# Homework Help: Scale Transformation of Density Function

1. Aug 24, 2010

### kipfilet

1. The problem statement, all variables and given/known data

Two random variables, x and y. Density functions are $m(x)$ and $f(y)$, respectively. x is defined on [0,1] and y on $[0,1-\rho]$. I also know that

$$\int_{0}^{1} m(x) \,dx = 0.5 \int_{0}^{1-\rho} f(y) \,dy = 0.5$$

Knowing that f(y) is essentially a transformation of m(x) (distribution has the same mass, but a smaller lower bound), is it possible to find a direct relation between the two density functions?

2. Relevant equations

See above.

3. The attempt at a solution

This is not really a physics problem, but rather something that appeared in the course of designing an economic model :) Those two distributions relate to the same thing - capital assets - but refer to the way how they are distributed among a population composed of men (m(x)) and women (f(y)). It would be really helpful for comparative statics and other types of results if I could find a direct relation between m(x) and f(y), knowing that f(y) is, essentially, a transformation of the former. However, I do not even know if this is possible. The only thing (don't laugh at me) that I can be sure of is that $M(1) = F(1-\rho)$ (where upper case letters refer to the distribution function). Could anyone better versed in statistics than I am lend me a helpful hand? Thank you very much in advance!

PS: Those are two different integrals that should be in separate lines, but I could not manage to make the tex code I know work in here. I apologise for the mess!

2. Aug 24, 2010

### LCKurtz

I'm not sure this is what you are getting at, but I'll have a go. Suppose you have two random variables X with density f(x) and Y with density g(y), both on [0,1]. Let's say that the probability "piles up" more rapidly for f than for g. By this I mean that if F and G are the cumulative distribution functions for f and g, respectively, then F(x) ≥ G(x).

Now, you could define a transformation as follows. For each x in [0,1], let y = h(x) be the unique value of y such that G(y) = F(x), which is just another way of saying that P(Y≤y) = P(X≤x). Since F and G are increasing y = h(x) = G-1(F(x)).

If you differentiate G(y)=F(x) with respect to x you get G'(y)y' = F'(x) or g(y) y' = f(x).

So, remembering that y = h(x) as defined above, you have the following relation between the densities: g(h(x))h'(x) = f(x).

Disclaimer: I don't know if that is what you are wondering about, nor whether this is just a circular argument, nor whether it is useful. :uhh: