Distributionally robust expectation inequalities for structured distributions

Abstract

Quantifying the risk of unfortunate events occurring, despite limited distributional information, is a basic problem underlying many practical questions. Indeed, quantifying constraint violation probabilities in distributionally robust programming or judging the risk of financial positions can both be seen to involve risk quantification under distributional ambiguity. In this work we discuss worst-case probability and conditional value-at-risk problems, where the distributional information is limited to second-order moment information in conjunction with structural information such as unimodality and monotonicity of the distributions involved. We indicate how exact and tractable convex reformulations can be obtained using standard tools from Choquet and duality theory. We make our theoretical results concrete with a stock portfolio pricing problem and an insurance risk aggregation example.

Appendices

Equality constrained quadratic programs (QPs)

We will state here a relevant result concerning equality constrained QPs used throughout the rest of this paper. Assume we define a function $I:\mathbb {R}^d\rightarrow \mathbb {R}$ as follows

$$\begin{aligned} \begin{aligned} I(b) :=\min _{x\in \mathbb {R}^n}&\quad x^\top G x + 2 x^\top c + y\\ \mathrm {s.t.}&\quad A x = b, \end{aligned} \end{aligned}$$

with $A \in \mathbb {R}^{d \times n}$ having full row rank and G positive semidefinite. It is assumed that the function $x^\top G x + 2 x^\top c$ in bounded from below such that $I(b)>\infty$. We can now represent the quadratic function I using a dual representation as indicated in the following theorem.

Theorem 5

(Parametric representation of I) The function I is lower bounded by

$$\begin{aligned} I(b) \ge b^\top T_1 b + 2 b^\top T_2 + T_3 \end{aligned}$$

(9)

for all $T_1 \in \mathbb {S}^{d}$, $T_2 \in \mathbb {R}^d$ and $T_3 \in \mathbb {R}$ such that there exist $\varLambda _1 \in \mathbb {R}^{d\times d}$, $\varLambda _2 \in \mathbb {R}^d$ with

$$\begin{aligned} \begin{pmatrix} \varLambda _1 + \varLambda _1^\top - T_1 &{} \varLambda _2 - T_2 &{} - \varLambda _1^\top A \\ \varLambda _2^\top - T_2^\top &{} y - T_3 &{} c^\top - \varLambda _2^\top A \\ - A^\top \varLambda _1 &{} c - A^\top \varLambda _2 &{} G \end{pmatrix} \succeq 0. \end{aligned}$$

(10)

Moreover, inequality (9) is tight uniformly in $b\in \mathbb {R}^d$ for some $T_1$, $T_2$ and $T_3$ satisfying condition (10).

Proof

The Lagrangian of the optimization problem defining I(b) is given as

$$\begin{aligned} \mathcal L (x, \lambda ) :=x^\top G x + 2 x^\top \left( c + A^\top \lambda \right) - 2\lambda ^\top b + y. \end{aligned}$$

As $x^\top G x + 2 x^\top c$ is bounded from below on $\mathbb {R}^n$, we have that for all $b \in \mathbb {R}^d$ there exists a minimizer $x^\star$ such that $I(b) =(x^\star )^\top G x^\star + 2 (x^\star )^\top c + y$ and $Ax^\star =b$. From the first order optimality conditions for convex QPs [14, Lemma 16.1], we have that $\min _x \max _\lambda ~ \mathcal {L}(x, \lambda ) = \mathcal L(x^\star , \lambda ^\star ) = \max _\lambda \min _x ~ \mathcal {L}(x, \lambda )$ where the saddle point $(x^\star , \lambda ^\star )$ is any solution of the linear system

$$\begin{aligned} \begin{pmatrix} G &{} A^\top \\ A &{} 0 \end{pmatrix} \begin{pmatrix} x^\star \\ \lambda ^\star \end{pmatrix} = \begin{pmatrix} -c \\ b \end{pmatrix}. \end{aligned}$$

(11)

