Relationship between the Gini coefficient and the variance

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2021-04-25 10:06 +0800.

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First, define the Lorenz curve: it is the curve that consists of all points $(u,v)$ such that the poorest $u$ portion of population in the country owns $v$ portion of the total wealth.

The Gini coefficient $G/\mu$ is defined as the area between the Lorenz curve and the line $u=v$ divided by the area enclosed by the three lines $u=v$, $v=0$, and $u=1$.

Now, suppose the wealth distribution in the country is $p(X)$, where $p\!\left(x\right)\mathrm dx$ is the portion of population that has wealth in the range $[x,x+\mathrm dx]$.

Then, the Lorenz curve is the graph of the function $g$ defined as $$g(F(x))=\frac1\mu\int_{-\infty}^xtp\!\left(t\right)\mathrm dt,$$ where $$F\!\left(x\right)\coloneqq\int_{-\infty}^xp\!\left(t\right)\mathrm dt$$ is the cumulative distribution function of $p(X)$, and $$\mu\coloneqq\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt$$ $$(1)$$ is the average wealth of the population, which is just $\mathrm E[\mathrm X]$ ($X$ is a random variable such that $X\sim p(X)$).

Then, the Lorenz curve is $$v=g(u)\coloneqq\frac1\mu\int_{-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt.$$

According to the definition of the Gini coefficient, $$\begin{align*} G&\coloneqq2\mu\int_0^1\left(u-g(u)\right)\mathrm du\\ &=\mu-2\mu\int_0^1g\!\left(u\right)\mathrm du\\ &=\mu-2\int_{u=0}^1\int_{t=-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt\,\mathrm du. \end{align*}$$ Interchange the order of integration, and we have $$\begin{align*} G&=\mu-2\int_{t=-\infty}^{+\infty}\int_{u=F(t)}^1tp\!\left(t\right)\mathrm dt\,\mathrm du\\ &=\mu-2\int_{-\infty}^{+\infty}\left(1-F(t)\right)tp\!\left(t\right)\mathrm dt. \end{align*}$$ Substitute Equation 1 into the above equation, and we have $$\begin{align*} G&=\int_{-\infty}^{+\infty}2tF\!\left(t\right)p\!\left(t\right)\mathrm dt-\mu\\ &=\int_{-\infty}^{+\infty}\left(2tF\!\left(t\right)-1\right)tp\!\left(t\right)\mathrm dt\\ &=\int_0^1\left(2u-1\right)F^{-1}\!\left(u\right)\mathrm du. \end{align*}$$ Now here is the neat part. Separate it into two parts, and write them in double integrals: $$\begin{align*} G&=\int_0^1uF^{-1}\!\left(u\right)\mathrm du-\int_0^1\left(1-u\right)F^{-1}\!\left(u\right)\mathrm du\\ &=\int_{u_2=0}^1\int_{u_1=0}^{u_2}F^{-1}\!\left(u_2\right)\mathrm du_1\,\mathrm du_2 -\int_{u_1=0}^1\int_{u_2=u_1}^1F^{-1}\!\left(u_1\right)\mathrm du_1\,\mathrm du_2. \end{align*}$$ Interchange the order of integration of the second term, and we have $$\begin{align*} G&=\int_{u_2=0}^1\int_{u_1=0}^{u_2}\left(F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right)\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{u_2=0}^1\int_{u_1=0}^1\left|F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right|\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\left|x_2-x_1\right|p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left|X_2-X_1\right|\right], \end{align*}$$ where $X_1$ and $X_2$ are two independent random variables with $p$ being their respective distribution functions: $\left(X_1,X_2\right)\sim p\!\left(X_1\right)p\!\left(X_2\right)$.

By this result, we can easily see how the Gini coefficient represents the statistical dispersion.

We can apply similar tricks to the variance $\sigma_X^2$. $$\begin{align*} \sigma_X^2&=\mathrm E\!\left[X^2\right]-\mathrm E\!\left[X\right]^2\\ &=\int_{-\infty}^{+\infty}t^2p\!\left(t\right)\mathrm dt -\left(\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt\right)^2\\ &=\int_0^1F^{-1}\!\left(u\right)^2\,\mathrm du -\left(\int_0^1F^{-1}\!\left(u\right)\mathrm du\right)^2. \end{align*}$$ Separate the first into two halves, and write the altogether three terms in double integrals: $$\begin{align*} \sigma_X^2&=\frac12\int_0^1F^{-1}\!\left(u_2\right)^2\,\mathrm du_2\int_0^1\mathrm du_1\\ &\phantom{=~}{}-\int_0^1F^{-1}\!\left(u_1\right)\mathrm du_1\int_0^1F^{-1}\!\left(u_2\right)\mathrm du_2\\ &\phantom{=~}{}+\frac12\int_0^1F^{-1}\!\left(u_1\right)^2\,\mathrm du_1\int_0^1\mathrm du_2\\ &=\frac12\int_0^1\int_0^1 \left(F^{-1}\!\left(u_2\right)^2-2F^{-1}\!\left(u_1\right)F^{-1}\!\left(u_2\right)+F^{-1}\!\left(u_1\right)^2\right) \mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \left(x_2-x_1\right)^2p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left(X_2-X_1\right)^2\right]. \end{align*}$$ Then we can derive the relationship between the Gini coefficient and the variance: $$2\sigma_X^2-4G^2=\sigma_{\left|X_2-X_2\right|}^2.$$