I understand the analytical proof given here.


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.

I fail to understand. Could someone please help?

  • 1
    $\begingroup$ 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). $\endgroup$
    – user83346
    Aug 27, 2017 at 6:46
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    $\begingroup$ 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). $\endgroup$
    – user83346
    Aug 27, 2017 at 6:51
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    $\begingroup$ en.wikipedia.org/wiki/Probability_integral_transform $\endgroup$
    – Glen_b
    Aug 27, 2017 at 7:56