The quadratic optimization problem $\max _x \, \mathcal L(x, \lambda ^\star )$ admits a maximizer if and only if $(c + A^\top \lambda ^\star )$ is in the range of G. It must thus hold that

$$\begin{aligned} \left( \mathbb {I}_{d} - G G^{\dagger }\right) \left( c+A^\top \lambda ^\star \right) = 0. \end{aligned}$$

(12)

Hence when dualizing the problem defining I(b), we get its dual representation $I(b) = \max _\lambda ~ - \left( c + A^\top \lambda \right) ^\top G^{\dagger } \left( c + A^\top \lambda \right) - 2 \lambda ^\top b + y.$ From equation (11) it follows that $\lambda ^\star$ is any solution of the linear equation $b + A G^{\dagger } A^\top \lambda ^\star + A G^{\dagger } c =0$. Therefore there exists an affine $\lambda ^\star (b) = - \varLambda _1^\star b - \varLambda ^\star _2$ with $\varLambda _1^\star \in \mathbb {R}^{d\times d}$ and $\varLambda _2^\star \in \mathbb {R}^d$ such that

$$\begin{aligned} I(b)= & {} -\left( c - A^\top \varLambda _1^\star b - A^\top \varLambda _2^\star \right) ^\top G^{\dagger } \left( c - A^\top \varLambda _1^\star b - A^\top \varLambda _2^\star \right) \nonumber \\&+\,2 b^\top {\varLambda _1^\star }^\top b +2{\varLambda _2^\star }^\top b + y. \end{aligned}$$

(13)

From equation (12) it follows that for all b in $\mathbb {R}^d$ it holds that $\left( \mathbb {I}_{d} - G G^{\dagger }\right) \left( c-A^\top \varLambda _1^\star b - A^\top \varLambda _2^\star \right) \text{= } \text{0 }.$ We must hence also have that

$$\begin{aligned} \left( \mathbb {I}_{d} - G G^{\dagger }\right) \left( -A^\top \varLambda _1^\star ,\, c - A^\top \varLambda _2^\star \right) = 0. \end{aligned}$$

(14)

The dual reprenstation of I(b) guarantees that for all $\lambda (b) = - \varLambda _1 b - \varLambda _2$ with $\varLambda _1 \in \mathbb {R}^{d\times d}$ and $\varLambda _2 \in \mathbb {R}^d$

$$\begin{aligned} I(b)\ge & {} -\left( c - A^\top \varLambda _1 b - A^\top \varLambda _2\right) ^\top G^{\dagger } \left( c - A^\top \varLambda _1 b - A^\top \varLambda _2 \right) \\&+\,2 b^\top \varLambda _1^\top b + 2 \varLambda _2^\top b \end{aligned}$$

Lower bounding the right hand side of the previous inequality with $b^\top T_1 b + 2 T_2^\top b + T_3$ yields $I(b) \ge b^\top T_1 b + 2 T_2^\top b + T_3$ if for all b in $\mathbb {R}^d$ it holds that

$$\begin{aligned}&\begin{pmatrix} b \\ 1 \end{pmatrix}^\top \left[ \begin{pmatrix} \varLambda _1+\varLambda _1^\top - T_1 &{} \varLambda _2-T_2 \\ \varLambda _2^\top -T_2^\top &{} y-T_3\end{pmatrix} \right. \\&\quad \left. - \begin{pmatrix} - \varLambda _1^\top A \\ c^\top - \varLambda _2^\top A \end{pmatrix} G^{\dagger } \begin{pmatrix} -A^\top \varLambda _1&c - A^\top \varLambda _2 \end{pmatrix}\right] \begin{pmatrix} b \\ 1 \end{pmatrix} \ge 0 \end{aligned}$$

and

$$\begin{aligned} \left( \mathbb {I}_{d} - G G^{\dagger }\right) \left( -A^\top \varLambda _1, \,c - A^\top \varLambda _2\right) =0. \end{aligned}$$

