A Tail Measure With Variable Risk Tolerance: Application in Dynamic Portfolio Insurance Strategy

Abstract

Risk measures for tail risk have an important application in the dynamic portfolio insurance strategies. We propose a new risk measure called SlideVaR which overcome the limitation of traditional measures like VaR and ES, and can sufficiently reflect the market changes. Several important properties of SlideVaR and its generalized risk measure have been investigated. Then, we further apply SlideVaR into constructing dynamic portfolio insurance strategy. Our numerical analysis shows that SlideVaR-based portfolio insurance strategy has advantage especially in markets where the state changes frequently.

Appendices

Preliminaries of Risk Measures

1.1 Coherent Risk Measure

Coherent risk measure was proposed by Artzner et al. (1999). Define a probability space $(\Omega , \mathcal {F},\mathbb {P})$, denote $\mathcal {X}$ as the set of loss random variables defined on $(\Omega , \mathcal {F},\mathbb {P})$, then

Definition 3

(Coherent) A risk measure satisfying the four axioms of translation invariance, sub-additivity, positive homogeneity and monotonicity, is called coherent risk measure.

• (a) Translation invariance:

  $$\begin{aligned} \rho (X + a)=\rho (X)+a, \,\, \forall X \in \mathcal {X}, \,\, \forall a \in \mathbb {R}. \end{aligned}$$

  (37)

• (b) Sub-additivity:

  $$\begin{aligned} \rho (X_1+X_2) \leqslant \rho (X_1)+\rho (X_2), \,\, \forall X_1, \,\, X_2 \in \mathcal {X}. \end{aligned}$$

  (38)

• (c) Positive homogeneity:

  $$\begin{aligned} \rho (\lambda X)=\lambda \rho (X), \,\, \forall \lambda \geqslant 0, \,\, \forall X \in \mathcal {X}. \end{aligned}$$

  (39)

• (d) Monotonicity:

  $$\begin{aligned} \text{ if } X \leqslant Y, \text{ then } \rho (X) \leqslant \rho (Y), \,\, \forall X, \,\, Y \in \mathcal {X}. \end{aligned}$$

  (40)

1.2 Spectral Risk Measure

As we mentioned in Sect. 3.1, $U_{\beta }^{\phi }$ is a spectral risk measure and $\phi _{\beta }$ is the corresponding risk aversion function. Spectral risk measure was proposed by Acerbi (2002). First, Acerbi (2002) define the risk aversion function $\phi$ in the normed space $L^1([0,1])$ where every element is represented by a class of functions which differ at most on a subset of [0, 1] of zero measure. The norm in given by

$$\begin{aligned} ||\phi || = \int _0^1|\phi (p)|dp. \end{aligned}$$

(41)

Second, let $\phi$ satisfies

• Monotonicity: $\phi$ is decreasing if $\forall q \in (a,b)$ and $\forall \varepsilon >0$ such that $[q-\varepsilon ,q+\varepsilon ] \subset [a,b]$,

  $$\begin{aligned} \int _{q-\varepsilon }^{q}\phi (p)dp \geqslant \int _{q}^{q+\varepsilon }\phi (p)dp. \end{aligned}$$

  (42)

  $\phi$ is increasing if $\forall q \in (a,b)$ and $\forall \varepsilon >0$ such that $[q-\varepsilon ,q+\varepsilon ] \subset [a,b]$,

  $$\begin{aligned} \int _{q-\varepsilon }^{q}\phi (p)dp \leqslant \int _{q}^{q+\varepsilon }\phi (p)dp. \end{aligned}$$

  (43)

• Positivity: $\phi$ is positive if $\forall I \subset [a,b]$,

  $$\begin{aligned} \int _I\phi (p)dp \geqslant 0. \end{aligned}$$

  (44)

Then, Acerbi (2002) give the following,

Definition 4

(Spectral risk measure) If a risk aversion function $\phi \in L^1([0,1])$ satisfies: (1) $\phi$ is positive, (2) $\phi$ is increasing, (3) $|| \phi ||=1$. Then we call the function

$$\begin{aligned} M_{\phi }(X):=\int _0^1F^{-1}_X(p)\phi (p)dp, X\in \mathcal {X}, \end{aligned}$$

(45)

is the spectral risk measure generated by $\phi$.

By some examples, Acerbi (2002) and Adam et al. (2008) show that, in a real-world risk management application the integral 45 will always be well defined and finite. For instance, if $\phi (p)$ is bounded and X is a finite integrable random variable, then $U_{\beta }^{\phi }$ is well-defined and finite.

1.3 Distortion Risk Measures

Distortion risk measures was proposed by Wang et al. (1997). Distortion function $g:[0,1] \rightarrow [0,1]$ is the non-decreasing function such that: $g(0) = 0$, $g(1) = 1$.

Definition 5

Given a distortion function g, the distortion risk measure $\rho _{g}(X)$ is:

$$\begin{aligned} \rho _{g}(X) = \int _{-\infty }^{0}[g(P[X> x])-1]dx + \int _{0}^{+\infty }g(P[X > x])dx. \end{aligned}$$

(46)

Distortion risk measure is a general framework that can be expressed as a special Chouqet Integral (Choquet 1954), and a distortion risk measure is coherent if and only if the distortion function is concave (Balbás et al. 2009; Wirch and Hardy 2001; Dhaene et al. 2012). Several widely used risk measures are special cases of $\rho _{g}(X)$. For example, the distortion function of $VaR_{\alpha }$ is

$$\begin{aligned} \psi _{\alpha }(u) = \left\{ \begin{aligned}&0 \,,&u&\in [0, 1-\alpha ),\\&1 \,,&u&\in [1-\alpha ,1]. \end{aligned} \right. \end{aligned}$$

(47)

The distortion function of $ES_{\alpha }$ is:

$$\begin{aligned} \gamma _{\alpha }(u) = \left\{ \begin{aligned}&\frac{u}{1-\alpha } \,,&u&\in [0, 1-\alpha ),\\&1 \,,&u&\in [1-\alpha ,1]. \end{aligned} \right. \end{aligned}$$

