# How does one prove probability integral transform?

How does one prove probability integral transform? So when $Y = F_X(X)$ where $X$ has a continuous distribution for which the cumulative distribution function is $F_X$, why does $Y$ have a uniform distribution? And what would Y's relationship with $\text{uniform}(0,1)$ distribution?

• Why minus vote? – rrr Mar 8 '13 at 10:16
• See the page "How to ask a question" and enumerate the ways in which yours differ. – Did Mar 9 '13 at 9:01

Write for $t \in (0;1)$, $$P(Y \leq t) = P(F_X(X) \leq t) = P(X \leq F_X^{-1}(t)) = F_X(F_X^{-1}(t)) = t.$$ It is the cdf of the U(0;1).
• @Peter, we are not inverting $X$ but instead $F_{X}$. Certainly it could be that this function is not invertible when we define the inverse function in the usual way for functions. But here we typically define the inverse of a CDF in a different way to guarantee that an inverse always exists, namely, for a CDF we usually define $F^{-1}_{X}(p) = \inf \{x|P(X\leq x) \geq p)$. – Addem Jan 4 '15 at 15:26
• From the answer it is not clear why $F_X$ must be continuous. – Sergey Zykov Jul 27 '15 at 15:26
• @Addem How can we be sure that $F_x^{-1}(p)$, in this particular definition, is always a real number? Can't it be infinite? – Amontillado Aug 13 '16 at 17:25