Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a probability distribution, its quantile function is defined in terms of its distribution function as

$$ Q(p)=F^{-1}(p) = \inf \{ x\in R : p \le F(x) \} $$

I was wondering if, conversely, a quantile function can uniquely determine a distribution and therefore fully describe the probability distribution just as a distribution function does?

Thanks and regards!


Please let me be more specific. Because a CDF is nondecreasing, right-continuous and limit is $0$ when $x \to -\infty$ and $1$ when $x \to \infty$, its quantile function is nondecreasing, left-continuous and a map from $(0,1)$ into $R$. If a function is nondecreasing, left-continuous and a map from $(0,1)$ into $R$, can it become a quantile function of some CDF? When it can, is there a way to represent the CDF in terms of the quantile function using infimum or supremum similar as quantile function in terms of CDF?

share|cite|improve this question
how to show "{" and "}"? Is it same as in Latex? "\{" and "\}" seem don't work. – Tim Oct 20 '10 at 10:33
I believe \\{ and \\} should work instead. I had a similar problem at first. – yunone Oct 20 '10 at 10:55
up vote 3 down vote accepted

Well, if $Q(p)$ is well-defined and monotonic in the interval $(0,1)$, then certainly.

share|cite|improve this answer
But don't we get that from the fact that $F$ is non-decreasing and $\lim_{x\to -\infty} F(x)=0$. – Jyotirmoy Bhattacharya Oct 20 '10 at 10:39
@Jyotirmoy, I interpreted his question as him not knowing $F$, but knowing $Q$. If $Q$ satisfies those conditions, then its inverse would be a valid CDF. – J. M. Oct 20 '10 at 10:42
Thanks! Can you be specific when the quantile function isn't? I mean when a quantile function cannot uniquely determine a distribution function? – Tim Oct 20 '10 at 10:42
Also how to represent cdf in terms of quantile function? – Tim Oct 20 '10 at 10:44
If your proposed $Q(p)$ has a singularity within the unit interval (it can be singular at 0 or 1 though), or oscillates, or other such wonky behavior, then it certainly cannot be the inverse of some CDF. – J. M. Oct 20 '10 at 10:45

I believe the following definition for a CDF is consistent with the definition of a quantile function in your original post:

$F(x) = \sup \{ p\in (0,1) : x \ge Q(p) \}$

This definition indeed makes the quantile function left-continuous as you proposed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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