Superquantile/CVaR risk measures: second-order theory

Abstract

Superquantiles, which refer to conditional value-at-risk in the same way that quantiles refer to value-at-risk, have many advantages in the modeling of risk in finance and engineering. However, some applications may benefit from a further step, from superquantiles to second-order superquantiles. Measures of risk based on second-order superquantiles have recently been explored in some settings, but key parts of the theory have been lacking: descriptions of the associated risk envelopes and risk identifiers. Those missing ingredients are supplied in this paper, and moreover not just for second-order superquantiles, but also for a much broader class of mixed superquantile measures of risk. Such dualizing expressions facilitate the development of dual methods for mixed and second-order superquantile risk minimization as well as superquantile regression, a proposed second-order version of quantile regression.

Appendix

As support for proving Proposition 2.3 in Sect. 2, we need the following consequence of the Fubini–Tonelli’s Theorem.

Proposition 4.3

Suppose that $(\mathcal{{X}}, \mathcal{{A}}, \mu )$ and $(\mathcal{{Y}},\mathcal{{B}},\nu )$ are sigma-finite measure spaces. If $f:\mathcal{{X}}\times \mathcal{{Y}}\rightarrow \overline{\mathbb {R}}$ is measurable with respect to the product sigma-algebra on $\mathcal{{X}}\times \mathcal{{Y}}$ and $g:\mathcal{{X}}\times \mathcal{{Y}}\rightarrow \overline{\mathbb {R}}$ is integrable with respect to the product measure $\mu \times \nu$, with $f(x,y)\ge g(x,y)$ for $(\mu \times \nu )$-a.e. $(x,y)\in \mathcal{{X}}\times \mathcal{{Y}}$, then the following hold:

• (i) the function $h_1 = \int f(x,\cdot ) d\mu (x)$ is $\mathcal{{B}}$-measurable,

• (ii) the function $h_2 = \int f(\cdot ,y) d\nu (y)$ is $\mathcal{{A}}$-measurable,

• (iii) and

  $$\begin{aligned} \int f d(\mu \times \nu ) = \int \left[ \int f(x,y) d\mu (x)\right] d\nu (y) = \int \left[ \int f(x,y) d\nu (y)\right] d\mu (x). \end{aligned}$$

Proof

We recall that the integral of the sum of a nonnegative measurable function and an integrable function equates the sum of the individual integrals under the usual rules for handling addition with infinity. Then,

$$\begin{aligned} h_1 = \int f(x,\cdot ) d\mu (x) = \int (f-g)(x,\cdot ) d\mu (x) + \int g(x,\cdot ) d\mu (x) \end{aligned}$$

is $\mathcal{{B}}$-measurable since both terms on the right-hand side are $\mathcal{{B}}$-measurable by the Fubini–Tonelli Theorem. A similar argument yields the conclusion for $h_2$. The final assertion follows by applying the Fubini–Tonelli Theorem to $f-g$ and g, and the above rule about interchange of summation and integration. $\square$

Proof of Proposition 2.3

For every $X\in \mathcal{{L}}^2$, $\bar{q}_X$ is continuous and finite on [0, 1) and therefore $\bar{\mathcal{{B}}}_{[0,1)}$-measurable. Moreover, $\bar{q}_X \ge E[X]$ and therefore $\mathcal{{R}}(X) \ge E[X] > -\infty$. Consequently, $\mathcal{{R}}$ is well-defined with values in $[E[X], \infty ]$. Its regularity and positive homogeneity follow directly from those of $\mathcal{{R}}_\alpha$; see Rockafellar and Uryasev (2013). Since $\bar{q}_X$ is strictly increasing on [0, 1) for nonconstant X, we have that if $\lambda (\{0\})<1$, then

$$\begin{aligned} \mathcal{{R}}(X) = E[X]\lambda (\{0\}) + \int _{1>\beta >0} \bar{q}_X(\beta )~d\lambda (\beta ) > E[X]\lambda (\{0\}) + E[X](1-\lambda (\{0\}) = E[X] \end{aligned}$$

and the strict lower bound follows. From (6),

$$\begin{aligned} \mathcal{{R}}(X) \le \int _{0}^1 E[X] + \frac{\sigma (X)}{\sqrt{1-\beta }} d\lambda (\beta ) = E[X] + \sigma (X)\int _{0}^1 \frac{1}{\sqrt{1-\beta }} d\lambda (\beta )<\infty \end{aligned}$$

under the stated assumption, which establishes the corresponding finiteness on $\mathcal{{L}}^2$. In the case of $\sup X <\infty$, finiteness of $\mathcal{{R}}(X)$ follows trivially.

