Why Allocation at a Kink Needs Special Care
Marginal capital contributions are often introduced as derivatives: increase one business unit’s exposure slightly and measure the resulting change in aggregate capital. That interpretation is straightforward when the capital function is differentiable. It needs more care when the portfolio lies at a kink, or when the loss distribution has an atom at the VaR threshold. In those cases, the change can depend on which direction the portfolio moves, and a single gradient may not exist.
This matters in practice because an allocation may be used for pricing, performance assessment, or risk limits. A precise calculation should identify the risk measure, the direction of change, and any convention used to resolve ties. In particular, VaR is a quantile, not generally a maximum of scenario losses; the active-scenario subgradient rule for a maximum of affine functions cannot automatically be applied to VaR.
VaR with Discrete Scenario Losses
Suppose there are finitely many scenarios \(s\), with fixed probabilities \(p_s\), and scenario loss \(L_s(w)\). The loss distribution at weights \(w\) is discrete. At confidence level \(\alpha\), define \(\operatorname{VaR}_{\alpha}(w)=\inf\{x:P(L(w)\le x)\ge\alpha\}\). This is the lower quantile convention. At an atom, several scenarios can have the same loss as VaR, and their combined probability may straddle the quantile level.
For a useful directional calculation, assume scenario losses are affine in the weights near the portfolio: \(L_s(w+th)=L_s(w)+t a_s\), where \(a_s\) is the scenario’s loss change per unit \(t\) in direction \(h\). Let \(q\) be the VaR, \(p_<\) the probability of losses strictly below \(q\), and \(p_=\) the probability of losses equal to \(q\). If \(p_< < \alpha \le p_<+p_=\), the quantile lies within the mass at \(q\). For small positive \(t\), scenarios initially at \(q\) separate according to their slopes \(a_s\). The right directional derivative is the weighted quantile of those slopes at conditional rank \(r=(\alpha-p_<)/p_=\). This result assumes fixed scenario probabilities and that scenarios not tied at \(q\) remain separated from \(q\) for sufficiently small \(t\). It does not say that VaR is globally convex or that its derivative is a subgradient.
Worked Example: VaR Directional Derivatives Inside a Tied Atom
A hypothetical portfolio has two weights \(w_1,w_2\), each initially 1. Its four scenario losses, in thousands of dollars, are \(L_A=20w_1+20w_2\), \(L_B=60w_1+40w_2\), \(L_C=20w_1+80w_2\), and \(L_D=90w_1+90w_2\). Their probabilities are respectively 0.20, 0.25, 0.25, and 0.30. Find the 60% VaR and the right-sided marginal contributions for each weight.
Method. First identify the quantile and its tied scenarios; then rank the slopes of those tied scenarios in the direction being tested.
At \((w_1,w_2)=(1,1)\), losses are \(40,100,100,180\). Cumulative probability is 0.20 below 100 and 0.70 at or below 100. Therefore \(\operatorname{VaR}_{0.60}=100\) thousand dollars. The tied mass is \(p_=0.25+0.25=0.50\), so the conditional rank within that mass is \(r=(0.60-0.20)/0.50=0.80\).
For an increase in \(w_1\), the slopes of tied scenarios B and C are 60 and 20. In increasing order, slope 20 has probability 0.25 of the tied mass, and slope 60 brings the cumulative tied probability to 0.50. The 80th-percentile slope is therefore 60. The right-sided marginal contribution for \(w_1\) is 60 thousand dollars per unit weight.
For an increase in \(w_2\), the tied-scenario slopes are 40 for B and 80 for C. Slope 40 accounts for 0.25 of tied probability, and slope 80 brings it to 0.50. The 80th-percentile slope is 80. The right-sided marginal contribution for \(w_2\) is 80 thousand dollars per unit weight.
Check. For a small \(t>0\), increasing \(w_1\) gives tied losses \(100+60t\) and \(100+20t\). After scenario A’s probability of 0.20, the 60% quantile falls in scenario B, so its change per unit \(t\) is 60. Increasing \(w_2\) instead gives \(100+40t\) and \(100+80t\); the quantile falls in scenario C, giving 80. The calculations agree with the weighted-slope method.
