1

I am trying to understand transforming random variables into a different distribution. I don't get how this works. Let's say X is normally distributed. We have some function Y = g(X) that transforms X into a variable that is distributed differently.

How would this happen? Unless we square X or something like that, each X would lead to a unique Y, with densities equal to whatever the density of X was. I am not understanding how we could transform X into another distributed random variable, and one that is known.

I'm doing homework and it wants me to transform X to a uniform distribution. Conceptually, I don't understand how this works. Let's say X comes from norm(0, 1). No matter what function g(X) is, I'd imagine Y = g(0) will occur more times than Y = g(5). I'm not understanding what exactly needs to be transformed. Are we trying to transform the density itself - like flatten the density function out?

Thanks!

confused
  • 2,453
  • 6
  • 26
  • 4
    [How to transform gaussian(normal) distribution to uniform distribution?](https://math.stackexchange.com/questions/2343952/how-to-transform-gaussiannormal-distribution-to-uniform-distribution) – user2974951 Oct 22 '20 at 12:18
  • If $g(y)=3$ for all $y$'s, would you support that $Y=3$ is Normal? – Xi'an Oct 22 '20 at 12:26
  • I don't see how that tells me anything about Y – confused Oct 22 '20 at 12:31
  • @user2974951 I see. We can use the CDF of the random variable, which converts it to a new random variable that exists between 0 and 1. And then somehow we can normalize it? – confused Oct 22 '20 at 12:32
  • Wow that's crazy, I just simulated it in R and using CDF converts it to uniform - seems like it holds true for any distribution. – confused Oct 22 '20 at 12:39
  • Your question assumes that *densities* are conserved under a transformation, but that's just not so. The distinction between a probability and a probability density is fundamental. This leads me to suspect that https://stats.stackexchange.com/questions/14483 answers your question. – whuber Oct 22 '20 at 12:53
  • Thanks. I think I've found what I needed. – confused Oct 22 '20 at 12:58

0 Answers0