After a Schur complement [7, Thm 4.3], we obtain the first part of the theorem

$$\begin{aligned}&\exists \varLambda _1, \varLambda _2 : \quad \begin{pmatrix} \varLambda _1 + \varLambda _1^\top - T_1 &{} \varLambda _2 - T_2 &{} - \varLambda _1^\top A \\ \varLambda _2^\top - T_2^\top &{} y - T_3 &{} c^\top - \varLambda _2^\top A \\ - A^\top \varLambda _1 &{} c - A^\top \varLambda _2 &{} G \end{pmatrix} \succeq 0 \implies I(b) \\&\quad \ge b^\top T_1 b + 2 T_2^\top b + T_3. \end{aligned}$$

As I(b) is a quadratic function there exist $T_1^\star$, $T_2^\star$ and $T_3^\star$ such that $I(b)=b^\top T_1^\star b + 2{T_2^\star }^\top b + T_3^\star$. The equations (13) and (14) guarantee [7, Thm 4.3] that

$$\begin{aligned} \begin{pmatrix} \varLambda _1^\star + {\varLambda _1^\star }^\top - T_1^\star &{} \varLambda _2^\star - T_2^\star &{} - {\varLambda _1^\star }^\top A \\ {\varLambda _2^\star }^\top - {T_2^\star }^\top &{} y - T_3^\star &{} c^\top - {\varLambda _2^\star }^\top A \\ - A^\top \varLambda _1^\star &{} c - A^\top \varLambda _2^\star &{} G \end{pmatrix} \succeq 0 \end{aligned}$$

completing the proof. $\square$

Proofs

Proposition 2:

Proof

The statement can be proved almost immediately from the definition of convexity. For all $\theta \in [0, 1]$

$$\begin{aligned} \begin{aligned} L_s(\theta a + (1-\theta ) b)&= \int _{0}^\infty L(\lambda (\theta a + (1-\theta ) b)) ~T(\mathrm {d}\lambda ) \\&= \int _{0}^\infty L(\theta (\lambda a) + (1-\theta ) (\lambda b)) ~T(\mathrm {d}\lambda ) \\&\le \int _{0}^\infty \theta L( \lambda a) + (1-\theta ) L(\lambda b) ~T(\mathrm {d}\lambda ) \\&= \theta L_s(a) + (1-\theta ) L_s(b) \end{aligned} \end{aligned}$$

showing convexity of $L_s$. $\square$

Corollary 1:

From Theorem 2, we have that the generating distribution T for $\alpha$-unimodal ambiguity sets satisfies

$$\begin{aligned} T([0, t]) = \alpha \, \int _{0}^t \lambda ^{\alpha -1} ~\mathrm {d}\lambda , \quad \forall t\in [0,1]. \end{aligned}$$

The moment transformations from Theorem 1 become

$$\begin{aligned} \begin{aligned} \mu _\alpha&:= \left[ \int _0^\infty \lambda \,T(\mathrm {d}\lambda )\right] ^{{-1}}\,\, \mu&= \left[ \alpha \int _0^1 \lambda ^\alpha (\mathrm {d}\lambda )\right] ^{{-1}}\,\, \mu&= \frac{\alpha +1}{\alpha } \mu \\ S_\alpha&:= \left[ \int _0^\infty \lambda ^2\,T(\mathrm {d}\lambda )\right] ^{{-1}}\,\, S&= \left[ \alpha \int _0^1 \lambda ^{\alpha +1}(\mathrm {d}\lambda )\right] ^{{-1}}\,\, S&=\frac{\alpha +2}{\alpha } S. \end{aligned} \end{aligned}$$

From Proposition 1, the transformed loss function $L_s$ required in Theorem 1 can be found as

$$\begin{aligned} \begin{aligned} L_s(y)&= \max _{i\in \mathcal I} ~ T\left( \left[ b_i/ a_i^\top y, \infty \right) \right) \cdot {\mathbb {1}}_{a_i^\top x \ge b_i}(y) \\&=: \max _{i\in \mathcal I}f_i(a_i^\top y), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} f_i(q) = {\left\{ \begin{array}{ll} \displaystyle \alpha \int _{b_i/q}^1 \lambda ^{\alpha -1}~\mathrm {d}\lambda ,\, &{} q \ge b_i,\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

