Calculating CVaR and bPOE for common probability distributions with application to portfolio optimization and density estimation

Abstract

Conditional value-at-risk (CVaR) and value-at-risk, also called the superquantile and quantile, are frequently used to characterize the tails of probability distributions and are popular measures of risk in applications where the distribution represents the magnitude of a potential loss. buffered probability of exceedance (bPOE) is a recently introduced characterization of the tail which is the inverse of CVaR, much like the CDF is the inverse of the quantile. These quantities can prove very useful as the basis for a variety of risk-averse parametric engineering approaches. Their use, however, is often made difficult by the lack of well-known closed-form equations for calculating these quantities for commonly used probability distributions. In this paper, we derive formulas for the superquantile and bPOE for a variety of common univariate probability distributions. Besides providing a useful collection within a single reference, we use these formulas to incorporate the superquantile and bPOE into parametric procedures. In particular, we consider two: portfolio optimization and density estimation. First, when portfolio returns are assumed to follow particular distribution families, we show that finding the optimal portfolio via minimization of bPOE has advantages over superquantile minimization. We show that, given a fixed threshold, a single portfolio is the minimal bPOE portfolio for an entire class of distributions simultaneously. Second, we apply our formulas to parametric density estimation and propose the method of superquantiles (MOS), a simple variation of the method of moments where moments are replaced by superquantiles at different confidence levels. With the freedom to select various combinations of confidence levels, MOS allows the user to focus the fitting procedure on different portions of the distribution, such as the tail when fitting heavy-tailed asymmetric data.

Proofs

1.1 Proposition 1

Proof

First, note that $q_\alpha (X) = \frac{-\ln (1 - \alpha ) }{\lambda }$ for exponential RVs with rate parameter $\lambda$. We then have,

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} q_p (X) dp \\&= \frac{-1}{\lambda (1-\alpha )} \int _{\alpha }^{1} \ln (1 - p ) dp \\&= \frac{-1}{\lambda (1-\alpha )} \int _{1-\alpha }^{0} -\ln (y ) dy = \frac{-1}{\lambda (1-\alpha )} \int _{0}^{1 - \alpha } \ln (y ) dy. \end{aligned}$$

Since $\int \ln (y)dy = y \ln (y) - y + C$, we have

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{-1}{\lambda (1-\alpha )} \int _{0}^{1 - \alpha } \ln (y ) dy \\&= \frac{-1}{\lambda (1-\alpha )} \left[ ( 1-\alpha ) \ln (1-\alpha ) - (1-\alpha ) \right] = \frac{ - \ln (1-\alpha ) + 1 }{ \lambda } . \end{aligned}$$

We can then see that

$$\begin{aligned} {\bar{p}}_x (X)&= \{ 1 - \alpha | {\bar{q}}_\alpha (X) = x \} \\&= \{ 1 - \alpha |\frac{ - \ln (1-\alpha ) + 1 }{ \lambda } = x \} \\&= \{ 1 - \alpha | \ln (1-\alpha ) = 1-\lambda x \} \\&= \{ 1 - \alpha | e^{\ln (1-\alpha )} = e^{1-\lambda x} \} = \{ 1 - \alpha | 1-\alpha = e^{1-\lambda x} \} = e^{1-\lambda x} . \end{aligned}$$

$\square$

1.2 Corollary 1

Proof

We know that X, being exponential, has CDF given by $P(X \ge x) = 1 - e^{-\lambda x}$. From Proposition 1, we know that

$$\begin{aligned} {\bar{p}}_x (X) = e^{(1 - \lambda x)} = e^{-\lambda (\frac{-1}{\lambda } + x)}. \end{aligned}$$

Then, since $\mu =\frac{1}{\lambda }$, it follows that ${\bar{p}}_x (X) = e^{-\lambda ( x - \mu )} = 1 - P(X \le x - \mu ) = P(X > x - \mu )$. The equality for CVaR follows easily from Proposition 1 since $q_\alpha (X) = \frac{-\ln (1 - \alpha ) }{\lambda }$.

