mcnalu's musings

This post describes the mathematics of inequality in income distributions. It is sufficiently abstract that it's applicable to wealth and many other distributions. You can find the ideas explained without mathematics in this post over on the Rational Intuition blog. The essential point is that commonly stated income fractions, such as

The top 30% of the population in terms of income receive 51% of the total income.

make for poor measures of inequality because they are insensitive to changes in the distribution. The mathematical reasons for this are given at the end of this post.

Definitions

Consider an income distribution $f(x)$. The fraction of the population in income bracket $(x,x+dx)$ is given by $f(x)dx$ and so we have

\begin{equation} \int_{-\infty}^\infty f(x) dx = 1 \label{population} \end{equation}

This admits negative incomes, but doesn't require them as $f(x)$ can be set to zero for $x<0$ if desired.

The mean income of this population is given by $\mu$, defined as

\begin{equation} \mu = \int_{-\infty}^\infty f(x)x dx \label{mean} \end{equation}

The standard deviation of the population is given by $\sigma$:

\begin{equation} \sigma^2 = \int_{-\infty}^\infty f(x) (x-\mu)^2 dx \label{sigma} \end{equation}

Let $x_n$ be the upper boundary of decile $n$, then we have

\begin{equation} \int_{-\infty}^{x_n} f(x) dx = \frac{n}{10} \label{decile-n} \end{equation}

and also define $\mu_n$ as

\begin{equation} \mu_n = \int_{-\infty}^{x_n} f(x)x dx \label{mean-n} \end{equation}

This is the mean of the group with $x<x_n$, i.e. all deciles up to and including $n$. The fraction of income of this group out of the total population's income will therefore be $p_n$

\begin{equation} p_n=\frac{\mu_n}{\mu} \label{fraction-n} \end{equation}

Income groups

We can divide the population into three groups in terms of income. The low group will have $x<x_3$, i.e. the bottom 30%, and the high group will have $x>x_7$, i.e. the top 30%. The remainder form the middle 40%. We can define the fraction of income going to these groups as follows:

\begin{eqnarray} p_L &=& p_3 \nonumber \\ p_M &=& p_7 - p_3 \label{p-groups}\\ p_H &=& 1 - p_7 \nonumber \end{eqnarray}

Note that by definition

$$ p_L+p_M+p_H=1 $$

Uniform distribution

For a uniform distribution which is zero except in $(a,b)$ we have from (\ref{population}) that

\begin{equation} f(x) = \frac{1}{b-a} \end{equation}

and from (\ref{mean}) we have the mean

\begin{equation} \mu = \frac{a+b}{2} \label{uniform-mean} \end{equation}

and from (\ref{sigma}) we have the standard deviation

\begin{equation} \sigma = \frac{b-a}{2\sqrt{3}} \label{uniform-sigma} \end{equation}

The upper bound for decile $n$ from (\ref{decile-n}) is

\begin{equation} x_n = a + \frac{n}{10} (b-a) \end{equation}

and $\mu_n$ from (\ref{mean-n}) is

\begin{equation} \mu_n = \frac{n}{20} \left[\left(2-\frac{n}{10}\right)a + \frac{n}{10}b\right] \label{uniform-mean-n} \end{equation}

From this and (\ref{uniform-mean}) and after a little algebraic reduction we can write out the fractions of incomes going to the three income groups in a form convenient for calculating percentages:

\begin{eqnarray} p_L &=& \frac{51a + 9b}{100(a+b)} \nonumber \\ p_M &=& \frac{40}{100} \label{uniform-p-groups}\\ p_H &=& \frac{9a + 51b}{100(a+b)} \nonumber \end{eqnarray}

Notice that $p_M$ is always 40%, independent of $a$ and $b$. The most extreme equality occurs for $a=0$ for which the trio of percentages will be 9%/40%/51%, independent of $b$.

We can re-express these in terms of the mean and standard deviation using (\ref{uniform-mean}) and (\ref{uniform-sigma}):

\begin{eqnarray} p_L &=& 0.3 - \frac{21\sqrt{3}}{100} \frac{\sigma}{\mu} \nonumber \\ p_M &=& 0.4 \label{uniform-p-musigma}\\ p_H &=& 0.3 + \frac{21\sqrt{3}}{100} \frac{\sigma}{\mu} \nonumber \end{eqnarray}

Notice that as $\sigma/\mu \rightarrow 0$, $p_L$ and $p_H$ both tend to 0.3.

