Why Convexity Follows From Two Coherent-Risk Axioms
When an insurer combines or reallocates loss exposures, it is useful to compare the risk of the resulting portfolio with the risks of its component positions. Convexity provides a precise bound for a portfolio formed as a weighted average: its risk cannot exceed the same weighted average of the component risks. This is not an additional assumption when a risk measure is coherent. It follows directly from subadditivity and positive homogeneity.
The result depends on the weights being nonnegative and summing to one. Those conditions make the portfolio a convex combination of positions. The inequality is an upper bound; it does not say that the portfolio risk must equal the bound, or that it must be less than the risk of every component.
The four standard coherence properties for losses are monotonicity, translation invariance, positive homogeneity, and subadditivity. Convexity uses only the last two. Thus, the implication below does not require a separate argument involving monotonicity or cash additions. The positions must belong to a domain on which their weighted sums are defined and on which the two axioms hold.
Deriving the Convexity Inequality
Let \(X\) and \(Y\) be loss positions and choose \(0\leq\lambda\leq1\). The weights \(\lambda\) and \(1-\lambda\) are nonnegative, so positive homogeneity can be applied to each scaled position. Subadditivity can then be applied to their sum:
The first line is exactly subadditivity. The second line follows by applying positive homogeneity separately with scale factors \(\lambda\) and \(1-\lambda\). Together, these steps prove convexity. The endpoints also work: when \(\lambda=0\), the expression is \(\rho(Y)\), and when \(\lambda=1\), it is \(\rho(X)\).
For more than two positions, the same result follows by repeated use of the two-position inequality. One can also see the structure directly: repeated subadditivity bounds the risk of a sum by the sum of the individual risks, and positive homogeneity pulls each nonnegative weight outside its risk measure. The weights must sum to one for this to be a convex combination, although subadditivity and positive homogeneity can also give a bound for nonnegative weights whose sum is not one.
For the convexity bound, verify \(w_j\geq0\) for every position and \(\sum_j w_j=1\).
Apply the specified risk measure to each loss position, using the same units and risk convention.
Multiply each component risk by its portfolio weight and add the weighted amounts.
Calculate the weighted loss in each scenario and then apply the risk measure to that portfolio loss. The bound need not be attained.
Worked Portfolio Calculations
The following examples use a finite set of stress scenarios and the maximum-scenario-loss risk measure. For a vector \(X=(x_1,\ldots,x_m)\), define \(\rho(X)=\max_i x_i\). This measure is positively homogeneous and subadditive: scaling all scenario losses by a nonnegative amount scales their maximum, and the maximum of the scenario-by-scenario sums cannot exceed the sum of the separate maxima. Accordingly, it is convex. Scenario losses below are stated in thousands of dollars; no scenario probabilities are needed for this particular risk measure.
Worked Example: Two Positions With Different Worst Scenarios
Two hypothetical loss positions have scenario losses \(M=(9,2)\) and \(N=(1,7)\), in thousands of dollars. A portfolio assigns weight \(0.30\) to \(M\) and \(0.70\) to \(N\). Find the component risks, the convexity bound, and the exact portfolio risk.
Method and component risks. Under the maximum-scenario-loss measure, take the largest entry of each vector. Thus, \(\rho(M)=\max(9,2)=9\), and \(\rho(N)=\max(1,7)=7\).
Convexity bound. The weights are nonnegative and \(0.30+0.70=1\), so the convexity inequality applies:
Exact portfolio risk. Combine the losses within each scenario before taking the maximum. Scenario 1 gives \(0.30(9)+0.70(1)=2.70+0.70=3.40\). Scenario 2 gives \(0.30(2)+0.70(7)=0.60+4.90=5.50\). Therefore, the portfolio risk is \(\max(3.40,5.50)=5.50\) thousand dollars, in scenario 2.
Check and trap avoided. The exact risk \(5.50\) is no greater than the bound \(7.60\), as convexity requires. A common mistake is to combine losses across scenarios incorrectly or to report the bound as though it were the exact portfolio risk. Here the exact portfolio risk is \(5.50\), not \(7.60\).
Worked Example: Finding the Worst Scenario After Weighting
Two hypothetical positions have losses \(P=(4,10,3)\) and \(Q=(8,2,6)\), in thousands of dollars. The portfolio weights are \(0.60\) for \(P\) and \(0.40\) for \(Q\). Calculate the weighted risk bound and the exact risk, identifying the scenario that determines the exact risk.
Method and component risks. The maximum losses are \(\rho(P)=10\) and \(\rho(Q)=8\). The weights are nonnegative and sum to one, so the bound is \(0.60(10)+0.40(8)=6.00+3.20=9.20\) thousand dollars.
Exact portfolio losses by scenario. In scenario 1, the portfolio loss is \(0.60(4)+0.40(8)=2.40+3.20=5.60\). In scenario 2, it is \(0.60(10)+0.40(2)=6.00+0.80=6.80\). In scenario 3, it is \(0.60(3)+0.40(6)=1.80+2.40=4.20\).
The largest portfolio loss is \(\max(5.60,6.80,4.20)=6.80\) thousand dollars, attained in scenario 2. It satisfies \(6.80\leq9.20\).