$\square$

1.3 Proposition 2

Proof

First, note that the conditional distribution of a Pareto, conditioned on the event that the random value is larger than some $\gamma$, is simply another Pareto with parameters $a,\gamma$. This implies that $E[ X | X > \gamma ] = \frac{ a \gamma }{ a - 1}$ if $a\ge 1$; otherwise the expectation is $\infty$. Also, $1- F(\gamma ) = \left( \frac{x_m}{\gamma } \right)^a$. Since,

$$\begin{aligned} E[ X - \gamma ]^+ = (E[ X | X > \gamma ] - \gamma )(1- F(\gamma )) , \end{aligned}$$

we will have that,

$$\begin{aligned} E[ X - \gamma ]^+ = \left( \frac{ a \gamma }{ a - 1}- \gamma \right) \left( \frac{x_m}{\gamma } \right)^a . \end{aligned}$$

This gives us bPOE formula,

$$\begin{aligned} {\bar{p}}_x(X)&= \min _{x_m\le \gamma<x} \frac{ \left( \frac{ a \gamma }{ a - 1}- \gamma \right) x_m^a }{ \gamma ^a (x-\gamma ) } \\&= \min _{x_m\le \gamma<x} \frac{ \left( \frac{ a }{ a - 1}- 1 \right) x_m^a }{ \gamma ^{a-1} (x-\gamma ) } = \left( \max _{x_m\le \gamma <x} \frac{\gamma ^{a-1} (x-\gamma )(a-1) }{x_m^a} \right)^{-1} \end{aligned}$$

Since $a>1$, the maximization objective is concave over the range $\gamma \in (0,\infty )$ which contains the range $[x_m,x)$, so we just need to take the gradient of function $g(\gamma )=\frac{\gamma ^{a-1} (x-\gamma )(a-1) }{x_m^a}$ and set it to zero to find the optimal $\gamma$ as follows:

$$\begin{aligned} \frac{\partial g}{\partial \gamma } = \frac{ x(a-1)^2 \gamma ^{a-2} - (a-1)a\gamma ^{a-1} }{x_m^a} =0&\implies x(a-1)^2 \gamma ^{a-2} = (a-1)a\gamma ^{a-1} \\&\implies \frac{x(a-1)}{a} = \gamma \\ \end{aligned}$$

Plugging this value of $\gamma$ into the objective of our bPOE formula yields,

$$\begin{aligned} {\bar{p}}_x(X)&=\frac{ \left( \frac{ \frac{ax(a-1)}{a} }{a-1} - \frac{x(a-1)}{a} \right) x_m^a }{ \left(\frac{x(a-1)}{a}\right)^a \left( x - \frac{x(a-1)}{a} \right) } \\&= \left( \frac{x_m a}{x(a-1) } \right)^a \end{aligned}$$

CVaR is then equal to the value of x which solves the equation $1-\alpha = {\bar{p}}_x(X)$ or,

$$\begin{aligned} 1-\alpha = \left( \frac{x_m a}{x(a-1) } \right)^a , \end{aligned}$$

which has solution,

$$\begin{aligned} {\bar{q}}_\alpha (X) = \frac{ x_m a }{ (1-\alpha )^{\frac{1}{a}} (a-1)} . \end{aligned}$$

$\square$

1.4 Corollary 2

Proof

It follows by simply comparing the formulas from Proposition 1 and the CDF and quantile formulas for a Pareto RV. $\square$

1.5 Proposition 3

Proof

For these results, we rely on the fact that if $X \sim GPD(\mu ,s,\xi )$, then $X-\gamma | X> \gamma \sim GPD(0,s+\xi (\gamma -\mu ),\xi )$, meaning that the excess distribution of a GPD random variable is also GPD. Now, note also that if $\xi <1$, then $E[X]=\mu + \frac{s}{1-\xi }$. This gives us,

$$\begin{aligned} E[X-\gamma | X>\gamma ] = E[ GPD(0,s+\xi (\gamma -\mu ),\xi )] = \frac{s+\xi (\gamma - \mu )}{1-\xi } \end{aligned}$$

which further implies that,

$$\begin{aligned} {\bar{q}}_\alpha (X)&= E[X-q_\alpha (X) | X>q_\alpha (X)] + q_\alpha (X) \\&= \frac{s + \xi ( q_\alpha (X) - \mu )}{1-\xi } +q_\alpha (X) . \end{aligned}$$

Plugging in the values of the quantile functions yields the final formulas. Using the formulas we just found for ${\bar{q}}_\alpha (X)$, it is straightforward to solve for ${\bar{p}}_x(X)$ which equals $1-\alpha$ such that $\alpha$ solves the equation $x={\bar{q}}_\alpha (X)$. $\square$

1.6 Proposition 4

Proof

To get the superquantile, we begin with the integral representation:

