I'm attempting to understand one line of the proof for the probability integral transform as found on wikipedia:

Suppose that a random variable $X$ has a continuous distribution for which the cdf is $F_X$. Then the random variable $Y = F_X(X)$ has a standard uniform distribution.


$ F_Y(y) = P(Y \leq y) = P(F_X(X) \leq y) = P(X \leq F^{-1}_X(y)) = F_X(F^{-1}_X(y)) = y $

What I do not understand is the definition of the random variable $Y$, namely why is there a capital $X$ in parentheses, $F_X(X)$, instead of lower-case, $F_X(x)$. More importantly, what does this mean?

I have looked at this post already, and my updated understanding is that $F_X(X)$ represents the distribution of the probabilities of $X$, not the variable itself. So, I believe that $Y$ is the distribution of probabilities of $X$. Is this correct? Or, if not, can someone explain what this difference in notation means?

  • $\begingroup$ Your questions are puzzling because they are identical to the ones asked at the post you reference and they are explicitly answered there in several ways. It is therefore unclear what you are seeking in terms of an answer. $\endgroup$
    – whuber
    Jun 11, 2020 at 14:41

1 Answer 1


Because, take $F_X(x)=G(x)$ as a function and we apply this transformation over the random variable $X$, to obtain $Y$. So, in general, if the input is a RV, the output is a RV, i.e. $Y=G(X)$, not $Y=G(x)$. You could say $y=G(x)$, for a specific pair of $(x,y)$ by the way. Therefore, the notation confusion you have doesn't have anything to do with the meaning of $F_X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.