Weibull-MBUR {Y} A New Family of Generalized Unit Distributions with Correlated Parameters: Is the Correlation, Distribution or Data Driven or Interaction Between Them

1. Weibull-MBUR {Lomax} Distribution

1.1. Lomax Link Distribution

The Lomax Distribution has the following PDF, CDF, and quantile functions $f_Y\left(y\right)={\displaystyle \frac{\psi }{\lambda }}{\left[1+{\displaystyle \frac{y}{\lambda }}\right]}^{-\left(\psi +1\right)}$ , $F_Y\left(y\right)=1-{\left[1+{\displaystyle \frac{y}{\lambda }}\right]}^{-\psi }$ , $Q_Y\left(p\right)=\lambda \left[{\left(1-p\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]$ , respectively, where; $y>0$, $0<p<1$, the shape and scale parameters are $\psi \&\lambda >0$ respectively

Use the approach discussed by Aljarrah 2014 as shown in the following steps to derive the PDF.

Step 1: substitute the CDF of MBUR into the quantile function of the link distribution (quantile function of Y), to obtain the t variable.

$T=Q_Y\left[F_X\left(x\right)\right]=\lambda \left[{\left(1-p\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]=\lambda \left[{\left(1-F_X\left(x\right)\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]$$T=\lambda \left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]$

Step 2: substitute this T variable into the PDF of the Weibull distribution

$f_T\left(t\right)={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{t}{\xi }}\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}$$f_T\left(x\right)$$={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^{c-1}exp\left[-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right]=$$=c{{\lambda }^{c-1}{\xi }^{-c}\left({\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right)}^{c-1}exp\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\}$

Step 3: get the derivative of the quantile function which is the reciprocal of the PDF of the quantile distribution:

${\displaystyle \frac{d}{dx}}\left(Q_Y\left(p\right)\right)={\displaystyle \frac{d}{dx}}\left(Q_Y\left(F_X\left(x\right)\right)\right)={\displaystyle \frac{1}{f_Y\left(t\right)}}={\displaystyle \frac{1}{\frac{\psi }{\lambda }{\left[1+\frac{\lambda \left[{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}^{\frac{-1}{\psi }}-1\right]}{\lambda }\right]}^{-\left(\psi +1\right)}}}$$={\displaystyle \frac{\lambda }{\psi {\left[1+{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}^{\frac{-1}{\psi }}-1\right]}^{-\left(\psi +1\right)}}}={\displaystyle \frac{\lambda }{\psi {\left[1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right]}^{\frac{\left(\psi +1\right)}{\psi }}}}$

Step 4: substitute into the equation of $g\left(x\right)$ :

$g\left(x\right)=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(Q_Y\left(F_X\left(x\right)\right)\right)}}=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(t\right)}}$, where $r\left(Q_Y\left(F_X\left(x\right)\right)\right)=f_T\left(t\right)$$g\left(x\right)$$={\displaystyle \frac{6}{b^2}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\displaystyle \frac{c{{\lambda }^{c-1}{\xi }^{-c}\left({\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}^{\frac{-1}{\psi }}-1\right)}^{c-1}exp\left\{-{\left(\frac{\lambda }{\xi }\left[{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}^{\frac{-1}{\psi }}-1\right]\right)}^c\right\}}{\frac{\psi }{\lambda }{\left[1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right]}^{\frac{\left(\psi +1\right)}{\psi }}}}$${\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}$is bigger than one and so ${\left({\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right)}^{c-1}$can be written as ${\left\{{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]\right\}}^{c-1}={\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1-c}{\psi }}}{\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{c-1}$.

And when ${\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1-c}{\psi }}}$ is multiplied by the denominator it can be written as

${\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1-c}{\psi }}}{\left[1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{{\displaystyle \frac{-\left(\psi +1\right)}{\psi }}}={\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-c-\psi }{\psi }}}={\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{-\left({\displaystyle \frac{c}{\psi }}+1\right)}$After rearrangement and for convenient the $x_g=x$ in the following PDF and CDF where x is defined on the unit interval:

${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\psi },\mathit{\xi },\mathit{\lambda }\right)=$$={\displaystyle \frac{6c}{{\psi b}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{-\left({\displaystyle \frac{c}{\psi }}+1\right)}{\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{c-1}$$exp\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\}$

To derive the ${\mathit{G}}_{{\mathit{X}}_{\mathit{g}}}$ substitute the t variable into the CDF of the generator Weibull distribution

${\mathit{G}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\psi },\mathit{\xi },\mathit{\lambda }\right)=F_T\left(t\right)=1-exp\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\};0<x<1$

Quantile function: to derive the quantile function follow the steps

Step 1: Take the quantile of the generator, or $Q_T\left(p\right)=\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}$

Step 2: Substitute this quantile in the CDF of the link distribution, $F_Y\left(y\right)=1-{\left[1+{\displaystyle \frac{y}{\lambda }}\right]}^{-\psi }$, $F_X\left(x\right)=F_Y\left(p\right)=1-{\left[1+{\displaystyle \frac{\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}{\lambda }}\right]}^{-\psi }$

Step 3: Substitute the result and call it, Z; $1-{\left[1+{\displaystyle \frac{\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}{\lambda }}\right]}^{-\psi }$into the quantile function of the baseline distribution MBUR distribution. This will give the same result as taking the inverse of the $G_{X_g}$

Another approach that can be deployed to obtain the quantile function of$G_{X_g}$is by taking the inverse of the$G_{X_g}$:

$p=1-exp\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\}$, solve for x yields the following equation $3{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^2-2{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^3=1-{\left\{1+{\displaystyle \frac{\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}{\lambda }}\right\}}^{-\psi }$ , the left hand side is the CDF of the baseline distribution MBUR and the right hand side is a constant, which is the z, and z is $0<z<1$. Solving the third degree polynomial for its root yields the $x_g$ variable or the quantile function for the new distribution is

$Q_X\left(p\right)=Q_{X_g}\left(z\right)={\left\{-.5\left(cos\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]-\sqrt{3}sin\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]\right)+.5\right\}}^{b^2}$$0<z<1,b,c,\psi ,\xi ,\lambda >0$

The hazard function is defined as $h\left(X_g\right)={\displaystyle \frac{g_{X_g}}{1-G_{X_g}}}$$h\left(X_g\right)={\displaystyle \frac{6c}{{\psi b}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{-\left({\displaystyle \frac{c}{\psi }}+1\right)}{\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{c-1}$

Theorem 1: The ^rth moment for the new distribution can be expressed in the formulas according to the scale parameter of the link function or the Lomax distribution as will be shown in the proof

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\left(-1\right)}^s{\left(\lambda \right)}^{-u}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _u^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\lambda }^{s\psi +u}$${\xi }^{-s\psi -u}\mathsf{\Gamma }\left(1-{\displaystyle \frac{1}{c}}\left[s\psi +u\right]\right)$

Proof: start with the t variable and isolate the x at one side

$t=\lambda \left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]$ so $3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}=1-{\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }$$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)=1-{\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }$this gives $x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}$$x{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{b^2}{2}}}={\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$use the generalized binomial theory on the left hand side

$x{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^m={\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$${\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}\right)}^m{x}^{{\displaystyle \frac{m}{b^2}}+1}={\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$$W_m.x^{{\displaystyle \frac{m}{b^2}}+1}={\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ this gives $x^{{\displaystyle \frac{m+b^2}{b^2}}}={\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}$${\left\{x^{\left({\displaystyle \frac{m+b^2}{b^2}}\right)}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}={\left\{{\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$$x={\left\{{\left\{{\displaystyle \frac{1-{\left(\frac{t}{\lambda }+1\right)}^{-\psi }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$

Take the r^th power of both sides

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left(1-{\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }\right)}^{\left({\displaystyle \frac{r{b}^4}{2m+2b^2}}\right)}$, 0 < ${\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }<1$

Use the generalized binomial theory to expand the ${\left(1-{\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }\right)}^{\left({\displaystyle \frac{r{b}^4}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\left[{\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }\right]}^s$

Use the generalized binomial theory to expand the ${\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-s\psi }$, here there are two cases

First case: $0<{\displaystyle \frac{t}{\lambda }}<1$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\left({\displaystyle \frac{t}{\lambda }}\right)}^u$

Take expectation of both sides

$E\left(x^r\right)={\left(3\right)}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{\left(W_m\right)}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\lambda }^{-u}E\left[{\left(t\right)}^u\right]$

$E\left[{\left(t\right)}^u\right]={\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$ is the u^th raw moment of the Weibull variable (the t variable)

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\left(-1\right)}^s{\left(\lambda \right)}^{-u}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$

The proof of the $E\left[{\left(t\right)}^u\right]={\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$ :

${\int }_0^{\infty }{t}^u{\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{t}{\xi }}\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}dt={\displaystyle \frac{c}{{\xi }^c}}{\int }_0^{\infty }{\left(t\right)}^u{\left(t\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}dt$

${\displaystyle \frac{c}{{\xi }^c}}{\int }_0^{\infty }{t}^u{\left(t\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}dt$ , use change of variable, $w={\left({\displaystyle \frac{t}{\xi }}\right)}^c$ so $t=w^{1/c}\xi$ and $dt={\displaystyle \frac{1}{c}}{w}^{{\displaystyle \frac{1}{c}}-1}\xi$

$\left({\displaystyle \frac{c}{{\xi }^c}}\right)\left({\displaystyle \frac{\xi }{c}}\right){\int }_0^{\infty }{\left(w^{1/c}\xi \right)}^{u+c-1}{e}^{-w}{w}^{{\displaystyle \frac{1}{c}}-1}dw={\xi }^{1-c+u+c-1}{\int }_0^{\infty }{\left(w^{1/c}\right)}^{u+c-1}{e}^{-w}{w}^{{\displaystyle \frac{1}{c}}-1}dw$

${\xi }^{1-c+u+c-1}{\int }_0^{\infty }{w}^{u/c}{e}^{-w}dw={\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$ is the raw moment of the Weibull variable (the t variable).

Second case: ${\displaystyle \frac{t}{\lambda }}>1$,

factor out the $\left({\displaystyle \frac{t}{\lambda }}\right)$ so ${\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-s\psi }={\left[\left({\displaystyle \frac{t}{\lambda }}\right)\left(1+{\left[{\displaystyle \frac{t}{\lambda }}\right]}^{-1}\right)\right]}^{-s\psi }={\left({\displaystyle \frac{t}{\lambda }}\right)}^{-s\psi }{\left(1+{\left({\displaystyle \frac{t}{\lambda }}\right)}^{-1}\right)}^{-s\psi }$, then expand the ${\left(1+{\left({\displaystyle \frac{t}{\lambda }}\right)}^{-1}\right)}^{-s\psi }$using the generalized binomial theorem then take the expectation.

