I am trying to wrap my head around a certain topic in my notes, but it seems very confusing.

Let $X$ a continuous random variable whose distribution function $F_X$ is strictly increasing on the possible values of $X$. Then $F_X$ has an inverse function. [Agreed]

Let $U = F_X(X)$, then for $u \in [0,1]$ we wish to find $F_U(u)$: Since $F_U(u) = P[U\leq u]:$ $$P[U \leq u] = P[F_X(X)\leq u] \; \; \; \; \ \;\;\;(1)$$ $$= P[X \leq F_X^{-1}(u)]\; \; \; \; \ \;\;\;(2)$$ $$= F_X(F_X^{-1}(u))=u\; \; \; \; \ \;\;\; (3)$$

I am probably missing something super obvious, but I am confused by the above 3 steps.

From $(1)$ to $(2)$ - I recognise that we defined $U = F_X(X)$, so the LHS is fine. But the right hand side of the inequality in $(2)$, how exactly does that make sense?

From $(2)$ to $(3)$, I think it is saying this is just the CDF of $U<F_X^{-1}(u)$? IS that correct? And the inverse cancels with $F_X$ to leave $u$.

I.e., $F_U(u) = u$

  • 1
    $\begingroup$ Eq 2 should be P(X<= ...) $\endgroup$
    – TPArrow
    Oct 23, 2019 at 4:32
  • 3
    $\begingroup$ I'm quite certain this is covered by answers already on site. $\endgroup$
    – Glen_b
    Oct 23, 2019 at 5:06
  • 1
    $\begingroup$ Search for 'inverse CDF method'. Many hits. $\endgroup$
    – BruceET
    Oct 23, 2019 at 8:18
  • 1
    $\begingroup$ Perhaps a more effective search is Probability Integral Transform. $\endgroup$
    – whuber
    Oct 23, 2019 at 15:42

1 Answer 1


First, for $F_X^{-1}$ to exist, $F_X$ must be strictly increasing and continuous.

Second, if $F_X(X)\le u$, then, $F_X$ being increasing, $F_X^{-1}$ is also increasing, hence $$F_X^{-1}(F_X(X))\le F_X^{-1}(u)$$ by applying $F_X^{-1}$ to both sides of the inequality.

Third, since $F_X^{-1}$ is the inverse function, $$F_X^{-1}(F_X(X))=X$$ hence the event $$U\le u$$ is the same as the event $$X\le F_X^{-1}(u)$$ which has probability$$F_X(F_X^{-1}(u))=u$$

In conclusion, $U=F_X(X)$ is thus distributed as $\mathcal U(0,1)$ when $X\sim F_X$.

  • $\begingroup$ Thanks - a typo in the source was leading to some confusion. makes sense now! $\endgroup$
    – codenoob
    Oct 24, 2019 at 5:55

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