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} q_p (X) dp \\&= \frac{1}{1-\alpha } \int _{\alpha }^{1} \mu - b\,sign(p-.5)\,\ln (1 - 2|p-.5|) dp \\&= \mu - \frac{b}{1-\alpha } \int _{\alpha }^{1} sign(p-.5)\,\ln (1 - 2|p-.5|) dp \\&= \mu - \frac{b}{1-\alpha } \left( \int _{\min \{\alpha ,.5\}}^{.5} -\ln (2p) dp + \int _{\max \{\alpha ,.5\}}^{1} \ln (2(1-p)) dp \right) . \end{aligned}$$

To evaluate the integral, we use simple substitution as well as the identity $\int \ln (y)dy = y \ln (y) - y + C$. After simplifying, we see that with $\alpha < .5$ the integral evaluates to,

$$\begin{aligned} {\bar{q}}_\alpha (X)= \mu +b \left(\frac{\alpha }{1-\alpha } \right)(1 - \ln (2 \alpha )) . \end{aligned}$$

Similarly, we find that with $\alpha \ge .5$ the integral evaluates to,

$$\begin{aligned} {\bar{q}}_\alpha (X) =\mu + b\left(1 - \ln \left(2(1- \alpha )\right) \right). \end{aligned}$$

For bPOE, first assume that threshold $x \ge \mu +b$. Using our formula for CVaR, we see that ${\bar{q}}_{.5} (X) = \mu + b$. Thus, $x \ge \mu +b$ implies that $1-{\bar{p}}_x(X) \ge .5$ implying that

$$\begin{aligned} {\bar{p}}_x (X)&= \{ 1 - \alpha | {\bar{q}}_\alpha (X) = x , \alpha \ge .5\} \\&= \{ 1 - \alpha |\mu + b\left(1 - \ln \left(2(1- \alpha )\right) \right) = x \} \\&= \frac{1}{2}e^{1-\left(\frac{x-\mu }{b} \right)} . \end{aligned}$$

Assume contrarily that $x< \mu + b$. Since ${\bar{q}}_{.5} (X) = \mu + b$, we have that $1-{\bar{p}}_x(X) < .5$ which implies that

$$\begin{aligned} {\bar{p}}_x (X)&= \{ 1 - \alpha | {\bar{q}}_\alpha (X) = x , \alpha < .5\} \\&= \left\{ 1 - \alpha |\mu +b \left(\frac{\alpha }{1-\alpha } \right)(1 - \ln (2 \alpha ))= x \right\} . \end{aligned}$$

Letting $z=\frac{x-\mu }{b}$, we must now find $\alpha$ which solves the equation $\left(\frac{\alpha }{1-\alpha } \right)(1 - \ln (2 \alpha ))= z$. We do so as follows:

$$\begin{aligned} \left(\frac{\alpha }{1-\alpha } \right)(1 - \ln (2 \alpha ))= z&\implies \frac{-z}{\alpha } = \frac{( \ln (2 \alpha ) - 1)}{1-\alpha } \\&\implies e^{\frac{-z}{\alpha }} = e^{\frac{(\ln (2 \alpha ) - 1 )}{1-\alpha }} = \left( \frac{2\alpha }{e} \right)^\frac{1}{1-\alpha } \\&\implies e^{\frac{-z(1-\alpha )}{\alpha }} = \left( \frac{2\alpha }{e} \right)\\&\implies \frac{-z}{\alpha } e^{-z(\frac{1}{\alpha } -1)} = -2ze^{-1} \\&\implies \frac{-z}{\alpha } e^{\frac{-z}{\alpha } } = -2ze^{-z-1} \\&\implies \frac{-z}{\alpha } = {\mathcal {W}}(-2ze^{-z-1}) . \end{aligned}$$

where the final step follows from the definition of the Lambert-${\mathcal {W}}$ function which is given by the relation $xe^x=y \iff {\mathcal {W}}(y)=x$. Thus, $\frac{-z}{\alpha } = {\mathcal {W}}(-2ze^{-z-1}) \implies {\bar{p}}_x(X) = 1-\alpha =1 + \frac{z}{{\mathcal {W}}( -2e^{-z-1}z)}$. $\square$

1.7 Proposition 5

Proof

It is well known that if $X \sim {\mathcal {N}}(0,1)$, then the conditional expectation is given by the inverse Mills Ratio, $E[ X | X > \gamma ] = \frac{f(\gamma )}{1 - F(\gamma ) }$. It follows then that ${\bar{q}}_\alpha (X) = E[ X | X > q_\alpha (X) ] = \frac{f(q_\alpha (X))}{1 - F(q_\alpha (X)) }= \frac{f(q_\alpha (X))}{1 - \alpha }$. $\square$

