# Dimensional analysis for the qdf/quantile function corresponding to the pdf/CDF for the size distribution of income

I have previously posted a very similar question on Stackoverflow, but based on responses there I have decided that the real nub of my question is economic. I will give a longish introduction, mainly to distinguish what I know (introduction) from what I am asking (last 3 paragraphs, skip to it if you like).

Suppose I have a pdf f(x) or a CDF F(x) that represents the size distribution of income. Income is a flow of money (dollars or some unit of currency), per some unit of time, to particular individuals. I therefor think the proper unit of measurement is dollars per person-year.

In order for the probability to come out in dimensionless units, f(x) must be measured in person-years per dollar. Integrating this over x also yields a dimensionless share for the CDF because the units of an integral are the product of the units of the integrand y = f(x), here measured in person-years/dollar, and of the differential dx, here measured in dollars/person-year.

Integrating xf(x) therefore puts the units on the mean income in $/person-year; this seems reasonable. From this I have drawn conclusions on the units and interpretation of the quantile density function q(y), quantile function Q(y), and the integral of yq(y)dy (i.e. the the inverse function analogue of the partial moment function of order 1) which if correct would be very helpful to me, but in which I have limited confidence. I’d be grateful if someone could confirm or deny these conclusions, perhaps in the context of some discussion/explanation of interpretation of dimensional analysis in the the context of distribution functions. I believe the units of the y axis and of q(y) are necessarily the same as those in the previous pdf/CDF analysis, i.e. the y values are still measured in person-years per dollar and x = q(y) is still measured in$/person-year. This implies that integrals of q(y) over some interval in general, and Q(y) (which is equal to the integral of q(y) from 0 to y) in particular, are both measured in unitless shares. However (uncertain interpretation #1) F(x) is in unitless shares of total income, while Q(y) is in unitless shares of the population for a year.

Finally, (uncertain interpretation #2) this would have the integral of yq(y)dy in units of person-years per dollar. But I am having trouble interpreting this. The corresponding x quantity is the mean (or for a bounded x, the partial mean) of income per person-year. So this would be the mean number of person-years--for what? for a single dollar? for the mean income?

Consider the truncated version of this function, where the integral of yq(y)dy from zero to q* is divided by F(q*). The corresponding truncated function of x yields the mean for those with incomes of x* or less, and $/person-year seems like reasonable units for a mean income. But I want the the y version to come out in number of people, or number of person-years, corresponding to the share of person-years over the same interval, and the "per dollar" in the denominator, which I need to make the CDF(x) and Q(y) unitless, does not have an interpretation which is obvious to me. • Your last-last sentence needs clarification (from "But I want the y version...: ). What do you mean by the "y version"? Please clarify. Apr 19, 2018 at 20:31 • I want this analysis yo produce consistent values for the income share and the population share of am income range, as well as th he mean income and the number of people. That is why I am trying to make one graph do such work. Because the quantile function is the inverse of the distribution function, I often think of it as if I could just trade axises. Jun 15, 2023 at 2:01 ## 1 Answer Your random variable is$X$= "Yearly Income per capita measured in dollars". So$X$is a real-valued function. Its domain is usually left unspecified, while its range is measured in "dollars per person-year", as the OP writes it. The density and the distribution (pdf/cdf) functions of$X$,say$f_X,F_X$have as their domain the range of$X$. So the domain of$f_X,F_X$is a set that carries the unit of measurement "dollars per person-year". Being strict in dimensional analysis, indeed we could say that the density must then be measured in "person-years per dollar" (irrespective of whether this is meaningful or not). So the expected value that is the integral$\int_0^{\infty} x f_X(x) dx$keeps having "dollars" as a measurement unit (I just assumed$X$is non-negative), applying the rule the OP states (the one unit is the reciprocal of the other, and we are left with the third which is again "dollars per person-year"). Moreover the integral that is the cdf$F_x(x) = \int_0^x f_X(s) ds$has a range that comes out "dimensionless", since probability is a relative measure. (Digression: Note that by definition, the distribution function of a random variable measures the probability of the random variable being smaller or equal than a threshold,$F_X(x): = Prob(X \leq x)$. Additional assumptions are required to "tranlsate" this probability as representing also "income shares", meaning I guess "percentage of population with per capita income below or equal to$x$") Turning to the Quantile function, it is defined as the inverse of the distribution function (generalized inverse to cover all cases, but let's keep it simple). Being the inverse, it uses as its domain the range of$F_X(x)$, so the domain of$Q(p)$carries "probability" as unit of measurement, while its range is the domain of$F_X$. And the domain of$F_x$is the range of$X$: the Quantile function has the same range as$X$and so it is measured in "dollars per person-year" too (i.e. its range caries the "dollars per person-year" unit of measurement). it follows that the derivative of$Q(p)$",$q(p) = dQ(p)/dp$has the same units of measurement as$Q(p)$, while the integral$\int_0^1 pq(p)dp\$ is measured in "dollars per person-year" also.

• This is a very helpful & thoughtful answer. I got pulled away from this project for several years due to a relationship breakup, organization bankruptcy, and cross-country move, all at once, plus a spot of open-heart surgery just to make the transition complete. I hope I accepted your answer at the tome--I should have--but I do not now see the "accept" checkmark. If I did not, please let me know and I will figure out how to get it back and accept it. I am actively returning to this project no, and this answer is still entirely relevant (and a great review of some issues I had forgotten. Jun 15, 2023 at 1:47
• @andrewH Thanks for the positive feedback. My answer appears accepted (green checkmark). Jun 19, 2023 at 14:13