# Inverse function for a non-decreasing CDF

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.

• 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... – Elvis Nov 27 '13 at 12:55
• Is this for some subject? – Glen_b -Reinstate Monica Nov 27 '13 at 17:44
• @Glen_b Yes indeed. It is part of the chapter on Monte Carlo techniques. – JohnK Nov 27 '13 at 18:15
• @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. – JohnK Nov 27 '13 at 18:25
• 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. – Glen_b -Reinstate Monica Nov 27 '13 at 18:29

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