From Gini to Bonferroni to Tsallis: an inequality-indices trek

Abstract

Arguably, the Gini index is the best known and the most widely applied inequality index in socioeconomics in particular, and across the sciences in general. On the other hand, far less known and less applied is the Bonferroni index. Addressed via Lorenz curves, the Gini index can be formulated as the average of two continuums of inequality indices that, respectively, stem from two sets of Lorenz-based distances: vertical and horizontal. These Gini-index formulations use one natural type of Lorenz-based distances. However, there is another natural type of Lorenz-based distances, and when using this type: (1) averaging the vertical continuum of inequality indices yields the known Bonferroni index; (2) averaging the horizontal continuum of inequality indices yields a new Bonferroni index. This paper explores comprehensively the two Bonferroni indices, and presents the many analogies between these indices and the Gini index. This paper also unifies the Bonferroni indices and the Gini index via two families of inequality induces: (1) a “vertical family” of which the known Bonferroni index and the Gini index are special cases; (2) a “horizontal family” of which the new Bonferroni index and the Gini index are special cases. These two families are shown to be counterparts of the Tsallis family of entropy measures, and the two Bonferroni indices are shown to be counterparts of the Shannon entropy. Written in an entirely self-contained manner, this paper is accessible also to audiences with no prior knowledge of inequality indices.

Appendix

1.1 Preliminaries

• A) Consider the general random variable $\xi$, as described at the beginning of Sect. 2.1. In terms of its density function $f_{\xi }\left( t\right)$, as well as in terms of its survival function $\bar{F}_{\xi }\left( t\right)$ and distribution function $F_{\xi }\left( t\right)$, the mean of the random variable is given by

  $$\begin{aligned} \mathbf {E}\left[ \xi \right]= & {} \int _{0}^{\infty }tf_{\xi }\left( t\right) dt \nonumber \\= & {} \int _{0}^{\infty }\bar{F}_{\xi }\left( t\right) dt=\int _{0}^{\infty }\left[ 1-F_{\xi }\left( t\right) \right] dt. \end{aligned}$$

  (51)

• B) Consider n IID copies $\left\{ \xi _{1},\ldots ,\xi _{n}\right\}$ of the random variable $\xi$. The distribution function of the maximum $\xi _{1}\vee \cdots \vee \xi _{n}=\max \left\{ \xi _{1},\ldots ,\xi _{n}\right\}$ is given by

  $$\begin{aligned} F_{\xi _{1}\vee \cdots \vee \xi _{n}}\left( t\right)= & {} \Pr \left( \xi _{1}\vee \cdots \vee \xi _{n}\le t\right) \nonumber \\= & {} \Pr \left( \xi _{1}\le t\right) \cdots \Pr \left( \xi _{n}\le t\right) =F_{\xi }\left( t\right) ^{n}. \end{aligned}$$

  (52)

  Differentiating Eq. (52) implies that the density function of the maximum $\xi _{1}\vee \cdots \vee \xi _{n}$ is given by

  $$\begin{aligned} f_{\xi _{1}\vee \cdots \vee \xi _{n}}\left( t\right) =nF_{\xi }\left( t\right) ^{n-1}f_{\xi }\left( t\right) . \end{aligned}$$

  (53)

  The survival function of the minimum $\xi _{1}\wedge \cdots \wedge \xi _{n}=\min \left\{ \xi _{1},\ldots ,\xi _{n}\right\}$ is given by

  $$\begin{aligned} \bar{F}_{\xi _{1}\wedge \cdots \wedge \xi _{n}}\left( t\right)= & {} \Pr \left( \xi _{1}\wedge \cdots \wedge \xi _{n}>t\right) \nonumber \\= & {} \Pr \left( \xi _{1}>t\right) \cdots \Pr \left( \xi _{n}>t\right) =\bar{F} _{\xi }\left( t\right) ^{n}. \end{aligned}$$

  (54)

  Differentiating Eq. (54) implies that the density function of the minimum $\xi _{1}\wedge \cdots \wedge \xi _{n}$ is given by

  $$\begin{aligned} f_{\xi _{1}\wedge \cdots \wedge \xi _{n}}\left( t\right) =n\bar{F}_{\xi }\left( t\right) ^{n-1}f_{\xi }\left( t\right) \text { .} \end{aligned}$$

  (55)

• C) Set the focus now on the minimum $\xi _{1}\wedge \xi _{2}$. Conditioning implies that

  $$\begin{aligned} \mathbf {E}\left[ \xi _{1}\wedge \xi _{2}\right]= & {} \int _{0}^{\infty }\mathbf {E} \left[ \xi _{1}\wedge \xi _{2}{{\vert } }\xi _{2}=t\right] f_{\xi }\left( t\right) dt \nonumber \\= & {} \int _{0}^{\infty }\mathbf {E}\left[ \min \left\{ \xi _{1},t\right\} \right] f_{\xi }\left( t\right) dt=\int _{0}^{\infty }\mathbf {E}\left[ \min \left\{ \xi ,t\right\} \right] f_{\xi }\left( t\right) dt. \end{aligned}$$

  (56)

  From Eq. (56) we obtain that

  $$\begin{aligned} \frac{\mathbf {E}\left[ \xi _{1}\wedge \xi _{2}\right] }{\mathbf {E}\left[ \xi \right] }=\int _{0}^{\infty }\frac{\mathbf {E}\left[ \min \left\{ \xi ,t\right\} \right] }{\mathbf {E}\left[ \xi \right] }f_{\xi }\left( t\right) dt . \end{aligned}$$

  (57)

