Deriving Top-Income Shares from a Power-Law Exponent

# Taleb's Black-Swan: interpretation of the exponent


# Taleb's Black-Swan: interpretation of the exponent












I am reading Taleb's "Black Swan" (revised 2020th edition). In chapter 16 "The Aesthetics of Randomness" he describes the meaning of the exponent in the context of extrapolation. On page 265 there is a table (table 3) which depicts that if the exponent attains the value of $1$ then the share of the top $1$% will almost represent the whole set ($\approx99.99$%).

I don't understand how this is possible? Let's set up a simple illustratrive example:

$30$ m people earn $50,000$ USD

$15$ m people earn $100,000$ USD

$7.5$ m people earn $200,000$ USD

$\cdots$

$1$ person earns $\approx1.6$ trillion USD.

Now if you consider the top $1$%, namely $300,000$ people, and sum up their income then this share isn't even close to $90$%.

What am I missing? Or, is Taleb's example so vague that one can't set up a proper example?

## Answer by Philipp (score 4, accepted)

https://quant.stackexchange.com/a/61648

I finally got the idea behind the example. To illustrate it in a more general setting I will present a rigorous proof:

Let $x_k$ denote the salary and $b_k$ the number of persons that earn $x_k$ or more. Following the proposed power law by Taleb, we have:

$x_{k}:=x_02^k$ and $b_k:=b_0\left(\frac{1}{2}\right)^{ka}$, where $a\geq1$ and $k\in \mathbb{N}$.

The sum of all earned incomes is given by:

$$ W_0:=\sum\limits_{k=0}^nx_k\left(b_k -b_{k+1}\right)=\sum\limits_{k=0}^nx_02^k\left(b_0 \left(\frac{1}{2}\right)^{ka}-b_0\left(\frac{1}{2}\right)^{(k+1)a}\right)=\sum\limits_{k=0}^nb_0x_02^{k(1-a)}\left(1-\left(\frac{1}{2}\right)^a\right)=\left(1-\left(\frac{1}{2}\right)^a\right)b_0x_0\sum\limits_{k=0}^n2^{k(1-a)}=\left(1-\left(\frac{1}{2}\right)^a\right)\frac{b_0x_0}{1-2^{1-a}}\left(1-2^{n(1-a)}\right). $$ Then the sum of all incomes that is earned by the top $q$% is simply represented by the last terms of the above sum. So the sum starts at some index $k=m$. Hence, $$ W_q:=\sum\limits_{k=m}^nx_k\left(b_k -b_{k+1}\right)=\cdots=\left(1-\left(\frac{1}{2}\right)^a\right)\frac{b_0x_0}{1-2^{1-a}}\left(2^{m(1-a)}-2^{n(1-a)}\right). $$ Assuming $a>1$, we get the share of the top $q$% by dividing: $\frac{W_q}{W_0}=\frac{\left(2^{m(1-a)}-2^{n(1-a)}\right)}{1-2^{n(1-a)}}$. If $n\to\infty$, which means we keep up increasing the salary and look how many persons will earn it we get: $\frac{W_q}{W_0}=2^{m(1-a)}$ (note that terms disappear because of $a>1$). To find the relevant index $m$ we simply take the logarithm: $m=\frac{\log_2(q)}{a}$ and plug it into $\frac{W_q}{W_0}=2^{m(1-a)}$. So it is actually irrelavant wether we assume halving or any other method to reduce quantities because $2$ cancels out: $\frac{W_q}{W_0}=2^{\frac{\log_2(q)}{a}(1-a)}=q^{\frac{a-1}{a}}$.

Examples:

1.) Share of the top $1$% and $a=1.1$. This yields: $\frac{W_{0.01}}{W_0}=0.01^{\frac{1.1-1}{1.1}}\approx0.66$.

2.) Share of the top $1$% and $a=1.3$. This yields: $\frac{W_{0.01}}{W_0}=0.01^{\frac{1.3-1}{1.3}}\approx0.35$.

3.) If $a=1$ then the share of the top $1$% is $100$%.

## Answer by Sergei Rodionov (score 2)

https://quant.stackexchange.com/a/61595

I don't know how to interpret the above example, but wealth distribution, of which inequality is one of the measures, is frequently described by the Pareto distribution.

Also, the IRS publishes annual income distribution statistics which is a good source to check the narrative. Top 1% earns some 20% of all income.