<p>A Lorenz curve is a P-P plot, so the shape of the curve depends on the underlying probability distribution. Depending on what probability distribution you're using for your model (uniform, normal, exponential, etc.), you'll end up with a different functional form for the Lorenz curve.</p>
<p>Given an underlying cumulative distribution function <span class="math-container">$ F(x) $</span> with inverse (if it exists) <span class="math-container">$ x(F) $</span>, the Lorenz curve is given by</p>
<p><span class="math-container">$$ L\big(F(x)\big) = \frac{\int_0^F{x(G)}dG}{\int_0^1{x(G)}dG} $$</span></p>
<p>In practice, when you're looking at Lorenz curves and Gini coefficients, it's common to use a <a href="https://en.wikipedia.org/wiki/Pareto_distribution" rel="nofollow noreferrer">Pareto distribution</a> as the underlying probability distribution.</p>
<p>Here's a derivation for the functional form of the Lorenz curve for the Pareto distribution as an example of the process. Starting from the CDF of the Pareto distribution:</p>
<p><span class="math-container">$$ F(x) = 1 - \bigg(\frac{x_m}{x}\bigg)^\alpha , $$</span></p>
<p>we can invert it to</p>
<p><span class="math-container">$$ x(F) = x_m (1-F)^{-\frac{1}{\alpha}} . $$</span></p>
<p>Then, inserting the inverse of the CDF into the definition of the Lorenz curve above, we have</p>
<p><span class="math-container">$$ L(F) = \frac{\int_0^F{x_m (1-G)^{-\frac{1}{\alpha}}}dG}{\int_0^1{x_m (1-G)^{-\frac{1}{\alpha}}}dG} $$</span></p>
<p><span class="math-container">$$ = 
\frac{\bigg[-\frac{x_m}{1-\frac{1}{\alpha}}\big(1-G\big)^{1-\frac{1}{\alpha}}\bigg]^F_0}{\bigg[-\frac{x_m}{1-\frac{1}{\alpha}}\big(1-G\big)^{1-\frac{1}{\alpha}}\bigg]^1_0} $$</span></p>
<p><span class="math-container">$$ = 
\frac{\frac{x_m}{1-\frac{1}{\alpha}}\bigg[1-\big(1-F\big)^{1-\frac{1}{\alpha}}\bigg]}{\frac{x_m}{1-\frac{1}{\alpha}}} . $$</span></p>
<p>So, for the Pareto distribution with parameters <span class="math-container">$x_m$</span> and <span class="math-container">$\alpha$</span>, the corresponding Lorenz curve has the functional form</p>
<p><span class="math-container">$$ L(F) = 1 - (1 - F)^{1-\frac{1}{\alpha}} . $$</span></p>
<p>Now, if you want to find the functional form of the Lorenz curve given data on some measure, like wealth or income, you would first have to fit the distribution of your data to the Pareto distribution first and estimate its two parameters, <span class="math-container">$x_m$</span> and <span class="math-container">$\alpha$</span> (<a href="https://en.wikipedia.org/wiki/Pareto_distribution#Estimation_of_parameters" rel="nofollow noreferrer">see here</a>).</p>
<p>Once you have your estimate of <span class="math-container">$\alpha$</span> (<span class="math-container">$\hat{\alpha}$</span>), you can simply plug that into the Lorenz curve for the Pareto distribution we found above:</p>
<p><span class="math-container">$$ L(F) = 1 - (1 - F)^{1-1/\hat{\alpha}} . $$</span></p>

CC BY-SA 4.0
2020-12-23T23:37:05.333Z
Thank you for mentioning this. Would you perhaps be able to provide a more explicit example to work from? It seems you are suggesting one could interpret the curves as either Bell curves or Beta functions and so on.
PhiEarl
CC BY-SA 4.0
2020-12-24T02:06:21.613Z
I added a bit more of context on the relationship between the probability distribution of your measure and the Lorenz curve. Different probability densities will have different functional forms for their Lorenz curves. The example I provide assumes that the distribution of your underlying measure is a Pareto distribution (so wealth or income or whatever has a Pareto distribution). Once you make that assumption, you can use the process I lay out above to get to the functional form of the Lorenz curve that works best for your purposes.
Amaan M
CC BY-SA 4.0
2020-12-23T05:40:24.223Z
2020-12-24T02:02:00.710Z
2020-12-24T02:02:00.710Z
Amaan M
<p>Is there a conventional curve used to interpolate the Lorenz curves that model the distribution for calculating the Gini coefficient? Is it polynomial? But why? Which degree?</p>

<mathematical-economics><statistics><gini>
Interpolation of Gini Index?
PhiEarl
# Interpolation of Gini Index?