(48)

Proof of Propositions and Corollaries

1.1 Proof of Proposition 1

Proof

Due to that $U_{\beta }^{\phi }(X)$ is a spectral risk measure and S(x) is a non-decreasing function, if $X \leqslant _{st} Y$, we have

$$\begin{aligned} S\circ U_{\beta }^{\phi }(Y) \geqslant S\circ U_{\beta }^{\phi }(X). \end{aligned}$$

Then we can have

$$\begin{aligned}&SlideVaR_{\alpha ,\beta }^{\phi }(Y) - SlideVaR_{\alpha ,\beta }^{\phi }(X) \\ =&S\circ U_{\beta }^{\phi }(Y) \left[ ES_{\alpha }(Y)- VaR_{\beta }(Y)\right] -S\circ U_{\beta }^{\phi }(X) \left[ ES_{\alpha }(X)- VaR_{\beta }(X)\right] +VaR_{\beta }(Y) \\ -&VaR_{\beta }(X) \\ \geqslant&S\circ U_{\beta }^{\phi }(X) \left[ ES_{\alpha }(Y)- VaR_{\beta }(Y)\right] -S\circ U_{\beta }^{\phi }(X) \left[ ES_{\alpha }(X)- VaR_{\beta }(X)\right] +VaR_{\beta }(Y) \\ -&VaR_{\beta }(X) \\ =&S\circ U_{\beta }^{\phi }(X)\left[ ES_{\alpha }(Y)- VaR_{\beta }(Y)\right] + \left[ 1-S\circ U_{\beta }^{\phi }(X) \right] \left[ VaR_{\beta }(Y)-VaR_{\beta }(X) \right] . \end{aligned}$$

Due to that $ES_{\alpha }(Y) \geqslant VaR_{\beta }(Y)$ and $VaR_{\beta }(Y) \geqslant VaR_{\beta }(X)$, we can have

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(Y) - SlideVaR_{\alpha ,\beta }^{\phi }(X)\geqslant 0. \end{aligned}$$

1.2 Proof of Proposition 2

Proof

$\forall X \in \tilde{\mathcal {X}}$, we have $SlideVaR_{\alpha ,\beta }^{\phi }(X) = ES_{\alpha }(X)$. Therefore, $\forall X,Y \in \tilde{\mathcal {X}}$, we have

$$\begin{aligned}&SlideVaR_{\alpha ,\beta }^{\phi }(X+Y) \leqslant ES_{\alpha }(X+Y)\leqslant ES_{\alpha }(X)+ES_{\alpha }(Y) \\ =&SlideVaR_{\alpha ,\beta }^{\phi }(X)+SlideVaR_{\alpha ,\beta }^{\phi }(Y). \end{aligned}$$

Then (a) is proved. $\forall X^{*}\in \mathcal {X}^{*}$, if $X^{*} \leqslant _{st} X$ (and Y), we have

$$\begin{aligned} 1 \geqslant S\circ U_{\beta }^{\phi }(Y) \geqslant S\circ U_{\beta }^{\phi }(X) \geqslant S\circ U_{\beta }^{\phi }(X^{*}) = 1, \end{aligned}$$

that is

$$\begin{aligned} S\circ U_{\beta }^{\phi }(Y) = S\circ U_{\beta }^{\phi }(X) = S\circ U_{\beta }^{\phi }(X^{*}) = 1, \end{aligned}$$

According to (a), (b) can be proved.

1.3 Proof of Corollary 2

Proof

$\forall X, Y$ which satisfy the conditions in Proposition 2, we have $SlideVaR_{\alpha ,\beta }^{\phi }(X) = ES_{\alpha }(X)$, $SlideVaR_{\alpha ,\beta }^{\phi }(Y) = ES_{\alpha }(Y)$. Then $X \leqslant _{icx} Y$ implies that $SlideVaR_{\alpha ,\beta }^{\phi }(X) \leqslant SlideVaR_{\alpha ,\beta }^{\phi }(Y)$.

1.4 Proof of Corollary 1

Proof

$\forall X\in \tilde{\mathcal {X}}$, if $X \leqslant _{st} Y$, we have

$$\begin{aligned} 1 \geqslant S\circ U_{\beta }^{\phi }(Y) \geqslant S\circ U_{\beta }^{\phi }(X)= 1, \end{aligned}$$

which implies that $S\circ U_{\beta }^{\phi }(Y)= 1$ and then $Y \in \tilde{\mathcal {X}}$.

1.5 Proof of Proposition 3

Proof

$\forall a \in \mathcal {R}$, We have

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(X+a)&= S\circ U_{\beta }^{\phi }(X+a) \left[ ES_{\alpha }(X+a)-VaR_{\beta }(X+a)\right] +VaR_{\beta }(X+a) \\&=S\circ (U_{\beta }^{\phi }(X)+a) \left[ ES_{\alpha }(X)-VaR_{\beta }(X)\right] +VaR_{\beta }(X)+a. \end{aligned}$$

If $a \geqslant 0$, we have $S\circ (U_{\beta }^{\phi }(X)+a) \geqslant S\circ (U_{\beta }^{\phi }(X))$, then

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(X+a)&\geqslant S\circ U_{\beta }^{\phi }(X) \left[ ES_{\alpha }(X)-VaR_{\beta }(X)\right] +VaR_{\beta }(X)+a \\&=SlideVaR_{\alpha ,\beta }^{\phi }(X)+a. \end{aligned}$$

The equal sign holds when $a=0$. Similarly, if $a \leqslant 0$, we have $S\circ (U_{\beta }^{\phi }(X)+a) \leqslant S\circ U_{\beta }^{\phi }(X)$, then

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(X+a)&\leqslant S\circ U_{\beta }^{\phi }(X) \left[ ES_{\alpha }(X)-VaR_{\beta }(X)\right] +VaR_{\beta }(X)+a \\&=SlideVaR_{\alpha ,\beta }^{\phi }(X)+a. \end{aligned}$$

The equal sign holds when $a=0$.