  Equations (51) and (54) imply that

  $$\begin{aligned} \mathbf {E}\left[ \xi _{1}\wedge \xi _{2}\right] =\int _{0}^{\infty }\bar{F} _{\xi _{1}\wedge \xi _{2}}\left( t\right) dt=\int _{0}^{\infty }\bar{F}_{\xi }\left( t\right) ^{2}dt. \end{aligned}$$

  (58)

  In turn, Eqs. (51) and (58) imply that

  $$\begin{aligned} \mathbf {E}\left[ \xi \right] -\mathbf {E}\left[ \xi _{1}\wedge \xi _{2}\right]= & {} \int _{0}^{\infty }\bar{F}_{\xi }\left( t\right) dt-\int _{0}^{\infty }\bar{F} _{\xi }\left( t\right) ^{2}dt \nonumber \\= & {} \int _{0}^{\infty }\left[ \bar{F}_{\xi }\left( t\right) -\bar{F}_{\xi }\left( t\right) ^{2}\right] dt=\int _{0}^{\infty }\bar{F}_{\xi }\left( t\right) \left[ 1-\bar{F}_{\xi }\left( t\right) \right] dt \nonumber \\= & {} \int _{0}^{\infty }\left[ 1-F_{\xi }\left( t\right) \right] F_{\xi }\left( t\right) dt. \end{aligned}$$

  (59)

  From Eq. (59) we obtain that

  $$\begin{aligned} 1-\frac{\mathbf {E}\left[ \xi _{1}\wedge \xi _{2}\right] }{\mathbf {E}\left[ \xi \right] }=\frac{1}{\mathbf {E}\left[ \xi \right] }\int _{0}^{\infty }F_{\xi }\left( t\right) \left[ 1-F_{\xi }\left( t\right) \right] dt. \end{aligned}$$

  (60)

• D) Note that

  $$\begin{aligned}&\int _{0}^{\infty }sf_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] ds \nonumber \\&\quad =\int _{0}^{\infty }\left( \int _{0}^{s}dt\right) \left\{ f_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] \right\} ds \nonumber \\&\quad =\int _{0}^{\infty }\left\{ \int _{t}^{\infty }f_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] ds\right\} dt. \end{aligned}$$

  (61)

  Integration-by-parts implies that

  $$\begin{aligned} \int _{t}^{\infty }f_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] ds=-\left[ 1-F_{\xi }\left( t\right) \right] -F_{\xi }\left( t\right) \ln \left[ F_{\xi }\left( t\right) \right] . \end{aligned}$$

  (62)

  Integrating Eq. (62) over the non-negative half-line, and using Eq. (51), we have

  $$\begin{aligned}&\int _{0}^{\infty }\left\{ \int _{t}^{\infty }f_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] ds\right\} dt \nonumber \\&\int _{0}^{\infty }\left\{ -\left[ 1-F_{\xi }\left( t\right) \right] -F_{\xi }\left( t\right) \ln \left[ F_{\xi }\left( t\right) \right] \right\} dx \nonumber \\&\quad =-\int _{0}^{\infty }\left[ 1-F_{\xi }\left( t\right) \right] dt-\int _{0}^{\infty }F_{\xi }\left( t\right) \ln \left[ F_{\xi }\left( t\right) \right] dt \nonumber \\&\quad =-\mathbf {E}\left[ \xi \right] -\int _{0}^{\infty }F_{\xi }\left( t\right) \ln \left[ F_{\xi }\left( t\right) \right] dt. \end{aligned}$$

  (63)

  Combined together, Eqs. (61) and (63) imply that

  $$\begin{aligned} \int _{0}^{\infty }sf_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] ds=-\mathbf {E}\left[ \xi \right] -\int _{0}^{\infty }F_{\xi }\left( t\right) \ln \left[ F_{\xi }\left( t\right) \right] dt. \end{aligned}$$

  (64)

  In turn, we can re-write Eq. (64) as follows:

  $$\begin{aligned} -\int _{0}^{\infty }sf_{\xi }\left( s\right) \ln \left[ F_{\xi }\left( s\right) \right] ds=\mathbf {E}\left[ \xi \right] -\int _{0}^{\infty }F_{\xi }\left( t\right) \ln \left[ 1/F_{\xi }\left( t\right) \right] dt. \end{aligned}$$

  (65)

1.2 Eqs. (12)–(15)

Equations (11) and (7) imply that

$$\begin{aligned} I_{Gini}= & {} \int _{0}^{1}\left[ \bar{L}\left( u\right) -L\left( u\right) \right] du \nonumber \\= & {} \int _{0}^{1}\left\{ \left[ 1-L\left( 1-u\right) \right] -L\left( u\right) \right\} du \nonumber \\= & {} 1-\int _{0}^{1}L\left( 1-u\right) du-\int _{0}^{1}L\left( u\right) du \nonumber \\= & {} 1-2\int _{0}^{1}L\left( u\right) du. \end{aligned}$$

(66)

Equation (66) yields Eq. (12).

Using the change-of-variables $u=F_{X}\left( t\right)$ and Eq. (3) we have

$$\begin{aligned} 2\int _{0}^{1}L\left( u\right) du= & {} 2\int _{0}^{\infty }L\left[ F_{X}\left( t\right) \right] f_{X}\left( t\right) dt \nonumber \\= & {} 2\int _{0}^{\infty }F_{Y}\left( t\right) f_{X}\left( t\right) dt. \end{aligned}$$

(67)

In turn, using Eq. (53) (for the random variable $\xi =X$ and $n=2$) we have

$$\begin{aligned} 2\int _{0}^{\infty }F_{Y}\left( t\right) f_{X}\left( t\right) dt= & {} \int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) } \left[ 2F_{X}\left( t\right) f_{X}\left( t\right) \right] dt \nonumber \\= & {} \int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) } f_{X_{1}\vee X_{2}}\left( t\right) dt. \end{aligned}$$

(68)

Equations (67)–(68) imply that

$$\begin{aligned} 2\int _{0}^{1}L\left( u\right) du=\int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) }f_{X_{1}\vee X_{2}}\left( t\right) dt. \end{aligned}$$

(69)

Substituting Eq. (69) into Eq. (66) yields Eq. (15).

