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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?

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  • $\begingroup$ Try this formula: ${\rm Share \, top\, }q=\frac{1}{\left(1-q\right)^{\frac{1}{a}-1}}$ where a is the exponent. Work only for $a>0$ so for the first entry please try a=1.00001 for example. $\endgroup$ Mar 9 at 17:31
  • $\begingroup$ @Magicisinthechain, don't know where you got this formula from, but it's wrong. For sake of simplicity try for example $a=1.1$, a group of $100$ people and start doubling from a lowest salary, say $10$USD. You will quickly see that the share of the top $1$% is equal to $\approx 66$% whereas "your" formula yields $\approx 99.9$%. $\endgroup$
    – Philipp
    Mar 10 at 23:41
  • $\begingroup$ Thanks - this is power law, so not sure it work for discrete group of 60, doesn’t it reproduce Taleb’s table? $\endgroup$ Mar 10 at 23:44
  • $\begingroup$ @Magicisinthechain, I figured out the formula. Please see my answer. If you plug in $a=1.1$ then you will more or less receive all the values which Taleb presents in table 3. However, I am not sure if Taleb had this formula in mind or if he was simply playing around with an excel spread sheet to produce some numbers. $\endgroup$
    – Philipp
    Mar 11 at 0:56
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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 $100$%.

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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.

enter image description here

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  • $\begingroup$ My question is not about a specific example but rather the general situation which is described by Taleb. Again, how is it possible that the top $1$% represents almost the whole set? Yes, there are examples where the top $1$% represent a large share of the set (your income example is one of those). However, if you obey the power law as described by Taleb you are not able to get an example where the top $1$% represents $99.99$% of the set. $\endgroup$
    – Philipp
    Mar 9 at 12:05
  • $\begingroup$ @Philipp - Is there a formula that describes the relationship (or the law as you call it) in the book? I couldn't understand it from the numbers. $\endgroup$ Mar 9 at 12:34
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    $\begingroup$ Thanks for following up :) Yes, if you double a quantity it's occurence is halved. So if we stick with my above income example it means, that if $30$m people earn $50,000$USD then $15$m people earn $100,000$USD, e.g. $y=50,000\left(\frac{30m}{15m}\right)^1$. So the formula is $y=x\left(\frac{q_x}{q_y}\right)^1$, where $q_x$ and $q_y$ denote the respective quantities und $x$ and $y$ represent the occurences of that quantity in the sample. $\endgroup$
    – Philipp
    Mar 9 at 12:50
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    $\begingroup$ You can certainly reach 99.99% and actually any number of nines, but you have to allow fractional people, i.e. 0.5 people making $3.2T etc. $\endgroup$ Mar 9 at 13:19
  • $\begingroup$ Oh yes that's true. Didn't see it :D If I also take into account that those $0.5$ people only contribute $0.5\cdot 3.2tn=1.6tn$ to the overall wealth then you realise that you need a very large "number" of those fractional people to achieve $99.99$%. This makes the whole example in the book a little bit odd. But I think I got the point $\endgroup$
    – Philipp
    Mar 9 at 14:43

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