1.6 Proof of Proposition 4

Proof

Due to that

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(\lambda X)&= S\circ U_{\beta }^{\phi }(\lambda X) \left[ ES_{\alpha }( \lambda X)-VaR_{\beta }(\lambda X)\right] +VaR_{\beta }(\lambda X) \\&= \lambda S\circ ( \lambda U_{\beta }^{\phi }( X)) \left[ ES_{\alpha }( X)-VaR_{\beta }(X)\right] +\lambda VaR_{\beta }( X), \end{aligned}$$

$$\begin{aligned} \lambda SlideVaR_{\alpha ,\beta }^{\phi }( X)&= \lambda S\circ U_{\beta }^{\phi }(X) \left[ ES_{\alpha }(X)-VaR_{\beta }(X)\right] +\lambda VaR_{\beta }( X)), \end{aligned}$$

we can have

$$\begin{aligned}&SlideVaR_{\alpha ,\beta }^{\phi }(\lambda X) - \lambda SlideVaR_{\alpha ,\beta }^{\phi }( X) \\ =&\lambda \left[ ES_{\alpha }( X)-VaR_{\beta }(X)\right] \left[ S\circ ( \lambda U_{\beta }^{\phi }( X))- S\circ U_{\beta }^{\phi }( X)\right] . \end{aligned}$$

From $U_{\beta }^{\phi }(X)>0$, we know that, if $\lambda \geqslant 1$, we have $S\circ ( \lambda U_{\beta }^{\phi }( X))- S\circ U_{\beta }^{\phi }( X) \geqslant 0$ which implies that $SlideVaR_{\alpha ,\beta }^{\phi }(\lambda X) - \lambda SlideVaR_{\alpha ,\beta }^{\phi }( X) \geqslant 0$. The equal sign holds when $\lambda =1$. Similarly, if $\lambda \leqslant 1$, we have $S\circ ( \lambda U_{\beta }^{\phi }( X))- S\circ U_{\beta }^{\phi }( X) \leqslant 0$ which implies that $SlideVaR_{\alpha ,\beta }^{\phi }(\lambda X) - \lambda SlideVaR_{\alpha ,\beta }^{\phi }( X) \leqslant 0$. The equal sign holds when $\lambda =1$.

1.7 Proof of Proposition 5

Proof

$\forall X \in \tilde{\mathcal {X}}$, we have $SlideVaR_{\alpha ,\beta }^{\phi }(X) = ES_{\alpha }(X)$. Therefore, $\forall X,Y \in \tilde{\mathcal {X}}$, we have

$$\begin{aligned}&SlideVaR_{\alpha ,\beta }^{\phi }(\lambda X+(1-\lambda ) Y) \leqslant ES_{\alpha }(\lambda X+(1-\lambda ) Y)\leqslant ES_{\alpha }(\lambda X)+ES_{\alpha }((1-\lambda ) Y) \\ =&\lambda ES_{\alpha }(X)+(1-\lambda )ES_{\alpha }( Y) = \lambda SlideVaR_{\alpha ,\beta }^{\phi }(X)+(1-\lambda ) SlideVaR_{\alpha ,\beta }^{\phi }(Y). \end{aligned}$$

Then (a) is proved. $\forall X^{*}\in \mathcal {X}^{*}$, if $X^{*} \leqslant _{st} X$ (and Y) (or $X^{*} \leqslant _{lr} X$ (and Y) or $X^{*} \leqslant _{hr} X$ (and Y)), we have

$$\begin{aligned} S\circ U_{\beta }^{\phi }(Y) = S\circ U_{\beta }^{\phi }(X) = S\circ U_{\beta }^{\phi }(X^{*}) = 1, \end{aligned}$$

which means that $X,Y \in \tilde{\mathcal {X}}$. According to (a), (b) can be proved.

1.8 Proof of Proposition 6

Proof

From Dhaene et al. (2006) we know that VaR and distortion risk measure are additive of for sums of comonotonic risks. Then we can have

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(X+Y)&= S\circ U_{\beta }^{\phi }(X+Y) \left[ ES_{\alpha }(X+Y)-VaR_{\beta }(X+Y)\right] +VaR_{\beta }(X+Y) \\&= S\circ ( U_{\beta }^{\phi }( X)+U_{\beta }^{\phi }( Y)) \left[ ES_{\alpha }( X)-VaR_{\beta }(X)\right] +VaR_{\beta }( X) \\&+S\circ ( U_{\beta }^{\phi }( X)+U_{\beta }^{\phi }( Y)) \left[ ES_{\alpha }( Y)-VaR_{\beta }(Y)\right] +VaR_{\beta }( Y). \end{aligned}$$

Due to that $U_{\beta }^{\phi }( X)>0$, $U_{\beta }^{\phi }(Y)>0$, we can have

$$\begin{aligned} SlideVaR_{\alpha ,\beta }^{\phi }(X+Y)&\geqslant S\circ U_{\beta }^{\phi }( X) \left[ ES_{\alpha }( X)-VaR_{\beta }(X)\right] +VaR_{\beta }( X) \\&+S\circ U_{\beta }^{\phi }( Y) \left[ ES_{\alpha }( Y)-VaR_{\beta }(Y)\right] +VaR_{\beta }( Y)\\&=SlideVaR_{\alpha ,\beta }^{\phi }(X)+SlideVaR_{\alpha ,\beta }^{\phi }(Y). \end{aligned}$$

1.9 Proof of Proposition 7

Proof

• 1. Due to that $\forall u \in [0,1]$, $\sum \limits _{i=1}^N S_i(x)\cdot g_i(u)$ is non-decreasing with respect to $x \in \mathcal {R}$, we can have that if $X \leqslant _{st} Y$ (or $X \leqslant _{lr} Y$, or $X \leqslant _{hr} Y$),

  $$\begin{aligned} \kappa _{X}(u) = \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot g_i(u) \leqslant \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(Y)\cdot g_i(u) = \kappa _{Y}(u). \end{aligned}$$