Equation (1) implies that

$$\begin{aligned} 2\int _{0}^{\infty }F_{Y}\left( t\right) f_{X}\left( t\right) dt= & {} 2\int _{0}^{\infty }F_{Y}\left( t\right) \left[ \frac{\mu }{t}f_{Y}\left( t\right) \right] dt \nonumber \\= & {} \mu \int _{0}^{\infty }\frac{1}{t}\left[ 2F_{Y}\left( t\right) f_{Y}\left( t\right) \right] dt. \end{aligned}$$

(70)

In turn, using Eq. (53) (for the random variable $\xi =Y$ and $n=2$), and then the reciprocation $Z=1/Y$, we have:

$$\begin{aligned} \mu \int _{0}^{\infty }\frac{1}{t}\left[ 2F_{Y}\left( t\right) f_{Y}\left( t\right) \right] dt= & {} \mu \int _{0}^{\infty }\frac{1}{t}f_{Y_{1}\vee Y_{2}}\left( t\right) dt \nonumber \\= & {} \mu \mathbf {E}\left[ \frac{1}{Y_{1}\vee Y_{2}}\right] =\mu \mathbf {E}\left[ \min \left\{ \frac{1}{Y_{1}},\frac{1}{Y_{2}}\right\} \right] \nonumber \\= & {} \mu \mathbf {E}\left[ Z_{1}\wedge Z_{2}\right] , \end{aligned}$$

(71)

where $Z_{1}$ and $Z_{2}$ are IID copies of the random variable Z. Equation (67) and Eqs. (70)–(71), together with the fact that $\mathbf {E}\left[ Z\right] =1/\mu$, imply that

$$\begin{aligned} 2\int _{0}^{1}L\left( u\right) du=\frac{\mathbf {E}\left[ Z_{1}\wedge Z_{2} \right] }{\mathbf {E}\left[ Z\right] }. \end{aligned}$$

(72)

In turn, Eqs. (66) and (72) imply that

$$\begin{aligned} I_{Gini}=1-\frac{\mathbf {E}\left[ Z_{1}\wedge Z_{2}\right] }{\mathbf {E}\left[ Z\right] }. \end{aligned}$$

(73)

Equations (73) and (57) (for $\xi =Z$) yield Eq. (14). Equations (73) and (60) (for $\xi =Z$) yield Eq. (13).

1.3 Eqs. (17)–(20)

Equations (16) and (8) imply that

$$\begin{aligned} I_{Gini}= & {} \int _{0}^{1}\left[ L^{-1}\left( u\right) -\bar{L}^{-1}\left( u\right) \right] du \nonumber \\= & {} \int _{0}^{1}\left\{ \left[ 1-\bar{L}^{-1}\left( 1-u\right) \right] -\bar{L} ^{-1}\left( u\right) \right\} du \nonumber \\= & {} 1-\int _{0}^{1}\bar{L}^{-1}\left( 1-u\right) du-\int _{0}^{1}\bar{L} ^{-1}\left( u\right) du \nonumber \\= & {} 1-2\int _{0}^{1}\bar{L}^{-1}\left( u\right) du. \end{aligned}$$

(74)

Equation (74) yields Eq. (17).

Using the change-of-variables $u=\bar{F}_{Y}\left( t\right)$ and Eq. (6) we have

$$\begin{aligned} 2\int _{0}^{1}\bar{L}^{-1}\left( u\right) du= & {} 2\int _{0}^{\infty }\bar{L}^{-1} \left[ \bar{F}_{Y}\left( t\right) \right] f_{Y}\left( t\right) dt \nonumber \\= & {} 2\int _{0}^{\infty }\bar{F}_{X}\left( t\right) f_{Y}\left( t\right) dt. \end{aligned}$$

(75)

In turn, using Eq. (55) (for the random variable $\xi =Y$ and $n=2$) we have

$$\begin{aligned} 2\int _{0}^{\infty }\bar{F}_{X}\left( t\right) f_{Y}\left( t\right) dt= & {} \int _{0}^{\infty }\frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }\left[ 2\bar{F}_{Y}\left( t\right) f_{Y}\left( t\right) \right] dt \nonumber \\= & {} \int _{0}^{\infty }\frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }f_{Y_{1}\wedge Y_{2}}\left( t\right) dt. \end{aligned}$$

(76)

Equations (75)–(76) imply that

$$\begin{aligned} 2\int _{0}^{1}\bar{L}^{-1}\left( u\right) du=\int _{0}^{\infty }\frac{\bar{F} _{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }f_{Y_{1}\wedge Y_{2}}\left( t\right) dt. \end{aligned}$$

(77)

Substituting Eq. (77) into Eq. (74) yields Eq. (20).

