For a CDF that is not strictly increasing, i.e. its inverse is not defined, define the quantile function

$$F^{-1} (u) =\inf \{x: F(x) \geq u \},\quad 0<u<1. $$

Where U has a uniform $(0,1)$ distribution. Prove that the random variable $F^{-1} (u)$ has cdf $F(x)$.

In case of a strictly increasing CDF the proof is quite easy because the inverse is defined. Define $X=F^{-1} (u)$

$$ P\left[X<x \right]= P \left [F^{-1} (U) \leq x \right]= P \left[U \leq F(x) \right] =F(x) $$

But how do I accomondate the nondecreasing CDF whose inverse is given by the quantile function? I am a begginer so any help is welcome. Thank you.

  • $\begingroup$ If $F$ is not increasing, the definition of $F^{-1}$ you give on the second line is still ok – it’s not the inverse but it’s well defined. You can use it... $\endgroup$
    – Elvis
    Commented Nov 27, 2013 at 12:55
  • $\begingroup$ Is this for some subject? $\endgroup$
    – Glen_b
    Commented Nov 27, 2013 at 17:44
  • $\begingroup$ @Glen_b Yes indeed. It is part of the chapter on Monte Carlo techniques. $\endgroup$
    – JohnK
    Commented Nov 27, 2013 at 18:15
  • $\begingroup$ @Glen_b Not a textbook question. My textbook simply states that the strictly increasing CDF assumption can be relaxed by using the quantile function. It was not a textbook exercise but a general query. I will nevertheless use the tag from now on. $\endgroup$
    – JohnK
    Commented Nov 27, 2013 at 18:25
  • $\begingroup$ It doesn't have to be specifically an exercise from a textbook; indeed the instruction "Prove ..." rather than a question 'How do I prove ...' makes it look like assigned work* rather than 'a general question'. * (but it's not restricted to assigned work either.) In any case, thanks for adding the tag. $\endgroup$
    – Glen_b
    Commented Nov 27, 2013 at 18:29

1 Answer 1


Let $U$ be a $\mathrm{U}[0,1]$ r.v. Let $F$ be a distribution function. Remember that every distribution function is non decreasing and right continuous . Define the quantile function $$ F^{-1}(u) = \inf\,\{x:u \leq F(x)\}. $$ Drawing a picture

enter image description here

we see that $F^{-1}(u)\leq x$ if and only if $u\leq F(x)$. Please, make sure that you understand both implications. Therefore, if $X=F^{-1}(U)$, then $$ P(X\leq x)=P(F^{-1}(U)\leq x)=P(U\leq F(x))=F(x) \, . $$


Your Answer

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

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