1.8 Proposition 6

Proof

Note that for a standard normal random variable, the tail expectation beyond any threshold $\gamma$ is given by the inverse Mills Ratio,

$$\begin{aligned} E[ X | X > \gamma ] = \frac{f(\gamma )}{1 - F(\gamma ) } . \end{aligned}$$

Note also that for any threshold $\gamma$ and any random variable we have,

$$\begin{aligned} E[ X - \gamma ]^+ = (E[ X | X > \gamma ] - \gamma )(1- F(\gamma )) . \end{aligned}$$

Using the Mills ratio gives us,

$$\begin{aligned} E[ X - \gamma ]^+ = \left( \frac{f(\gamma )}{1 - F(\gamma ) } - \gamma \right) (1- F(\gamma )) = f(\gamma ) - \gamma (1 - F(\gamma )) . \end{aligned}$$

Plugging this result into the minimization formula for bPOE yields the final formula. $\square$

1.9 Proposition 7

Proof

This follows from the fact that ${\bar{q}}_\alpha (X) = E[ X | X > q_\alpha (X) ] = \frac{f(q_\alpha (X))}{1 - F(q_\alpha (X)) }$ and the optimization formula of bPOE given in the previous proposition for normally distributed variables. $\square$

1.10 Proposition 8

Proof

To derive the integral representation, simply plug in the formula for $E[ X - \gamma ]^+$, then utilize the definition of the PDF and CDF. The gradient calculation is a standard calculus exercise. $\square$

1.11 Proposition 9

Proof

We simply evaluate the integral of the quantile function as follows.

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} q_p (X) dp \\&= \frac{1}{1-\alpha }\int _{\alpha }^{1} e^{\mu + s\sqrt{2} \text {erf}^{-1}(2p -1)} dp \\&= \frac{e^{\mu }}{1-\alpha }\int _{\alpha }^{1} e^{ s\sqrt{2} \text {erf}^{-1}(2p -1)} dp \\&= \frac{e^{\mu }}{1-\alpha } \left[ -\frac{1}{2} e^{ \frac{s^2}{2}} \left( 1 + \text {erf}\left( \frac{s}{\sqrt{2}} - \text {erf}^{-1}(2p -1) \right) \right) \right]_{p=\alpha }^1 \\&= \frac{e^{\mu }}{1-\alpha } \left[ \frac{1}{2} e^{ \frac{s^2}{2}} + \frac{1}{2} e^{ \frac{s^2}{2}} \left( 1 + \text {erf}\left( \frac{s}{\sqrt{2}} - \text {erf}^{-1}(2p -1) \right) \right) \right] \\&= \frac{1}{2} e^{\mu + \frac{s^2}{2}} \frac{ \left[ 1 + \text {erf}\left( \frac{s}{\sqrt{2}} - \text {erf}^{-1}(2\alpha -1) \right) \right] }{ 1-\alpha }. \end{aligned}$$

$\square$

1.12 Proposition 10

Proof

To obtain the superquantile, we have

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} q_p (X) dp \\&= \frac{1}{1-\alpha } \int _{\alpha }^{1} \mu + s \ln \left( \frac{\alpha }{1-\alpha } \right) dp \\&= \mu + \frac{s}{1-\alpha } \int _{\alpha }^{1} \ln (p) - \ln (1-p) dp \\&= \mu + \frac{s}{1-\alpha }\left( \int _{\alpha }^{1} \ln (p) dp + \int _{\alpha }^{1} - \ln (1-p) dp \right) \end{aligned}$$

Utilizing simple substitution as well as the identity $\int \ln (y)dy = y \ln (y) - y + C$, we get

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \mu + \frac{s}{1-\alpha } \left( -1 - \alpha \ln \alpha + \alpha - (1-\alpha ) \ln (1-\alpha ) + (1-\alpha ) \right) \\&= \mu + \frac{s}{1-\alpha } \left( - \alpha \ln \alpha - (1-\alpha ) \ln (1-\alpha ) \right) \\&= \mu + \frac{s}{1-\alpha } H( \alpha ) . \end{aligned}$$