Equation (1) implies that

$$\begin{aligned} 2\int _{0}^{\infty }\bar{F}_{X}\left( t\right) f_{Y}\left( t\right) dt= & {} 2\int _{0}^{\infty }\bar{F}_{X}\left( t\right) \left[ \frac{t}{\mu } f_{X}\left( t\right) \right] dt \nonumber \\= & {} \frac{1}{\mu }\int _{0}^{\infty }t\left[ 2\bar{F}_{X}\left( t\right) f_{X}\left( t\right) \right] dt. \end{aligned}$$

(78)

In turn, using Eq. (55) (for the random variable $\xi =X$ and $n=2$) we have:

$$\begin{aligned} \frac{1}{\mu }\int _{0}^{\infty }t\left[ 2\bar{F}_{X}\left( t\right) f_{X}\left( t\right) \right] dt= & {} \frac{1}{\mu }\int _{0}^{\infty }tf_{X_{1}\wedge X_{2}}\left( t\right) dt \nonumber \\= & {} \frac{1}{\mu }\mathbf {E}\left[ X_{1}\wedge X_{2}\right] . \end{aligned}$$

(79)

Equation (75) and Eqs. (78)–(79), together with the fact that $\mathbf {E}\left[ X\right] =\mu$, imply that

$$\begin{aligned} 2\int _{0}^{1}\bar{L}^{-1}\left( u\right) du=\frac{\mathbf {E}\left[ X_{1}\wedge X_{2}\right] }{\mathbf {E}\left[ X\right] }. \end{aligned}$$

(80)

In turn, Eqs. (74) and (80) imply that

$$\begin{aligned} I_{Gini}=1-\frac{\mathbf {E}\left[ X_{1}\wedge X_{2}\right] }{\mathbf {E}\left[ X\right] }. \end{aligned}$$

(81)

Equations (81) and (57) (for $\xi =X$) yield Eq. (19). Equations (81) and (60) (for $\xi =X$) yield Eq. (18).

1.4 Eq. (21) and Eqs. (23)–(25)

Equation (7) implies the coincidence of Eq. (21):

$$\begin{aligned} \int _{0}^{1}\frac{\bar{L}\left( u\right) -u}{1-u}du= & {} \int _{0}^{1}\frac{\left[ 1-L\left( 1-u\right) \right] -u}{1-u}du \nonumber \\= & {} \int _{0}^{1}\frac{\left( 1-u\right) -L\left( 1-u\right) }{\left( 1-u\right) }du=\int _{0}^{1}\frac{u-L\left( u\right) }{u}du. \end{aligned}$$

(82)

Using the change-of-variables $u=F_{X}\left( t\right)$ and Eq. (3) we have

$$\begin{aligned} \int _{0}^{1}\frac{L\left( u\right) }{u}du= & {} \int _{0}^{\infty }\frac{L\left[ F_{X}\left( t\right) \right] }{F_{X}\left( t\right) }f_{X}\left( t\right) dt\nonumber \\= & {} \int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) } f_{X}\left( t\right) dt. \end{aligned}$$

(83)

Equations (22) and (83) yield Eq. (25).

Using Eq. (1) we have

$$\begin{aligned} \frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) }= & {} \frac{1}{F_{X}\left( t\right) }\int _{0}^{t}f_{Y}\left( x\right) dx \nonumber \\= & {} \frac{1}{F_{X}\left( t\right) }\int _{0}^{t}\frac{x}{\mu }f_{X}\left( x\right) dx \nonumber \\= & {} \frac{1}{\mu }\int _{0}^{t}x\frac{f_{X}\left( x\right) }{F_{X}\left( t\right) }dx. \end{aligned}$$

(84)

The integral appearing on the bottom line of Eq. (84) is the conditional mean of the random variable X, given the information that $X\le t$. Hence, Eq. (84) implies that

$$\begin{aligned} \frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) }=\frac{\mathbf {E}[X{{\vert } }X\le t]}{\mathbf {E}\left[ X\right] }. \end{aligned}$$

(85)

Substituting Eq. (85) into Eq. (83) yields

$$\begin{aligned} \int _{0}^{1}\frac{L\left( u\right) }{u}du=\int _{0}^{\infty }\frac{\mathbf {E} [X{{\vert } }X\le t]}{\mathbf {E}\left[ X\right] } f_{X}\left( t\right) dt. \end{aligned}$$

(86)

In turn, Eqs. (22) and (86) yield Eq. (24).

Note that

$$\begin{aligned} \int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) } f_{X}\left( t\right) dt= & {} \int _{0}^{\infty }F_{Y}\left( t\right) \frac{ f_{X}\left( t\right) }{F_{X}\left( t\right) }dt \nonumber \\= & {} \int _{0}^{\infty }\left[ \int _{0}^{t}f_{Y}\left( y\right) dy\right] \left\{ \ln \left[ F_{X}\left( t\right) \right] ^{\prime }\right\} dt \nonumber \\= & {} \int _{0}^{\infty }f_{Y}\left( y\right) \left( \int _{y}^{\infty }\left\{ \ln \left[ F_{X}\left( t\right) \right] ^{\prime }\right\} dt\right) dy \nonumber \\= & {} \int _{0}^{\infty }f_{Y}\left( y\right) \left\{ -\ln \left[ F_{X}\left( y\right) \right] \right\} dy \nonumber \\= & {} -\int _{0}^{\infty }f_{Y}\left( y\right) \ln \left[ F_{X}\left( y\right) \right] dy \end{aligned}$$

(87)

Using Eq. (1), and then Eq. (65) (for $\xi =X$), we have

$$\begin{aligned} -\int _{0}^{\infty }f_{Y}\left( s\right) \ln \left[ F_{X}\left( s\right) \right] ds= & {} -\int _{0}^{\infty }\frac{s}{\mu }f_{X}\left( s\right) \ln \left[ F_{X}\left( s\right) \right] ds \nonumber \\= & {} \frac{1}{\mathbf {E}\left[ X\right] }\left\{ -\int _{0}^{\infty }sf_{X}\left( s\right) \ln \left[ F_{X}\left( s\right) \right] ds\right\} \nonumber \\= & {} \frac{1}{\mathbf {E}\left[ X\right] }\left\{ \mathbf {E}\left[ X\right] -\int _{0}^{\infty }F_{X}\left( t\right) \ln \left[ 1/F_{X}\left( t\right) \right] dt\right\} \nonumber \\= & {} 1-\frac{1}{\mathbf {E}\left[ X\right] }\int _{0}^{\infty }F_{X}\left( t\right) \ln \left[ 1/F_{X}\left( t\right) \right] dt. \end{aligned}$$