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\left[{\left({\displaystyle \frac{t}{\lambda }}+1\right)}^{-\psi }\right]}^s$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\left({\displaystyle \frac{t}{\lambda }}\right)}^{-s\psi }{\left(1+{\left({\displaystyle \frac{t}{\lambda }}\right)}^{-1}\right)}^{-s\psi }$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\left({\displaystyle \frac{t}{\lambda }}\right)}^{-s\psi }{\displaystyle \sum _u^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\left({\left({\displaystyle \frac{t}{\lambda }}\right)}^{-1}\right)}^u$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _u^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\lambda }^{s\psi +u}E\left[t^{-s\psi -u}\right]$

$E\left[t^{-s\psi -u}\right]={\xi }^{-s\psi -u}\mathsf{\Gamma }\left(1-{\displaystyle \frac{1}{c}}\left[s\psi +u\right]\right)$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _u^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-s\psi }{u}}\right){\lambda }^{s\psi +u}$

${\xi }^{-s\psi -u}\mathsf{\Gamma }\left(1-{\displaystyle \frac{1}{c}}\left[s\psi +u\right]\right)$ where the $0<{\displaystyle \frac{1}{c}}\left[s\psi +u\right]<1$

Theorem 2: linear series expansion of this new pdf ${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}$, is applied to the exponent term and each of the power terms, using the generalized binomial expansion and Pochhammer falling factorial

Proof:

$F_X\left(x\right)=3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}$, is the CDF for MBUR and it is monotonically increasing function and it is $0<F_X\left(x\right)<1$ , so ${\left[S\left(x\right)\right]}^{-1/\psi }={\left[1-\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)\right]}^{-1/\psi }$can be expanded using the generalized binomial theory. Also to be noticed that $3x^{{\displaystyle \frac{2}{b^2}}}>2x^{{\displaystyle \frac{3}{b^2}}}$so

${\left[1-\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)\right]}^{-1/\psi }={\left[1-3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right]}^{-1/\psi }$

$={\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-1}{\psi }\right)}{n}}\right){\left[3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right]}^n={\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-1}{\psi }\right)}{n}}\right)3^n{x}^{{\displaystyle \frac{2n}{b^2}}}{\left[1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right]}^n$ and ${\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}$ is less than 1 so using the generalized binomial theory

${\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-1}{\psi }}{n}}\right)3^n{x}^{{\displaystyle \frac{2n}{b^2}}}{\left[1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right]}^n={\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-1}{\psi }\right)}{n}}\right)3^n{x}^{{\displaystyle \frac{2n}{b^2}}}{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{n}{m}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^m$

${\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^{n+m}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-1}{\psi }}{n}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n}{m}}\right)3}^n{{\left({\displaystyle \frac{2}{3}}\right)}^{m}x}^{{\displaystyle \frac{2n}{b^2}}}{x}^{{\displaystyle \frac{m}{b^2}}}={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^{n+m}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-1}{\psi }}{n}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n}{m}}\right)2}^m{3^{n-m}x}^{{\displaystyle \frac{2n+m}{b^2}}}$

For the power term ${\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{c-1}$ , ${\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}$ is less than 1 so generalized binomial theory can be applied on

${\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{c-1}={\displaystyle \sum _{k=0}^{\infty }}{\left(-1\right)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right){\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{k}{\psi }}}$ another generalized binomial expansion be applied on the ${\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{k}{\psi }}}$

$={\displaystyle \sum _{k-0}^{\infty }}{\left(-1\right)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right){\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\left(-1\right)}^{f+g}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right){3^{f-g}{2}^{g}x}^{{\displaystyle \frac{2f+g}{b^2}}}$

$={\displaystyle \sum _{k-0}^{\infty }}{\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\left(-1\right)}^{k+f+g}\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right){3^{f-g}{2}^{g}x}^{{\displaystyle \frac{2f+g}{b^2}}}$

For the term ${\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{-\left({\displaystyle \frac{c}{\psi }}+1\right)}$applying binomial expansion gives

${\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}{\left(-1\right)}^{p+q}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{c}{\psi }+1\right)}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{q}}\right){3^{p-q}{2}^{q}x}^{{\displaystyle \frac{2p+q}{b^2}}}$

Assembling both power terms together

$={\displaystyle \sum _{k-0}^{\infty }}{\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\left(-1\right)}^{k+f+g}\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right){3^{f-g}{2}^{g}x}^{{\displaystyle \frac{2f+g}{b^2}}}$

${\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}{\left(-1\right)}^{p+q}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{c}{\psi }+1\right)}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{q}}\right){3^{p-q}{2}^{q}x}^{{\displaystyle \frac{2p+q}{b^2}}}$

$={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}{\left(-1\right)}^{k+f+g+p+q}\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{c}{\psi }+1\right)}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{q}}\right)3^{f-g+p-q}{2}^{g+q}{x}^{{\displaystyle \frac{2f+g+2p+q}{b^2}}}$

The exponent term can be expanded as

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{l}^r;calll=\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\};$

$exp\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\}=e^{-l}$

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left[{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right]}^r$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}{\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]}^{cr}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}{{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-cr}{\psi }}}\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{cr}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^{s+t}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right){3^{s-t}{2}^{t}x}^{{\displaystyle \frac{2s+t}{b^2}}}{\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{cr}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^{s+t}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right){3^{s-t}{2}^{t}x}^{{\displaystyle \frac{2s+t}{b^2}}}{\displaystyle \sum _{v=0}^{\infty }}{\left(-1\right)}^v\left({\displaystyle \genfrac{}{}{0pt}{}{cr}{v}}\right){\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{v}{\psi }}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^{s+t}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right){3^{s-t}{2}^{t}x}^{{\displaystyle \frac{2s+t}{b^2}}}{\displaystyle \sum _{v=0}^{\infty }}{\left(-1\right)}^v\left({\displaystyle \genfrac{}{}{0pt}{}{cr}{v}}\right)$

${\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\left(-1\right)}^{u+w}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{v}{\psi }}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{u}{w}}\right){3^{u-w}{2}^{w}x}^{{\displaystyle \frac{2u+w}{b^2}}}$

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{v=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{r+s+t+v+u+w}}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{cr}{v}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{v}{\psi }}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{u}{w}}\right){3^{s-t+u-w}{2}^{t+w}x}^{{\displaystyle \frac{2s+t+2u+w}{b^2}}}$

${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\psi },\mathit{\xi },\mathit{\lambda }\right)=$

$={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{-\left({\displaystyle \frac{c}{\psi }}+1\right)}{\left[1-{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\psi }}}\right]}^{c-1}$

$exp\left\{-{\left({\displaystyle \frac{\lambda }{\xi }}\left[{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\psi }}}-1\right]\right)}^c\right\}$

Now put the all the parts together $g_{X_g}\left(x\right)$

$={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}$

${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}{\left(-1\right)}^{k+f+g+p+q}\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{c}{\psi }+1\right)}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{q}}\right)3^{f-g+p-q}{2}^{g+q}{x}^{{\displaystyle \frac{2f+g+2p+q}{b^2}}}$

${\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{v=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{r+s+t+v+u+w}}{r!}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{cr}{v}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{v}{\psi }}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{u}{w}}\right){3^{s-t+u-w}{2}^{t+w}x}^{{\displaystyle \frac{2s+t+2u+w}{b^2}}}$

$={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{v=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}\begin{array}{c}{\displaystyle \frac{{\left(-1\right)}^{p+q+k+f+g+r+s+t+v+u+w}}{r!}}\\ \end{array}$

$\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{c}{\psi }+1\right)}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{q}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right){\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{cr}{v}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{v}{\psi }}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{u}{w}}\right)3^{p-q+f-g+s-t+u-w}{2}^{q+g+t+w}$

$x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w}{b^2}}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}$

${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\psi },\mathit{\xi },\mathit{\lambda }\right)={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}{x}^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w}{b^2}}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}$

${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\psi },\mathit{\xi },\mathit{\lambda }\right)={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}$

Where $W_{p,q,k,f,g,r,s,t,v,u,w}=$

${\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \sum _{f=0}^{\infty }}{\displaystyle \sum _{g=0}^{\infty }}{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{v=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}\begin{array}{c}{\displaystyle \frac{{\left(-1\right)}^{p+q+k+f+g+r+s+t+v+u+w}}{r!}}\\ \end{array}$

$\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{c}{\psi }+1\right)}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{q}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{c-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{k}{\psi }}{f}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{f}{g}}\right){\left({\displaystyle \frac{\lambda }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{\psi }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{cr}{v}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{v}{\psi }}{u}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{u}{w}}\right)\left(3^{p-q+f-g+s-t+u-w}\right)\left(2^{q+g+t+w}\right)$

Integrate this series expansion by integrating the x from 0 to 1 because the support of the distribution is the support of the MBUR so the integral will pass to the x

${\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}dx$

This gives ${\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}dx=$

${\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}dx=$

${\int }_0^1\left(x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}-x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+3}{b^2}}-1}\right)dx$

$={\left.{\displaystyle \frac{x^{\frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}}{\frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}}\right|}_0^1-{\left.{\displaystyle \frac{x^{\frac{2p+q+2f+g+2s+t+2u+w+3}{b^2}}}{\frac{2p+q+2f+g+2s+t+2u+w+3}{b^2}}}\right|}_0^1$

${\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}dx$

$={\displaystyle \frac{b^2}{2p+q+2f+g+2s+t+2u+w+2}}-{\displaystyle \frac{b^2}{2p+q+2f+g+2s+t+2u+w+3}}$

$=b^2\left\{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+3}{2p+q+2f+g+2s+t+2u+w+2}}-{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{2p+q+2f+g+2s+t+2u+w+3}}\right\}$

${\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}dx$

$={\displaystyle \frac{\frac{6c}{\psi }{\left(\frac{\lambda }{\xi }\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}}{\left(2p+q+2f+g+2s+t+2u+w+2\right)\left(2p+q+2f+g+2s+t+2u+w+3\right)}}$

$0<x<1,b,c,\xi ,\psi ,\lambda >0$

Theorem 3: the h^th non-central moments are given by the following formula:

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\displaystyle \frac{\frac{6c}{\psi }{\left(\frac{\lambda }{\xi }\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}}{\left(2p+q+2f+g+2s+t+2u+w+h{b}^2+2\right)\left(2p+q+2f+g+2s+t+2u+w+h{b}^2+3\right)}}$

Proof:

$E\left(x^h\right)={\int }_0^{1}g\left(x\right)x^{h}dx={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}{\int }_0^1{x}^h\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+2}{b^2}}-1}dx.$

$E\left(x^h\right)={\int }_0^{1}g\left(x\right)x^{h}dx={\displaystyle \frac{6c}{\psi {\left(b\right)}^2}}{\left({\displaystyle \frac{\lambda }{\xi }}\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2p+q+2f+g+2s+t+2u+w+h{b}^2+2}{b^2}}-1}dx.$

$E\left(x^h\right)$

$={\displaystyle \frac{\frac{6c}{\psi }{\left(\frac{\lambda }{\xi }\right)}^c{W}_{p,q,k,f,g,r,s,t,v,u,w}}{\left(2p+q+2f+g+2s+t+2u+w+h{b}^2+2\right)\left(2p+q+2f+g+2s+t+2u+w+h{b}^2+3\right)}}$

In the supplementary material, the author depicts different figures illustrating the behavior of the distribution as regards the PDF, CDF and hazard functions. The PDF can exhibit bathtub, increasing and decreasing, right skewed and left skewed shapes and other different shapes. The hazard rate is mainly increasing and J shaped.

2. Weibull-MBUR {Dagum} Distribution

2.1. Dagum Link Distribution

The Dagum Distribution has the following PDF, CDF, and quantile functions $f_Y\left(y\right)=a\gamma {\psi }^a{y}^{-a-1}{\left[1+{\left({\displaystyle \frac{y}{\psi }}\right)}^{-a}\right]}^{-\gamma -1}$ , $F_Y\left(y\right)={\left[1+{\left({\displaystyle \frac{y}{\psi }}\right)}^{-a}\right]}^{-\gamma }$ , $Q_Y\left(p\right)=\psi {\left(p^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}$ , respectively, where; $y>0$, $0<p<1$, the two shape and scale parameters are $a,\gamma ,\psi >0$ respectively.

