Auditing a Risk Formula
Knowing the names of the coherence axioms is not enough when a proposed risk formula is placed in front of you. The practical task is to substitute losses that isolate each condition, keep the loss convention consistent, and compare the resulting quantities. One failed condition is sufficient to show that the formula is not coherent; a few successful numerical checks do not establish that it is coherent for every admissible loss.
Use \(X\) and \(Y\) for random losses, so a larger value means a worse financial outcome. Write \(\rho(X)\) for the risk assigned by a candidate functional \(\rho\). Unless a formula requires more, take the loss distributions to have the moments used in the calculation. For finite examples, every probability and loss amount will be specified.
The tests are deliberately different. For monotonicity, compare ordered losses outcome by outcome, not just their averages. For translation invariance, add the same constant in every outcome. For positive homogeneity, multiply all outcomes by the same nonnegative factor. For subadditivity, retain the joint outcomes of both losses when forming the portfolio sum; the separate marginal distributions alone may not determine its risk.
Check what the formula takes as input and what moments or other quantities it requires. Do not apply a variance formula to a distribution without a finite variance.
Use coordinatewise ordered losses for monotonicity, a constant shift for translation invariance, a nonnegative multiplier for homogeneity, and a jointly specified pair for subadditivity.
Compare \(\rho(X)\) with \(\rho(Y)\), \(\rho(X+c)\) with \(\rho(X)+c\), \( \rho(aX)\) with \(a\rho(X)\), or \(\rho(X+Y)\) with \(\rho(X)+\rho(Y)\), as appropriate.
A strict violation disproves the relevant axiom. Equality in one calculation, or success on a few examples, only confirms those cases; a general claim requires a general argument.
Cash Shifts and Scaling in a Formula
A formula can behave correctly under one operation and fail under another. Variance is unchanged by adding a sure amount, but it scales by the square of an exposure multiplier. That distinction makes mean-plus-variance formulas useful for practicing direct substitution.
Worked Example: Auditing a Mean-Plus-Variance Formula
Suppose losses are measured in thousands of dollars and a candidate formula is \(\rho(X)=E[X]+0.1\operatorname{Var}(X)\). Loss \(X\) is 2 or 6 with probability \(1/2\) each. Test translation invariance for a sure addition of 3, and positive homogeneity for a scaling factor of 2. The variance term is an illustrative formula component; the calculation does not assume this candidate is coherent.
Method. Calculate the mean and variance of the original loss, then recalculate the formula after each transformation. For an equally likely pair, the mean is the midpoint and the variance is the average squared deviation from that midpoint.
For \(X=(2,6)\), the mean is \((2+6)/2=4\). The deviations from 4 are \(-2\) and 2, so \(\operatorname{Var}(X)=[(-2)^2+2^2]/2=4\). Therefore, \(\rho(X)=4+0.1(4)=4.4\).
After adding 3 in every outcome, \(X+3=(5,9)\). Its mean is 7 and its variance is \([(-2)^2+2^2]/2=4\). Thus \(\rho(X+3)=7+0.1(4)=7.4\). The translation-invariance comparison is \(\rho(X)+3=4.4+3=7.4\), so this test passes.
After scaling by 2, \(2X=(4,12)\). Its mean is 8, and its deviations from 8 are \(-4\) and 4, giving variance \([(-4)^2+4^2]/2=16\). Hence \(\rho(2X)=8+0.1(16)=9.6\). But \(2\rho(X)=2(4.4)=8.8\), which is not 9.6. Positive homogeneity fails.
Check. The cash shift leaves the variance at 4 and increases the mean by exactly 3. Doubling the losses multiplies the variance by 4, not by 2; that is why the formula’s variance term prevents the scaling comparison from holding. A common mistake is to scale the variance linearly when exposure doubles.
Ordered Losses and Monotonicity
Monotonicity compares losses on the same outcomes. The requirement \(X\le Y\) almost surely means that \(X\) is no larger than \(Y\) in every outcome except possibly on an event of probability zero. It does not mean only that \(E[X]\le E[Y]\). A risk formula that includes a measure of dispersion can assign a smaller value to a less variable loss even when that loss is larger in every outcome.