  From the definition, we know that

  $$\begin{aligned} SlideM(X)&= \int _{-\infty }^{0}\left[ \kappa _{X}\left( P\left[ X> x\right] \right) -1\right] dx + \int _{0}^{+\infty }\kappa _{X}\left( P\left[ X> x\right] \right) dx \\&\leqslant \int _{-\infty }^{0}[\kappa _{Y}(P[Y> y])-1]dy + \int _{0}^{+\infty }\kappa _{Y}(P[Y > y])dy \\&= SlideM(Y). \end{aligned}$$

• 2. $\forall X \in \tilde{\mathcal {X}}_G$, we have $SlideM(X) = \rho _{g_1}(X)$. If $g_1(u)$ is concave, we have $\rho _{g_1}(X+Y)\leqslant \rho _{g_1}(X) +\rho _{g_1}(Y)$ (Dhaene et al. 2012). Therefore, $\forall X,Y \in \tilde{\mathcal {X}}_G$, we have

  $$\begin{aligned} SlideM(X+Y) \leqslant \rho _{g_1}(X+Y)\leqslant \rho _{g_1}(X)+\rho _{g_1}(Y) =SlideM(X)+SlideM(Y). \end{aligned}$$

  Then (a) is proved. Due to that $g_1(u) \geqslant g_2(u) \geqslant \cdots \geqslant g_N(u)$, and $\sum \limits _{i=1}^N S_i(x)\cdot g_i(u)$ is non-decreasing with respect to x, $\forall X^{*}\in \mathcal {X}^{*}$, if $X^{*} \leqslant _{st} X$ (and Y) (or $X^{*} \leqslant _{lr} X$ (and Y) or $X^{*} \leqslant _{hr} X$ (and Y)), we have

  $$\begin{aligned} S_1 \circ U_{\beta }^{\phi }(Y) = S_1 \circ U_{\beta }^{\phi }(X) = S_1 \circ U_{\beta }^{\phi }(X^{*}) = 1, \end{aligned}$$

  According to (a), (b) can be proved.

• 3. $\forall X, Y$ which satisfy (a) $\forall$ $X,Y\in \tilde{\mathcal {X}}_G$, or (b) if ${\exists }X^{*}\in \mathcal {X}^{*}_G$ s.t. $X^{*} \leqslant _{st} X$ (and Y), we have $SlideM(X) = \rho _{g_1}(X)$, $SlideM(Y) = \rho _{g_1}(Y)$. Then $X \leqslant _{icx} Y$ implies that $SlideM(X) \leqslant SlideM(Y)$.

• 4. $\forall X\in \tilde{\mathcal {X}}_G$, we have $S_1\circ U_{\beta }^{\phi }(X)= 1$. If $X \leqslant _{st} Y$ (or $X \leqslant _{lr} Y$ or $X \leqslant _{hr} Y$), due to that $g_1(u) \geqslant g_2(u) \geqslant \cdots \geqslant g_N(u)$, and $\sum \limits _{i=1}^N S_i(x)\cdot g_i(u)$ is non-decreasing with respect to x, we have $S\circ U_{\beta }^{\phi }(Y) = S\circ U_{\beta }^{\phi }(X)= 1$, which implies that $Y \in \tilde{\mathcal {X}}_G$.

• 5. $\forall a \in \mathcal {R}$, we have

  $$\begin{aligned} SlideM(X+a) = \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X+a)\cdot \rho _{g_i}(X+a) = \sum \limits _{i=1}^N S_i \circ \left( U_{\beta }^{\phi }(X)+a \right) \cdot \rho _{g_i}(X) +a. \end{aligned}$$

  If $a \geqslant 0$, we know that $\sum \limits _{i=1}^N S_i \circ \left( U_{\beta }^{\phi }(X)+a \right) \geqslant \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)$ which implies that

  $$\begin{aligned} SlideM(X+a) \geqslant \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X)+a = SlideM(X)+a. \end{aligned}$$

  The equal sign holds when $a=0$. Similarly, if $a \leqslant 0$, we know that $\sum \limits _{i=1}^N S_i \circ \left( U_{\beta }^{\phi }(X)+a \right) \leqslant \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)$ which implies that

  $$\begin{aligned} SlideM(X+a) \leqslant \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X)+a = SlideM(X)+a. \end{aligned}$$

  The equal sign holds when $a=0$.

• 6. Due to that

  $$\begin{aligned} SlideM(\lambda X)&= \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(\lambda X)\cdot \rho _{g_i}(\lambda X) = \lambda \sum \limits _{i=1}^N S_i \circ (\lambda U_{\beta }^{\phi }( X))\cdot \rho _{g_i}(X), \\ \lambda SlideM(X)&= \lambda \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X), \end{aligned}$$

  we can have

  $$\begin{aligned} SlideM(\lambda X) - \lambda SlideM(X) = \lambda \left[ \sum \limits _{i=1}^N S_i \circ (\lambda U_{\beta }^{\phi }( X)) \cdot \rho _{g_i}(X) - \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X) \right] . \end{aligned}$$

  From $U_{\beta }^{\phi }(X)>0$, we know that, if $\lambda \geqslant 1$, we have $\sum \limits _{i=1}^N S_i \circ (\lambda U_{\beta }^{\phi }( X)) \cdot \rho _{g_i}(X) - \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X) \geqslant 0$ which implies that $SlideM(\lambda X) - \lambda SlideM(X) \geqslant 0$. The equal sign holds when $\lambda =1$. Similarly, if $\lambda \leqslant 1$, we have $\sum \limits _{i=1}^N S_i \circ (\lambda U_{\beta }^{\phi }( X)) \cdot \rho _{g_i}(X) - \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X) \leqslant 0$ which implies that $SlideM(\lambda X) - \lambda SlideM(X) \leqslant 0$. The equal sign holds when $\lambda =1$.