Use the approach discussed by Aljarrah 2014 as shown in the following steps to derive the PDF.

Step 1: substitute the CDF of MBUR into the quantile function of the link distribution (quantile function of Y), to obtain the t variable.

$T=Q_Y\left[F_X\left(x\right)\right]=\psi {\left(p^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}=\psi {\left({\left[F_X\left(x\right)\right]}^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}$

$T=\psi {\left({\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}$

Step 2: substitute this T variable into the PDF of the Weibull distribution

$f_T\left(t\right)={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{t}{\xi }}\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}$

$f_T\left(x\right)$

$={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{\psi {\left({\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{-1/\gamma }-1\right)}^{\frac{-1}{a}}}{\xi }}\right)}^{c-1}exp\left[-{\left({\displaystyle \frac{\left[\psi {\left({\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{-1/\gamma }-1\right)}^{\frac{-1}{a}}\right]}{\xi }}\right)}^c\right]$

$=c{\psi }^{c-1}{\xi }^{-c}{\left({\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-1/\gamma }-1\right)}^{\left({\displaystyle \frac{1-c}{a}}\right)}exp\left\{-{\left({\displaystyle \frac{\psi }{\xi }}\left[{\left({\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}\right]\right)}^c\right\}$

Step 3: get the derivative of the quantile function which is the reciprocal of the PDF of the quantile distribution:

${\displaystyle \frac{d}{dx}}\left(Q_Y\left(p\right)\right)={\displaystyle \frac{d}{dx}}\left(Q_Y\left(F_X\left(x\right)\right)\right)={\displaystyle \frac{1}{f_Y\left(t\right)}}$

$={\displaystyle \frac{1}{a\gamma {\psi }^a{\left(\psi {\left({\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{-1/\gamma }-1\right)}^{\frac{-1}{a}}\right)}^{-a-1}{\left[1+{\left(\frac{\psi {\left({\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{-1/\gamma }-1\right)}^{\frac{-1}{a}}}{\psi }\right)}^{-a}\right]}^{-\gamma -1}}}$

Step 4: substitute into the equation of $g\left(x\right)$ :

$g\left(x\right)=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(Q_Y\left(F_X\left(x\right)\right)\right)}}=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(t\right)}}$, where $r\left(Q_Y\left(F_X\left(x\right)\right)\right)=f_T\left(t\right)$

$g\left(x\right)={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\displaystyle \frac{{\left({\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{-1}{\gamma }}-1\right)}^{-\left(\frac{a+c}{a}\right)}exp\left\{-{\left(\frac{\psi }{\xi }{\left[{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{-1}{\psi }}-1\right]}^{\frac{-1}{a}}\right)}^c\right\}}{{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{\left(\frac{\gamma +1}{\gamma }\right)}}}$

As long as ${\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}>1$ for $\gamma >0$ the following term can be written as

${\left({\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}={\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}\right]}^{-\left({\displaystyle \frac{a+c}{a}}\right)}{\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$

After rearrangement and for convenient the $x_g=x$ in the following PDF and CDF where x is defined on the unit interval:

$g_{X_g}\left(x_g;b,c,,\xi ,a,\gamma ,\psi \right)=$

${\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$

$exp\left\{-{\left({\displaystyle \frac{\psi }{\xi }}{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right]}^{{\displaystyle \frac{-1}{a}}}\right)}^c\right\}$

$={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$

$exp\left\{-{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{c}{\gamma a}}}\left[1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right]}^{{\displaystyle \frac{-c}{a}}}\right\}$

To derive the $G_{X_g}$, substitute the t variable into the CDF of the Weibull distribution

$G_{X_g}\left(x_g;b,c,,\xi ,a,\gamma ,\psi \right)=F_T\left(t\right)=1-exp\left\{-{\left({\displaystyle \frac{\psi }{\xi }}{\left({\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}\right)}^c\right\};0<x<1$

To find the quantile function apply the following steps.

Step 1: Take the quantile of the generator, $Q_T\left(p\right)=\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}$

Step 2: Substitute this quantile in the CDF of the link distribution. $F_Y\left(y\right)={\left[1+{\left({\displaystyle \frac{y}{\psi }}\right)}^{-a}\right]}^{-\gamma }$,

$F_X\left(x\right)=F_Y\left(p\right)={\left[1+{\left({\displaystyle \frac{\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}{\psi }}\right)}^{-a}\right]}^{-\gamma }$

Step 3: Substitute the result and call it, Z; ${\left[1+{\left({\displaystyle \frac{\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}{\psi }}\right)}^{-a}\right]}^{-\gamma }$into the quantile function of the baseline distribution MBUR distribution. This will give the same result as taking the inverse of the $G_{X_g}$ .

Another approach that can be deployed to obtain the quantile function of $G_{X_g}$ is by taking the inverse of the $G_{X_g}$: $p=1-exp\left\{-{\left({\displaystyle \frac{\psi }{\xi }}{\left({\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}\right)}^c\right\}$, solve for x yields the following equation $3{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^2-2{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^3={\left[1+{\left({\displaystyle \frac{\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}{\psi }}\right)}^{-a}\right]}^{-\gamma }$ , the left hand side is the CDF of the baseline distribution MBUR and the right hand side is a constant, which is the z, and z is $0<z<1$. Solving the third degree polynomial for its root yields the ${\mathit{x}}_{\mathit{g}}$ variable or the quantile function for the new distribution is

$Q_X\left(p\right)=Q_{X_g}\left(z;b,c,,\xi ,a,\gamma ,\psi \right)={\left\{-.5\left(cos\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]-\sqrt{3}sin\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]\right)+.5\right\}}^{b^2}$

$0<z<1,b,c,,\xi ,a,\gamma ,\psi >0$

The hazard function is defined as $h\left(X_g\right)={\displaystyle \frac{g_{X_g}}{1-G_{X_g}}}=$

$h\left(X_g\right)={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$

Theorem 4: the r^th moment for the new distribution is presented in the formulas according to the shape $a$ and scale $\psi$ parameters of the link function or the Dagum random variable, this will be illustrated in the proof.

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{sa}{\xi }^{-sa}\mathsf{\Gamma }\left(1-{\displaystyle \frac{sa}{c}}\right)$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{-au-\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}{\xi }^{au+{\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}}$

$\mathsf{\Gamma }\left(1+{\displaystyle \frac{1}{c}}\left(au+{\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)\right)$

Proof: start with the t variable and isolate the x on one side

$t=\psi {\left({\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-1/\gamma }-1\right)}^{{\displaystyle \frac{-1}{a}}}$ this gives ${\left\{{\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}+1\right\}}^{-\gamma }=3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\left\{{\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}+1\right\}}^{-\gamma }$ this gives $x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}$

$x{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{b^2}{2}}}={\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ use generalized binomial expansion

$x{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^m={\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

${\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}\right)}^m{x}^{{\displaystyle \frac{m}{b^2}}+1}={\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

$W_m.x^{{\displaystyle \frac{m}{b^2}}+1}={\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ the following step is $x^{{\displaystyle \frac{m+b^2}{b^2}}}={\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}$

${\left\{x^{\left({\displaystyle \frac{m+b^2}{b^2}}\right)}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}={\left\{{\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$ so $x={\left\{{\left\{{\displaystyle \frac{{\left\{{\left(\frac{t}{\psi }\right)}^{-a}+1\right\}}^{-\gamma }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$

Take the r^th power of both sides

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left({\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}+1\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}$ there are two cases:

First case: if ${0<\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}<1$

Use the generalized binomial theory to expand the ${\left({\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}+1\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{s}}\right){\left({\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}\right)}^s$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{sa}{\left(t\right)}^{-sa}$

Take the expectation of both sides

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{sa}E\left[{\left(t\right)}^{-sa}\right]$

$E\left[{\left(t^{-a}\right)}^s\right]={\xi }^{-sa}\mathsf{\Gamma }\left(1-{\displaystyle \frac{sa}{c}}\right)$ is the s^th raw moment of the Weibull variable (the t variable), where $0<{\displaystyle \frac{sa}{c}}<1$ in other words , $0<s<{\displaystyle \frac{c}{a}}$ . That is to say s^th moment index whether 1 ,2 3, or 4 should lie between the 0 and the ratio between the $c$shape parameter of the Weibull random variable and $a$ shape parameter of the Dagum random variable

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{sa}{\xi }^{-sa}\mathsf{\Gamma }\left(1-{\displaystyle \frac{sa}{c}}\right)$

second case: if ${\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}>1$ , factor out the ${\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}$ then distribute the power then apply the generalized binomial theorem

${\left({\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}+1\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}={\left({\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}{\left(1+{\left({\displaystyle \frac{t}{\psi }}\right)}^a\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}$ then expand ${\left(1+{\left({\displaystyle \frac{t}{\psi }}\right)}^a\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}$ by applying the generalized binomial theorem and then taking the expectation

${\left(1+{\left({\displaystyle \frac{t}{\psi }}\right)}^a\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}={\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{u}}\right){\left({\displaystyle \frac{t}{\psi }}\right)}^{au}$

$\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left({\left({\displaystyle \frac{t}{\psi }}\right)}^{-a}\right)}^{\left({\displaystyle \frac{-\gamma r{b}^4}{2m+2b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{-au}\left[{\left(t\right)}^{au}\right]$

$\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{-au-\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}{\left(t\right)}^{au+\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{-au-\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}E\left[{\left(t\right)}^{au+\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}\right]$

$E\left[{\left(t\right)}^{au+\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}\right]={\xi }^{au+{\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}}\mathsf{\Gamma }\left(1+{\displaystyle \frac{1}{c}}\left(au+{\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)\right)$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-\gamma r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{-au-\left({\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)}{\xi }^{au+{\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}}$

$\mathsf{\Gamma }\left(1+{\displaystyle \frac{1}{c}}\left(au+{\displaystyle \frac{a\gamma r{b}^4}{2m+2b^2}}\right)\right)$

Theorem 5: linear series expansion of this new pdf ${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}$, is applied to the exponent term and each of the power terms, using the generalized binomial expansion and Pochhammer falling factorial

Proof: using the series expansion for the power terms; the first power term

$F_X\left(x\right)=3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}$, is the CDF for MBUR and it is monotonically increasing function can be expanded using the generalized binomial theory. Also to be noticed that $3x^{{\displaystyle \frac{2}{b^2}}}>2x^{{\displaystyle \frac{3}{b^2}}}$ so the first power term can be expanded using generalized binomial theorem as

${\left[3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right]}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}={\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{a\gamma }-1\right)}{n}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^n$

$=3^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{x}^{\left({\displaystyle \frac{2c-2a\gamma }{b^{2}a\gamma }}\right)}{\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{a\gamma }-1\right)}{n}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^n$

$={\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{a\gamma }-1\right)}{n}}\right){\left({\displaystyle \frac{2}{3}}\right)}^n{3}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{x}^{{\displaystyle \frac{n}{b^2}}}{x}^{\left({\displaystyle \frac{2c-2a\gamma }{b^{2}a\gamma }}\right)}$

$={\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{a\gamma }-1\right)}{n}}\right){\left(2\right)}^n{3}^{\left({\displaystyle \frac{c}{a\gamma }}-1-n\right)}{x}^{{\displaystyle \frac{n}{b^2}}}{x}^{\left({\displaystyle \frac{2c-2a\gamma }{b^{2}a\gamma }}\right)}$