(88)

Equation (83) and Eqs. (87)–(88) imply that

$$\begin{aligned} \int _{0}^{1}\frac{L\left( u\right) }{u}du=1-\frac{1}{\mathbf {E}\left[ X \right] }\int _{0}^{\infty }F_{X}\left( t\right) \ln \left[ 1/F_{X}\left( t\right) \right] dt. \end{aligned}$$

(89)

In turn, Eqs. (22) and (89) yield Eq. (23).

1.5 Eq. (26) and Eqs. (28)–(30)

Equation (8) implies the coincidence of Eq. (26):

$$\begin{aligned} \int _{0}^{1}\frac{L^{-1}\left( u\right) -u}{1-u}du= & {} \int _{0}^{1}\frac{\left[ 1-\bar{L}^{-1}\left( 1-u\right) \right] -u}{1-u}du \nonumber \\= & {} \int _{0}^{1}\frac{\left( 1-u\right) -\bar{L}^{-1}\left( 1-u\right) }{\left( 1-u\right) }du=\int _{0}^{1}\frac{u-\bar{L}^{-1}\left( u\right) }{u}du\text { . } \end{aligned}$$

(90)

Using the change-of-variables $u=\bar{F}_{Y}\left( t\right)$ and Eq. (6) we have

$$\begin{aligned} \int _{0}^{1}\frac{\bar{L}^{-1}\left( u\right) }{u}du= & {} \int _{0}^{\infty }\frac{ \bar{L}^{-1}\left[ \bar{F}_{Y}\left( t\right) \right] }{\bar{F}_{Y}\left( t\right) }f_{Y}\left( t\right) dt \nonumber \\= & {} \int _{0}^{\infty }\frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }f_{Y}\left( t\right) dt. \end{aligned}$$

(91)

Equations (27) and (91) yield Eq. (30).

Using Eq. (1) we have

$$\begin{aligned} \frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }= & {} \frac{1}{ \bar{F}_{Y}\left( t\right) }\int _{t}^{\infty }f_{X}\left( y\right) dy \nonumber \\= & {} \frac{1}{\bar{F}_{Y}\left( t\right) }\int _{t}^{\infty }\frac{\mu }{y} f_{Y}\left( y\right) dy \nonumber \\= & {} \mu \int _{t}^{\infty }\frac{1}{y}\frac{f_{Y}\left( y\right) }{\bar{F} _{Y}\left( t\right) }dy. \end{aligned}$$

(92)

The integral appearing on the bottom line of Eq. (92) is the conditional mean of the random variable 1/Y, given the information that $Y>t$. Note that, as Y is an absolutely continuous random variable, the information $Y>t$ is effectively identical to the information $Y\ge t$. Hence, using the fact that $Z=1/Y$ and $\mathbf {E}\left[ Z\right] =1/\mu$, Eq. (92) implies that

$$\begin{aligned} \frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }=\frac{ \mathbf {E}[1/Y{{\vert } }Y\ge t]}{\left( 1/\mu \right) }= \frac{\mathbf {E}[Z{{\vert } }Z\le 1/t]}{\mathbf {E}\left[ Z \right] }. \end{aligned}$$

(93)

Substituting Eq. (93) into Eq. (91), and using the change of variables $s=1/t$ and the fact that $Z=1/Y$, we have

$$\begin{aligned} \int _{0}^{1}\frac{\bar{L}^{-1}\left( u\right) }{u}du= & {} \int _{0}^{\infty }\frac{ \mathbf {E}[Z{{\vert } }Z\le 1/t]}{\mathbf {E}\left[ Z\right] }f_{Y}\left( t\right) dt \nonumber \\= & {} \int _{0}^{\infty }\frac{\mathbf {E}[Z{{\vert } }Z\le s]}{ \mathbf {E}\left[ Z\right] }\left[ f_{Y}\left( \frac{1}{s}\right) \frac{1}{ s^{2}}\right] ds \nonumber \\= & {} \int _{0}^{\infty }\frac{\mathbf {E}[Z{{\vert } }Z\le s]}{ \mathbf {E}\left[ Z\right] }f_{Z}\left( s\right) ds\text { .} \end{aligned}$$

(94)

In turn, Eqs. (27) and (94) yield Eq. (29).

Note that

$$\begin{aligned} \int _{0}^{\infty }\frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }f_{Y}\left( t\right) dt= & {} \int _{0}^{\infty }\bar{F}_{X}\left( t\right) \frac{f_{Y}\left( t\right) }{\bar{F}_{Y}\left( t\right) }dt \nonumber \\= & {} \int _{0}^{\infty }\left[ \int _{t}^{\infty }f_{X}\left( x\right) dx\right] \left\{ -\ln \left[ \bar{F}_{Y}\left( t\right) \right] ^{\prime }\right\} dt\nonumber \\= & {} -\int _{0}^{\infty }f_{X}\left( x\right) \left( \int _{0}^{x}\left\{ \ln \left[ \bar{F}_{Y}\left( t\right) \right] ^{\prime }\right\} dt\right) dx \nonumber \\= & {} -\int _{0}^{\infty }f_{X}\left( x\right) \ln \left[ \bar{F}_{Y}\left( x\right) \right] dx. \end{aligned}$$