• 7. $\forall X \in \tilde{\mathcal {X}}_G$, we have $SlideM(X) = \rho _{g_1}(X)$. Therefore, $\forall X,Y \in \tilde{\mathcal {X}}_G$, we have

  $$\begin{aligned}&SlideM(\lambda X+(1-\lambda ) Y) \leqslant \rho _{g_1}(\lambda X+(1-\lambda ) Y)\leqslant \rho _{g_1}(\lambda X)+\rho _{g_1}((1-\lambda ) Y) \\ =&\lambda \rho _{g_1}(X)+(1-\lambda )\rho _{g_1}( Y) = \lambda SlideM(X)+(1-\lambda ) SlideM(Y). \end{aligned}$$

  Then (a) is proved. $\forall X^{*}\in \mathcal {X}^{*}_G$, if $X^{*} \leqslant _{st} X$ (and Y) (or $X^{*} \leqslant _{lr} X$ (and Y) or $X^{*} \leqslant _{hr} X$ (and Y)), we have

  $$\begin{aligned} S_1\circ U_{\beta }^{\phi }(Y) = S_1\circ U_{\beta }^{\phi }(X) = S_1\circ U_{\beta }^{\phi }(X^{*}) = 1, \end{aligned}$$

  which means that $X,Y \in \tilde{\mathcal {X}}_G$. According to (a), (b) can be proved.

• 8. From Dhaene et al. (2006) we know that distortion risk measure is additive of for sums of comonotonic risks. Then we can have

  $$\begin{aligned} SlideM(X+Y)&= \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X+Y)\cdot \rho _{g_i}(X+Y) \\&= \sum \limits _{i=1}^N S_i \circ (U_{\beta }^{\phi }(X)+U_{\beta }^{\phi }(Y))\cdot \rho _{g_i}(X)+\sum \limits _{i=1}^N S_i \circ (U_{\beta }^{\phi }(X)+U_{\beta }^{\phi }(Y))\cdot \rho _{g_i}(Y). \end{aligned}$$

  Due to that $U_{\beta }^{\phi }( X)>0$, $U_{\beta }^{\phi }(Y)>0$, we can have

  $$\begin{aligned} SlideM(X+Y)&\geqslant \sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(X)\cdot \rho _{g_i}(X)+\sum \limits _{i=1}^N S_i \circ U_{\beta }^{\phi }(Y)\cdot \rho _{g_i}(Y) \\&= SlideM(X)+SlideM(Y). \end{aligned}$$

Other Examples of Analytical SlideVaR Expressions

1.1 Exponential Distribution

Let X be an exponential distributed random variable with parameters $\lambda$, i.e. $X \sim Exp(\lambda )$, then we know that $E[X]=SD[X]=1/\lambda$. Because the exponential is

$$\begin{aligned} F(x)=\left\{ \begin{aligned}&1-e^{-\lambda x},&\,\, \text {if }\,\,&x \geqslant 0,\\&0,&\,\, \text {if }\,\,&x <0. \end{aligned} \right. \end{aligned}$$

we can have

$$\begin{aligned} VaR_{\alpha }(X) = \frac{-\ln {(1-\alpha )}}{\lambda }. \end{aligned}$$

Thanks to integration by parts, we can get

$$\begin{aligned} ES_{\alpha }(X) = \frac{-\ln {(1-\alpha )}+1}{\lambda }. \end{aligned}$$

For the function $\phi ^e_{\beta }(p)$ in Eq. (11), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^e}(X)&= \frac{1}{\gamma (1-e^{\frac{\beta -1}{\gamma }})}\sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) (1-\beta _i) \cdot \frac{-\ln {(1-\beta _i)}+1}{\lambda } \\&+ \frac{e^{\frac{\beta -1}{\gamma }}(1-\beta )}{\gamma (1-e^{\frac{\beta -1}{\gamma }})}\cdot \frac{-\ln {(1-\beta )}+1}{\lambda }. \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^e}(X) < a$, $SlideVaR(X) = VaR_{\beta }(X) = \dfrac{-\ln {(1-\beta )}}{\lambda }$;

• When $a \leqslant U_{\beta ,N}^{\phi ^e}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \frac{1}{\lambda (b-a)} \left( \ln \frac{1-\beta }{1-\alpha }+1\right) \left[ \frac{1}{\gamma (1-e^{\frac{\beta -1}{\gamma }})}\sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) (1-\beta _i) \right. \\&\left. \frac{-\ln {(1-\beta _i)}+1}{\lambda } + \frac{e^{\frac{\beta -1}{\gamma }}(1-\beta )}{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \cdot \frac{-\ln {(1-\beta )}+1}{\lambda }-a \right] + \frac{-\ln {(1-\beta )}}{\lambda }; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^e}(X) \geqslant b$, $SlideVaR(X) = ES_{\alpha }(X)=\dfrac{-\ln {(1-\alpha )}+1}{\lambda }$.

For the function $\phi ^p_{\beta }(p)$ in Eq. (12), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^p}(X)&= \frac{1-\gamma }{(1-\beta )^{1-\gamma }}\sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] (1-\beta _i) \cdot \frac{-\ln {(1-\beta _i)}+1}{\lambda } \\&+ (1-\gamma )\cdot \frac{-\ln {(1-\beta )}+1}{\lambda }. \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^p}(X) < a$, $SlideVaR(X) = VaR_{\beta }(X) = \dfrac{-\ln {(1-\beta )}}{\lambda }$;

• When $a \leqslant U_{\beta ,N}^{\phi ^p}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \frac{1}{\lambda (b-a)} \left( \ln \frac{1-\beta }{1-\alpha }+1\right) \left[ \frac{1-\gamma }{(1-\beta )^{1-\gamma }}\sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] \right. \\&\left. (1-\beta _i)\frac{-\ln {(1-\beta _i)}+1}{\lambda } + (1-\gamma )\cdot \frac{-\ln {(1-\beta )}+1}{\lambda } - a \right] + \frac{-\ln {(1-\beta )}}{\lambda }; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^p}(X) \geqslant b$, $SlideVaR(X) =ES_{\alpha }(X)= \dfrac{-\ln {(1-\alpha )}+1}{\lambda }$.

