# If $Y=F(X)$ is $U[0,1]$ where $F$ is cdf of continuous $X$ then shouldn't its plot be rectangular? [duplicate]

I understand the analytical proof given here.

https://math.stackexchange.com/questions/868400/showing-that-y-has-a-uniform-distribution-if-y-fx-where-f-is-the-cdf-of-contin

But in that case since $Y=F(X)$, say where $F(X)$ is the cdf of a standard normal distribution $N(0,1)$ the plot of which is not rectangular but rather S shaped.

• If you define $Y=F(X)$, where $F$ is the CDF of $X$ (continuous) then you have another random variable $Y$ with another cdf than the cdf of $X$, let's name if $F_Y$. In the answer on math.stackexchange it is shown that for this other random variable the cdf is $F_Y(Y)=P(y \le Y)=Y$, so you have transformed the variable $X$ into a variable $Y$ with a cdf of an U(0,1).
• It means that $Y$, the new random variable has a cdf that is a line (the bisector) between 0 and 1, and if you look at the cdf of a uniform random variable you have exactly this. Other way to see this: you have shown that $F_Y(y)=y$ as cdf, after derivation you get the density $f_Y(y)=1$ which is the density of a U(0,1).