(95)

Using Eq. (1), the change of variables $s=1/x$, the fact that $Z=1/Y$ and $\mathbf {E}\left[ Z\right] =1/\mu$, and Eq. (65) (for $\xi =Z$), we have

$$\begin{aligned} -\int _{0}^{\infty }f_{X}\left( x\right) \ln \left[ \bar{F}_{Y}\left( x\right) \right] dx= & {} -\int _{0}^{\infty }\frac{\mu }{x}f_{Y}\left( x\right) \ln \left[ \bar{F}_{Y}\left( x\right) \right] dx \nonumber \\= & {} -\mu \int _{0}^{\infty }s\left[ f_{Y}\left( \frac{1}{s}\right) \frac{1}{s^{2} }\right] \ln \left[ \bar{F}_{Y}\left( \frac{1}{s}\right) \right] ds \nonumber \\= & {} \frac{1}{\mathbf {E}\left[ Z\right] }\left\{ -\int _{0}^{\infty }sf_{Z}\left( s\right) \ln \left[ F_{Z}\left( s\right) \right] ds\right\} \nonumber \\= & {} \frac{1}{\mathbf {E}\left[ Z\right] }\left\{ \mathbf {E}\left[ Z\right] -\int _{0}^{\infty }F_{Z}\left( t\right) \ln \left[ 1/F_{Z}\left( t\right) \right] dt\right\} \nonumber \\= & {} 1-\frac{1}{\mathbf {E}\left[ Z\right] }\int _{0}^{\infty }F_{Z}\left( t\right) \ln \left[ 1/F_{Z}\left( t\right) \right] dt. \end{aligned}$$

(96)

Equation (91) and Eqs. (95)–(96) imply that

$$\begin{aligned} \int _{0}^{1}\frac{\bar{L}^{-1}\left( u\right) }{u}du=1-\frac{1}{\mathbf {E} \left[ Z\right] }\int _{0}^{\infty }F_{Z}\left( t\right) \ln \left[ 1/F_{Z}\left( t\right) \right] dt. \end{aligned}$$

(97)

In turn, Eqs. (27) and (97) yield Eq. (28).

1.6 Eq. (32), Eq. (33), and Eq. (35)

For $n=2$ note that

$$\begin{aligned} \max \left\{ \xi _{1},\xi _{2}\right\} +\min \left\{ \xi _{1},\xi _{2}\right\} =\xi _{1}+\xi _{2} \end{aligned}$$

(98)

and

$$\begin{aligned} \max \left\{ \xi _{1},\xi _{2}\right\} -\min \left\{ \xi _{1},\xi _{2}\right\} =\left| \xi _{1}-\xi _{2}\right| . \end{aligned}$$

(99)

Summing Equations (98) and (99) up we have

$$\begin{aligned} 2\max \left\{ \xi _{1},\xi _{2}\right\} =\xi _{1}+\xi _{2}+\left| \xi _{1}-\xi _{2}\right| . \end{aligned}$$

(100)

Applying expectation to Eq. (100) implies that

$$\begin{aligned} 2\mathbf {E}\left[ \xi _{1}\vee \xi _{2}\right]= & {} 2\mathbf {E}\left[ \max \left\{ \xi _{1},\xi _{2}\right\} \right] \nonumber \\= & {} \mathbf {E}\left[ \xi _{1}\right] +\mathbf {E}\left[ \xi _{2}\right] +\mathbf { E}\left[ \left| \xi _{1}-\xi _{2}\right| \right] \nonumber \\= & {} 2\mathbf {E}\left[ \xi \right] +\mathbf {E}\left[ \left| \xi _{1}-\xi _{2}\right| \right] , \end{aligned}$$

(101)

and hence

$$\begin{aligned} \mathbf {E}\left[ \xi _{1}\vee \xi _{2}\right] -\mathbf {E}\left[ \xi \right] = \frac{1}{2}\mathbf {E}\left[ \left| \xi _{1}-\xi _{2}\right| \right] . \end{aligned}$$

(102)

Substituting Eq. (102) into Eq. (31) yields Eq. (32).

Using Eqs. (51) and (52) we have

$$\begin{aligned} \mathbf {E}\left[ \xi _{1}\vee \cdots \vee \xi _{n}\right]= & {} \int _{0}^{\infty } \left[ 1-F_{\xi _{1}\vee \cdots \vee \xi _{n}}\left( t\right) \right] dt \nonumber \\= & {} \int _{0}^{\infty }\left[ 1-F_{\xi }\left( t\right) ^{n}\right] dt. \end{aligned}$$

(103)

In turn, using Equations (103) and (51) we have

$$\begin{aligned}&\mathbf {E}\left[ \xi _{1}\vee \cdots \vee \xi _{n}\right] -\mathbf {E}\left[ \xi \right] \nonumber \\&\quad =\int _{0}^{\infty }\left[ 1-F_{\xi }\left( t\right) ^{n}\right] dt-\int _{0}^{\infty }\left[ 1-F_{\xi }\left( t\right) \right] dt \nonumber \\&\quad =\int _{0}^{\infty }\left\{ \left[ 1-F_{\xi }\left( t\right) ^{n}\right] - \left[ 1-F_{\xi }\left( t\right) \right] \right\} dt \nonumber \\&\quad =\int _{0}^{\infty }\left[ F_{\xi }\left( t\right) -F_{\xi }\left( t\right) ^{n}\right] dt. \end{aligned}$$

(104)

Substituting Eq. (104) into Eq. (31) yields Eq. (33).