The scenario selected at the quantile can change with the direction. Consequently, “the VaR scenario” need not specify a unique marginal contribution for every business unit. If a scenario-based allocation is required, one transparent convention is to choose a quantile scenario and allocate its loss across units using that scenario’s component losses. At a tie, the choice or weighting of scenarios must be stated. In this example, scenario B’s losses at the base weights are 60 and 40, which sum to its VaR loss of 100; scenario C’s losses are 20 and 80, which also sum to 100. Both reconcile to VaR, but they give different unit allocations. Neither allocation is implied uniquely by the VaR value alone.
Subgradients for Convex Kinks
A different situation arises when the aggregate capital function itself is a convex, piecewise-linear function on the full weight space. For example, a maximum of finitely many affine functions is convex. At a point where several affine functions attain the maximum, there may be no unique gradient, but the convex subdifferential describes the supporting slopes.
The full-space condition matters. If weights are restricted to a feasible convex domain \(C\), the subgradient inequality is required only for \(y\in C\). At a boundary point, this restricted subdifferential can be larger than the convex hull of active gradients. Under the usual convex-analysis convention it includes the domain’s normal cone, defined here as \(N_C(w)=\{n:n\mathbin{\cdot}(y-w)\le 0\text{ for all }y\in C\}\). For instance, on \(C=[0,\infty)\), \(K(x)=x\), and \(w=0\), the inequality \(y\ge gy\) for every \(y\ge0\) holds for all \(g\le1\); thus the restricted subdifferential is \((-\infty,1]\), not just \(\{1\}\).
Worked Example: Allocating a Convex Capital Function at a Kink
Let \(K(w)=\max\{40w_1+20w_2,\;20w_1+40w_2\}\) on the full weight space, with \(w_1=w_2=1\). Find the capital, the one-sided marginal contributions for \(w_1\), and one subgradient allocation.
At the stated weights, both affine functions equal \(40+20=60\) and \(20+40=60\), so \(K(1,1)=60\). Their gradients are \(g^{(1)}=(40,20)\) and \(g^{(2)}=(20,40)\).
Increasing \(w_1\) by \(t>0\) gives the two candidate values \(60+40t\) and \(60+20t\). Their maximum is \(60+40t\), so the right-sided marginal contribution is \((60+40t-60)/t=40\). For a decrease \(t>0\), set \(w_1=1-t\). The candidates become \(60-40t\) and \(60-20t\), whose maximum is \(60-20t\). The left-sided derivative with respect to \(w_1\) is therefore \([60-(60-20t)]/t=20\). The distinct one-sided values confirm the kink.
Every vector in the convex hull of the two active gradients is a subgradient. Taking their equal-weight average gives \(g=(30,30)\). The corresponding Euler contributions at \(w=(1,1)\) are \(w_1g_1=30\) and \(w_2g_2=30\), which sum to 60.
Check. The two one-sided slopes, 20 and 40, are the slopes of the two active affine pieces in the appropriate direction. The chosen subgradient allocation sums to \(K(w)=60\). It is one valid allocation under this rule, not the only possible one.
The active-gradient result follows from the supporting-slope condition. Each active affine function \(l_j\) satisfies \(l_j(w)=K(w)\), and \(K(y)\ge l_j(y)=K(w)+\nabla l_j\mathbin{\cdot}(y-w)\). Thus each active gradient is a subgradient, as is any convex combination of them. Conversely, for a finite maximum of affine functions on the full space, a vector outside their active-gradient convex hull cannot support the function at \(w\): a separating direction would make its proposed linear support exceed the maximum’s directional slope. This gives precisely the convex hull. The conclusion depends on the full-space setup; a restricted feasible domain changes the supporting inequalities as described above.
For a positively homogeneous convex capital function on the full space, a subgradient \(g\) at \(w\) satisfies \(g\mathbin{\cdot}w=K(w)\). Indeed, applying the subgradient inequality at \(y=0\) gives \(g\mathbin{\cdot}w\ge K(w)\), while applying it at \(y=2w\) gives \(g\mathbin{\cdot}w\le K(w)\). This explains why a subgradient can produce a full allocation in this setting. It does not establish that a subgradient exists for a nonconvex measure such as VaR, nor that an arbitrary finite-difference estimate must fully allocate capital.
