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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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    how to show "{" and "}"? Is it same as in Latex? "\{" and "\}" seem don't work.2010-10-20
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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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    But don't we get that from the fact that $F$ is non-decreasing and $\lim_{x\to -\infty} F(x)=0$.2010-10-20
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    @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.2010-10-20
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    Thanks! Can you be specific when the quantile function isn't? I mean when a quantile function cannot uniquely determine a distribution function?2010-10-20
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    Also how to represent cdf in terms of quantile function?2010-10-20
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    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.2010-10-20
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    Your second question is the difficult part. The [inverse function theorem](http://mathworld.wolfram.com/InverseFunctionTheorem.html) guarantees the existence, but construction might require defining a new function. It's a bit similar to the situation with the Lambert function...2010-10-20
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    @J.M.: Thanks! Please let me be more specific. Because a CDF is nondecreasing, right-continous and limit is 0 when $x \rightarrow -\infty$ and 1 when $x \rightarrow \infty$, its quantile function is nondecreasing, left-continous and a map from (0,1) into $R$. If a function is nondecreasing, left-continous and a map from (0,1) into $R$, can it become a quantile function of some CDF? When it can, as you mentioned the inverse function theorem, there is no way to represent the CDF in terms of the quantile function using infimum or supremum similar as quantile function in terms of CDF?2010-10-20
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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.