Equation (31) implies that

$$\begin{aligned} 1-I_{n}\left( \xi \right)= & {} 1-\frac{\mathbf {E}\left[ \xi _{1}\vee \cdots \vee \xi _{n}\right] -\mathbf {E}\left[ \xi \right] }{n\mathbf {E}\left[ \xi \right] -\mathbf {E}\left[ \xi \right] } \nonumber \\= & {} \frac{n\mathbf {E}\left[ \xi \right] -\mathbf {E}\left[ \xi \right] -\mathbf {E }\left[ \xi _{1}\vee \cdots \vee \xi _{n}\right] +\mathbf {E}\left[ \xi \right] }{n\mathbf {E}\left[ \xi \right] -\mathbf {E}\left[ \xi \right] } \nonumber \\= & {} \frac{n\mathbf {E}\left[ \xi \right] -\mathbf {E}\left[ \xi _{1}\vee \cdots \vee \xi _{n}\right] }{\left( n-1\right) \mathbf {E}\left[ \xi \right] }. \end{aligned}$$

(105)

Using Eqs. (51) and (53) we have

$$\begin{aligned}&n\mathbf {E}\left[ \xi \right] -\mathbf {E}\left[ \xi _{1}\vee \cdots \vee \xi _{n}\right] \nonumber \\&\quad =n\int _{0}^{\infty }sf_{\xi }\left( s\right) ds-\int _{0}^{\infty }sf_{\xi _{1}\vee \cdots \vee \xi _{n}}\left( s\right) ds \nonumber \\&\quad =n\int _{0}^{\infty }sf_{\xi }\left( s\right) ds-\int _{0}^{\infty }s\left[ nF_{\xi }\left( s\right) ^{n-1}f_{\xi }\left( s\right) \right] ds \nonumber \\&\quad =n\int _{0}^{\infty }sf_{\xi }\left( s\right) \left[ 1-F_{\xi }\left( s\right) ^{n-1}\right] ds \nonumber \\&\quad =n\int _{0}^{\infty }sf_{\xi }\left( s\right) \left[ \int _{s}^{\infty }\left( n-1\right) F_{\xi }\left( t\right) ^{n-2}f_{\xi }\left( t\right) dt\right] ds \nonumber \\&\quad =\left( n-1\right) \int _{0}^{\infty }\left[ \int _{0}^{t}s\frac{f_{\xi }\left( s\right) }{F_{\xi }\left( t\right) }ds\right] \left[ nF_{\xi }\left( t\right) ^{n-1}f_{\xi }\left( t\right) \right] dt \nonumber \\&\quad =\left( n-1\right) \int _{0}^{\infty }\left[ \int _{0}^{t}s\frac{f_{\xi }\left( s\right) }{F_{\xi }\left( t\right) }ds\right] f_{\xi _{1}\vee \cdots \vee \xi _{n}}\left( t\right) dt \nonumber \\&\quad =\left( n-1\right) \int _{0}^{\infty }\mathbf {E}[\xi {{\vert } }\xi \le t]f_{\xi _{1}\vee \cdots \vee \xi _{n}}\left( t\right) dt. \end{aligned}$$

(106)

Substituting Eq. (106) into Eq. (105) yields Eq. (35).

1.7 Eqs. (36)–(39)

Differentiating Eq. (3) with respect to the variable t implies that

$$\begin{aligned} L^{\prime }\left[ F_{X}\left( t\right) \right] f_{X}\left( t\right) =f_{Y}\left( t\right) . \end{aligned}$$

(107)

Equations (107) and (1) imply that

$$\begin{aligned} L^{\prime }\left[ F_{X}\left( t\right) \right] =\frac{t}{\mu }\text { .} \end{aligned}$$

(108)

Using Eqs. (51) and (53) (for $\xi =X$), then Eq. (108), and then the change-of-variables $u=F_{X}\left( t\right)$, we have

$$\begin{aligned} \mathbf {E}\left[ X_{1}\vee \cdots \vee X_{n}\right]= & {} \int _{0}^{\infty }tf_{X_{1}\vee \cdots \vee X_{n}}\left( t\right) dt \nonumber \\= & {} \mathbf {E}\left[ X\right] \int _{0}^{\infty }\frac{t}{\mu }\left[ nF_{X}\left( t\right) ^{n-1}f_{X}\left( t\right) \right] dt \nonumber \\= & {} \mathbf {E}\left[ X\right] \int _{0}^{\infty }L^{\prime }\left[ F_{X}\left( t\right) \right] \left[ nF_{X}\left( t\right) ^{n-1}f_{X}\left( t\right) \right] dt \nonumber \\= & {} \mathbf {E}\left[ X\right] \int _{0}^{1}L^{\prime }\left( u\right) nu^{n-1}du \nonumber \\= & {} \mathbf {E}\left[ X\right] n\left\{ \left[ 1-0\right] -\int _{0}^{1}L\left( u\right) \left( n-1\right) u^{n-2}du\right\} \nonumber \\= & {} n\mathbf {E}\left[ X\right] \left\{ 1-\left( n-1\right) \int _{0}^{1}u^{n-2}L\left( u\right) du\right\} \text { , } \end{aligned}$$

(109)

and hence

$$\begin{aligned}&n\mathbf {E}\left[ X\right] -\mathbf {E}\left[ X_{1}\vee \cdots \vee X_{n} \right] \nonumber \\&\quad =\left( n-1\right) \mathbf {E}\left[ X\right] \cdot n\int _{0}^{1}u^{n-2}L\left( u\right) du. \end{aligned}$$

(110)