We next consider the alternative expression. By definition,

$$\begin{aligned} \mathcal{{R}}(X) = \int _0^1 \left[ \int _0^1 q_X(\beta ) \psi (\alpha ,\beta ) d\beta \right] d\lambda (\alpha ), \end{aligned}$$

(20)

with $\psi (\alpha ,\beta ) = \frac{1}{1-\alpha }$ if $0\le \alpha < \beta < 1$ and $\psi (\alpha ,\beta ) = 0$ otherwise. We equip $[0,1)\times (0,1)$ with the product measure $\lambda \times m$ defined on the product sigma-algebra $\bar{\mathcal{{B}}}_{[0,1)}\otimes \mathcal{{B}}_{(0,1)}$. It is obvious that $\psi :[0,1)\times (0,1)\rightarrow \mathbb {R}$ is $(\bar{\mathcal{{B}}}_{[0,1)}\otimes \mathcal{{B}}_{(0,1)})$-measurable and likewise $q_X$, viewed as a function on $[0,1)\times (0,1)$ that is constant in its first argument, due its monotonicity. Consequently, the function $(\alpha ,\beta )\mapsto q_X(\beta ) \psi (\alpha ,\beta )$ is measurable in the same sense. Then, we look toward the interchange of integration order in (20).

We consider three cases. (i) Suppose that $X\ge 0$ a.e. Then, $q_X\ge 0$ and $q_X\psi \ge 0$, and the interchange of integration order is permitted by Tonelli–Fubini’s Theorem. (ii) Suppose that $X\le 0$ a.e. Then, $-q_X\ge 0$ and $-q_X\psi \ge 0$, and the interchange of integration order is again permitted by Tonelli-Fubini’s Theorem. (iii) Suppose that neither (i) nor (ii) holds. Then, there exists a $\beta _X \in (0,1)$ such that $q_X(\beta ) \ge 0$ for $\beta \ge \beta _X$ and $q_X(\beta ) \le 0$ for $\beta \le \beta _X$. In view of Proposition 4.3, it suffices to find an integrable, lower-bounding function of $q_X\psi$. Let $g:[0,1)\times (0,1) \rightarrow \mathbb {R}$ be given by

$$\begin{aligned} g(\alpha ,\beta ) = {\left\{ \begin{array}{ll} q_X(\beta ){/}(1-\beta _X) &{}\quad \text{ if }\; 0\le \alpha < \beta \le \beta _X\\ q_X(\beta ) &{}\quad \text{ if }\; 0\le \alpha < \beta <1, \beta _X<\beta \\ 0 &{}\quad \text{ otherwise. }\end{array}\right. } \end{aligned}$$

Clearly, $q_X\psi \ge g$ and

$$\begin{aligned} \int |g| d(\lambda \times m) \le \frac{1}{1-\beta _X}\int |q_X| d(\lambda \times m) = \frac{1}{1-\beta _X} \int _0^1 \left[ \int _0^1 |q_X(\beta )| d\beta \right] d\lambda (\alpha ), \end{aligned}$$

(21)

where the equality follows by Tonelli–Fubini’s Theorem. The inner integral simplifies to

$$\begin{aligned} \int _0^1 |q_X(\beta )| d\beta = \int _{\beta _X}^1 q_X(\beta ) d\beta - \int _0^{\beta _X} q_X(\beta ) d\beta = (1-\beta _X)\bar{q}_X(\beta _X) - \int _0^{\beta _X} q_X(\beta ) d\beta . \end{aligned}$$

The last term requires further simplification. Recall that for $\alpha \in (0,1)$,

$$\begin{aligned} \frac{1}{\alpha }\int _0^\alpha q_X(\beta )d\beta = -\frac{1}{\alpha }\int _{1-\alpha }^1 q_{-X}(\beta )d\beta = -\bar{q}_{-X}(1-\alpha ). \end{aligned}$$