In addition, if $U_{\beta ,N}^{\phi ^e}(X)$ or $U_{\beta ,N}^{\phi ^p}(X) \equiv h(b-a)+a$ where h is a constant in [0, 1]. Then

$$\begin{aligned} SlideVaR(X) = h\cdot \dfrac{-\ln {(1-\alpha )}+1}{\lambda } + (1-h)\cdot \dfrac{-\ln {(1-\beta )}}{\lambda }, \end{aligned}$$

which becomes $GlueVaR_{\alpha ,\beta }^{h_1,h_2}(X)$ with $h_1=h_2=h$ (Belles-Sampera et al. 2014).

1.2 Pareto Distribution

Let X be a Pareto distributed random variable with parameters $(k, \sigma )$ where $k>0$, $\sigma >0$, i.e. $X \sim Pareto(k, \sigma )$, then we know that

$$\begin{aligned} E(X)=\left\{ \begin{aligned}&\infty ,&\,\, \text {if }\,\,&k \leqslant 1,\\&\frac{k\sigma }{k-1},&\,\, \text {if }\,\,&k >1. \end{aligned} \right. \end{aligned}$$

$$\begin{aligned} D(X)=\left\{ \begin{aligned}&\quad \quad \infty ,&\,\, \text {if }\,\,&0< k \leqslant 2,\\&\frac{k\sigma ^2}{(k-1)^2(k-2)},&\,\, \text {if }\,\,&k >2. \end{aligned} \right. \end{aligned}$$

Because the Pareto CDF is

$$\begin{aligned} F(x)=\left\{ \begin{aligned}&1-\left( \frac{\sigma }{x}\right) ^k,&\,\, \text {if }\,\,&x \geqslant \sigma ,\\&0,&\,\, \text {if }\,\,&x<\sigma , \end{aligned} \right. \end{aligned}$$

we can have

$$\begin{aligned} VaR_{\alpha }(X) = \frac{\sigma }{(1-\alpha )^{\frac{1}{k}}}. \end{aligned}$$

From Norton et al. (2021) we know that

$$\begin{aligned} ES_{\alpha }(X) = \frac{k\sigma }{(1-\alpha )^{\frac{1}{k}}(k-1)}. \end{aligned}$$

For the function $\phi ^e_{\beta }(p)$ in Eq. (11), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^e}(X)&= \frac{k\sigma }{\gamma (1-e^{\frac{\beta -1}{\gamma }})(k-1)}\sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) (1-\beta _i)^{1-\frac{1}{k}}+ \frac{k\sigma e^{\frac{\beta -1}{\gamma }}(1-\beta )^{1-\frac{1}{k}}}{\gamma (1-e^{\frac{\beta -1}{\gamma }})(k-1)}. \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^e}(X) < a$, $SlideVaR(X) = VaR_{\beta }(X) =\dfrac{\sigma }{(1-\beta )^{\frac{1}{k}}}$;

• When $a \leqslant U_{\beta ,N}^{\phi ^e}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \frac{k (1-\beta )^{\frac{1}{k}} - (1-\alpha )^{\frac{1}{k}}(k-1)}{(1-\alpha )^{\frac{1}{k}}(1-\beta )^{\frac{1}{k}}(k-1)} \cdot \frac{\sigma }{b-a}\left[ \frac{k\sigma }{\gamma (1-e^{\frac{\beta -1}{\gamma }})(k-1)} \right. \\&\left. \sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) (1-\beta _i)^{1-\frac{1}{k}}+ \frac{k\sigma e^{\frac{\beta -1}{\gamma }}(1-\beta )^{1-\frac{1}{k}}}{\gamma (1-e^{\frac{\beta -1}{\gamma }})(k-1)}-a \right] \\&+ \dfrac{\sigma }{(1-\beta )^{\frac{1}{k}}}; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^e}(X) \geqslant b$, $SlideVaR(X) = ES_{\alpha }(X)=\dfrac{k\sigma }{(1-\alpha )^{\frac{1}{k}}(k-1)}$.

For the function $\phi ^p_{\beta }(p)$ in Eq. (12), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^p}(X)&= \frac{(1-\gamma )k\sigma }{(1-\beta )^{1-\gamma }(k-1)}\sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] (1-\beta _i)^{1-\frac{1}{k}}+ \frac{(1-\gamma )k\sigma }{(1-\beta )^{\frac{1}{k}}(k-1)}. \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^p}(X) < a$, $SlideVaR(X) = VaR_{\beta }(X) =\dfrac{\sigma }{(1-\beta )^{\frac{1}{k}}}$;

• When $a \leqslant U_{\beta ,N}^{\phi ^p}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \frac{k (1-\beta )^{\frac{1}{k}} - (1-\alpha )^{\frac{1}{k}}(k-1)}{(1-\alpha )^{\frac{1}{k}}(1-\beta )^{\frac{1}{k}}(k-1)} \cdot \frac{\sigma }{b-a}\left[ \frac{(1-\gamma )k\sigma }{(1-\beta )^{1-\gamma }(k-1)} \right. \\&\left. \sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] (1-\beta _i)^{1-\frac{1}{k}}+ \frac{(1-\gamma )k\sigma }{(1-\beta )^{\frac{1}{k}}(k-1)} - a \right] \\&+ \dfrac{\sigma }{(1-\beta )^{\frac{1}{k}}}; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^p}(X) \geqslant b$, $SlideVaR(X) = ES_{\alpha }(X)=\dfrac{k\sigma }{(1-\alpha )^{\frac{1}{k}}(k-1)}$.

In addition, if $U_{\beta ,N}^{\phi ^e}(X)$ or $U_{\beta ,N}^{\phi ^p}(X) \equiv h(b-a)+a$ where h is a constant in [0, 1]. Then

$$\begin{aligned} SlideVaR(X) = \frac{h k (1-\beta )^{\frac{1}{k}} - h (1-\alpha )^{\frac{1}{k}}(k-1) + (1-\alpha )^{\frac{1}{k}}(k-1) }{(1-\alpha )^{\frac{1}{k}}(1-\beta )^{\frac{1}{k}}(k-1)} \cdot \sigma , \end{aligned}$$

which equals to $GlueVaR_{\alpha ,\beta }^{h_1,h_2}(X)$ with $h_1=h_2=h$ (Belles-Sampera et al. 2014).