To get bPOE, we simply follow the bPOE definition, needing to find $\alpha$ which solves $\mu + \frac{s}{1-\alpha } H( \alpha )=x$. The transformed system arises from combining logarithms within the superquantile formula and applying exponential transformations. $\square$

1.13 Proposition 11

Proof

This follows from the fact that $E[X-\gamma ]^+ = \int _{\gamma }^{\infty } (1-F(t))dt$. Evaluating this integral for $X \sim Logistic(\mu ,s)$ yields, $E[X-\gamma ]^+= s\ln ( 1+e^{-(\frac{\gamma -\mu }{s}) } )$ which can then be plugged into the minimization formula for bPOE. The second part of the proposition follows from the fact that the gradient of the objective function w.r.t. $\gamma$ is given by,

$$\begin{aligned} \frac{s\ln ( 1+e^{-(\frac{\gamma -\mu }{s}) } )}{ (x-\gamma )^2 } - \frac{ e^{-(\frac{\gamma -\mu }{s} )} }{ (x-\gamma )\left( 1+e^{-(\frac{\gamma -\mu }{s} )} \right) } . \end{aligned}$$

Setting this gradient to zero and simplifying yields the stated optimality condition. $\square$

1.14 Proposition 12

Proof

Since there is no closed-form expression for the quantile, we utilize the representation of the superquantile given by $\frac{1}{1-\alpha } \int _{q_\alpha (X)}^\infty t f(t) dt$. To evaluate this integral, we first take the derivative of the PDF, giving

$$\begin{aligned} \frac{df(x)}{dx} = \frac{-f(x)(x-\mu )(\nu +1)}{\nu s^2 + (x-\mu )}. \end{aligned}$$

Rearranging yields,

$$\begin{aligned} xf(x)dx = \frac{-\nu s^2 df(x) }{(\nu +1)} - \frac{(x-\mu )^2df(x)}{(\nu +1)} + \mu f(x) dx . \end{aligned}$$

We can then integrate both sides,

$$\begin{aligned} \int xf(x)dx = \frac{-\nu s^2 f(x) }{(\nu +1)} - \frac{1}{(\nu +1)} \int (x-\mu )^2df(x)+ \mu F(x). \end{aligned}$$

Integrating by parts gives us the following form of the middle term;

$$\begin{aligned} \int (x-\mu )^2df(x) = (x-\mu )^2 f(x) - 2 \int x f(x) dx + 2 \mu F(x) . \end{aligned}$$

Then, finally, after substituting this new expression for the middle term and simplifying, we get

$$\begin{aligned} \int xf(x)dx = - \frac{(\nu s^2 + (x- \mu )^2 )}{(\nu -1)}f(x) + \mu F(x). \end{aligned}$$

Taking the definite integral yields,

$$\begin{aligned} \int _{q_\alpha (X)}^\infty xf(x)dx&= \left( - \lim _{x \rightarrow \infty }\frac{(\nu s^2 + (x- \mu )^2) }{(\nu -1)}f(x) + \lim _{x \rightarrow \infty } \mu F(x) \right) \\&\qquad \qquad \quad - \left(- \frac{(\nu s^2 + (q_\alpha (X)- \mu )^2) }{(\nu -1)}f(q_\alpha (X)) + \mu F(q_\alpha (X)) \right). \end{aligned}$$

It is easy to see that the second limit goes to one and, after applying l’Hôpital’s rule where necessary, that the first limit goes to zero. This leaves

$$\begin{aligned} \int _{q_\alpha (X)}^\infty xf(x)dx&= \mu - \left(- \frac{(\nu s^2 + (q_\alpha (X)- \mu )^2) }{(\nu -1)}f(q_\alpha (X)) + \mu F(q_\alpha (X)) \right)\\&= \mu ( 1 - \alpha ) + \left( \frac{\nu s^2 + (q_\alpha (X)- \mu )^2 }{(\nu -1)} \right) f(q_\alpha (X)) \\&= \mu ( 1 - \alpha ) + s \left( \frac{\nu + T^{-1}(\alpha )^2 }{(\nu -1)} \right) \tau (T^{-1}(\alpha ) ), \end{aligned}$$

where the final step comes from writing the non-standardized quantile $q_\alpha (X)$ and PDF f(x) in their standardized form. Then, finally, dividing by $1-\alpha$ yields the formula

$$\begin{aligned} {\bar{q}}_\alpha (X) =\frac{1}{1-\alpha } \int _{q_\alpha (X)}^\infty xf(x)dx = \mu + s \left( \frac{\nu + T^{-1}(\alpha )^2 }{(\nu -1)(1-\alpha )} \right) \tau (T^{-1}(\alpha ) ) . \end{aligned}$$