Applying this result, the inner integral from above simplifies further to

$$\begin{aligned} \int _0^1 |q_X(\beta )| d\beta = (1-\beta _X)\bar{q}_X(\beta _X) + \beta _X\bar{q}_{-X}(1-\beta _X)<\infty . \end{aligned}$$

Consequently in view of (21), g is integrable and therefore furnishes the necessary lower-bounding, integrable function in Proposition 4.3, which completes part (iii). We are therefore permitted to interchange the order of integration in (20) and get

$$\begin{aligned} \mathcal{{R}}(X)= & {} \int _0^1 \left[ \int _0^1 q_X(\beta ) \psi (\alpha ,\beta ) d\beta \right] d\lambda (\alpha ) = \int _0^1 q_X(\beta ) \left[ \int _0^1 \psi (\alpha ,\beta ) d\lambda (\alpha )\right] d\beta \\= & {} \int _0^1 q_X(\beta ) \varphi (\beta ) d\beta , \end{aligned}$$

where the last equality follows from the definition of $\varphi$.

The final assertions follow from recognizing that the Lebesgue–Stieltjes measure $d\varphi$ associated with a function $\varphi$ has $d\varphi (\alpha ) = \frac{1}{1-\alpha } d\lambda (\alpha )$ for a weighting measure $\lambda$ on [0, 1). $\square$

Now we articulate other definitions and technical results required in the paper.

Definition 4.4

Let $(T,\mathcal{{A}},\mu )$ be a complete measure space, with $\mu$ sigma-finite, $\mathcal{{X}}$ a separable reflexive Banach space, and $\mathcal{{M}}$ a linear subspace of the linear space of all $(\mathcal{{A}},\mathcal{{B}}_{\mathcal{{X}}})$-measurable functions $x:T\rightarrow \mathcal{{X}}$. The set $\mathcal{{M}}$ is $(\mathcal{{A}},\mathcal{{B}}_{\mathcal{{X}}})$-decomposable if, whenever $x\in \mathcal{{M}}$ and $x_0:S\rightarrow \mathcal{{X}}$ is a bounded $(\mathcal{{A}},\mathcal{{B}}_{\mathcal{{X}}})$-measurable function on a set $S\in \mathcal{{A}}$, with $\mu (S)<\infty$, then the function $y:T\rightarrow \mathcal{{X}}$ given by

$$\begin{aligned} y(t) = {\left\{ \begin{array}{ll} x_0(t) &{}\quad \text{ if }\; t\in S\\ x(t) &{}\quad \text{ if }\; t \in T{\setminus } S \end{array}\right. } \end{aligned}$$

also belongs to $\mathcal{{M}}$.

Definition 4.5

In the notation of Definition 4.4, we say that a function $f:T\times \mathcal{{X}}\rightarrow (-\infty , \infty ]$ is a normal integrand if the following hold:

• (i) f is $(\mathcal{{A}}\otimes \mathcal{{B}}_{\mathcal{{X}}})$-measurable and

• (ii) for every $t\in T$, $f(t, \cdot )$ is lower semicontinuous on $\mathcal{{X}}$ and not identical to $\infty$.

Proposition 4.6

Suppose that the conditions and notation of Definition 4.4 hold and $f:T\times \mathcal{{X}}\rightarrow (-\infty , \infty ]$ is a normal integrand. Then, the following hold:

• (i) the functions $t\mapsto \inf _{\xi \in \mathcal{{X}}} f(t,\xi )$ and $t\mapsto f(t,x(t))$, with $x:T\rightarrow \mathcal{{X}}$ $(\mathcal{{A}},\mathcal{{B}}_{\mathcal{{X}}})$-measurable, are $\mathcal{{A}}$-measurable and

• (ii) if $\mathcal{{M}}$ is $(\mathcal{{A}},\mathcal{{B}}_{\mathcal{{X}}})$-decomposable and there exists an $x\in \mathcal{{M}}$ such that $\int f(t,x(t)) ~d\mu (t) <\infty$, then

  $$\begin{aligned} \inf _{x\in \mathcal{{M}}} \int f(t,x(t)) ~d\mu (t) = \int \varphi (t) ~d\mu (t),\quad \text{ where }\quad \varphi (t) = \inf _{\xi \in \mathcal{{X}}} f(t,\xi ). \end{aligned}$$