1.3 Generalized Pareto Distribution

Let X be a Generalized Pareto distributed random variable with parameters $(\mu , \sigma , \xi )$ where $\sigma >0$, i.e. $X \sim GP(\mu , \sigma , \xi )$, then we know that

$$\begin{aligned} E(X)= \mu + \frac{\sigma }{1-\xi }, \quad if \quad \xi <1, \end{aligned}$$

$$\begin{aligned} D(X)=\frac{\sigma ^2}{(1-\xi )^2(1-2\xi )}, \quad if \quad \xi <0.5. \end{aligned}$$

Because the Pareto CDF is

$$\begin{aligned} F(x)=\left\{ \begin{aligned}&1-\left( 1+ \frac{\xi (x-\mu )}{\sigma }\right) ^{-\frac{1}{\xi }},&\,\, \text {if }\,\,&\xi \ne 0,\\&1-e^{-\frac{x-\mu }{\sigma }},&\,\, \text {if }\,\,&\xi = 0, \end{aligned} \right. \end{aligned}$$

for $x \geqslant \mu$ when $\xi \geqslant 0$ and $\mu \leqslant x \leqslant \mu -\frac{\sigma }{\xi }$ when $\xi <0$, we can have

$$\begin{aligned} VaR_{\alpha }(X) =\left\{ \begin{aligned}&\mu + \sigma \frac{(1-\alpha )^{-\xi }-1}{\xi },&\,\, \text {if }\,\,&\xi \ne 0,\\&\mu -\sigma \ln (1-\alpha ),&\,\, \text {if }\,\,&\xi = 0, \end{aligned} \right. \end{aligned}$$

From Norton et al. (2021) we know that, with $-1< \xi <1$,

$$\begin{aligned} ES_{\alpha }(X) =\left\{ \begin{aligned}&\mu + \sigma \frac{(1-\alpha )^{-\xi }-1}{\xi }+ \sigma \frac{(1-\alpha )^{-\xi }}{1-\xi },&\,\, \text {if }\,\,&\xi \ne 0,\\&\mu - \sigma \ln (1-\alpha )+\sigma ,&\,\, \text {if }\,\,&\xi = 0, \end{aligned} \right. \end{aligned}$$

We know that, if $\xi =0$, X would become an exponential distributed random variable which we have discussed in Sect. 1. If $\xi <0$, X would become a Pareto distributed random variable which we have discussed in Sect. 2. Therefore, in this section, we only consider the case where $\xi \in (0,1)$.

For the function $\phi ^e_{\beta }(p)$ in Eq. (11), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^e}(X)&= \frac{1}{\gamma (1-e^{\frac{\beta -1}{\gamma }})}\sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) (1-\beta _i)\left( \mu + \sigma \frac{(1-\beta _i)^{-\xi }-1}{\xi }+ \sigma \frac{(1-\beta _i)^{-\xi }}{1-\xi } \right) \\&+ \frac{e^{\frac{\beta -1}{\gamma }}(1-\beta )}{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \left( \mu + \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }+ \sigma \frac{(1-\beta )^{-\xi }}{1-\xi } \right) . \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^e}(X) < a$, $SlideVaR(X) = VaR_{\beta }(X) = \mu + \sigma \dfrac{(1-\beta )^{-\xi }-1}{\xi }$;

• When $a \leqslant U_{\beta ,N}^{\phi ^p}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \left[ \frac{(1-\alpha )^{-\xi }-(1-\beta )^{-\xi }}{\xi }+ \frac{(1-\alpha )^{-\xi }}{1-\xi } \right] \frac{\sigma }{b-a}\left[ \frac{1}{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \right. \\&\left. \sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) (1-\beta _i)\left( \mu + \sigma \frac{(1-\beta _i)^{-\xi }-1}{\xi }+ \sigma \frac{(1-\beta _i)^{-\xi }}{1-\xi } \right) \right. \\&\left. + \frac{e^{\frac{\beta -1}{\gamma }}(1-\beta )}{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \left( \mu + \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }+ \sigma \frac{(1-\beta )^{-\xi }}{1-\xi } \right) -a \right] + \mu \\&+ \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^e}(X) \geqslant b$, $SlideVaR(X)=ES_{\alpha }(X)=\mu + \sigma \dfrac{(1-\alpha )^{-\xi }-1}{\xi }+ \sigma \dfrac{(1-\alpha )^{-\xi }}{1-\xi }$.

For the function $\phi ^p_{\beta }(p)$ in Eq. (12), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^e}(X)&= \frac{1-\gamma }{(1-\beta )^{1-\gamma }}\sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] (1-\beta _i)\left( \mu + \sigma \frac{(1-\beta _i)^{-\xi }-1}{\xi } \right. \\&\left. + \sigma \frac{(1-\beta _i)^{-\xi }}{1-\xi } \right) + (1-\gamma ) \left( \mu + \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }+ \sigma \frac{(1-\beta )^{-\xi }}{1-\xi } \right) . \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^e}(X) < a$, $SlideVaR(X) = VaR_{\beta }(X) = \mu + \sigma \dfrac{(1-\beta )^{-\xi }-1}{\xi }$;

• When $a \leqslant U_{\beta ,N}^{\phi ^p}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \left[ \frac{(1-\alpha )^{-\xi }-(1-\beta )^{-\xi }}{\xi }+ \frac{(1-\alpha )^{-\xi }}{1-\xi } \right] \frac{\sigma }{b-a}\left[ \frac{1-\gamma }{(1-\beta )^{1-\gamma }} \right. \\&\left. \sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma } -(1-\beta _{i-1})^{-\gamma }\right] (1-\beta _i)\left( \mu + \sigma \frac{(1-\beta _i)^{-\xi }-1}{\xi } \right. \right. \\&\left. \left. +\sigma \frac{(1-\beta _i)^{-\xi }}{1-\xi } \right) + (1-\gamma ) \left( \mu + \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }+ \sigma \frac{(1-\beta )^{-\xi }}{1-\xi } \right) \right. \\&\left. -a \right] + \mu + \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^e}(X) \geqslant b$, $SlideVaR(X)=ES_{\alpha }(X)=\mu + \sigma \dfrac{(1-\alpha )^{-\xi }-1}{\xi }+ \sigma \dfrac{(1-\alpha )^{-\xi }}{1-\xi }$.

