I've tried to derive a proof myself and have looked through a lot of sources but can't seem to find an understandable mathematical explanation/proof for why this is the case. I found one at(https://www.dannybugingo.com/en/math/gini/ginius.pdf) but this proof seems to be wrong in how the integrals for the expected absolute difference is calculated.

Does anyone have any mathematical derivation for this?

Edit: Basically you would start with this... enter image description here ...where g() is the absolute value function and try to prove that this divided by the expected value of the variable X is equivalent to the GINI index as defined by:

enter image description here

PS: I really wish I knew how to write LATEX so I could explain the mathematics I've tried but I don't have time to do so now. Hopefully someone still understands the question.

  • 1
    $\begingroup$ Does this help with your Latex woes? $\endgroup$
    – Giskard
    Aug 11, 2020 at 20:41
  • $\begingroup$ Thx. I couldn't make it work quickly now so since I'm in a bit of a stress I added some images that hopefully conveys everything well enough :-) $\endgroup$
    – E.K
    Aug 11, 2020 at 21:11
  • $\begingroup$ @E.K you should typeset those images following the guide in the link giskard provided. That is this site policy especially for those who need help with homework question. Typesetting equation makes it searchable whereas having picture does not which then hurts other people who might have similar question $\endgroup$
    – 1muflon1
    Aug 11, 2020 at 22:15
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    $\begingroup$ @1muflon1 I used to be of the same position, but it was pointed out to me that it is enough if the surrounding text is typed and searchable. No one actually googles integrals, people google text. $\endgroup$
    – Giskard
    Aug 12, 2020 at 3:46
  • $\begingroup$ @E.K How is $g(x_1,x_2)$ the absolute value function? That is a single variable function. What is $f_1$ and $f_2$? $\endgroup$
    – Giskard
    Aug 12, 2020 at 3:47

1 Answer 1


The proof provided in my opinion is right, maybe some of the definition is not clear and fully explained. However, the idea of the proof is exactly right.

Let's start defining Lorentz curve from probability density function (PDF) representing the corresponding proportion of population for each income. We denote this PDF. as $f(x)$, where $x$ is the amount of income. In other words, $f(x)dx$ proportion of population has the income $x$.

Then, we have the CDF(Cumulative Distribution Function) $F(x)$. The is the cumulative population.

Lorentz curve essentially should dependent on $F$, the cumulative population. However, it's easier to first describe it with the independent variable being income $x$. Then $$L(x) = \frac{\int_{-\infty}^x f(y)y dy}{\int_{-\infty}^\infty f(y)y dy} = \frac{\int_{-\infty}^x f(y)y dy}{E(X) \text{ or } \mu}. $$

With these functions in hand, we can now write down mathematically the 2 definitions of gini coefficient, $$G_1 = 1 - 2 \int_0^1 L(F) dF$$ $$G_2 = \frac 1 {2\mu} \int \int |y-x| f(x)f(y) dx dy$$.

I personally prefer to derive from $G_1$ to $G_2$, but the logic works both way.

To work out $G_1$, we need to evaluate $\int_0^1 L(F) dF$. $$\begin{eqnarray} \\ &&\int_0^1 L(F) dF \\ & =& \int_{-\infty}^{\infty} L(x) \frac{dF}{dx} dx \\ & = & \frac{1}{\mu}\int_{-\infty}^{\infty} \int_{-\infty}^x f(y)y dy \, \frac{dF}{dx} dx \end{eqnarray}$$ Note that the integrant contains an integral, so we can apply integration by parts. $$ \begin{eqnarray} \\&&\int_0^1 L(F) dF \\ &=& 1 - 1/\mu \int_{-\infty}^{\infty} f(x)F(x)xdx \\ \end{eqnarray} $$ $G_1$ thus become $$\frac{1}{\mu}\int_{-\infty}^{\infty} f(x)F(x)xdx - \int_0^1 L(F) dF \\ =\frac{1}{\mu} \int_{-\infty}^{\infty} f(x)F(x)xdx - \frac{1}{\mu}\int_{-\infty}^{\infty} \int_{-\infty}^x f(y)y dy \, \frac{dF}{dx} dx \\ = \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^x f(x)x f(y) dydx - \frac{1}{\mu}\int_{-\infty}^{\infty} \int_{-\infty}^x f(x) f(y)y dy dx \\ = \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^x (x-y)f(x)f(y) dydx $$ By symmetry, $\displaystyle \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^x (x-y)f(x)f(y) dydx =\frac 1 2 \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |x-y|f(x)f(y) dydx$, which is just $G_2$.


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