CDF and uniformly distributed random variable

probability distributionsprobability theory

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 $y\in(0,1)$, $G(y)=\sup\{x\in\mathbb{R};F(x)<y\}$ and show that the randon variable $X=G(Y)$ satisfies $F_X=F$.

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:
$$F_X(x)=\mathbb{P}(X\le x) = \mathbb{P}(G(Y)\le x).$$

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

Best Answer

Let's assume for simplicity that $Y$ is a continuous random variable. Then $G(Y) \leq x$ iff $Y \leq F(x)$ 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.