]> Pareto's principle

Pareto's principle

Resource → 0 1 Population → 0 1 x In 1896, Vilfredo Pareto observed that wealth inequality tends to put around 80% of the wealth in the hands of 20% of the population. His actual data related to wealth in England with a 70-30 split, but the 80-20 split is the common example. Of course, the complement of the split equally arises: if the richest 30% hold 70% of the wealth, then the poorest 70% holds 30% of the wealth.

Some such split into complementary fractions is always sure to happen: the proportion of the wealth held by the poorest fraction x of the population increases from 0 at x = 0 to 1 at x = 1, while the complement 1 −x decreases from 1 at 0 to 0 at 1; as both functions are continuous, there is necessarily some point at which the proportion of the wealth held by the poorest x of the population is equal to 1 −x. The richest proportion 1 −x of the population then hold proportion x of the total wealth. The same happens for any quantity distributed among a population, be it the wealth of people, the number of Nobel prizes won by citizens of countries, the distribution of thermal energy among modes of vibration in a solid or anything else; a fraction (less than half, as we'll see below) of those with a high score have between them the complement of that fraction (so more than half) of the total score.

Given a population among whom some resource is distributed, we can define a function F from amount of that resource to proportion of the population with at most the given amount of the resource; this function is necessarily monotonically non-decreasing (i.e. it just keeps going up). Consequently, we can invert it to get a function G that maps proportion of the population to the amount of resource for which the given proportion of the population has at most that amount; this is also necessarily non-decreasing. The partial integral of G up to some proportion x of the population is then (aside from a constant factor of the total population size) the amount of resource held by those with at most G(x) of the resource; dividing by the integral of G over its full range, this gives us the proportion of the resource held by those with at most G(x). This increases continuously from 0 at 0 to 1 at 1. It consequently necessarily meets the diagonal straight line from 0 at 1 to 1 at 0; say it does so at x; then a fraction 1 −x of the total resource is held by a fraction x of the population.

Since G is non-decreasing, its integral's slope never decreases, so its average slope after the cross-over with 1 −x ←x is necessarily greater than its average slope before there. Consequently, the meeting must be on the lower-right half of this diagonal line, at some x ≥ half. The proportion of the population with less than G(x) is x and it owns a proportion 1 −x ≤ half ≤ x of the resource. To attain equality, we would need G to be constant, which means that every member of the population has exactly the same amount of the resource. If there is any inequality, then there is a majority of the population that owns a minority of the resource (and, conversely, a minority that owns a majority of the resource). If we split the population at the median, those above have more than those below; as there are equally many on either side, the upper half of the population has more than half the resource; as we move our cut upwards from the median, we reduce the proportion of the population above and of resource they hold; the sum of these two proportions was above 1 at the median (and shall come down to zero eventually) so their sum must sooner or latter come down to 1 as we move the cut upwards.

Since G's integral is always increasing, there is only one intersection point, so there is only one proportion x of the population for which the poorest x hold only 1 −x of the resource (while the richest 1 −x hold x of the total); I'll describe this as the complementary split point of the distribution. The interesting discussion then comes down to asking where this complementary split happens in a distribution, which I'll parameterise by the size 1 −x of the minority (of the resource, that a majority of the population holds; or of the population, that holds a majority of the resource); I'll call this the Pareto parameter for the distribution of resources across the population. For Pareto's data, this is 0.3 (a.k.a. 30%), for the usual case that gets cited it's 0.2 (one fifth or 20%), in the perfectly egalitarian distribution it would be 0.5 (i.e. 50%), which is as high as it can get. So we can think of this Pareto parameter as an equality-index; the higher it is, the more equal is the distribution of the resource.


We can represent our distribution by a measure P on reals, for which P({real x: x < k}) is the proportion of our population with less than k of the resource we're studying. I'll write f@P for the induced measure that integrates (or sums) a function f using P to do the integration. (Typically, P itself shall be p@Σ for some standard measure Σ, typically Lesbesgue integration or summation; f@P is then (f.p)@Σ using the pointwise product f.p = (: f(x).p(x) ←x :), so we use Σ to integrate f scaled by P's density p; however, I want to abstract away the mess of just exactly which Σ we're using, and doing so means the analysis applies equally to cases where neither of the usual suspects for Σ will do.) We can think of P(S), for a subset S of {reals}, as (: 1 ←x |S)@P and we have P({reals}) = 1. Let X = (: x ← x :{reals}), so that the average of the resource over the whole population is then E = X@P({reals}); for any k, we can split this into two parts L(k) = X@P({real x: x < k}) and H(k) = X@P({real x: x ≥ k}). If we also split P({reals}) as M(k) = P({real x: x < k}) and K(k) = P({real x: x ≥ k}), we get K(k) +M(k) = 1, H(k) +L(k) = E and the condition for K(k) to be our Pareto parameter is just that K(k) = L(k)/E. Substituting for E, this becomes L(k) = K(k).(L(k) +H(k)) whence K(k).H(k) = L(k).(1 −K(k)) = L(k).M(k). If we now define a measure Q by Q(S) = P(S).X@P(S), our constraint on k just becomes Q({real x: x < k}) = Q({real x: x ≥ k}) so k is just Q's median, with P({real x: x ≥ k}) giving our Pareto parameter.

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