  (22)

Proof

First, we consider $t\mapsto \inf _{\xi \in \mathcal{{X}}} f(t,\xi )$. For measurable spaces $(\mathcal{{X}}_1,\mathcal{{A}}_1)$ and $(\mathcal{{X}}_2,\mathcal{{A}}_2)$, we recall that a set-valued mapping $S:\mathcal{{X}}_1\rightrightarrows \mathcal{{X}}_2$ is $(\mathcal{{A}}_1,\mathcal{{A}}_2)$-measurable if its graph is measurable in the sense that

$$\begin{aligned} \left\{ (x_1,x_2)\in \mathcal{{X}}_1\times \mathcal{{X}}_2|x_2 \in S(x_1)\right\} \in \mathcal{{A}}_1 \otimes \mathcal{{A}}_2, \end{aligned}$$

where $\mathcal{{A}}_1 \otimes \mathcal{{A}}_2$ is the product sigma-algebra generated by $\mathcal{{A}}_1$ and $\mathcal{{A}}_2$. Since f is a normal integrand, the set-valued mapping $t\mapsto \mathop {\mathrm{epi}}f(t,\cdot )$ is $\mathcal{{A}}$-measurable and closed-valued; see for example Rockafellar (1971, Proposition 1). By Rockafellar (1971, Theorem 1(f)), there exists a countable collection $\{g_i\}_{i\in I}$ of $\mathcal{{A}}$-measurable functions $g_i:T\rightarrow \mathcal{{X}}\times \mathbb {R}$ of the form $g_i(t) = (x_i(t),\alpha _i(t))$, $x_i(t)\in \mathcal{{X}}$ and $\alpha _i(t)\in \mathbb {R}$, such that

$$\begin{aligned} \mathop {\mathrm{epi}}f(t,\cdot ) = \mathop {\mathrm{cl}}\{ g_i(t)\}_{i\in I}\quad \text{ for } \text{ all }\quad t\in T, \end{aligned}$$

where $\mathop {\mathrm{cl}}$ denotes closure. The mapping $t\mapsto \alpha _i(t)$ is also $\mathcal{{A}}$-measurable. Consequently,

$$\begin{aligned} \inf _{\xi \in \mathcal{{X}}} f(t,\xi ) = \inf _{i\in I} \alpha _i(t)\quad \text{ for } \text{ all }\quad t\in T \end{aligned}$$

and the conclusion follows from the fact that the pointwise infimum of a countable collection of measurable functions is a measurable function.

Second, we consider $t\mapsto f(t,x(t))$, which is a composition of f with the measurable mapping $t\mapsto (t,x(t))$ and therefore measurable.

Third, we establish part (ii) by following the arguments in the proof of Theorem 2 in Rockafellar (1971). By assumption there exists a function $x_1\in \mathcal{{M}}$ and a $\mu$-integrable function $\alpha _1:T\rightarrow \mathbb {R}$ such that

$$\begin{aligned} f(t,x_1(t)) \le \alpha _1(t)\quad \text{ for } \text{ every }\quad t\in T. \end{aligned}$$

Since $\varphi (t) \le f(t,x(t))$ for every function $x\in \mathcal{{M}}$ and $t\in T$ by definition and $\varphi$ is $\mathcal{{A}}$-measurable by part (i), the integral of $\varphi$ is well-defined and either finite or equals $-\infty$. Consequently, the inequality $\ge$ holds in (22). Now, let $\gamma \in \mathbb {R}$ be such that

$$\begin{aligned} \int \varphi (t) d\mu (t) < \gamma . \end{aligned}$$

(23)

We will prove the existence of a function $x\in \mathcal{{M}}$ such that

$$\begin{aligned} \int f(t,x(t)) d\mu (t) < \gamma , \end{aligned}$$

(24)

thereby establishing part (ii). From (23) and the properties of $(T,\mathcal{{A}},\mu )$, there exists a $\mu$-integrable function $\alpha _0:T\rightarrow \mathbb {R}$ such that $\varphi (t) < \alpha _0(t)$ for every $t\in T$ and

$$\begin{aligned} \int \alpha _0(t) ~d\mu (t) < \gamma . \end{aligned}$$

(25)

