means gives $E[X]=8$ and $E[Y]=8.8$, then might tempt someone to infer that variance must also increase. Expected loss ordering is not a proof about variance. Calculate the measure being tested.

Worked Example: A Monotonicity Failure from a Variability Penalty

Consider the proposed score $\rho(X)=E[X]-0.5\sqrt{\operatorname{Var}(X)}$, with losses measured in dollars. State $A$ has probability $0.90$, and state $B$ has probability $0.10$. Let $X=0$ in both states. Let $Y=0$ in state $A$ and $Y=12$ in state $B$. Does this score satisfy monotonicity for this pair?

Method. Compare the losses state by state. They are equal in state $A$, and $0\leq12$ in state $B$; hence $X\leq Y$. Monotonicity would require $\rho(X)\leq\rho(Y)$.

Because $X$ is constant at zero, $E[X]=0$, $\operatorname{Var}(X)=0$, and $\rho(X)=0-0.5(0)=0$.

For $Y$, $E[Y]=0.90(0)+0.10(12)=1.20$. Its variance is $\operatorname{Var}(Y)=0.90(0.10)(12-0)^2=0.09(144)=12.96$, so its standard deviation is $\sqrt{12.96}=3.60$. Therefore $\rho(Y)=1.20-0.5(3.60)=1.20-1.80=-0.60$.

The result is $\rho(X)=0>-0.60=\rho(Y)$, which violates monotonicity. The occasional positive loss raises the mean but raises the variability penalty enough to lower this score. This pair isolates how a formula that subtracts variability can assign a lower score to a statewise larger loss.

Check. The standard deviation is $3.60$, not $12.96$; $12.96$ is the variance. Using the variance in place of the standard deviation would change the formula and give the wrong result.

Building a Subadditivity Counterexample

For subadditivity, the critical step is to specify which losses occur together. A convenient construction makes the two losses move together in both states. If they are identical, the aggregate has twice each outcome. That can make a nonlinear risk score grow faster than the sum of the separate scores.

Worked Example: Failure of Subadditivity for Mean Plus Variance

Suppose a proposed score is $\rho(X)=E[X]+0.10\operatorname{Var}(X)$, where losses are in dollars and the coefficient $0.10$ has units of inverse dollars. There are two equally likely states. In state $A$, two hypothetical portfolios each lose $0$; in state $B$, each loses $\$10$. Test subadditivity for the aggregate $X+Y$.

Method. Calculate the score for each portfolio, then calculate the aggregate in each state and apply the same formula. The losses are perfectly aligned: $X=Y=0$ in state $A$ and $X=Y=10$ in state $B$.

For either portfolio, $E[X]=0.50(0)+0.50(10)=5$. Its variance is $0.50(0.50)(10-0)^2=0.25(100)=25$. Thus $\rho(X)=5+0.10(25)=5+2.5=7.5$, and the same calculation gives $\rho(Y)=7.5$. The sum of the separate scores is $15$.

The aggregate loss is $0$ in state $A$ and $20$ in state $B$. Hence $E[X+Y]=0.50(0)+0.50(20)=10$, and $\operatorname{Var}(X+Y)=0.50(0.50)(20-0)^2=0.25(400)=100$. Therefore $\rho(X+Y)=10+0.10(100)=20$.

Subadditivity would require $20\leq15$, which is false. The variance term grows by a factor of four when the losses are doubled, while the mean only doubles. The common state-by-state movement preserves the full variability in the aggregate rather than diversifying it away.

Check. The calculation uses the same joint states for both portfolios. The example disproves subadditivity for this proposed formula under this pairing; it does not say that every pair of losses violates subadditivity.

Common Traps When Constructing Counterexamples

Using mean order instead of statewise order. Monotonicity requires $x_A\leq y_A$ and $x_B\leq y_B$. A comparison of $E[X]$ and $E[Y]$ alone does not establish that condition.
Changing probabilities without tracking them. The same outcome values can have very different variances under different state probabilities. State the probability attached to each outcome before calculating.
Ignoring the pairing of losses. Subadditivity concerns the actual aggregate $X+Y$. Its outcomes depend on which values of $X$ and $Y$ occur in the same state.
Showing only a numerical reversal. A valid counterexample must first satisfy the axiom’s premise, then show that its conclusion fails. State both parts explicitly.
Confusing a counterexample with a general theorem. One pair disproves a universal claim for the tested formula. It does not establish how the formula behaves for every distribution or prove that a different measure has the same defect.

Two-state examples are particularly useful as diagnostic tests: they make it possible to separate a risk measure’s response to the loss ordering from its response to probability spread or dependence. This is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus, where axiom claims must be tested against the exact mathematical conditions they assert.

Check Your Understanding

For each problem, specify the relevant states and show enough arithmetic to verify the stated comparison.

States $A$ and $B$ have probabilities $0.25$ and $0.75$. Let $X=(2,8)$ and $Y=(5,8)$, where each ordered pair lists the losses in states $A$ and $B$. For $\rho(Z)=\operatorname{Var}(Z)$, test monotonicity. (Answer: $X\leq Y$; $\operatorname{Var}(X)=6.75$, $\operatorname{Var}(Y)=1.6875$, so monotonicity fails.)
Let $X=0$ in both states. Let $Y=0$ with probability $0.80$ and $Y=15$ with probability $0.20$. For $\rho(Z)=E[Z]-0.60\sqrt{\operatorname{Var}(Z)}$, test monotonicity. (Answer: $E[Y]=3$, $\operatorname{Var}(Y)=36$, $\rho(Y)=3-0.60(6)=-0.60<\rho(X)=0$; monotonicity fails.)
Two equally likely states have losses $X=Y=0$ in state $A$ and $X=Y=8$ in state $B$. For $\rho(Z)=E[Z]+0.05\operatorname{Var}(Z)$, test subadditivity. (Answer: $\rho(X)=\rho(Y)=4.80$; $X+Y$ is $0$ or $16$, so $\rho(X+Y)=8+0.05(64)=11.20>9.60$; subadditivity fails.)
States $A$ and $B$ are equally likely. For $\rho(Z)=\operatorname{Var}(Z)$, let $X=(0,10)$ and $Y=(10,0)$. Calculate the aggregate variance and test subadditivity. (Answer: $\operatorname{Var}(X)=\operatorname{Var}(Y)=25$, while $X+Y=(10,10)$ has variance $0$; $0\leq50$, so this pair satisfies subadditivity.)
State $A$ has probability $0.10$, and state $B$ has probability $0.90$. For $\rho(Z)=\operatorname{Var}(Z)$, let $X=(0,6)$ and $Y=(2,6)$. Check the statewise ordering and compare the variances. (Answer: $X\leq Y$; $\operatorname{Var}(X)=0.10(0.90)(6^2)=3.24$, and $\operatorname{Var}(Y)=0.10(0.90)(4^2)=1.44$; monotonicity fails.)

What You'll Learn