In order to apply Theorem 4, we now need only reformulate the semi-infinite constraint ($\mathcal C_2$), i.e. the constraint

$$\begin{aligned} T_{3, i} + 2 q T_{2, i} + q^2 T_{1, i} \ge f_i(q) \quad ~\forall q\in \mathbb {R}, \quad \forall i\in \mathcal I. \end{aligned}$$

Because $0\in \varXi$ and hence $b_i > 0$, we have equivalently, for each $i\in \mathcal I$, and for all $q \in \mathbb {R}^+$

$$\begin{aligned} T_{3, i} + 2 q T_{2, i} + q^2 T_{1, i} \ge {\left\{ \begin{array}{ll} \displaystyle 1 - (b_i/q)^\alpha &{} q \ge b_i,\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

which can be seen to reduce to

$$\begin{aligned} T_{3, i} + 2 q T_{2, i} + q^2 T_{1, i} \ge 1 - \frac{b_i^\alpha }{q^\alpha }, \quad \forall q\ge 0. \end{aligned}$$

Defining a new scalar variable ${\tilde{q}}$ and applying the variable substitution ${\tilde{q}}^{\,w} = q$, this can be rewritten as

$$\begin{aligned} {\tilde{q}}^{2 \, w+ v} T_{1, i} + 2 {\tilde{q}}^{w+ v} T_{2, i} + {\tilde{q}}^{v} (T_{3, i} - 1) + b_i^\alpha \ge 0, \quad \forall \tilde{q}\ge 0 \end{aligned}$$

after multiplying both sides with ${\tilde{q}}^v> 0$. The final result is obtained after the substitution $b_i^{1/w} \bar{q} = \tilde{q}$.

Corollary   2: The method of proof follows that of Corollary 1, except that we now apply Proposition 3 to generate the transformed loss function $L_s$.

In this case the loss function L is equivalent to $L = \ell \circ {\kappa }_{\varXi }$ with $\ell (t) = \max \{0,t-1\}$. Recalling from Theorem 2 the generating distribution T for $\alpha$-unimodal distributions, we set

$$\begin{aligned} \begin{aligned} \ell _s(t)&= \int _0^{\infty }\ell (\lambda t)T(\mathrm {d}\lambda )\\&= \alpha \int _0^{1}\max \{0,(\lambda t-1)\}\lambda ^{\alpha -1}\mathrm {d}\lambda , \end{aligned} \end{aligned}$$

which is zero for any $t \le 1$. For $t\ge 1$, we can evaluate the integral to get

$$\begin{aligned} \begin{aligned} \forall t \ge 1{:}\,\ell _s(t)&= \alpha \int _{1/t}^{1}(t\lambda ^\alpha -\lambda ^{\alpha -1})\mathrm {d}\lambda \\&= \frac{\alpha }{\alpha +1} t - 1 + \frac{1}{\alpha +1}\left( \frac{1}{t}\right) ^\alpha \end{aligned} \end{aligned}$$

and then set $L_s(x) = \max _{i\in \mathcal I} f_i(a_i^\top x)$ where each $f_i(q) := \ell _s(q/b_i)$.

We can now apply Theorem 4 by reformulating the constraint ($\mathcal C_2$) for this choice of $f_i$ for each $i\in \mathcal I$, resulting in the constraint

$$\begin{aligned} T_{3, i} + 2 q T_{2, i} + q^2 T_{1, i} \ge \frac{\alpha }{\alpha +1} \frac{q}{b_i} - 1 + \frac{1}{\alpha +1} \frac{b_i^\alpha }{q^\alpha } \quad ~\forall q\ge b_i \end{aligned}$$

because $0\in \varXi$ and hence $b_i > 0$. We define a new scalar variable ${\tilde{q}}$ and apply the variable substitution ${\tilde{q}}^{\,w} = q$, resulting in the constraint

$$\begin{aligned} {\tilde{q}}^{2 w+ v} T_{1, i} + {\tilde{q}}^{w+ v} \left( 2 T_{2, i} - \frac{\alpha }{(\alpha +1) b_i} \right) + {\tilde{q}}^{v} \left( 1 + T_{3, i} \right) - \frac{b_i^\alpha }{\alpha +1} \ge 0, \quad \forall {\tilde{q}}\ge b_i^{1/w} \end{aligned}$$

after multiplying both sides by ${\tilde{q}}^v> 0$. The final result is obtained after the substitution $b_i^{1/w} \bar{q} = \tilde{q}$.

