Let $X$ be a non-negative random variable with positive finite expectation. The Lorenzcurve $L_X$ is defined by $$ L_X(u) = \frac{\int_0^u F_X^{-1}(y) dy}{E(X)}, \quad 0 \leq u \leq 1,$$ where $$ F_X^{-1}(y) = \begin{cases} \sup\{x : F_X(x) \leq y\}, \quad 0 \leq y < 1 \\ \sup\{x : F_X(x) \leq y\}, \quad y = 1 \end{cases} $$ is the right continuous inverse distribution function of the random variable X. The Gini-Index $G$ is defined by $G = 1 - 2\int_0^1 L_X(u) du$.
I've stepped across the following equation $$ G = \frac{E(X_{2:2}) - E( X_{1:2} )}{E(X_{2:2}) + E(X_{1:2})}, $$ where $X_{1:2} < X_{2:2}$ are the order statistics of a random sample of size 2 drawn from X.
Is this a true statement for any random variable X? If so, why?
In addition, I have calculated the Gini-Index in the case of $X \sim \exp(\lambda)$ for both expressions above and came to the same result, but I do fail to see any justification for the general case.
First, notice that $$ \max(x,y)+\min(x,y)=x+y, $$ and $$ \max(x,y)-\min(x,y)=|x-y|=(x+y)-2\min(x,y). $$ Then, if $X$ and $Y$ are i.i.d. with cdf $F$, \begin{align} &\frac{\mathsf{E}\max(X,Y)-\mathsf{E}\min(X,Y)}{\mathsf{E}\max(X,Y)+\mathsf{E}\min(X,Y)}=1-\frac{\mathsf{E}\min(X,Y)}{\mathsf{E}X}\\ &\qquad=1-\frac{1}{\mathsf{E}X}\int_0^\infty[1-F(x)]^2\,dx. \end{align} It remains to show that the RHS equals the first definition.