# Does the quantile function uniquely determine the distribution function?

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!

UPDATE:

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?

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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

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

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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. –  Guess who it is. 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. –  Guess who it is. 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.

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