The second power term ${\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$ where ${\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}$ is <1 for $\gamma >0$

${\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}={\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{1/\gamma }{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$

$={\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right){\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{1/\gamma }{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right]}^m$

$={\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right){\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{m}{\gamma }}}{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{m}{\gamma }}}$

$={\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right){\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{m}{\gamma }}}{\displaystyle \sum _w^{\infty }}{\left(-1\right)}^w\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{m}{\gamma }\right)}{w}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^w$

$={\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\left(-1\right)}^{m+w}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{m}{\gamma }\right)}{w}}\right){\left(3\right)}^{{\displaystyle \frac{m}{\gamma }}}{\left({\displaystyle \frac{2}{3}}\right)}^w{x}^{{\displaystyle \frac{w}{b^2}}}{x}^{{\displaystyle \frac{2m}{{\gamma b}^2}}}$

$={\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\left(-1\right)}^{m+w}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{m}{\gamma }\right)}{w}}\right){\left(3\right)}^{{\displaystyle \frac{m}{\gamma }}-w}{\left(2\right)}^w{x}^{\left({\displaystyle \frac{2m}{{\gamma b}^2}}+{\displaystyle \frac{\gamma w}{{\gamma b}^2}}\right)}$

The exponent term can be expanded as

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{l}^r;calll=\left\{-{\left({\displaystyle \frac{\psi }{\xi }}{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right]}^{{\displaystyle \frac{-1}{a}}}\right)}^c\right\};$

$exp\left\{-\left[{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right]}^{{\displaystyle \frac{-c}{a}}}\right]\right\}=e^{-l}$

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left[{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right]}^{{\displaystyle \frac{-c}{a}}}\right]}^r={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right]}^{{\displaystyle \frac{-cr}{a}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}\right]}^{{\displaystyle \frac{-cr}{a}}}\left[1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right]}^{{\displaystyle \frac{-cr}{a}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{cr}{a\gamma }}}\left[1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right]}^{{\displaystyle \frac{-cr}{a}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{a\gamma }}}{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{cr}{a\gamma }}}\left[1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right]}^{{\displaystyle \frac{-cr}{a}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{a\gamma }}}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{a\gamma }\right)}{s}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^s{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{a}\right)}{t}}\right)\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right]}^t$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{a\gamma }}}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{a\gamma }\right)}{s}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^s{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{a}\right)}{t}}\right){\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{t}{\gamma }}}{\displaystyle \sum _{u=0}^{\infty }}{\left(-1\right)}^u\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{t}{\gamma }\right)}{u}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^u$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(x\right)}^{{\displaystyle \frac{2cr}{b^{2}a\gamma }}}{\left(3\right)}^{{\displaystyle \frac{cr}{a\gamma }}}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{a\gamma }\right)}{s}}\right){\left({\displaystyle \frac{2}{3}}\right)}^s{x}^{{\displaystyle \frac{s}{b^2}}}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{a}\right)}{t}}\right){\left(3\right)}^{{\displaystyle \frac{t}{\gamma }}}{x}^{{\displaystyle \frac{2t}{{\gamma b}^2}}}{\displaystyle \sum _{u=0}^{\infty }}{\left(-1\right)}^u\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{t}{\gamma }\right)}{u}}\right){\left({\displaystyle \frac{2}{3}}\right)}^u{x}^{{\displaystyle \frac{u}{b^2}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{r+s+t+u}}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{a\gamma }\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{a}\right)}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{t}{\gamma }\right)}{u}}\right){\left(3\right)}^{{\displaystyle \frac{cr}{a\gamma }}+{\displaystyle \frac{t}{\gamma }}-s-u}{\left(2\right)}^{s+u}{\left(x\right)}^{{\displaystyle \frac{2cr}{b^{2}a\gamma }}}{\left(x\right)}^{{\displaystyle \frac{s}{b^2}}}{\left(x\right)}^{{\displaystyle \frac{2t}{{\gamma b}^2}}}{\left(x\right)}^{{\displaystyle \frac{u}{b^2}}}$

${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{a},\mathit{b},\mathit{c},\mathit{\psi },\mathit{\xi },\mathit{\psi }\right)=$

$={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{\left({\displaystyle \frac{c}{a\gamma }}-1\right)}{\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{1}{\gamma }}}\right)}^{-\left({\displaystyle \frac{a+c}{a}}\right)}$

$exp\left\{-{\left({\displaystyle \frac{\psi }{\xi }}{\left[{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-1}{\gamma }}}-1\right]}^{{\displaystyle \frac{-1}{a}}}\right)}^c\right\}$

Now put the three parts together $g_{X_g}\left(x\right)$

$={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\displaystyle \sum _{n=0}^{\infty }}{\left(-1\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{a\gamma }-1\right)}{n}}\right){\left(2\right)}^n{3}^{\left({\displaystyle \frac{c-a\gamma }{a\gamma }}-n\right)}{x}^{{\displaystyle \frac{n}{b^2}}}{x}^{\left({\displaystyle \frac{2c-2a\gamma }{b^{2}a\gamma }}\right)}$

${\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\left(-1\right)}^{m+w}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{m}{\gamma }\right)}{w}}\right){\left(3\right)}^{{\displaystyle \frac{m}{\gamma }}-w}{\left(2\right)}^w{x}^{\left({\displaystyle \frac{2m}{{\gamma b}^2}}+{\displaystyle \frac{\gamma w}{{\gamma b}^2}}\right)}$

${\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{r+s+t+u}}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{a\gamma }\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{a}\right)}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{t}{\gamma }\right)}{u}}\right){\left(3\right)}^{{\displaystyle \frac{cr}{a\gamma }}+{\displaystyle \frac{t}{\gamma }}-s-u}{\left(2\right)}^{s+u}{\left(x\right)}^{{\displaystyle \frac{2cr}{b^{2}a\gamma }}}{\left(x\right)}^{{\displaystyle \frac{s}{b^2}}}{\left(x\right)}^{{\displaystyle \frac{2t}{{\gamma b}^2}}}{\left(x\right)}^{{\displaystyle \frac{u}{b^2}}}$

$={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{n+m+w+r+s+t+u}}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3\right)}^{{\displaystyle \frac{cr}{a\gamma }}+{\displaystyle \frac{t}{\gamma }}+{\displaystyle \frac{c-a\gamma }{a\gamma }}+{\displaystyle \frac{m}{\gamma }}-w-n-s-u}{\left(2\right)}^{n+w+s+u}$

$\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{c}{a\gamma }-1}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{m}{\gamma }}{w}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{cr}{a\gamma }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{a}}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{t}{\gamma }}{u}}\right)$

$\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{x}^{{\displaystyle \frac{n}{b^2}}}{x}^{\left({\displaystyle \frac{2c-2a\gamma }{b^{2}a\gamma }}\right)}{x}^{\left({\displaystyle \frac{2m}{{\gamma b}^2}}+{\displaystyle \frac{\gamma w}{{\gamma b}^2}}\right)}{\left(x\right)}^{{\displaystyle \frac{2cr}{b^{2}a\gamma }}}{\left(x\right)}^{{\displaystyle \frac{s}{b^2}}}{\left(x\right)}^{{\displaystyle \frac{2t}{{\gamma b}^2}}}{\left(x\right)}^{{\displaystyle \frac{u}{b^2}}}$

$={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{w=0}^{\infty }}{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{n+m+w+r+s+t+u}}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3\right)}^{{\displaystyle \frac{cr}{a\gamma }}+{\displaystyle \frac{t}{\gamma }}+{\displaystyle \frac{c-a\gamma }{a\gamma }}+{\displaystyle \frac{m}{\gamma }}-w-n-s-u}{\left(2\right)}^{n+w+s+u}$

$\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{c}{a\gamma }-1}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(\frac{a+c}{a}\right)}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{m}{\gamma }}{w}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{cr}{a\gamma }}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{-cr}{a}}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{t}{\gamma }}{u}}\right)$

$\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}$

Integrate this series expansion by integrating the x from 0 to 1 because the support of the distribution is the support of the MBUR so the integral will pass to the x

${\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,w,r,s,t,u}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}dx$

This gives ${\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}dx=$

${\int }_0^1\left(x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}-x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}\right)dx$

$={\left.{\displaystyle \frac{x^{\left[\frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }\right]}}{\left[\frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }\right]}}\right|}_0^1-{\left.{\displaystyle \frac{x^{\left[\frac{a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }\right]}}{\left[\frac{a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }\right]}}\right|}_0^1$

$=b^{2}a\gamma \left\{{\displaystyle \frac{1}{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}}-{\displaystyle \frac{1}{a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}}\right\}$

$=b^{2}a\gamma \left\{{\displaystyle \frac{a\gamma }{\left[a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]\left[a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]}}\right\}$

${\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,w,r,s,t,u}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}dx$

$={\displaystyle \frac{\frac{6c}{b^{2}a\gamma }{\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,w,r,s,t,u}{\left(ba\gamma \right)}^2}{\left[a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]\left[a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]}}$

$={\displaystyle \frac{6ca\gamma {\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,w,r,s,t,u}}{\left[a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]\left[a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]}}$

Theorem 6: the r^th non-central moments are given by the following formula:

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\displaystyle \frac{6ca\gamma {\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,w,r,s,t,u}}{\left(a\gamma \left[m+n+s+2t+u+h{b}^2\right]+2c\left[1+r\right]+2a\left(m+t\right)\right)\left(a\gamma \left[m+n+s+2t+u+h{b}^2+1\right]+2c\left[1+r\right]+2a\left(m+t\right)\right)}}$

Proof:

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\int }_0^1\mathit{g}\left(\mathit{x}\right){\mathit{x}}^{\mathit{h}}\mathit{d}\mathit{x}={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,w,r,s,t,u}{\int }_0^1{\mathit{x}}^{\mathit{h}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}\mathit{d}\mathit{x}.$

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\int }_0^1\mathit{g}\left(\mathit{x}\right){\mathit{x}}^{\mathit{h}}\mathit{d}\mathit{x}={\displaystyle \frac{6c}{b^{2}a\gamma }}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,w,r,s,t,u}{\int }_0^1{\mathit{x}}^{\mathit{h}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}\mathit{d}\mathit{x}.$

${\int }_0^1{\mathit{x}}^{\mathit{h}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left[{\displaystyle \frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }}\right]-1}dx$

$={\left.{\displaystyle \frac{x^{\left[\frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)+h{b}^{2}a\gamma }{b^{2}a\gamma }\right]}}{\left[\frac{a\gamma \left(w+n+s+u\right)+2c\left(1+r\right)+2a\left(m+t\right)+h{b}^{2}a\gamma }{b^{2}a\gamma }\right]}}\right|}_0^1$

$-{\left.{\displaystyle \frac{x^{\left[\frac{a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)+h{b}^{2}a\gamma }{b^{2}a\gamma }\right]}}{\left[\frac{a\gamma \left(w+n+s+u+1\right)+2c\left(1+r\right)+2a\left(m+t\right)+h{b}^{2}a\gamma }{b^{2}a\gamma }\right]-1}}\right|}_0^1$

$={\displaystyle \frac{1}{\left[\frac{a\gamma \left(w+n+s+u+h{b}^2\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }\right]}}$

$-{\displaystyle \frac{1}{\left[\frac{a\gamma \left(w+n+s+u+h{b}^2+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}{b^{2}a\gamma }\right]}}$

$=b^{2}a\gamma \left\{{\displaystyle \frac{1}{\left[a\gamma \left(w+n+s+u+h{b}^2\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]}}-{\displaystyle \frac{1}{a\gamma \left(w+n+s+u+h{b}^2+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}}\right\}$

$=b^{2}a\gamma \left\{{\displaystyle \frac{1}{\left[a\gamma \left(m+n+s+u+h{b}^2\right)+2c\left(1+r\right)+2a\left(m+t\right)\right]}}-{\displaystyle \frac{1}{a\gamma \left(m+n+s+u+h{b}^2+1\right)+2c\left(1+r\right)+2a\left(m+t\right)}}\right\}$

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\displaystyle \frac{6ca\gamma {\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,w,r,s,t,u}}{\left(a\gamma \left[m+n+s+2t+u+h{b}^2\right]+2c\left[1+r\right]+2a\left(m+t\right)\right)\left(a\gamma \left[m+n+s+2t+u+h{b}^2+1\right]+2c\left[1+r\right]+2a\left(m+t\right)\right)}}$

In the supplementary material, the author depicts different figures illustrating the behavior of the distribution as regards the PDF, CDF and hazard functions. The PDF can exhibit reversed J shaped, increasing and decreasing, N shape, left skewed and right skewed and other different shapes. The hazard rate is mainly increasing and J shaped.