Worked Example: A VaR Atom Exactly at the Quantile Boundary
Consider three hypothetical scenarios with probabilities 0.25, 0.25, and 0.50. Their losses are \(L_A=30w_1+20w_2\), \(L_B=10w_1+40w_2\), and \(L_C=60w_1+60w_2\), with weights initially \((1,1)\). Find the 50% VaR and the right- and left-sided marginal contributions for \(w_1\).
At the base weights, the losses are 50, 50, and 120. The tied mass at 50 is 0.50, and the 50% quantile is exactly at the end of that mass: the within-tie rank is \(r=0.50/0.50=1\). For a small increase \(t>0\) in \(w_1\), the tied losses become \(50+30t\) in A and \(50+10t\) in B. The 50% quantile is the larger of these two tied losses, because the quantile rank is the end of their combined mass. It is \(50+30t\), giving a right-sided marginal contribution of 30.
For a small decrease \(t>0\) in \(w_1\), the tied losses are \(50-30t\) and \(50-10t\). Scenario B is now the larger tied loss, so the 50% VaR is \(50-10t\). The left-sided derivative is \([50-(50-10t)]/t=10\). Thus the two one-sided contributions are 30 and 10.
Check. At the exact boundary rank, the perturbed quantile selects the largest loss among the scenarios in the tied mass. The selected scenario is A for an increase and B for a decrease, matching the two computed derivatives.
Common Mistakes and Practical Interpretation
- Using one tied scenario for every direction. As the first example shows, increasing different weights can select different tied scenarios and produce different VaR derivatives.
- Calling a directional derivative a subgradient. A directional derivative is a slope along a specified direction. A convex subgradient must satisfy a supporting inequality for all points in its domain. VaR need not be convex.
- Applying the active-gradient formula on a restricted domain without adjustment. The convex-hull formula stated here is for the full ambient space; boundary constraints can add normal-cone directions.
- Assuming every marginal contribution is unique or that the components automatically sum to aggregate capital. At a kink, one may select a subgradient or a scenario convention, but the choice and its reconciliation should be explicit.
A sound allocation analysis therefore reports the capital measure and confidence level, identifies any atom or active pieces, specifies the direction or allocation convention, and checks the resulting contributions against aggregate capital when the chosen method supports full allocation. These distinctions are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
Check Your Understanding
For each problem, state the relevant direction or convention and check how the tie or kink affects the result.
- Let \(K(w)=\max\{25w_1+15w_2,\;10w_1+30w_2\}\) at \(w=(1,1)\). Find the right- and left-sided marginal contributions for \(w_1\). (Answer: right-sided \(25\); left-sided \(10\).)
- A discrete loss has probability 0.30 strictly below its VaR atom and two scenarios tied at the atom with probabilities 0.40 and 0.30. At confidence level 0.60, their directional slopes are 12 and 20, respectively. Find the right directional derivative. (Answer: \(12\), since the required within-atom rank is \((0.60-0.30)/0.70=3/7\), reached in the first tied scenario.)
- On domain \([0,\infty)\), let \(K(x)=3x\). Using the subgradient inequality restricted to this domain, find \(\partial K(0)\). (Answer: \((-\infty,3]\).)
- At \(w=(2,1)\), let \(K(w)=\max\{5w_1+2w_2,\;3w_1+6w_2\}\). Choose the subgradient that gives weight 0.25 to the first active gradient and 0.75 to the second. Find its two Euler contributions and verify their sum. (Answer: \(g=(3.5,5)\); contributions \(7\) and \(5\), totaling \(12=K(w)\).)
- Two scenarios, each with probability 0.20, tie at a loss of 50; a third scenario has probability 0.60 and loss greater than 50. At confidence level 0.40, the tied scenarios’ slopes for an increase in one weight are 7 and 4. Find the right directional VaR derivative. (Answer: \(7\), since the quantile rank is at the end of the tied mass.)