Gaussian distribution

For a Gaussian distribution with mean $\mu$ and standard deviation $\sigma$, (\ref{population}) gives

\begin{equation} f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \end{equation}

and $\mu_n$ from (\ref{mean-n}) is

\begin{equation} \mu_n = \frac{n}{10}\mu - \sigma^2 f(x_n) \end{equation}

which leads to the following income fractions:

\begin{eqnarray} p_L &=& 0.3 - \frac{\sigma}{\mu} g \nonumber \\ p_M &=& 0.4 \label{gaussian-p-musigma}\\ p_H &=& 0.3 + \frac{\sigma}{\mu} g \nonumber \end{eqnarray}

where $g=f(x_3)\sigma = f(x_7)\sigma$ is a constant independent of $\mu$ and $\sigma$. By solving (\ref{decile-n}) for $x_3$ (using the inverse of the error function erf) we find that $g=0.348$ to 3 decimal places.

Comparison

For both uniform and Gaussian distributions $p_M$ is 40%. For the uniform distribution, the low and high fractions are:

$$ p_\frac{H}{L} = 0.3 \pm 0.364 \frac{\sigma}{\mu} $$

and the equivalent for the Gaussian is very similar:

$$ p_\frac{H}{L} = 0.3 \pm 0.348 \frac{\sigma}{\mu} $$

For a uniform distribution starting at zero ($a=0$) we have $\sigma/\mu=1/\sqrt{3}$ which gives a trio of percentages for the Gaussian as 10%/40%/50% which is only slightly different for the uniform case we saw above with 9%/40%/51%.

Symmetry invariance

If $f(x)$ is symmetric about the mean then $p_M=0$. This can be proved as follows. First, rewrite (\ref{mean-n}) as

$$ \mu_n = \int_{-\infty}^{x_n} f(x) \mu dx + \int_{-\infty}^{x_n} f(x) (x-\mu) dx $$

Then, because $\mu$ is constant, we can use (\ref{decile-n}) to replace the first integral to get

$$ \mu_n = \frac{n}{10}\mu + \int_{-\infty}^{x_n} f(x) (x-\mu) dx $$

This is the general form of expressions derived in (\ref{uniform-p-musigma}) and (\ref{gaussian-p-musigma}).

\begin{eqnarray} \mu_7-\mu_3 &=& \frac{7-3}{10}\mu + \int_{-\infty}^{x_7} f(x) (x-\mu) dx - \int_{-\infty}^{x_3} f(x) (x-\mu) dx \nonumber \\ &=& 0.4\mu + \int_{x_3}^{x_7} f(x) (x-\mu) dx \nonumber \\ \end{eqnarray}

First, note that $f(x)$ is symmetric about $x=\mu$ and so $f(x)(x-\mu)$ is anti-symmetric about $\mu$. The symmetry also means that $x_7-\mu$=$\mu-x_3$ and so the integral must be zero. From (\ref{p-groups}) and (\ref{fraction-n}) we now arrive at the result

$$ p_M = 0.4 $$

Scale invariance

Suppose we create a new distribution $g(y)$ from scaling $f(x)$ as follows:

$$ g(y=\alpha x)\alpha = f(x) $$

then it follows that all income fractions for $g$ will be the same as those for $f$, i.e. $p_n$ and so $p_L$, $p_M$ and $p_H$.

The new distribution can be thought of as that obtained if everyone in the original population has their incomes increased by a factor of $\alpha$. Note that $\alpha$ appears on the LHS above so that $g(y)$ satisfies (\ref{population}).

It can be shown from (\ref{decile-n}) that the decile boundaries $y_n$ for this new distribution are given by

$$ y_n = \alpha x_n $$

and from (\ref{mean-n}) the means $\nu_n$ for this distribution are given by

$$ \nu_n = \alpha \mu_n $$

since this is true also for the mean of the whole distribution, i.e. $\nu = \alpha \mu$, then we have

$$ \frac{\nu_n}{\nu}=\frac{\mu_n}{\mu} $$

and from this it follows that all income fractions for the scaled distribution will be the same as the $p$ values for $f$.

Insensitivity to distribution shape

The income fractions involve integrals over $f(x)$ and so are only weakly dependent on how income is distributed within deciles or income groups. This is why the income percentages for the same $\sigma/\mu$ ratio are so similar for the Gaussian and the uniform distribution even though the two distributions have significantly different functional forms.

All symmetric distributions will have the form

\begin{eqnarray} p_L &=& 0.3 - \frac{\sigma}{\mu} g \nonumber \\ p_M &=& 0.4\\ p_H &=& 0.3 + \frac{\sigma}{\mu} g \nonumber \end{eqnarray}

where $g$ is typically around 0.36 for many realistic distributions. Note that $p_L+p_H=0.6$.

Real income distributions are asymmetric and so $p_M$ will not equal 0.4. Usually they are skewed so that the distribution peaks at incomes below the mean and has a long tail to large $x$. This can cause $p_M$ to be less than 0.4, which in turn means that $p_L+p_H$ must take on a higher value than the 0.6.