3. Weibull-MBUR {Exponential} Distribution

3.1. Exponential Link Distribution

The exponential Distribution has the following PDF, CDF, and quantile functions $f_Y\left(y\right)=\lambda e^{-\lambda y}$ , $F_Y\left(y\right)=1-e^{-\lambda y}$ , $Q_Y\left(p\right)={\displaystyle \frac{-\ln \left(1-p\right)}{\lambda }}$ , respectively, where; $y>0$, $0<p<1$, the scale parameter is $\lambda >0$ respectively.

Use the approach discussed by Aljarrah 2014 as shown in the following steps to derive the PDF.

Step 1: substitute the CDF of MBUR into the quantile function of the link distribution (quantile function of Y), to obtain the t variable.

$T=Q_Y\left[F_X\left(x\right)\right]={\displaystyle \frac{-\ln \left(1-p\right)}{\lambda }}={\displaystyle \frac{-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}{\lambda }}$

$T={\displaystyle \frac{-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}{\lambda }}$

Step 2: substitute this T variable into the PDF of the Weibull distribution

$f_T\left(t\right)={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{t}{\xi }}\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}$

$f_T\left(x\right)$

$={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}{\lambda \xi }}\right)}^{c-1}exp\left[-{\left({\displaystyle \frac{\left[-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)\right]}{\lambda \xi }}\right)}^c\right]$

$=c{\lambda }^{1-c}{\xi }^{-c}{\left(-\ln \left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)\right)}^{c-1}exp\left\{-{\left({\left(\lambda \xi \right)}^{-1}\left[-\ln \left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)\right]\right)}^c\right\}$

Step 3: get the derivative of the quantile function which is the reciprocal of the PDF of the quantile distribution:

${\displaystyle \frac{d}{dx}}\left(Q_Y\left(p\right)\right)={\displaystyle \frac{d}{dx}}\left(Q_Y\left(F_X\left(x\right)\right)\right)={\displaystyle \frac{1}{f_Y\left(t\right)}}={\displaystyle \frac{1}{\lambda exp\left[-\lambda \left\{\frac{-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}{\lambda }\right\}\right]}}={\displaystyle \frac{1}{\lambda \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}}$

Step 4: substitute into the equation of $g\left(x\right)$ :

$g\left(x\right)=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(Q_Y\left(F_X\left(x\right)\right)\right)}}=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(t\right)}}$, where $r\left(Q_Y\left(F_X\left(x\right)\right)\right)=f_T\left(t\right)$

The PDF is$\mathit{g}\left(\mathit{x}\right)=$

${\displaystyle \frac{6c}{{\left(\lambda \xi \right)}^c{b}^2}}$

$\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\displaystyle \frac{{\left(-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)\right)}^{c-1}exp\left\{-{\left({\left(\lambda \xi \right)}^{-1}\left[-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)\right]\right)}^c\right\}}{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}}$

The CDF is ${\mathit{G}}_{{\mathit{X}}_{\mathit{g}}}=1-exp\left\{-{\left({\left(\lambda \xi \right)}^{-1}\left[-\ln \left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)\right]\right)}^c\right\}$

To find the quantile follow the steps.

Step 1: take the quantile of the generator, or $Q_T\left(p\right)=\xi {\left[-\ln \left(1-p\right)\right]}^{1/c}$

Step 2: Substitute this quantile in the CDF of the link distribution. $F_Y\left(y\right)=F_Y\left(y\right)=1-e^{-\lambda y}$,

$F_X\left(x\right)=F_Y\left(p\right)=F_Y\left(y\right)=1-e^{-\lambda \xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}$

Step 3: Substitute the result and call it, Z; $1-e^{-\lambda \xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}$ into the quantile function of the baseline distribution MBUR distribution. This will give the same result as taking the inverse of the $G_{X_g}$ .

Another approach that can be deployed to obtain the quantile function of ${\mathit{G}}_{{\mathit{X}}_{\mathit{g}}}$ is by taking the inverse of the $G_{X_g}=p=1-exp\left\{-{\left({\left(\lambda \xi \right)}^{-1}\left[-\ln \left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)\right]\right)}^c\right\}$, solving for x yields the following equation, $3{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^2-2{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^3=1-e^{-\lambda \xi {\left[-\ln \left(1-p\right)\right]}^{1/c}}$ , the left hand side is the CDF of the baseline distribution MBUR and the right hand side is a constant, which is the z, and z is $0<z<1$. Solving the third degree polynomial for its root yields the ${\mathit{x}}_{\mathit{g}}$ variable or the quantile function for the new distribution which is

$Q_X\left(p\right)=Q_{X_g}\left(z;b,c,\xi ,\lambda \right)={\left\{-.5\left(cos\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]-\sqrt{3}sin\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]\right)+.5\right\}}^{b^2}$

$0<z<1,b,c,\xi ,\lambda >0$

The hazard function is defined as $h\left(X_g\right)={\displaystyle \frac{g_{X_g}}{1-G_{X_g}}}=$

$h\left(X_g\right)={\displaystyle \frac{6c}{{\left(\lambda \xi \right)}^c{b}^2}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\displaystyle \frac{{\left(-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)\right)}^{c-1}}{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}}$

Theorem 7: the r^th non-central moments are given by the following formula

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{s+u}{\left(\lambda s\right)}^u}{u!}}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$

Proof: Starting with the t variable

$t={\displaystyle \frac{-\ln \left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}{\lambda }}$ then isolating the x as shown in the following steps

$-t\lambda =\ln \left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)$ this gives $e^{-t\lambda }=\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)$

$3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}=\left(1-e^{-t\lambda }\right)$ this gives $3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)=\left(1-e^{-t\lambda }\right)$

$x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\displaystyle \frac{\left(1-e^{-t\lambda }\right)}{3}}$

${\left\{x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}={\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ this gives $x{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{\left({\displaystyle \frac{b^2}{2}}\right)}={\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

$x{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^m={\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ using the binomial expansion

${\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}\right)}^m{x}^{{\displaystyle \frac{m}{b^2}}+1}={\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ this gives $W_m.x^{{\displaystyle \frac{m}{b^2}}+1}={\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

$x^{{\displaystyle \frac{m+b^2}{b^2}}}={\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}$

${\left\{x^{\left({\displaystyle \frac{m+b^2}{b^2}}\right)}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}={\left\{{\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$

$x={\left\{{\left\{{\displaystyle \frac{1-e^{-t\lambda }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}={\left(1-e^{-t\lambda }\right)}^{\left({\displaystyle \frac{b^4}{2m+2b^2}}\right)}{3}^{\left({\displaystyle \frac{-b^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-b^2}{m+b^2}}\right)}$

Take the r^th power of both sides

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left(1-e^{-t\lambda }\right)}^{\left({\displaystyle \frac{r{b}^4}{2m+2b^2}}\right)}$

Use the generalized binomial theory to expand the ${\left(1-e^{-t\lambda }\right)}^{\left({\displaystyle \frac{r{b}^4}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right)e^{-t\lambda s}$

Then expand the exponent $e^{-t\lambda s}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^u}{u!}}{\left(t\lambda s\right)}^u$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^u}{u!}}{\left(\lambda s\right)}^u{\left(t\right)}^u$

Take expectation of both sides

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{s+u}{\left(\lambda s\right)}^u}{u!}}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right)E\left[{\left(t\right)}^u\right]$

$E\left[{\left(t\right)}^u\right]={\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$ is the u^th raw moment of the Weibull variable (the t variable)

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{s+u}{\left(\lambda s\right)}^u}{u!}}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r{b}^4}{2m+2b^2}\right)}{s}}\right){\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$

In the supplementary material, the author depicts different figures illustrating the behavior of the distribution as regards the PDF, CDF and hazard functions. The PDF can exhibit reversed J shaped, increasing and decreasing, N shape, left skewed and right skewed shapes and other different shapes. The hazard function is mainly increasing, bathtub or J shape.

4. Weibull-MBUR {Exponentiated-Exponential} Distribution

4.1. Exponentiated-Exponential Link Distribution

The exponentiated exponential Distribution has the following PDF, CDF, and quantile functions $f_Y\left(y\right)=\alpha \lambda e^{-\lambda y}{\left(1-e^{-\lambda y}\right)}^{\alpha -1}$ , $F_Y\left(y\right)={\left(1-e^{-\lambda y}\right)}^{\alpha }$ , $Q_Y\left(p\right)={\displaystyle \frac{-\ln \left(1-p^{1/\alpha }\right)}{\lambda }}$ , respectively, where; $y>0$, $0<p<1$, the scale and shape parameters are $\lambda ,\alpha >0$ respectively.

Use the approach discussed by Aljarrah 2014 as shown in the following steps to derive the PDF.

Step 1: substitute the CDF of MBUR into the quantile function of the link distribution (quantile function of Y), to obtain the t variable.

