Monday, 3 July 2017

probability theory - CDF and uniformly distributed random variable



I am trying to solve a basic exercise about random variables, but I am having some trouble.




Let Y be an uniformly distributed random variable on [0,1] and F an arbitrary CDF. Define, for every y(0,1), $G(y)=\sup\{x\in\mathbb{R};F(x) and show that the randon variable X=G(Y) satisfies FX=F.



I am really puzzled with this question. It is very basic, I suppose, but I am still struggling with the concepts, and I think if I solved it, I could unterstand the underlying definitions better.



Here is some of my thoughts:
FX(x)=P(Xx)=P(G(Y)x).



I also know that, since Y is uniformly distributed, FY(x)=x. (Is this correct?) But I am having some trouble to connect this to the CDF of G(Y). Any hints would be apprreciated.


Answer



Let's assume for simplicity that Y is a continuous random variable. Then G(Y)x iff YF(x) by definition of G, which happens with probability F(x) (over Y).




When Y is an arbitrary random variable, a similar reasoning should work (if the statement is true), but you have to be more careful since Y could have atoms.


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