(Asked by: username_0 on 2020-12-16)

Is there a conventional curve used to interpolate the Lorenz curves that model the distribution for calculating the Gini coefficient? Is it polynomial? But why? Which degree?

**Tags:** mathematical-economics, statistics, gini

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## 1 answers

### Answer by username_1 on 2020-12-23. Score: 2

A Lorenz curve is a P-P plot, so the shape of the curve depends on the underlying probability distribution. Depending on what probability distribution you're using for your model (uniform, normal, exponential, etc.), you'll end up with a different functional form for the Lorenz curve.

Given an underlying cumulative distribution function $ F(x) $ with inverse (if it exists) $ x(F) $, the Lorenz curve is given by

$$ L\big(F(x)\big) = \frac{\int\_0^F{x(G)}dG}{\int\_0^1{x(G)}dG} $$

In practice, when you're looking at Lorenz curves and Gini coefficients, it's common to use a [Pareto distribution](https://en.wikipedia.org/wiki/Pareto_distribution) as the underlying probability distribution.

Here's a derivation for the functional form of the Lorenz curve for the Pareto distribution as an example of the process. Starting from the CDF of the Pareto distribution:

$$ F(x) = 1 - \bigg(\frac{x\_m}{x}\bigg)^\alpha , $$

we can invert it to

$$ x(F) = x\_m (1-F)^{-\frac{1}{\alpha}} . $$

Then, inserting the inverse of the CDF into the definition of the Lorenz curve above, we have

$$ L(F) = \frac{\int\_0^F{x\_m (1-G)^{-\frac{1}{\alpha}}}dG}{\int\_0^1{x\_m (1-G)^{-\frac{1}{\alpha}}}dG} $$

$$ =
\frac{\bigg[-\frac{x\_m}{1-\frac{1}{\alpha}}\big(1-G\big)^{1-\frac{1}{\alpha}}\bigg]^F\_0}{\bigg[-\frac{x\_m}{1-\frac{1}{\alpha}}\big(1-G\big)^{1-\frac{1}{\alpha}}\bigg]^1\_0} $$

$$ =
\frac{\frac{x\_m}{1-\frac{1}{\alpha}}\bigg[1-\big(1-F\big)^{1-\frac{1}{\alpha}}\bigg]}{\frac{x\_m}{1-\frac{1}{\alpha}}} . $$

So, for the Pareto distribution with parameters $x\_m$ and $\alpha$, the corresponding Lorenz curve has the functional form

$$ L(F) = 1 - (1 - F)^{1-\frac{1}{\alpha}} . $$

Now, if you want to find the functional form of the Lorenz curve given data on some measure, like wealth or income, you would first have to fit the distribution of your data to the Pareto distribution first and estimate its two parameters, $x\_m$ and $\alpha$ ([see here](https://en.wikipedia.org/wiki/Pareto_distribution#Estimation_of_parameters)).

Once you have your estimate of $\alpha$ ($\hat{\alpha}$), you can simply plug that into the Lorenz curve for the Pareto distribution we found above:

$$ L(F) = 1 - (1 - F)^{1-1/\hat{\alpha}} . $$

> **Comments:**
> **username_0**: Thank you for mentioning this. Would you perhaps be able to provide a more explicit example to work from? It seems you are suggesting one could interpret the curves as either Bell curves or Beta functions and so on.  
> **username_1**: I added a bit more of context on the relationship between the probability distribution of your measure and the Lorenz curve. Different probability densities will have different functional forms for their Lorenz curves. The example I provide assumes that the distribution of your underlying measure is a Pareto distribution (so wealth or income or whatever has a Pareto distribution). Once you make that assumption, you can use the process I lay out above to get to the functional form of the Lorenz curve that works best for your purposes.  

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