Worked Example: Testing an Expectation-Plus-Standard-Deviation Formula
Consider \(\rho(Z)=E[Z]+2\sqrt{\operatorname{Var}(Z)}\), with losses in thousands of dollars. Two equally likely outcomes define \(X=(0,10)\) and \(Y=(9,10)\), with coordinates referring to the same two outcomes. Test monotonicity.
Method. First compare the losses outcome by outcome. Then calculate each mean and standard deviation using the population variance of the two equally likely outcomes.
The outcomes satisfy \(0\le9\) and \(10\le10\), so \(X\le Y\) in every outcome. For \(X\), the mean is 5. The deviations are \(-5\) and 5, so the variance is \([25+25]/2=25\), and the standard deviation is 5. Therefore, \(\rho(X)=5+2(5)=15\).
For \(Y\), the mean is \((9+10)/2=9.5\). The deviations are \(-0.5\) and 0.5, so the variance is \([0.25+0.25]/2=0.25\), and the standard deviation is 0.5. Therefore, \(\rho(Y)=9.5+2(0.5)=10.5\).
Monotonicity would require \(\rho(X)\le\rho(Y)\), but \(15>10.5\). Thus this functional fails monotonicity. The common trap is to see that \(E[X]<E[Y]\) and conclude that the formula must preserve the order; the standard-deviation term must also be included.
An ordered-pair calculation is especially effective as a disproof: once the outcome-by-outcome order and the strict reversal in risk are verified, no additional cases are needed to show failure. Conversely, a formula that preserves the order for this pair has not thereby passed monotonicity in general.
Portfolio Sums and Subadditivity
For subadditivity, the joint behavior of \(X\) and \(Y\) matters. In particular, variance of a sum includes a covariance term. Under finite second moments,
To see why, center each loss at its mean. The deviation of the sum from its mean is \((X-E[X])+(Y-E[Y])\). Squaring this expression produces the two squared deviations and the cross-product \(2(X-E[X])(Y-E[Y])\). Taking expectations gives the two variances and \(2\operatorname{Cov}(X,Y)\). This is why testing a portfolio formula requires joint outcomes or an explicitly stated dependence assumption.
Worked Example: A Variance Penalty Fails the Portfolio-Sum Test
Again measure losses in thousands of dollars and use \(\rho(Z)=E[Z]+0.1\operatorname{Var}(Z)\). In two equally likely states, let \(X=(0,2)\) and \(Y=(0,2)\). The coordinates represent the same states, so the losses move together. Test subadditivity.
Method. Calculate each stand-alone risk and the risk of the state-by-state portfolio sum. Keep the dependence aligned: in the first state both losses are 0, and in the second both are 2.
Each loss has mean \((0+2)/2=1\), variance \([(-1)^2+1^2]/2=1\), and risk \(1+0.1(1)=1.1\). Therefore, \(\rho(X)+\rho(Y)=1.1+1.1=2.2\).
The portfolio sum is \(X+Y=(0,4)\). Its mean is 2, and its variance is \([(-2)^2+2^2]/2=4\). Thus \(\rho(X+Y)=2+0.1(4)=2.4\). Since \(2.4>2.2\), subadditivity fails.
Check. Here \(\operatorname{Cov}(X,Y)=1\), so the portfolio variance is \(1+1+2(1)=4\), matching the direct calculation. The common mistake is to add the stand-alone variances and omit the covariance term. That shortcut would give 2 rather than the correct portfolio variance of 4.
When Direct Tests Are Enough
A finite calculation can settle a claim in one direction but not the other. If a candidate violates even one axiom for one admissible input, it is not coherent on a domain containing that input. To establish coherence, however, each axiom must hold throughout the stated domain. This usually calls for an algebraic argument or a result about the functional, not a collection of favorable examples.
For comparison, consider a finite set of scenarios indexed by \(j\) and the worst-scenario functional \(\rho(X)=\max_j X_j\). Its four checks can be expressed directly. If \(X_j\le Y_j\) for every \(j\), then \(\max_j X_j\le\max_j Y_j\). For a sure amount \(c\), \(\max_j(X_j+c)=\max_jX_j+c\). For \(a\ge0\), \(\max_j(aX_j)=a\max_jX_j\). Finally, for each scenario \(j\), \(X_j+Y_j\le\max_kX_k+\max_kY_k\); taking the maximum over \(j\) proves subadditivity. Here the same scenario set must be used in every comparison.