$T=Q_Y\left[F_X\left(x\right)\right]={\displaystyle \frac{-\ln \left(1-p^{1/\alpha }\right)}{\lambda }}=-{\displaystyle \frac{1}{\lambda }}\ln \left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{1/\alpha }\right)$

$T=-{\displaystyle \frac{1}{\lambda }}\ln \left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{1/\alpha }\right)$

Step 2: substitute this T variable into the PDF of the Weibull distribution

$f_T\left(t\right)={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{t}{\xi }}\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}$

$f_T\left(x\right)$

$={\displaystyle \frac{c}{\xi }}{\left\{{\displaystyle \frac{-\frac{1}{\lambda }\ln \left(1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{1/\alpha }\right)}{\xi }}\right\}}^{c-1}exp\left[-{\left\{{\displaystyle \frac{-\frac{1}{\lambda }\ln \left(1-{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{1/\alpha }\right)}{\xi }}\right\}}^c\right]$

$={\displaystyle \frac{c}{{\xi }^c}}{\left\{-{\displaystyle \frac{1}{\lambda }}\ln \left(1-{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{1/\alpha }\right)\right\}}^{c-1}exp\left[-{\left\{-{\displaystyle \frac{1}{\mathit{\xi \lambda }}}\ln \left(1-{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{1/\alpha }\right)\right\}}^c\right]$

Step 3: get the derivative of the quantile function which is the reciprocal of the PDF of the quantile distribution:

${\displaystyle \frac{d}{dx}}\left(Q_Y\left(p\right)\right)={\displaystyle \frac{d}{dx}}\left(Q_Y\left(F_X\left(x\right)\right)\right)={\displaystyle \frac{1}{f_Y\left(t\right)}}$

$={\displaystyle \frac{1}{\alpha \lambda exp\left(-\lambda \left\{-\frac{1}{\lambda }\ln \left(1-{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{1/\alpha }\right)\right\}\right){\left(1-exp\left(-\lambda \left\{-\frac{1}{\lambda }\ln \left(1-{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{1/\alpha }\right)\right\}\right)\right)}^{\alpha -1}}}$

$={\displaystyle \frac{1}{\alpha \lambda \left[1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{1}{\alpha }}\right]{\left(1-\left[1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{1}{\alpha }}\right]\right)}^{\alpha -1}}}$

$={\displaystyle \frac{1}{\alpha \lambda \left[1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{1/\alpha }\right]{\left(\left[{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{1/\alpha }\right]\right)}^{\alpha -1}}}={\displaystyle \frac{{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{1-\alpha }{\alpha }}}{\alpha \lambda \left[1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{1/\alpha }\right]}}$Step 4: substitute into the equation of $g\left(x\right)$ :

$g\left(x\right)=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(Q_Y\left(F_X\left(x\right)\right)\right)}}=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(t\right)}}$, where $r\left(Q_Y\left(F_X\left(x\right)\right)\right)=f_T\left(t\right)$

The PDF is$\mathit{g}\left(\mathit{x}\right)=$

$={\displaystyle \frac{6c}{\alpha \lambda b^2{\xi }^c}}{\displaystyle \frac{\left(1-x^{\frac{1}{b^2}}\right)x^{\frac{2}{b^2}-1}{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{1-\alpha }{\alpha }}}{\left[1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{1/\alpha }\right]}}$

${\left\{-{\displaystyle \frac{1}{\lambda }}\ln \left(1-{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{1/\alpha }\right)\right\}}^{c-1}exp\left[-{\left\{-{\displaystyle \frac{1}{\mathit{\xi \lambda }}}\ln \left(1-{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{1/\alpha }\right)\right\}}^c\right]$

The CDF is $G_{X_g}=1-exp\left[-{\left\{-{\displaystyle \frac{1}{\mathit{\xi \lambda }}}\ln \left(1-{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{1/\alpha }\right)\right\}}^c\right]$

To find the quantile follow the steps.

Step 1: take the quantile of the generator, $Q_{{\mathit{X}}_{\mathit{g}}}\left(p\right)$ or $Q_T\left(p\right)=\xi {\left[-\ln \left(1-p\right)\right]}^{{\displaystyle \frac{1}{c}}}$

Step 2: Substitute this quantile in the CDF of the link distribution. $F_Y\left(y\right)=F_Y\left(y\right)={\left(1-e^{-\lambda y}\right)}^{\alpha }={\left(1-\exp \left(-\left\{\lambda \left(\xi {\left[-\ln \left(1-p\right)\right]}^{{\displaystyle \frac{1}{c}}}\right)\right\}\right)\right)}^{\alpha }$

$F_X\left(x\right)=F_Y\left(p\right)=F_Y\left(y\right)={\left(1-\exp \left(-\left\{\lambda \left(\xi {\left[-\ln \left(1-p\right)\right]}^{{\displaystyle \frac{1}{c}}}\right)\right\}\right)\right)}^{\alpha }$

Step 3: Substitute the result and call it, Z; ${\left(1-\exp \left(-\left\{\lambda \left(\xi {\left[-\ln \left(1-p\right)\right]}^{{\displaystyle \frac{1}{c}}}\right)\right\}\right)\right)}^{\alpha }$ into the quantile function of the baseline distribution MBUR distribution. This will give the same result as taking the inverse of the $G_{X_g}$ .

Another approach that can be deployed to obtain the quantile function of $G_{X_g}$ is by taking the inverse of the $G_{X_g}=p=1-exp\left[-{\left\{-{\displaystyle \frac{1}{\mathit{\xi \lambda }}}\ln \left(1-{\left[3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{1/\alpha }\right)\right\}}^c\right]$, solving for x yields the following equation, $3{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^2-2{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^3={\left(1-\exp \left(-\left\{\lambda \left(\xi {\left[-\ln \left(1-p\right)\right]}^{{\displaystyle \frac{1}{c}}}\right)\right\}\right)\right)}^{\alpha }$ , the left hand side is the CDF of the baseline distribution MBUR and the right hand side is a constant, which is the z, and z is $0<z<1$. Solving the third degree polynomial for its root yields the $x_g$ variable or the quantile function for the new distribution which is

$Q_X\left(p\right)=Q_{X_g}\left(z\right)={\left\{-.5\left(cos\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]-\sqrt{3}sin\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]\right)+.5\right\}}^{b^2}$

$0<z<1,b,c,\xi ,\alpha ,\lambda >0$

The hazard function is defined as $h\left(X_g\right)={\displaystyle \frac{g_{X_g}}{1-G_{X_g}}}=$

$h\left(X_g\right)={\displaystyle \frac{6c}{\alpha \lambda b^{2{\xi }^c}}}{\displaystyle \frac{\left(1-x^{\frac{1}{b^2}}\right)x^{\frac{2}{b^2}-1}{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{1-\alpha }{\alpha }}{\left\{-\frac{1}{\lambda }\ln \left(1-{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{1/\alpha }\right)\right\}}^{c-1}}{\left[1-{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{1/\alpha }\right]}}$

Theorem 8: the r^th non-central moments are given by the following formula

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{u+s}}{u!}}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r\alpha b^4}{2m+2b^2}\right)}{s}}\right){\left(s\lambda \right)}^u{\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$Proof: Starting with the t variable then isolating the x as shown in the following steps

$t=-{\displaystyle \frac{1}{\lambda }}\ln \left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{1/\alpha }\right)$ this gives $-\lambda t=\ln \left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{1/\alpha }\right)$

$e^{-\lambda t}=\left(1-{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{1/\alpha }\right)$ this gives $1-e^{-\lambda t}={\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{1/\alpha }$

${\left(1-e^{-\lambda t}\right)}^{\alpha }=3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\left(1-e^{-\lambda t}\right)}^{\alpha }$ this gives ${\left\{x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

${\left\{x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ this gives ${x\left\{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

Expand the ${\left\{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}$ using the generalized binomial theory

$x{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^m={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

$x{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}\right)}^m{x}^{\left({\displaystyle \frac{m}{b^2}}\right)}={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$

$W_m.{\left(x\right)}^{\left({\displaystyle \frac{m}{b^2}}+1\right)}={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}$ this gives $x^{{\displaystyle \frac{m+b^2}{b^2}}}={\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}$

${\left\{x^{\left({\displaystyle \frac{m+b^2}{b^2}}\right)}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}={\left({\left\{{\displaystyle \frac{{\left(1-e^{-\lambda t}\right)}^{\alpha }}{3}}\right\}}^{{\displaystyle \frac{b^2}{2}}}\right)}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}{\left[{\left(W_m\right)}^{-1}\right]}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$

$x=3^{\left({\displaystyle \frac{-b^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-b^2}{m+b^2}}\right)}{\left[1-e^{-\lambda t}\right]}^{\left({\displaystyle \frac{\alpha b^4}{2m+2b^2}}\right)}$

Take the r^th power of both sides

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left[1-e^{-\lambda t}\right]}^{\left({\displaystyle \frac{r\alpha b^4}{2m+2b^2}}\right)}$

Use the generalized binomial theory to expand the ${\left[1-e^{-\lambda t}\right]}^{\left({\displaystyle \frac{r\alpha b^4}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r\alpha b^4}{2m+2b^2}\right)}{s}}\right){\left[e^{-\lambda t}\right]}^s$

Expand the exponent $e^{-s\lambda t}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r\alpha b^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^u}{u!}}{\left[s\lambda t\right]}^u$

Take expectation of both sides

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r\alpha b^4}{2m+2b^2}\right)}{s}}\right){\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^u}{u!}}E\left\{{\left[s\lambda t\right]}^u\right\}$

$E\left\{{\left[s\lambda t\right]}^u\right\}={\left(s\lambda \right)}^u{\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{u+s}}{u!}}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{r\alpha b^4}{2m+2b^2}\right)}{s}}\right){{\left(s\lambda \right)}^u\xi }^u\mathsf{\Gamma }\left(1+{\displaystyle \frac{u}{c}}\right)$

In the supplementary material, the author depicts different figures illustrating the behavior of the distribution as regards the PDF, CDF and hazard functions. The PDF can exhibit reversed J shaped, increasing and decreasing, N shape, left skewed and right skewed shapes and other different shapes. The hazard function is mainly increasing and J shaped.

5. Weibull-MBUR {Log Logistic} Distribution

5.1. Log Logistic Link Distribution

The log_logistic Distribution has the following PDF, CDF, and quantile functions $f_Y\left(y\right)={\displaystyle \frac{\frac{B}{\psi }{\left(\frac{y}{\psi }\right)}^{B-1}}{{\left[1+{\left(\frac{y}{\psi }\right)}^B\right]}^2}}$ , $F_Y\left(y\right)={\displaystyle \frac{{\left(\frac{y}{\psi }\right)}^B}{1+{\left(\frac{y}{\psi }\right)}^B}}$ , $Q_Y\left(p\right)={\psi \left[{\displaystyle \frac{p}{1-p}}\right]}^{{\displaystyle \frac{1}{B}}}$ , respectively, where; $y>0$, $0<p<1$, the scale and shape parameters are $\psi \&B>0$ respectively.

Use the approach discussed by Aljarrah 2014 as shown in the following steps to derive the PDF.

Step 1: substitute the CDF of MBUR into the quantile function of the link distribution (quantile function of Y), to obtain the t variable.

