Scale Transformation of Density Function

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SUMMARY

The discussion centers on the relationship between two density functions, m(x) and f(y), defined on the intervals [0,1] and [0,1-ρ], respectively. The integrals of both functions equal 0.5, indicating equal mass distribution. The user seeks to establish a direct relationship between these two functions, noting that f(y) is a transformation of m(x). A proposed transformation is defined as y = h(x), where G(y) = F(x), leading to the relation g(h(x))h'(x) = f(x) upon differentiation.

PREREQUISITES
  • Understanding of probability density functions
  • Knowledge of cumulative distribution functions (CDFs)
  • Familiarity with differentiation and inverse functions
  • Basic concepts of economic modeling related to capital assets
NEXT STEPS
  • Study the properties of transformations of random variables
  • Learn about the relationship between density functions and cumulative distribution functions
  • Explore comparative statics in economic models
  • Investigate the implications of mass distribution in statistical analysis
USEFUL FOR

Statisticians, economists, and students in advanced probability theory seeking to understand the relationships between different density functions and their applications in economic modeling.

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Homework Statement



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

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

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?

Homework Equations



See above.

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!
 
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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. :rolleyes:
 

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