<p>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.</p>
<p>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 <span class="math-container">$f(x)$</span>, where <span class="math-container">$x$</span> is the amount of income. In other words, <span class="math-container">$f(x)dx$</span> proportion of population has the income <span class="math-container">$x$</span>.</p>
<p>Then, we have the CDF(Cumulative Distribution Function) <span class="math-container">$F(x)$</span>. The is the cumulative population.</p>
<p>Lorentz curve essentially should dependent on <span class="math-container">$F$</span>, the cumulative population. However, it's easier to first describe it with the independent variable being income <span class="math-container">$x$</span>. Then <span class="math-container">$$L(x) = \frac{\int_{-\infty}^x f(y)y dy}{\int_{-\infty}^\infty f(y)y dy} =  \frac{\int_{-\infty}^x f(y)y dy}{E(X) \text{ or } \mu}.  $$</span></p>
<p>With these functions in hand, we can now write down mathematically the 2 definitions of gini coefficient,
<span class="math-container">$$G_1 = 1 - 2 \int_0^1  L(F) dF$$</span>
<span class="math-container">$$G_2 = \frac 1 {2\mu} \int \int |y-x| f(x)f(y) dx dy$$</span>.</p>
<p>I personally prefer to derive from <span class="math-container">$G_1$</span> to <span class="math-container">$G_2$</span>, but the logic works both way.</p>
<p>To work out <span class="math-container">$G_1$</span>, we need to evaluate <span class="math-container">$\int_0^1  L(F) dF$</span>.
<span class="math-container">$$\begin{eqnarray} 
\\
&amp;&amp;\int_0^1  L(F) dF  \\
&amp; =&amp; \int_{-\infty}^{\infty}  L(x) \frac{dF}{dx}
dx \\
&amp; = &amp; \frac{1}{\mu}\int_{-\infty}^{\infty} \int_{-\infty}^x f(y)y dy \, \frac{dF}{dx}
dx 
 \end{eqnarray}$$</span>
Note that the integrant contains an integral, so we can apply integration by parts.
<span class="math-container">$$
\begin{eqnarray} 
\\&amp;&amp;\int_0^1  L(F) dF \\
&amp;=&amp; 1 - 1/\mu \int_{-\infty}^{\infty} f(x)F(x)xdx  \\
\end{eqnarray}
$$</span>
<span class="math-container">$G_1$</span> thus become <span class="math-container">$$\frac{1}{\mu}\int_{-\infty}^{\infty} f(x)F(x)xdx - \int_0^1  L(F) dF \\ =\frac{1}{\mu} \int_{-\infty}^{\infty} f(x)F(x)xdx - \frac{1}{\mu}\int_{-\infty}^{\infty} \int_{-\infty}^x f(y)y dy \, \frac{dF}{dx} 
dx \\
=  \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^x f(x)x f(y) dydx - \frac{1}{\mu}\int_{-\infty}^{\infty} \int_{-\infty}^x f(x) f(y)y dy  
dx \\
= \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^x (x-y)f(x)f(y) dydx 
 $$</span>
By symmetry, <span class="math-container">$\displaystyle \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^x (x-y)f(x)f(y) dydx =\frac 1 2 \frac{1}{\mu} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |x-y|f(x)f(y) dydx$</span>, which is just <span class="math-container">$G_2$</span>.</p>

CC BY-SA 4.0
2023-09-11T02:16:59.393Z
2023-09-11T02:16:59.393Z
Abl grp
<p>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(<a href="https://www.dannybugingo.com/en/math/gini/ginius.pdf" rel="nofollow noreferrer">https://www.dannybugingo.com/en/math/gini/ginius.pdf</a>) but this proof seems to be wrong in how the integrals for the expected absolute difference is calculated.</p>
<p>Does anyone have any mathematical derivation for this?</p>
<p>Edit: Basically you would start with this...
<a href="https://i.sstatic.net/mwV7C.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mwV7C.png" alt="enter image description here" /></a>
...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:</p>
<p><a href="https://i.sstatic.net/1yENi.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1yENi.png" alt="enter image description here" /></a></p>
<p>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.</p>

CC BY-SA 4.0
2020-08-11T20:41:28.423Z
Does [this](https://economics.meta.stackexchange.com/questions/1417/mathjax-basic-tutorial-and-quick-reference) help with your Latex woes?
Giskard
CC BY-SA 4.0
2020-08-11T21:11:30.800Z
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 :-)
E.K
CC BY-SA 4.0
2020-08-11T22:15:45.367Z
@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
1muflon1
CC BY-SA 4.0
2020-08-12T03:46:33.183Z
@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.
Giskard
CC BY-SA 4.0
2020-08-12T03:47:12.820Z
@E.K How is $g(x_1,x_2)$ the absolute value function? That is a single variable function. What is $f_1$ and $f_2$?
Giskard
CC BY-SA 4.0
2020-08-12T07:56:44.833Z
Well the formula is just a general formula for the expected value of a function of two stochastic variables, hence g(x1,x2) is just an arbitrary bivariate function. f(x1) and f(x2) are density functions. In my case g(x1, x2) is the absolute value of the difference (x1 - x2) where x1 and x2 have the same distribution and hence the same prob. density function.
E.K
<macroeconomics><statistics>
How is the GINI coefficient half the relative mean absolute difference?
E.K
# How is the GINI coefficient half the relative mean absolute difference?

(Asked by: username_0 on 2020-08-11)

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](https://i.sstatic.net/mwV7C.png)](https://i.sstatic.net/mwV7C.png)
...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](https://i.sstatic.net/1yENi.png)](https://i.sstatic.net/1yENi.png)

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.

**Tags:** macroeconomics, statistics
> **Comments:**
> **username_1**: Does [this](https://economics.meta.stackexchange.com/questions/1417/mathjax-basic-tutorial-and-quick-reference) help with your Latex woes?  
> **username_0**: 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 :-)  
> **username_2**: @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  
> **username_1**: @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.  
> **username_1**: @E.K How is $g(x_1,x_2)$ the absolute value function? That is a single variable function. What is $f_1$ and $f_2$?  
> **username_0**: Well the formula is just a general formula for the expected value of a function of two stochastic variables, hence g(x1,x2) is just an arbitrary bivariate function. f(x1) and f(x2) are density functions. In my case g(x1, x2) is the absolute value of the difference (x1 - x2) where x1 and x2 have the same distribution and hence the same prob. density function.  

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## 1 answers

### Answer by username_3 on 2023-09-11. Score: 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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