Corollary 3:

We follow the same approach as the proof of Corollary 1, but this time use the generating distribution T for $\gamma$-monotone distributions from Theorem 3, i.e.

$$\begin{aligned} T([0, t]) = \frac{1}{B(n, \gamma )} \cdot \int _{0}^t \lambda ^{n-1} \cdot (1-\lambda )^{\gamma -1} ~\mathrm {d}\lambda , \quad \forall t \in [0,1]. \end{aligned}$$

In this case the moment transformations from Theorem 1 become

$$\begin{aligned} \begin{aligned} \mu _\gamma&:= \left[ \int _0^\infty \lambda \,T(\mathrm {d}\lambda )\right] ^{{-1}}\,\, \mu&= \left[ \frac{1}{B(n,\gamma )} \int _0^1 \lambda ^n(1-\lambda )^{\gamma -1}(\mathrm {d}\lambda )\right] ^{{-1}}\,\, \mu \\&\,\,= \frac{n+\gamma }{n}\mu \\ S_\gamma&:= \left[ \int _0^\infty \lambda ^2\,T(\mathrm {d}\lambda )\right] ^{{-1}}\,\, S&= \left[ \frac{1}{B(n,\gamma )} \int _0^1 \lambda ^{n+1}(1-\lambda )^{\gamma -1}(\mathrm {d}\lambda )\right] ^{{-1}}\,\, S\\&\,\,=\frac{n+\gamma }{n} \frac{n+\gamma +1}{n+1} S. \end{aligned} \end{aligned}$$

From Proposition 1, the transformed loss function $L_s$ required in Theorem 1 become

$$\begin{aligned} \begin{aligned} L_s(y)&= \max _{i\in \mathcal I} ~ T\left( \left[ b_i/ a_i^\top y, \infty \right) \right) \cdot {\mathbb {1}}_{a_i^\top x \ge b_i}(y) \\&=: \max _{i\in \mathcal I}f_i(a_i^\top y). \end{aligned} \end{aligned}$$

where

$$\begin{aligned} f_i(q) = {\left\{ \begin{array}{ll} \displaystyle \frac{1}{B(n,\gamma )} \int _{b_i/q}^1 \lambda ^{n-1}(1-\lambda )^{\gamma -1}~\mathrm {d}\lambda ,\, &{} q \ge b_i,\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

For $q\ge b_i$, we can use a binomial expansion to evaluate this integral,⁴ obtaining

$$\begin{aligned} B(n, \gamma ) f_i(q)&= B(n, \gamma ) - \int _0^{b_i/q} \lambda ^{n-1} \cdot (1-\lambda )^{\gamma -1} ~\mathrm {d}\lambda \\&= B(n, \gamma ) - \sum _{k=0}^{\gamma -1} \int _0^{b_i/q} ({-}1)^k \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) \lambda ^{n+k-1} ~\mathrm {d}\lambda \\&= B(n, \gamma )-{b_i^n} \sum _{k=0}^{\gamma -1} \frac{(-b_i)^k}{n+k} \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) \frac{1}{q^{n+k}}. \end{aligned}$$