We define the set-valued mapping $S:T\rightrightarrows \mathcal{{X}}$ by

$$\begin{aligned} S(t) = \left\{ \xi \in \mathcal{{X}}~|~f(t,\xi ) \le \alpha _0(t)\right\} \quad \text{ for }\quad t\in T. \end{aligned}$$

Since the function $(t,\xi )\mapsto f(t,\xi ) - \alpha _0(t)$ is $(\mathcal{{A}}\otimes \mathcal{{B}}_{\mathcal{{X}}})$-measurable, S is also $\mathcal{{A}}$-measurable. Moreover, S(t) is for each $t\in T$ closed and nonempty. Since S is $\mathcal{{A}}$-measurable, there exists a $\mathcal{{A}}$-measurable selection $x_0$, i.e., a $\mathcal{{A}}$-measurable function $x_0$ such that $x_0(t) \in S(t)$ for every $t\in T$; see for example the corollary of Theorem 1 in Rockafellar (1971). Since (25) holds, there exists a measurable set $T_0\subset T$, with $\mu (T_0) < \infty$, such that

$$\begin{aligned} \int _{T_0} \alpha _0(t) d\mu (t) + \int _{T{\setminus }T_0} \alpha _1(t) ~d\mu (t) < \gamma . \end{aligned}$$

(26)

By the construction of S in terms of $\alpha _0$, the measurable selection $x_0$ can be chosen to be bounded on $T_0$. Let $x:T\rightarrow \mathcal{{X}}$ be such that $x(t) = x_0(t)$ for $t\in T_0$ and $x(t) = x_1(t)$ for $t\in T{\setminus }T_0$. Then, $x\in \mathcal{{M}}$ by the assumption of decomposability, and we have that $f(t,x(t)) \le \alpha _0(t)$ for $t\in T_0$ and $f(t,x(t)) \le \alpha _1(t)$ for $t\in T{\setminus } T_0$. From (26) we then conclude (24), which establishes part (ii). $\square$

Lemma 4.7

If $q:[0,1)\rightarrow \mathcal{{L}}^2$ is $(\bar{\mathcal{{B}}}_{[0,1)}, \mathcal{{B}}_{\mathcal{{L}}^2})$-measurable, then

• (i) the function $f_1:[0,1)\times \Omega \rightarrow \overline{\mathbb {R}}$ given by $f_1(\beta ,\omega ) = q(\beta )(\omega )$ is $(\bar{\mathcal{{B}}}_{[0,1)}\otimes \mathcal{{F}})$-measurable, and

• (ii) the function $f_2:[0,1)\rightarrow \mathbb {R}$ given by $f_2(\beta ) = \Vert q(\beta )\Vert _2$ is $\bar{\mathcal{{B}}}_{[0,1)}$-measurable.

Proof

For part (i) simply observe that $f_1 = g \circ h$, where $h:[0,1)\times \Omega \rightarrow \mathcal{{L}}^2 \times \Omega$, with $h(\alpha ,\omega ) = (q(\alpha ),\omega )$, and $g:\mathcal{{L}}^2\times \Omega \rightarrow \overline{\mathbb {R}}$, with $g(Q,\omega ) = Q(\omega )$. The conclusion then follows from the measurability of q and elements of $\mathcal{{L}}^2$, and the fact that composition of measurable functions is measurable. Next, we consider part (ii). A trivial extension of part (i) establishes that the function $(\beta ,\omega )\mapsto [q(\beta )(\omega )]^2$ is $(\bar{\mathcal{{B}}}_{[0,1)}\otimes \mathcal{{F}})$-measurable. Since it is also nonnegative, it follows from Tonelli–Fubini’s Theorem that $[f_2(\cdot )]^2$ is $\bar{\mathcal{{B}}}_{[0,1)}$-measurable. $\square$

The following is a direct consequence of Definition 4.4.

Proposition 4.8

The set $\mathcal{{M}}$ is $(\bar{\mathcal{{B}}}_{[0,1)}, \mathcal{{B}}_{\mathcal{{L}}^2})$-decomposable.