Check and trap avoided. Scenario 1 is the worst scenario for \(Q\)'s combined contribution only in neither a way that determines the final result nor by inspection of the component maxima. The portfolio must be evaluated scenario by scenario. In particular, the exact risk is \(6.80\) in scenario 2, not scenario 1.
Worked Example: A Three-Position Portfolio
Three hypothetical positions have losses \(R=(6,1,4)\), \(S=(2,8,3)\), and \(T=(5,4,11)\), in thousands of dollars. Their respective portfolio weights are \(0.25\), \(0.25\), and \(0.50\). Find the weighted risk bound and the exact portfolio risk.
Method and component risks. Taking the maximum entry in each vector gives \(\rho(R)=6\), \(\rho(S)=8\), and \(\rho(T)=11\). The weights are nonnegative and \(0.25+0.25+0.50=1\). The weighted bound is
thousand dollars.
Exact portfolio risk. In scenario 1, the loss is \(0.25(6)+0.25(2)+0.50(5)=1.50+0.50+2.50=4.50\). In scenario 2, it is \(0.25(1)+0.25(8)+0.50(4)=0.25+2.00+2.00=4.25\). In scenario 3, it is \(0.25(4)+0.25(3)+0.50(11)=1.00+0.75+5.50=7.25\). Hence the exact risk is \(\max(4.50,4.25,7.25)=7.25\) thousand dollars, in scenario 3.
Check and trap avoided. The exact risk \(7.25\) is below the bound \(9.00\). A frequent arithmetic trap is to miscalculate one component maximum or a scenario-weighted loss. The component risks here are \(6\), \(8\), and \(11\), and scenario 3—not a component’s individual worst scenario taken on its own—determines the portfolio maximum.
What the Bound Does and Does Not Say
Convexity guarantees that the risk of a weighted-average position is at most the weighted average of the separate risks. It does not guarantee strict improvement, and equality is possible. For example, if the positions have losses that are identical in every scenario, taking a weighted average leaves that loss unchanged, so the bound is attained. More generally, whether the inequality is strict depends on the risk measure and the positions being combined.
Convexity also does not say that portfolio risk must be lower than each component risk. If one position has much higher risk than another, the weighted average of their risks can itself exceed the lower component risk. The meaningful comparison is with the weighted average of component risks, using the same weights as the portfolio.
A second trap is applying the convexity inequality when the weights are negative or do not sum to one. A negative weight is not a nonnegative scaling to which positive homogeneity in its standard form applies. Weights that do not sum to one do not describe a convex combination as written. Normalize them if the intended position is a weighted average, then calculate the bound using the normalized weights. If the weights represent unnormalized exposures instead, subadditivity and positive homogeneity may still supply a bound, but it is not the convex-combination statement unless the weights sum to one.
Worked Example: Normalizing Weights Before Using Convexity
A proposed weighted average uses weight \(0.20\) for position \(U\) and \(0.50\) for position \(V\). Their risks are \(10\) and \(6\) thousand dollars, respectively. Determine whether the weights can be used directly in the convexity inequality. Then normalize them and calculate the resulting bound.
Method. The proposed weights sum to \(0.20+0.50=0.70\), not \(1\). Therefore, they do not define a convex combination as written. Their normalized weights are \(0.20/0.70=2/7\) and \(0.50/0.70=5/7\), which are nonnegative and sum to \(1\).
Normalized convexity bound. The bound is
thousand dollars, rounded to two decimals. The calculation uses the normalized weights, not the original pair.
Check and trap avoided. The original weights fail the sum-to-one condition. A common arithmetic error is to add the weighted contributions incorrectly: \(20/7+30/7=50/7\), not \(58/7\). The normalized bound is approximately \(7.14\) thousand dollars.
Check Your Understanding
For each problem, use the maximum-scenario-loss measure where scenario vectors are given. All loss amounts are in thousands of dollars.
- Positions \(A=(3,9,2)\) and \(B=(7,4,5)\) receive weights \(0.40\) and \(0.60\). Find their component risks, the convexity bound, and the exact portfolio risk. (Answer: component risks \(9\) and \(7\); bound \(7.80\); exact risk \(6.00\) in scenario 2.)
- Positions \(C=(12,3)\) and \(D=(5,8)\) receive weights \(0.25\) and \(0.75\). Calculate the bound and the exact risk, and state whether the bound is attained. (Answer: bound \(9.00\); exact risk \(6.75\), tied across both scenarios; the bound is not attained.)
- Positions \(E=(2,8,1)\), \(F=(9,4,6)\), and \(G=(3,5,10)\) receive weights \(0.20\), \(0.30\), and \(0.50\). Calculate the weighted bound and the exact portfolio risk, including its scenario. (Answer: bound \(9.30\); exact risk \(7.00\) in scenario 3.)
- Positions \(H\) and \(J\) have risks \(14\) and \(4\). Proposed weights are \(0.30\) and \(0.50\). Can these weights be used as a convex combination as written? If normalized, what are the weights and risk bound? (Answer: no, they sum to \(0.80\); normalized weights \(3/8\) and \(5/8\); bound \(7.75\).)
- A candidate argues that convexity applies to weights \(1.20\) and \(-0.20\) because they sum to one. Is that sufficient? State which condition fails and why the convexity inequality cannot be invoked on that basis. (Answer: no; although the weights sum to one, \(-0.20\) is negative, so the weights are not a convex combination and positive homogeneity does not establish the stated bound for that negative scaling.)