In order to apply Theorem 4, we now need only reformulate the semi-infinite constraint ($\mathcal C_2$). We obtain, for each $i\in \mathcal I$, the constraint

$$\begin{aligned} T_{3, i} + 2 q T_{2, i} + q^2 T_{1, i} \ge 1-\frac{b_i^n}{B(n, \gamma )} \sum _{k=0}^{\gamma -1} \frac{(-b_i)^k}{n+k} \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) \frac{1}{q^{n+k}}, ~\forall q\ge b_i, \end{aligned}$$

recalling that $0\in \varXi$ and hence $b_i > 0$. We multiply both sides by $q^{n+\gamma -1}>0$ to produce, for each $i\in \mathcal I$ the constraint

$$\begin{aligned}&T_{1, i} q^{n+\gamma +1} + 2 T_{2, i} q^{n+\gamma } + \left( T_{3, i}-1\right) q^{n+\gamma -1} \\&\quad + \frac{b_i^{n}}{B(n, \gamma )} \sum _{k=0}^{\gamma -1} \frac{(-b_i)^k}{n+k} \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) q^{\gamma - k - 1 } \ge 0 , \quad \forall q\ge b_i. \end{aligned}$$

The final result is obtained after the substitution $b_i \bar{q} = q$.

Corollary 4:

The method of proof parallels that of Corollary 2, but this time using the generating distribution T for $\gamma$-monotone distributions from Theorem 3. In this case we set

$$\begin{aligned} \begin{aligned} \ell _s(t)&= \frac{1}{B(n,\gamma )}\int _0^{1}\max \{0,(\lambda t-1)\}\lambda ^{n-1}(1-\lambda )^{\gamma -1}\mathrm {d}\lambda , \end{aligned} \end{aligned}$$

which is zero for any $t \le 1$. For any $t\ge 1$, using a binomial expansion we can evaluate the integral to get

$$\begin{aligned} \forall t \ge 1 {:}\,B(n, \gamma ) \ell _s(t)&= t \int _{1/t}^1 \lambda ^n (1-\lambda )^{\gamma -1}~\mathrm {d}\lambda - \int _{1/t}^1 \lambda ^{n-1} (1-\lambda )^{\gamma -1}~\mathrm {d}\lambda \\&= tB(n+1, \gamma ) - B(n, \gamma ) + \int _0^{1/t} \lambda ^{n-1} (1-\lambda )^{\gamma -1} \mathrm {d}\lambda \\&\quad - t\int _{0}^{1/t} \lambda ^n (1-\lambda )^{\gamma -1} \mathrm {d}\lambda \\&= tB(n+1, \gamma ) - B(n, \gamma ) \\&\quad + \sum _{k=0}^{\gamma -1}\left[ (-1)^k \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) \int _0^{1/t}\left( \lambda ^{n-1}-t\lambda ^n\right) \lambda ^{k}~\mathrm {d}\lambda \right] \\&= tB(n+1, \gamma ) - B(n, \gamma ) \\&\quad + \sum _{k=0}^{\gamma -1} \frac{(-1)^{k}}{(n+k)(n+k+1)} \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) \left( \frac{1}{t}\right) ^{n+k} \end{aligned}$$

and then set $L_s(x) = \max _{i\in \mathcal I} f_i(a_i^\top x)$ where each $f_i(q) := \ell _s(q/b_i)$. In order to apply Theorem 4, we now need only reformulate the semi-infinite constraint ($\mathcal C_2$). We obtain, for each $i\in \mathcal I$, the constraint

$$\begin{aligned} \begin{aligned}&T_{3, i} + 2 q T_{2, i} + q^2 T_{1, i} \ge \frac{B(n+1, \gamma )}{b_i B(n, \gamma )} q - 1 \\&\quad +\frac{b^n_i}{B(n, \gamma )} \sum _{k=0}^{\gamma -1} \frac{(-b_i)^{k}}{(n+k)(n+k+1)} \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) \frac{1}{q^{n+k}} \quad ~\forall q\ge b_i \end{aligned} \end{aligned}$$

because $0\in \varXi$ and hence $b_i > 0$. We multiply both sides by $q^{n+\gamma -1}>0$ to produce the constraint

$$\begin{aligned} \begin{aligned}&T_{1, i} q^{n+\gamma +1} + \left( 2 T_{2, i} - \frac{B(n+1, \gamma )}{b_i B(n, \gamma )} \right) q^{n+\gamma } + \left( T_{3, i}+1\right) q^{n+\gamma -1} \\&\quad -\frac{b_i^{n}}{B(n, \gamma )} \sum _{k=0}^{\gamma -1} \frac{(-b_i)^k}{(n+k)(n+k+1)} \left( {\begin{array}{c}\gamma -1\\ k\end{array}}\right) q^{\gamma - k - 1 } \ge 0 , \quad \forall q\ge b_i. \end{aligned} \end{aligned}$$

The final result is obtained after the substitution $b_i \bar{q} = \tilde{q}$.