Substituting Eq. (110) into Eq. (105) (for $\xi =X$) yields Eq. (36). Using the change-of-variables $u=F_{X}\left( t\right)$ and Eq. (3), and then using Eq. (53) (for $\xi =X$), we have

$$\begin{aligned}&n\int _{0}^{1}u^{n-2}L\left( u\right) du \nonumber \\&\quad =n\int _{0}^{\infty }F_{X}\left( t\right) ^{n-2}L\left[ F_{X}\left( t\right) \right] f_{X}\left( t\right) dt \nonumber \\&\quad =n\int _{0}^{\infty }F_{X}\left( t\right) ^{n-2}F_{Y}\left( t\right) f_{X}\left( t\right) dt \nonumber \\&\quad =\int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) }\left[ nF_{X}\left( t\right) ^{n-1}f_{X}\left( t\right) \right] dt \nonumber \\&\quad =\int _{0}^{\infty }\frac{F_{Y}\left( t\right) }{F_{X}\left( t\right) } f_{X_{1}\vee \cdots \vee X_{n}}\left( t\right) dt. \end{aligned}$$

(111)

Substituting Eq. (111) into Eq. (36) yields Eq. (38).

Differentiating Eq. (6) with respect to the variable t implies that

$$\begin{aligned} \left( \bar{L}^{-1}\right) ^{\prime }\left[ \bar{F}_{Y}\left( t\right) \right] f_{Y}\left( t\right) =f_{X}\left( t\right) \text { .} \end{aligned}$$

(112)

Equations (112) and (1) imply that

$$\begin{aligned} \left( \bar{L}^{-1}\right) ^{\prime }\left[ \bar{F}_{Y}\left( t\right) \right] =\frac{\mu }{t}. \end{aligned}$$

(113)

As $Z=1/Y$, note that $Z_{1}\vee \cdots \vee Z_{n}=1/\left( Y_{1}\wedge \cdots \wedge Y_{n}\right)$. Consequently, using Eqs. (51) and (55) (for $\xi =Y$), then Eq. (113) and the fact that $\mathbf {E} \left[ Z\right] =1/\mu$, and then the change-of-variables $u=\bar{F} _{Y}\left( t\right)$, we have

$$\begin{aligned} \mathbf {E}\left[ Z_{1}\vee \cdots \vee Z_{n}\right]= & {} \mathbf {E}\left[ 1/\left( Y_{1}\wedge \cdots \wedge Y_{n}\right) \right] \nonumber \\= & {} \int _{0}^{\infty }\frac{1}{t}f_{Y_{1}\wedge \cdots \wedge Y_{n}}\left( t\right) dt=\int _{0}^{\infty }\frac{1}{t}\left[ n\bar{F}_{Y}\left( t\right) ^{n-1}f_{Y}\left( t\right) \right] dt \nonumber \\= & {} \mathbf {E}\left[ Z\right] \int _{0}^{\infty }\frac{\mu }{t}\left[ n\bar{F} _{Y}\left( t\right) ^{n-1}f_{Y}\left( t\right) \right] dt \nonumber \\= & {} \mathbf {E}\left[ Z\right] \int _{0}^{\infty }\left\{ \left( \bar{L} ^{-1}\right) ^{\prime }\left[ \bar{F}_{Y}\left( t\right) \right] \right\} \left[ n\bar{F}_{Y}\left( t\right) ^{n-1}f_{Y}\left( t\right) \right] dt \nonumber \\= & {} \mathbf {E}\left[ Z\right] \int _{0}^{1}\left[ \left( \bar{L}^{-1}\right) ^{\prime }\left( u\right) \right] nu^{n-1}du \nonumber \\= & {} \mathbf {E}\left[ Z\right] n\left\{ \left[ 1-0\right] -\int _{0}^{1}\bar{L} ^{-1}\left( u\right) \left( n-1\right) u^{n-2}du\right\} \nonumber \\= & {} n\mathbf {E}\left[ Z\right] \left\{ 1-\left( n-1\right) \int _{0}^{1}u^{n-2} \bar{L}^{-1}\left( u\right) du\right\} , \end{aligned}$$

(114)

and hence

$$\begin{aligned}&n\mathbf {E}\left[ Z\right] -\mathbf {E}\left[ Z_{1}\vee \cdots \vee Z_{n} \right] \nonumber \\&\quad =\left( n-1\right) \mathbf {E}\left[ Z\right] \cdot n\int _{0}^{1}u^{n-2}\bar{L }^{-1}\left( u\right) du. \end{aligned}$$

(115)

Substituting Eq. (115) into Eq. (105) (for $\xi =Z$) yields Eq. (37). Using the change-of-variables $u=\bar{F}_{Y}\left( t\right)$ and Eq. (6), and then using Eq. (55) (for $\xi =Y$), we have

$$\begin{aligned}&n\int _{0}^{1}u^{n-2}\bar{L}^{-1}\left( u\right) du \nonumber \\&\quad =n\int _{0}^{\infty }\bar{F}_{Y}\left( t\right) ^{n-2}\bar{L}^{-1}\left[ \bar{ F}_{Y}\left( t\right) \right] f_{Y}\left( t\right) dt \nonumber \\&\quad =n\int _{0}^{\infty }\bar{F}_{Y}\left( t\right) ^{n-2}\bar{F}_{X}\left( t\right) f_{Y}\left( t\right) dt \nonumber \\&\quad =\int _{0}^{\infty }\frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }\left[ n\bar{F}_{Y}\left( t\right) ^{n-1}f_{Y}\left( t\right) \right] dt \nonumber \\&\quad =\int _{0}^{\infty }\frac{\bar{F}_{X}\left( t\right) }{\bar{F}_{Y}\left( t\right) }f_{Y_{1}\wedge \cdots \wedge Y_{n}}\left( t\right) dt. \end{aligned}$$

(116)

Substituting Eq. (116) into Eq. (37) yields Eq. (39).