Weighted gamma Weibull distribution I want a formula of Gini's concentration ratio and the Lorenz concentration ratio  involving three parameters about gamma Weibull distribution. 
I have formula about gamma distribution that contains only one parameter. but I want it for three parameters. For the gamma distribution,the Gini concentration ratio is
$ G= (Γ(α+1/2))/(√π Γ(α+1)) $.
 A: The Gini coefficient of any distribution $F$ with mean $\alpha$ is
$$\frac{1}{\alpha}\int_{-\infty}^\infty F(x)\left(1-F(x)\right) dx.$$
The three-parameter Gamma family is the location-scale extension of the Gamma distribution.  That is, the location parameter $\mu$ and the scale parameter $\sigma\gt 0$ induce a new distribution
$$F_{\mu,\sigma}(z) = F\left(\frac{z-\mu}{\sigma}\right)$$
whose standard deviation has been multiplied by $\sigma$ and whose mean is $\sigma\alpha+\mu$.
Substitute $x = \frac{z-\mu}{\sigma}$ in the integral.  Noting that $dz = \sigma\, d\left(\frac{z-\mu}{\sigma}\right) = \sigma dx,$   it is immediate that the Gini coefficient of $F_{\mu,\sigma}$ is
$$\frac{1}{\sigma\alpha+\mu}\int_{-\infty}^\infty F_{\mu,\sigma}(z)\left(1-F_{\mu,\sigma}(z)\right) dz = \frac{\sigma \alpha}{\sigma\alpha+\mu}\left(\frac{1}{\alpha}\int_{-\infty}^\infty F(x)\left(1-F(x)\right) dx\right).$$
The parenthesized expression on the right hand side is the Gini coefficient of $F$.  This general result shows how the Gini coefficient responds to changes of location and scale.
For the Gamma distribution with shape parameter $\alpha$ the mean is $\alpha$ and its Gini coefficient is $1/(\alpha B(\alpha, 1/2))$ (equivalent to the expression in the question).  Therefore the Gini coefficient for the three-parameter version equals $$\frac{\sigma\alpha}{\sigma\alpha+\mu} \frac{1}{\alpha B(\alpha, 1/2)}$$
provided $\mu \ne -\alpha$.