$T=Q_Y\left[F_X\left(x\right)\right]={\psi \left[{\displaystyle \frac{p}{1-p}}\right]}^{{\displaystyle \frac{1}{B}}}=\psi {\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}$

$T=\psi {\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}$

Step 2: substitute this T variable into the PDF of the Weibull distribution

$f_T\left(t\right)={\displaystyle \frac{c}{\xi }}{\left({\displaystyle \frac{t}{\xi }}\right)}^{c-1}{e}^{-{\left({\displaystyle \frac{t}{\xi }}\right)}^c}$

$f_T\left(x\right)$

$={\displaystyle \frac{c}{\xi }}{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^{c-1}exp\left[-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right]$

$=c{\psi }^{c-1}{\xi }^{-c}{\displaystyle \frac{{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{c-1}{B}}}{{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}^{\frac{c-1}{B}}}}exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$

Step 3: get the derivative of the quantile function which is the reciprocal of the PDF of the quantile distribution:

${\displaystyle \frac{d}{dx}}\left(Q_Y\left(p\right)\right)={\displaystyle \frac{d}{dx}}\left(Q_Y\left(F_X\left(x\right)\right)\right)={\displaystyle \frac{1}{f_Y\left(t\right)}}={\displaystyle \frac{{\left[1+{\left(\frac{y}{\psi }\right)}^B\right]}^2}{\frac{B}{\psi }{\left(\frac{y}{\psi }\right)}^{B-1}}}={\displaystyle \frac{{\left[1+{\left(\frac{\psi {\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{1}{B}}}{\psi }\right)}^B\right]}^2}{\frac{B}{\psi }{\left(\frac{\psi {\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{1}{B}}}{\psi }\right)}^{B-1}}}$

$={\displaystyle \frac{{\left[1+{\left({\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{1}{B}}\right)}^B\right]}^2}{\frac{B}{\psi }{\left({\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{1}{B}}\right)}^{B-1}}}={\displaystyle \frac{{\left[1+\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]\right]}^2}{\frac{B}{\psi }{\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{B-1}{B}}}}$

$={\displaystyle \frac{{\left[\frac{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}+\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]\right]}^2}{\frac{B}{\psi }{\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{B-1}{B}}}}={\displaystyle \frac{{\left[\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)\right]}^{-2}}{\frac{B}{\psi }{\left[\frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}\right]}^{\frac{B-1}{B}}}}$

$={\displaystyle \frac{\psi {\left[1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right]}^{-2}{\left[1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right]}^{\frac{B-1}{B}}}{B{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{\frac{B-1}{B}}}}={\displaystyle \frac{\psi {\left[1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right]}^{-1-\frac{1}{B}}}{B{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{\frac{B-1}{B}}}}$

Step 4: substitute into the equation of $g\left(x\right)$ :

$g\left(x\right)=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(Q_Y\left(F_X\left(x\right)\right)\right)}}=f_X\left(x\right){\displaystyle \frac{r\left(Q_Y\left(F_X\left(x\right)\right)\right)}{f_Y\left(t\right)}}$, where $r\left(Q_Y\left(F_X\left(x\right)\right)\right)=f_T\left(t\right)$

The PDF is${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\xi },\mathit{B},\mathit{\psi }\right)=$

${\displaystyle \frac{6}{b^2}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\displaystyle \frac{\psi {\left[1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right]}^{-1-\frac{1}{B}}}{B{\left[3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right]}^{\frac{B-1}{B}}}}c{\psi }^{c-1}{\xi }^{-c}{\displaystyle \frac{{\left(3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}\right)}^{\frac{c-1}{B}}}{{\left(1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}\right)}^{\frac{c-1}{B}}}}exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$

$={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-\left(1+{\displaystyle \frac{c}{B}}\right)}{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{c}{B}}-1}exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$

The CDF is ${\mathit{G}}_{{\mathit{X}}_{\mathit{g}}}=1-exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$

To find the quantile follow the steps.

Step 1: take the quantile of the generator, $Q_{{\mathit{X}}_{\mathit{g}}}\left(p\right)$ or $Q_T\left(p\right)=\left[\xi {\left(-\ln \left(1-p\right)\right)}^{{\displaystyle \frac{1}{c}}}\right]$

Step 2: Substitute this quantile in the CDF of the link distribution.

$F_Y\left(y\right)=F_Y\left(y\right)={\displaystyle \frac{{\left(\frac{\xi {\left(-\ln \left(1-p\right)\right)}^{\frac{1}{c}}}{\psi }\right)}^B}{1+{\left(\frac{{\xi \left(-\ln \left(1-p\right)\right)}^{\frac{1}{c}}}{\psi }\right)}^B}}={\displaystyle \frac{\frac{{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{{\psi }^B}}{\frac{{\psi }^B}{{\psi }^B}+\frac{{{\xi }^B\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{{\psi }^B}}}={\displaystyle \frac{\left({\psi }^B\right)\frac{{{\xi }^B\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{\left({\psi }^B\right)}}{\left({\psi }^B\right)+{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}}={\displaystyle \frac{{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{\left({\psi }^B\right)+{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}}$

$F_X\left(x\right)=F_Y\left(p\right)=F_Y\left(y\right)={\displaystyle \frac{{{\xi }^B\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{\left({\psi }^B\right)+{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}}$

Step 3: Substitute the result and call it, Z; ${\displaystyle \frac{{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{\left({\psi }^B\right)+{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}}$ into the quantile function of the baseline distribution MBUR distribution. This will give the same result as taking the inverse of the $G_{X_g}$ .

Another approach that can be deployed to obtain the quantile function of $G_{X_g}$ is by taking the inverse of the $G_{X_g}=p=1-exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$, solving for x yields the following equation, $3{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^2-2{\left(x^{{\displaystyle \frac{1}{b^2}}}\right)}^3={\displaystyle \frac{{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}{\left({\psi }^B\right)+{\xi }^B{\left(-\ln \left(1-p\right)\right)}^{\frac{B}{c}}}}$ , the left hand side is the CDF of the baseline distribution MBUR and the right hand side is a constant, which is the z, and z is $0<z<1$. Solving the third degree polynomial for its root yields the $x_g$ variable or the quantile function for the new distribution which is

$Q_X\left(p\right)=Q_{X_g}\left(z\right)={\left\{-.5\left(cos\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]-\sqrt{3}sin\left[{\displaystyle \frac{{cos}^{-1}\left(1-2z\right)}{3}}\right]\right)+.5\right\}}^{b^2}$

$0<z<1,\left(x_g;b,c,\xi ,B,\psi \right)$

The hazard function is defined as $h\left(X_g\right)={\displaystyle \frac{g_{X_g}}{1-G_{X_g}}}=$

$={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-\left(1+{\displaystyle \frac{c}{B}}\right)}{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{c}{B}}-1}$

Theorem 9: the r^th non-central moments are given by the following formula depending on the scale and shape parameters of the link distribution function as shown in the proof

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{-\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}{\xi }^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}$

$\mathsf{\Gamma }\left(1+{\displaystyle \frac{\left(rB{b}^4+2mBs+2b^{2}Bs\right)}{c\left(2m+2b^2\right)}}\right)$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{uB}{\xi }^{-uB}\mathsf{\Gamma }\left(1-{\displaystyle \frac{uB}{c}}\right),0<{\displaystyle \frac{uB}{c}}<1$

Proof: Starting with the t variable

$t=\psi {\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}$ then isolating the x as shown in the following steps

${\displaystyle \frac{t}{\psi }}={\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}$ this gives ${\left({\displaystyle \frac{t}{\psi }}\right)}^B={\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}$

$\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right){\left({\displaystyle \frac{t}{\psi }}\right)}^B=3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)=\left(1-3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right){\left({\displaystyle \frac{t}{\psi }}\right)}^B$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\left({\displaystyle \frac{t}{\psi }}\right)}^B-3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right){\left({\displaystyle \frac{t}{\psi }}\right)}^B$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)+3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right){\left({\displaystyle \frac{t}{\psi }}\right)}^B={\left({\displaystyle \frac{t}{\psi }}\right)}^B$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\left\{1+{\left({\displaystyle \frac{t}{\psi }}\right)}^B\right\}={\left({\displaystyle \frac{t}{\psi }}\right)}^B$

$3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}$

$x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)={\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}$

${\left\{x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}={\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}$

$x{\left\{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right\}}^{{\displaystyle \frac{b^2}{2}}}={\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}$

$x{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^m={\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}$

${\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{b^2}{2}\right)}{m}}\right){\left({\displaystyle \frac{2}{3}}\right)}^m{x}^{{\displaystyle \frac{m}{b^2}}+1}={\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}$

$W_m.x^{{\displaystyle \frac{m}{b^2}}+1}={\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}$

$x^{{\displaystyle \frac{m+b^2}{b^2}}}={\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}{\left(W_m\right)}^{-1}$

${\left\{x^{\left({\displaystyle \frac{m+b^2}{b^2}}\right)}\right\}}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}={\left({\left[{\displaystyle \frac{{\left(\frac{t}{\psi }\right)}^B}{3\left\{1+{\left(\frac{t}{\psi }\right)}^B\right\}}}\right]}^{{\displaystyle \frac{b^2}{2}}}\right)}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}{\left[{\left(W_m\right)}^{-1}\right]}^{\left({\displaystyle \frac{b^2}{m+b^2}}\right)}$

$x=3^{\left({\displaystyle \frac{-b^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-b^2}{m+b^2}}\right)}{\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{B{b}^4}{2m+2b^2}}\right)}{\left\{1+{\left({\displaystyle \frac{t}{\psi }}\right)}^B\right\}}^{\left({\displaystyle \frac{-b^4}{2m+2b^2}}\right)}$

Take the r^th power of both sides

$x^r=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{rB{b}^4}{2m+2b^2}}\right)}{\left\{1+{\left({\displaystyle \frac{t}{\psi }}\right)}^B\right\}}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}$

Use the generalized binomial theory to expand the ${\left\{1+{\left({\displaystyle \frac{t}{\psi }}\right)}^B\right\}}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)},$ there are two cases

First case : if $0<{\left({\displaystyle \frac{t}{\psi }}\right)}^B<1$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{rB{b}^4}{2m+2b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\left[{\left({\displaystyle \frac{t}{\psi }}\right)}^B\right]}^s$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{rB{b}^4}{2m+2b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\left[{\left({\displaystyle \frac{t}{\psi }}\right)}^B\right]}^s$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{-\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}{\left(t\right)}^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}$

Take expectation of both sides

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{-\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}E\left[{\left(t\right)}^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}\right]$

$E\left[{\left(t\right)}^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}\right]={\xi }^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}\mathsf{\Gamma }\left(1+{\displaystyle \frac{\left(rB{b}^4+2mBs+2b^{2}Bs\right)}{c\left(2m+2b^2\right)}}\right)$ is the raw moment of the Weibull variable (the t variable)

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{s=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{s}}\right){\psi }^{-\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}{\xi }^{\left({\displaystyle \frac{rB{b}^4+2mBs+2b^{2}Bs}{2m+2b^2}}\right)}$

$\mathsf{\Gamma }\left(1+{\displaystyle \frac{\left(rB{b}^4+2mBs+2b^{2}Bs\right)}{c\left(2m+2b^2\right)}}\right)$

Second case :${\left({\displaystyle \frac{t}{\psi }}\right)}^B>1$ , factor out the ${\left({\displaystyle \frac{t}{\psi }}\right)}^B$ so that

${\left\{{\left({\displaystyle \frac{t}{\psi }}\right)}^B+1\right\}}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}={\left[{\left({\displaystyle \frac{t}{\psi }}\right)}^B\left(1+{\left({\displaystyle \frac{t}{\psi }}\right)}^{-B}\right)\right]}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}={\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{-rB{b}^4}{2m+2b^2}}\right)}{\left(1+{\left({\displaystyle \frac{t}{\psi }}\right)}^{-B}\right)}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}$ then use the generalized binomial theory, then take the expectation

$x^r=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{rB{b}^4}{2m+2b^2}}\right)}{\left({\displaystyle \frac{t}{\psi }}\right)}^{\left({\displaystyle \frac{-rB{b}^4}{2m+2b^2}}\right)}{\left(1+{\left({\displaystyle \frac{t}{\psi }}\right)}^{-B}\right)}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\left\{1+{\left({\displaystyle \frac{t}{\psi }}\right)}^{-B}\right\}}^{\left({\displaystyle \frac{-r{b}^4}{2m+2b^2}}\right)}$

$x^r=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{u}}\right){\left({\displaystyle \frac{t}{\psi }}\right)}^{-uB}$

Take expectation of both sides

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}{\displaystyle \sum _{u=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{uB}E\left[{\left(t\right)}^{-uB}\right]$

$E\left[{\left(t\right)}^{-uB}\right]={\xi }^{-uB}\mathsf{\Gamma }\left(1-{\displaystyle \frac{uB}{c}}\right)$

