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?

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

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    $\begingroup$ Worth mentioning that this is also just an inverse sample transform to a standard uniform distribution. $\endgroup$ – Emily Mar 8 '13 at 21:13
  • $\begingroup$ What if X is not invertible? $\endgroup$ – Peter Jul 4 '13 at 7:15
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    $\begingroup$ @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 $F^{-1}_{X}(p) = \inf \{x|P(X\leq x) \geq p)$. $\endgroup$ – Addem Jan 4 '15 at 15:26
  • $\begingroup$ From the answer it is not clear why $F_X$ must be continuous. $\endgroup$ – Sergey Zykov Jul 27 '15 at 15:26
  • $\begingroup$ @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? $\endgroup$ – Amontillado Aug 13 '16 at 17:25

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