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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.

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  • $\begingroup$ 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

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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 $$ P(X\leq x)=P(F^{-1}(U)\leq x)=P(U\leq F(x))=F(x) \, . $$

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