In addition, if $U_{\beta ,N}^{\phi ^e}(X)$ or $U_{\beta ,N}^{\phi ^p}(X) \equiv h(b-a)+a$ where h is a constant in [0, 1]. Then

$$\begin{aligned} SlideVaR(X) = \left[ \frac{(1-\alpha )^{-\xi }-(1-\beta )^{-\xi }}{\xi }+ \frac{(1-\alpha )^{-\xi }}{1-\xi } \right] \sigma + \mu + \sigma \frac{(1-\beta )^{-\xi }-1}{\xi }, \end{aligned}$$

which equals to $GlueVaR_{\alpha ,\beta }^{h_1,h_2}(X)$ with $h_1=h_2=h$ (Belles-Sampera et al. 2014).

1.4 Weibull Distribution

Let X be a Weibull distributed random variable with parameters $(\lambda , k)$, i.e. $X \sim Weibull(\lambda , k)$, then we know that $E[X]=\lambda \Gamma (1+\frac{1}{k})$, $D[X]=\lambda ^2 \left[ \Gamma (1+\frac{2}{k})-\Gamma (1+\frac{1}{k})^2 \right]$ where $\Gamma (a) = \int _0^\infty p^{a-1}e^{-p}dp$ is the gamma function. Because the Weibull CDF is

$$\begin{aligned} F(x) = 1-e^{-(\frac{x}{\lambda })^k}, \end{aligned}$$

we can have

$$\begin{aligned} VaR_{\alpha }(X) = \lambda [-\ln (1-\alpha )]^{\frac{1}{k}}. \end{aligned}$$

From Norton et al. (2021) we know that

$$\begin{aligned} ES_{\alpha }(X) = \frac{\lambda }{1-\alpha }\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\alpha ) \right) , \end{aligned}$$

where $\Gamma _U(a,b) = \int _b^\infty p^{a-1}e^{-p}dp$ is the upper incomplete gamma function. For the function $\phi ^e_{\beta }(p)$ in Eq. (11), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^e}(X)&= \frac{\lambda }{\gamma (1-e^{\frac{\beta -1}{\gamma }})}\sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) \Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta _i) \right) \\&+ \frac{\lambda e^{\frac{\beta -1}{\gamma }}}{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta ) \right) . \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^e}(X) < a$, $SlideVaR(X) = \lambda (-\ln (1-\beta ))^{\frac{1}{k}}$;

• When $a \leqslant U_{\beta ,N}^{\phi ^e}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \left[ \frac{\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\alpha ) \right) }{1-\alpha } - [-\ln (1-\beta )]^{\frac{1}{k}} \right] \frac{\lambda }{b-a}\left[ \frac{\lambda }{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \right. \\&\left. \sum _{i=2}^{N} \left( e^{\frac{\beta _i-1}{\gamma }}-e^{\frac{\beta _{i-1}-1}{\gamma }}\right) \Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta _i) \right) + \frac{\lambda e^{\frac{\beta -1}{\gamma }}}{\gamma (1-e^{\frac{\beta -1}{\gamma }})} \right. \\&\left. \Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta ) \right) -a \right] +\lambda (-\ln (1-\beta ))^{\frac{1}{k}}; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^e}(X) \geqslant b$, $SlideVaR(X) = \dfrac{\lambda }{1-\alpha }\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\alpha ) \right)$.

For the function $\phi ^p_{\beta }(p)$ in Eq. (12), we have

$$\begin{aligned} U_{\beta ,N}^{\phi ^p}(X)&= \frac{\lambda (1-\gamma )}{(1-\beta )^{1-\gamma }}\sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] \Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta _i) \right) \\&+ \frac{\lambda (1-\gamma )}{1-\beta }\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta ) \right) . \end{aligned}$$

Therefore, we have

• When $U_{\beta ,N}^{\phi ^e}(X) < a$, $SlideVaR(X) = \lambda (-\ln (1-\beta ))^{\frac{1}{k}}$;

• When $a \leqslant U_{\beta ,N}^{\phi ^e}(X) < b$,

  $$\begin{aligned} SlideVaR(X)&= \left[ \frac{\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\alpha ) \right) }{1-\alpha } - [-\ln (1-\beta )]^{\frac{1}{k}} \right] \frac{\lambda }{b-a}\left[ \frac{\lambda (1-\gamma )}{(1-\beta )^{1-\gamma }} \right. \\&\left. \sum _{i=2}^{N} \left[ (1-\beta _{i})^{-\gamma }-(1-\beta _{i-1})^{-\gamma }\right] \Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta _i) \right) \right. \\&\left. + \frac{\lambda (1-\gamma )}{1-\beta }\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\beta ) \right) -a \right] +\lambda (-\ln (1-\beta ))^{\frac{1}{k}}; \end{aligned}$$

• When $U_{\beta ,N}^{\phi ^e}(X) \geqslant b$, $SlideVaR(X) = \dfrac{\lambda }{1-\alpha }\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\alpha ) \right)$.

In addition, if $U_{\beta ,N}^{\phi ^e}(X)$ or $U_{\beta ,N}^{\phi ^p}(X) \equiv h(b-a)+a$ where h is a constant in [0, 1]. Then

$$\begin{aligned} SlideVaR(X)&= \left[ \frac{\Gamma _U\left( 1+\frac{1}{k},-\ln (1-\alpha ) \right) }{1-\alpha } - [-\ln (1-\beta )]^{\frac{1}{k}} \right] \lambda h +\lambda (-\ln (1-\beta ))^{\frac{1}{k}}, \end{aligned}$$

which equals to $GlueVaR_{\alpha ,\beta }^{h_1,h_2}(X)$ with $h_1=h_2=h$.