Factor models in insurance

As mentioned in Sect. 1.1, any worst-case CVaR problem can be reduced to a related worst-case expectation problem. We are therefore interested in loss functions of the form $L(S_d) = \min \left( \max \left( S_d, 0 \right) ,k\right) -\beta$ for $0\le \beta \le k$. We have that the loss function $L(S_d)$ can be written as the gauge function $L(S_d) = \ell \circ \kappa _\varXi (S_d)$ for $\varXi =\{x\in \mathbb {R}^d|\sum _{i=1}^d x_i \ge 1\}$ and

$$\begin{aligned} \ell = {\left\{ \begin{array}{ll} 0 &{}\quad \text{ if }\quad t \le \beta , \\ t-\beta &{}\quad \text{ if }\quad \beta \le t < k, \\ k-\beta &{} \quad \text{ if }\quad t \ge k. \end{array}\right. } \end{aligned}$$

Recalling from Theorem 2 the generating distribution T for $\alpha$-unimodal distributions, we set $\ell _s(t) = \int _0^{\infty }\ell (\lambda t)T(\mathrm {d}\lambda )$ which is zero for any $t \le \beta$. For $\beta \le t < k$, we can evaluate the integral to get

$$\begin{aligned} \begin{aligned} \beta \le \forall t < k {:}\,\ell _s(t)&= \alpha \int _{\beta /t}^{1}(\lambda t - \beta ) \lambda ^{\alpha -1} \mathrm {d}\lambda \\&= \frac{\alpha }{\alpha +1} t - \beta + \frac{\beta ^{\alpha +1}}{\alpha +1}\frac{1}{t^\alpha }. \end{aligned} \end{aligned}$$

Similarly for $t\ge k$, we get

$$\begin{aligned} \begin{aligned} \forall t \ge k {:}\,\ell _s(t)&= \alpha \int _{\beta /t}^{k/t}(\lambda t - \beta ) \lambda ^{\alpha -1} \mathrm {d}\lambda + \alpha \int _{k/t}^{1}(k-\beta ) \lambda ^{\alpha -1} \mathrm {d}\lambda \\&= k - \beta - \frac{k^{\alpha +1}-\beta ^{\alpha +1}}{\alpha +1} \frac{1}{t^\alpha } \end{aligned} \end{aligned}$$

and then set $L_s(x) = \ell _s(\sum _{i=1}^d x_i)$. In order to apply Theorem 4, we now need only reformulate the semi-infinite constraint ($\mathcal C_2$). This can be done using methods analogous to the method described in the proof of Corollary 2, but is omitted here for the sake of brevity. We get finally

$$\begin{aligned} \left\{ \begin{aligned} T_{1, i} q^{2} + 2 q T_{2, i} + T_{3, i} \ge 0,&\quad \forall q\in \mathbb {R}\\ T_{1, i} \beta ^2 q^{2 w{+} v} {+} q^{w{+}v} \beta \left( 2 T_{2, i} {-} \frac{\alpha }{\alpha +1} \right) {+} q^{v}\left( T_{3, i}+\beta \right) {-} \frac{\beta }{\alpha +1} \ge 0,&\quad 1 \le \forall q < \left( \frac{k}{\beta }\right) ^{1/w}\\ T_{1, i} k^2 q^{2 w+ v} + 2 k q^{w+v} T_{2, i} + q^{v}\left( T_{3, i}+\beta -k\right) + k \frac{1-(\beta /k)^{\alpha +1}}{\alpha +1} \ge 0,&\quad \forall q \ge 1 \end{aligned}\right\} (\mathcal C_2). \end{aligned}$$