$\square$

1.15 Proposition 13

Proof

To calculate the superquantile, we utilize the integral representation (1), which is

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} q_p (X) dp \\&= \frac{1}{1-\alpha } \int _{\alpha }^{1} \lambda {(-\ln (1-p))}^{1/k} dp . \end{aligned}$$

To put this integral into the form of the upper incomplete gamma function, make the change of variable $y= -\ln (1-p)$. This gives $e^y = \frac{1}{1-p}$ and $dp = (1-p) dy = e^{-y} dp$ with new lower limit of integration $-\ln (1-\alpha )$ and upper limit of integration $\infty$. Applying to the integral yields

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{\lambda }{1-\alpha } \int _{-\ln (1-\alpha )}^{\infty } {y}^{1/k} e^{-y} dy \\&= \frac{\lambda }{1-\alpha } \varGamma _U \left( 1 + \frac{1}{k} , -\ln (1-\alpha ) \right) . \end{aligned}$$

$\square$

1.16 Proposition 14

Proof

To calculate the superquantile, we utilize the integral representation as follows:

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} q_p (X) dp \\&= \frac{1}{1-\alpha } \left( \int _{0}^{1} q_p (X) dp - \int _{0}^{\alpha } q_p (X) dp \right)\\&= \frac{1}{1-\alpha } \left( E[X] - \int _{0}^{\alpha } q_p (X) dp \right)\\&= \frac{1}{1-\alpha } \left( E[X] - a \int _{0}^{\alpha } \left( \frac{p}{1-p} \right)^{\frac{1}{b}} dp \right) . \end{aligned}$$

Now, note first that for $X \sim LogLogistic(a,b)$, we have $E[X] =a\frac{\pi }{b} csc\left( \frac{\pi }{b} \right)$. Next, for the incomplete beta function, letting $A_1 = \frac{1}{b}+1$ and $A_2=1 - \frac{1}{b}$, we can see that

$$\begin{aligned} B_\alpha \left( \frac{1}{b}+1,1 - \frac{1}{b}\right) =\int _{0}^{\alpha }p^{\frac{1}{b}}\,(1-p)^{-\frac{1}{b}}\,dp . \end{aligned}$$

Using these two facts, we have,

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \left( E[X] - a \int _{0}^{\alpha } \left( \frac{p}{1-p} \right)^{\frac{1}{b}} dp \right) \\&= \frac{1}{1-\alpha } \left( a\frac{\pi }{b} csc\left( \frac{\pi }{b} \right) - a B_\alpha \left( \frac{1}{b}+1,1 - \frac{1}{b}\right) \right) \\&= \frac{a}{1-\alpha } \left( \frac{\pi }{b} csc\left( \frac{\pi }{b} \right) - B_\alpha \left( \frac{1}{b}+1,1 - \frac{1}{b}\right) \right) . \end{aligned}$$

$\square$

1.17 Proposition 15

Proof

Assume we have $\xi =0$. Then, we have

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} \mu - s \ln (-\ln (p))dp \\&= \mu - \frac{s}{1-\alpha } \left( \int _{0}^{1} \ln (-\ln (p))dp - \int _{0}^{\alpha } \ln (-\ln (p))dp \right)\\&= \mu - \frac{s}{1-\alpha } \left(y - \int _{0}^{\alpha } \ln (-\ln (p))dp \right)\\&= \mu - \frac{s}{1-\alpha } \left(y + \alpha \ln (-\ln (\alpha )) - {\mathrm {li}}(\alpha ) \right) .\\ \end{aligned}$$

Assume now that $\xi \ne 0$. Then, we have that,

$$\begin{aligned} {\bar{q}}_\alpha (X)&= \frac{1}{1-\alpha } \int _{\alpha }^{1} \mu + \frac{s}{\xi } \left( (\ln (\frac{1}{p})^{-\xi } -1 \right)dp \\&= \mu - \frac{s}{\xi (1-\alpha )} \int _{\alpha }^{1} \left( (\ln (\frac{1}{p})^{-\xi } -1 \right)dp\\&= \mu + \frac{s}{\xi (1-\alpha )} \left[ \varGamma _L(1-\xi ,\ln (\frac{1}{\alpha })) - (1-\alpha ) \right] . \end{aligned}$$

$\square$