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: $m=\frac{\log_2(q)}{a}$.

Examples:

1.) Share of the top $1$% and $a=1.1$, then we get $m\approx6$. This yields: $\frac{W_{0.01}}{W_0}=2^{6(-0.1)}\approx0.66$.

2.)Share of the top $1$% and $a=1.3$, then we get $m\approx5$. This yields: $\frac{W_{0.01}}{W_0}=2^{5(-0.3)}\approx0.35$.

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

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 $100$%.

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: $m=\frac{\log_2(q)}{a}$.

Examples:

1.) Share of the top $1$% and $a=1.1$, then we get $m\approx6$. This yields: $\frac{W_{0.01}}{W_0}=2^{6(-0.1)}\approx0.66$.

2.)Share of the top $1$% and $a=1.3$, then we get $m\approx5$. This yields: $\frac{W_{0.01}}{W_0}=2^{5(-0.3)}\approx0.35$.

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

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: $m=\frac{\log_2(q)}{a}$.

Examples:

1.) Share of the top $1$% and $a=1.1$, then we get $m\approx6$. This yields: $\frac{W_{0.01}}{W_0}=2^{6(-0.1)}\approx0.66$.

2.)Share of the top $1$% and $a=1.3$, then we get $m\approx5$. This yields: $\frac{W_{0.01}}{W_0}=2^{5(-0.3)}\approx0.35$.

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