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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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    I believe \\{ and \\} should work instead. I had a similar problem at first.2010-10-20

2 Answers 2

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

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    Yes to the first. For the second, it depends, but more often than not, you can't say anything more than "$p=F(y)$ is the function such that $Q(p)=y$".2010-10-20
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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.