$E\left(x^r\right)=3^{\left({\displaystyle \frac{-{rb}^4}{2m+2b^2}}\right)}{W_m}^{\left({\displaystyle \frac{-r{b}^2}{m+b^2}}\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-r{b}^4}{2m+2b^2}\right)}{u}}\right){\psi }^{uB}{\xi }^{-uB}\mathsf{\Gamma }\left(1-{\displaystyle \frac{uB}{c}}\right),0<{\displaystyle \frac{uB}{c}}<1$

Theorem 10: linear series expansion of this new pdf ${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}$, is applied to the exponent term and each of the power terms, using the generalized binomial expansion and Pochhammer falling factorial

Proof:

The PDF is${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\xi },\mathit{B},\mathit{\psi }\right)=$

${\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}{\left[1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-\left(1+{\displaystyle \frac{c}{B}}\right)}{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{c}{B}}-1}exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$

The power term ${\left[1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-\left(1+{\displaystyle \frac{c}{B}}\right)}$ is expanded as

${\left[1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right]}^{-\left(1+{\displaystyle \frac{c}{B}}\right)}={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^{n+m}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(1+\frac{c}{B}\right)}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{m}}\right){\left(3\right)}^{n-m}{\left(2\right)}^m{x}^{\left({\displaystyle \frac{2n+m}{b^2}}\right)}$

The power term ${\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{\left({\displaystyle \frac{c}{B}}-1\right)}$ is expanded as

${\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{\left({\displaystyle \frac{c}{B}}-1\right)}={\displaystyle \sum _{p=0}^{\infty }}{\left(-1\right)}^p\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{B}-1\right)}{P}}\right){\left(2\right)}^P{3}^{\left({\displaystyle \frac{c}{B}}-P-1\right)}{x}^{\left({\displaystyle \frac{pB+2c-2B}{b^{2}B}}\right)}$

The exponent term can be expanded as $exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}$

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{l}^r;calll=\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\};$

$exp\left\{-{\left(\left[{\displaystyle \frac{\psi }{\xi }}\right]{\left[{\displaystyle \frac{3x^{\frac{2}{b^2}}-2x^{\frac{3}{b^2}}}{1-3x^{\frac{2}{b^2}}+2x^{\frac{3}{b^2}}}}\right]}^{{\displaystyle \frac{1}{B}}}\right)}^c\right\}=e^{-l}$

$e^{-l}={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left[{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{c}{B}}}{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-c}{B}}}\right]}^r$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}-2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}+2x^{{\displaystyle \frac{3}{b^2}}}\right)}^{{\displaystyle \frac{-cr}{B}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right)}^{{\displaystyle \frac{-cr}{B}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^t{\left(1-3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right)}^{{\displaystyle \frac{-cr}{B}}}$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right){\left({\displaystyle \frac{2}{3}}\right)}^t\left(x^{{\displaystyle \frac{t}{b^2}}}\right){\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{B}\right)}{s}}\right){\left[3x^{{\displaystyle \frac{2}{b^2}}}\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)\right]}^s$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right){\left({\displaystyle \frac{2}{3}}\right)}^t\left(x^{{\displaystyle \frac{t}{b^2}}}\right){\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{B}\right)}{s}}\right)3^s{x}^{{\displaystyle \frac{2s}{b^2}}}{\left(1-{\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^s$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^r}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}{\left(3x^{{\displaystyle \frac{2}{b^2}}}\right)}^{{\displaystyle \frac{cr}{B}}}{\displaystyle \sum _{t=0}^{\infty }}{\left(-1\right)}^t\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right){\left({\displaystyle \frac{2}{3}}\right)}^t\left(x^{{\displaystyle \frac{t}{b^2}}}\right){\displaystyle \sum _{s=0}^{\infty }}{\left(-1\right)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{B}\right)}{s}}\right)3^s{x}^{{\displaystyle \frac{2s}{b^2}}}{\displaystyle \sum _{u=0}^{\infty }}{\left(-1\right)}^u\left({\displaystyle \genfrac{}{}{0pt}{}{s}{u}}\right){\left({\displaystyle \frac{2}{3}}{x}^{{\displaystyle \frac{1}{b^2}}}\right)}^u$

$={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{r+t+s+u}}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{B}\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{u}}\right)3^{\left({\displaystyle \frac{cr}{B}}-t+s-u\right)}{2}^{\left(t+u\right)}{x}^{\left({\displaystyle \frac{2cr}{b^{2}B}}+{\displaystyle \frac{t}{b^2}}+{\displaystyle \frac{2s+u}{b^2}}\right)}$

Assemble all the terms together so the PDF can be represented as ${\mathit{g}}_{{\mathit{X}}_{\mathit{g}}}\left({\mathit{x}}_{\mathit{g}};\mathit{b},\mathit{c},\mathit{\xi },\mathit{B},\mathit{\psi }\right)={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}$

${\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^{n+m}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(1+\frac{c}{B}\right)}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{m}}\right){\left(3\right)}^{n-m}{\left(2\right)}^m{x}^{\left({\displaystyle \frac{2n+m}{b^2}}\right)}$

${\displaystyle \sum _{p=0}^{\infty }}{\left(-1\right)}^p\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{B}-1\right)}{P}}\right){\left(2\right)}^P{3}^{\left({\displaystyle \frac{c}{B}}-P-1\right)}{x}^{\left({\displaystyle \frac{pB+2c-2B}{b^{2}B}}\right)}$

${\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{r+t+s+u}}{r!}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{B}\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{u}}\right)3^{\left({\displaystyle \frac{cr}{B}}-t+s-u\right)}{2}^{\left(t+u\right)}{x}^{\left({\displaystyle \frac{2cr}{b^{2}B}}+{\displaystyle \frac{t}{b^2}}+{\displaystyle \frac{2s+u}{b^2}}\right)}$

$={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \sum _{t=0}^{\infty }}{\displaystyle \sum _{s=0}^{\infty }}{\displaystyle \sum _{u=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^{n+m+p+r+t+s+u}}{r!}}\left({\displaystyle \genfrac{}{}{0pt}{}{-\left(1+\frac{c}{B}\right)}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{c}{B}-1\right)}{P}}\right){\left({\displaystyle \frac{\psi }{\xi }}\right)}^{cr}\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{cr}{B}\right)}{t}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(\frac{-cr}{B}\right)}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{u}}\right)$

$3^{\left({\displaystyle \frac{cr}{B}}-t+s-u+n-m+{\displaystyle \frac{c}{B}}-P-1\right)}{2}^{\left(t+u+m+p\right)}{x}^{\left({\displaystyle \frac{2cr}{b^{2}B}}+{\displaystyle \frac{t}{b^2}}+{\displaystyle \frac{2s+u}{b^2}}+{\displaystyle \frac{2n+m}{b^2}}+{\displaystyle \frac{pB+2c-2B}{b^{2}B}}\right)}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}$

$g_{X_g}\left(x_g;b,c,\xi ,B,\psi \right)={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,p,r,t,s,u}{x}^{\left({\displaystyle \frac{2cr}{b^{2}B}}+{\displaystyle \frac{t}{b^2}}+{\displaystyle \frac{2s+u}{b^2}}+{\displaystyle \frac{2n+m}{b^2}}+{\displaystyle \frac{pB+2c-2B}{b^{2}B}}\right)}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{{\displaystyle \frac{2}{b^2}}-1}$

$g_{X_g}\left(x_g;b,c,\xi ,B,\psi \right)={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,p,r,t,s,u}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left({\displaystyle \frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}}\right)-1}$

Integrate this series expansion by integrating the x from 0 to 1 because the support of the distribution is the support of the MBUR so the integral will pass to the x

${\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,p,r,t,s,u}{\int }_0^1\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left({\displaystyle \frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}}\right)-1}dx$

${\int }_0^1\left(x^{\left({\displaystyle \frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}}\right)-1}-x^{\left({\displaystyle \frac{B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)}{b^{2}B}}\right)-1}\right)dx$

$={\left.{\displaystyle \frac{x^{\left[\frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}\right]}}{\left[\frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}\right]}}\right|}_0^1-{\left.{\displaystyle \frac{x^{\left[\frac{B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)}{b^{2}B}\right]}}{\left[\frac{B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)}{b^{2}B}\right]}}\right|}_0^1$

$=b^{2}B\left\{{\displaystyle \frac{1}{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}}-{\displaystyle \frac{1}{B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)}}\right\}$

$=b^{2}B\left\{{\displaystyle \frac{1}{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}}-{\displaystyle \frac{1}{B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)}}\right\}$

$=\left\{{\displaystyle \frac{\frac{6c}{b^{2}B}{\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,p,r,t,s,u}{b}^2{B}^2}{\left[B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)\right]\left[B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)\right]}}\right\}$

$=\left\{{\displaystyle \frac{6cB{\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,p,r,t,s,u}}{\left[B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)\right]\left[B\left(t+2s+u+2n+m+p+1\right)+2c\left(r+1\right)\right]}}\right\}$

Theorem 11: the r^th non-central moments are given by the following formula:

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\displaystyle \frac{6cB{\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,p,r,t,s,u}}{\left[B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)\right]\left[B\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)\right]}}$Proof:

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\int }_0^1\mathit{g}\left(\mathit{x}\right){\mathit{x}}^{\mathit{h}}\mathit{d}\mathit{x}={\displaystyle \frac{6c}{b^{2}B}}{\left({\displaystyle \frac{\psi }{\xi }}\right)}^c{W}_{n,m,p,r,t,s,u}{\int }_0^1{\mathit{x}}^{\mathit{h}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left({\displaystyle \frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}}\right)-1}\mathit{d}\mathit{x}.$

${\int }_0^1{\mathit{x}}^{\mathit{h}}\left(1-x^{{\displaystyle \frac{1}{b^2}}}\right)x^{\left({\displaystyle \frac{B\left(t+2s+u+2n+m+p\right)+2c\left(r+1\right)}{b^{2}B}}\right)-1}dx$

$={\left.{\displaystyle \frac{x^{\left(\frac{B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)}{b^{2}B}\right)}}{\left[\frac{B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)}{b^{2}B}\right]}}\right|}_0^1$

$-{\left.{\displaystyle \frac{x^{\left(\frac{B\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)}{b^{2}B}\right)}}{\left[\frac{B\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)}{b^{2}B}\right]}}\right|}_0^1$

$=b^{2}B\left\{{\displaystyle \frac{1}{B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)}}-{\displaystyle \frac{1}{aB\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)}}\right\}$

$=b^{2}B\left\{{\displaystyle \frac{\left[B\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)\right]-\left[B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)\right]}{\left[B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)\right]\left[B\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)\right]}}\right\}$

$\mathit{E}\left({\mathit{x}}^{\mathit{h}}\right)={\displaystyle \frac{6cB{\left(\frac{\psi }{\xi }\right)}^c{W}_{n,m,p,r,t,s,u}}{\left[B\left(t+2s+u+2n+m+p+h{b}^2\right)+2c\left(r+1\right)\right]\left[B\left(t+2s+u+2n+m+p+h{b}^2+1\right)+2c\left(r+1\right)\right]}}$

In the supplementary material, the author depicts different figures illustrating the behavior of the distribution as regards the PDF, CDF and hazard functions. The PDF can exhibit, J shape and reversed J shaped, increasing and decreasing, M shape, left skewed and right skewed shape and other different shapes. The hazard function shapes show mainly increasing and J shaped appearance.