So I finally finished Thomas Piketty's book  Capital in the 21st Century, and I thought I'd write up an interesting result that Piketty mentions, but does not elaborate on. Given the book is already 700 pages, it's probably for the best that he drew the line somewhere. The result is specifically, that under basic models of the development of distribution of wealth in a society, it can be shown that when growth is equal to $g$, and return on capital is equal to $r$, then the distribution of wealth tends towards a Pareto distribution with parameter $rg$. That sounds pretty interesting right? My notes below are largely based on following paper by Charles I. Jones of Stanford Business School, my addition is to derive the assumption of an exponential distribution of income from more basic assumptions about labour and capital income. Link to Jones's paper [1] Pareto inequality of Income
The simplest version of the problem is to investigate the inequality of income rather than capital, this way we leave aside issues of inheritance, intergenerational effects, shocks to capital, etc. In this case, the model as described in the link above from Jones only requires two assumptions:
Where $\delta$ is the death rate. This is my kind of model! I’m pretty sure we derived this distribution in one of the actuarial exams (CT4 or CT5?), and this distribution can be derived just from the assumption of a constant death rate. It’s not without limitations (death rates have a miniincrease in adolescent for example), but the overall shape of the curve is pretty close to the empirical distribution. What’s our other assumption? 2. Income $y$, increases exponentially with age $x$: $$y = e^{\mu x}$$ I was happy following the paper until this point, is this a reasonable assumption? Empirically income does not tend to increase exponentially right? Salaries tends to increase up until middle age, but then peak around age 5055. Before falling off slightly. After more consideration, I realised I was considering the wrong type of income, the important distinction is we are talking about total income here, not just income from labour. We know that: $$ \text{Income(x)} = \text{Labour income(x)} + \text{Capital income(x)}$$ Below I give a demonstration of why we might be more willing to assume total income increases exponentially with age. Exponential distribution of Income Assume we have a constant savings rate $s$, capital earns interest at rate $r$, and let $C_x$ denote capital at time $x$, and $L_x$ denote labour income at time $x$. Then we can easily link capital at time $x$ to capital at time $x+1$ using the following: $$ C_x = C_{x1}(1+r)+ s*L_x$$ And this formula can then be used to examine individual’s capital over time: $$ C_1 = 0 + s * L_0 $$ $$ C_2 = C_1 (1+r) + s * L_1 = (s*L_0)(1+r) +s L_1 $$ $$ C_3 = C_2 (1+r) + s * L_2 = (s*L_0)(1+r)^2 +s L_1(1+r)+s*L_2 $$ And generalising (which we can prove by induction, but I’m just going to take as given here): $$C_n = s(L_0 *(1+r)^{n1} + L_1*(1+r)^{n2} + … + L_{n1})$$ Or in short hand: $$C_n = s \sum_{i=0}^{n1} L_{i}* (1+r)^{n1i} $$ We now need to commit to a particular form for $L_n$ to progress any further. It turns out even with a linearly increasing function of $L$, we still end up with an exponential aggregate income over time. Using a linear function for $L_n$ is weaker than using an exponential function (as we would just be begging the question if we already assumed $L_n$ increased exponentially.) Using $L_n = \alpha n$, gives us: $$C_n = s \sum_{i=0}^{n1} \alpha n * (1+r)^{n1i} $$ Taking $\alpha$, and a $(1+r)$ outside gives the following: $$C_n = \frac{\alpha s}{1+r} \sum_{i=0}^{n1} n * (1+r)^{ni} $$ The trick now is to think of this sum, which I’ll refer to as $S$ (i.e. ignoring the factors outside the sum), as the following: \begin{matrix} (1+r) & + & (1+r) & + & ... &...& ... & ...& +& (1+r)\\ (1+r)^2 & + & (1+r)^2 & + &...& ... & ... & + & (1+r)^2&\\ (1+r)^3 & + & (1+r)^3 & + &...& + &(1+r)^3 &\\ ... & & & & & & & &\\ (1+r)^{n1} & & & & & & & & \end{matrix} We now rewrite this as a series of sums of columns as follows: $$S = \sum_{i=1}^{1} (1+r)^i + \sum_{i=1}^{2} (1+r)^i + … + \sum_{i=1}^{n1} (1+r)^i$$ But each of these is now just a geometric series, that is to say our sum is equal to: $$S = \frac{(1+r)^21}{r} + \frac{(1+r)^31}{r} + … +\frac{(1+r)^n1}{r} – (n1)$$ And writing this with summation notation: $$S = \frac{\sum_{i=2}^{n} (1+r)^i – (n1)}{r} – (n1) $$ The trick now is to once again apply the formula for a geometric sum giving: $$S = \frac{ (1+r)^{n+1} – 1 2r – r^2}{r^2}  \frac{n1}{r}  \frac{r (n1)}{r}$$ Which we can simply slightly: $$S = \frac{ (1+r)^{n+1} – (1+r)^2}{r^2}  \frac{(n1)(1+r)}{r}$$ And then plugging this into the formula for $C_n$: $$C_n = \frac { s \alpha }{1+r} \left( \frac{ (1+r)^{n+1} – (1+r)^2}{r^2}  \frac{(n1)(1+r)}{r} \right)$$ And this is the result we require. We can see that capital, and hence income increases exponentially with age (though we are subtracting a term which increases linearly with age to slow it down). Pareto distribution Now that we've motivated the two assumptions, all that remains is to combine them, and show that we end up with a Pareto distribution. Inverting the assumption about income gives the age at which an individual earns a given level of income: $$y = e^{\mu x}$$ Gives: $$x(y) = \frac{1}{\mu} \text(log) y $$ And then, using this to evaluate the probability of income being greater than y: $$P(\text{Income} > y) = P(\text{Age} > x(y)) = e^{\delta x(y)} = y^{\frac{\delta}{\mu}}$$ As required. [1] Charles I. Jones's paper: web.stanford.edu/~chadj/SimpleParetoJEP.pdf 
AuthorI work as an actuary and underwriter at a global reinsurer in London. Categories
All
Archives
October 2022

Leave a Reply.