Worked Example: Checking a Worst-Scenario Portfolio Calculation
A hypothetical two-risk portfolio is represented across three common scenarios, with losses in thousands of dollars. Risk \(A\) has losses \((2,0,4)\), and risk \(B\) has losses \((1,3,2)\). For \(\rho(Z)=\max_j Z_j\), calculate the two stand-alone risks and the portfolio risk, then check the subadditivity inequality.
Method. Take the largest loss across the same three scenarios for each risk and for the scenario-by-scenario sum.
The stand-alone values are \(\rho(A)=\max(2,0,4)=4\) and \(\rho(B)=\max(1,3,2)=3\), so their sum is 7. The portfolio losses by scenario are \((2+1,0+3,4+2)=(3,3,6)\). Therefore, \(\rho(A+B)=\max(3,3,6)=6\), and \(6\le7\). Subadditivity holds for these inputs.
Check. The maximum of the combined scenario losses is 6, not the sum of the two worst losses unless those worst losses occur in the same scenario and align to produce that total. This calculation verifies this portfolio comparison; the preceding argument, rather than this single example, establishes the property for the stated finite-scenario functional.
Common Exam Traps
- Testing the wrong order. Monotonicity requires \(X\le Y\) outcome by outcome. Comparing only means does not test the condition.
- Changing more than one feature at once. Isolate the relevant operation: add a constant for translation invariance, or multiply by a nonnegative scalar for homogeneity.
- Dropping dependence in a portfolio. Calculate \(X+Y\) using joint outcomes. For variance-based formulas, ignoring covariance can change the result.
- Overclaiming from a successful example. One calculation can disprove an axiom, but success on one input does not prove it universally.
- Using a transformation outside its condition. Positive homogeneity is for \(a\ge0\); a negative multiplier reverses loss ordering and is not covered by that axiom.
Check Your Understanding
For each problem, use the stated loss convention and compare the relevant quantities directly.
- Loss \(Z\), in thousands of dollars, is equally likely to be 1 or 5. A candidate is \(\rho(Z)=E[Z]+0.05\operatorname{Var}(Z)\). Calculate \(\rho(Z+2)\) and \(\rho(Z)+2\), then compare \(\rho(3Z)\) with \(3\rho(Z)\). (Answer: \(\rho(Z+2)=5.20=\rho(Z)+2\); \(\rho(3Z)=10.80\), while \(3\rho(Z)=9.60\), so positive homogeneity fails.)
- For two equally likely states, let \(A=(1,13)\) and \(B=(10,13)\), and let \(\rho(W)=E[W]+1.5\sqrt{\operatorname{Var}(W)}\). Are the losses ordered, and does the functional satisfy monotonicity for this pair? (Answer: \(A\le B\) state by state, but \(\rho(A)=16\) and \(\rho(B)=13.75\); monotonicity fails.)
- Losses \(X\) and \(Y\), in thousands of dollars, both equal \((1,4)\) in the same two equally likely states. For \(\rho(W)=E[W]+0.2\operatorname{Var}(W)\), compare the combined risk with the sum of the stand-alone risks. (Answer: Each stand-alone risk is \(2.95\), so their sum is \(5.90\); \(X+Y=(2,8)\) has risk \(6.80\), so subadditivity fails.)
- Across three common scenarios, \(C=(4,1,3)\) and \(D=(0,5,2)\). For \(\rho(W)=\max_j W_j\), calculate \(\rho(C+D)\) and \(\rho(C)+\rho(D)\), and state whether the subadditivity inequality holds for this pair. (Answer: \(C+D=(4,6,5)\), so \(\rho(C+D)=6\); \(\rho(C)+\rho(D)=4+5=9\), and \(6\le9\).)
- Across the same three scenarios, \(C=(2,7,1)\). For \(\rho(W)=\max_j W_j\), calculate \(\rho(C+3)\) and \(2\rho(C)\), then say which equality the calculations test. (Answer: \(\rho(C+3)=10\) and \(2\rho(C)=14\); the first value tests translation invariance for \(c=3\), and the second is the scaled risk \(2\rho(C)\), not \(\rho(2C)\).)