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A quantile function Q is defined in terms of its distribution function F as:

$Q(p) = \inf \{ x\in R : p \le F(x) \}, p \in (0,1).$

I was wondering:

(1) if the definition of Q can be extended to $[0,1]$ by letting $ x \in R \cup \{\infty, -\infty \}$? if not, why?

(2) if it is correct that a distribution function can always be represented in terms of a quantile function defined on $(0,1)$ as:

$F(x) = \sup \{ p\in (0,1) : Q(p) \le x\}, x \in R-\{x|F(x)=0\text{ or }1\}.$

(3) if it is correct that a distribution function can always be represented in terms of a quantile function defined on $[0,1]$ (if (1) is correct) as:

$F(x) = \sup \{ p\in [0,1] : Q(p) \le x\}, x \in R \cup \{\infty, -\infty\}.$

Thanks and regards!


UPDATE:

One more question:

(4) if the followings are correct:

$F$ is discontiunous at $x_0$ with $F(x_0)=p_0$ and $lim_{x \rightarrow x_0-}F(x)= p_1$ $\Longleftrightarrow$ $Q$ is constant over $[p_1, p_0]$, taking value $x_0$.

$F$ is constant over $[x_1, x_0]$ taking value $p_0$ $\Longleftrightarrow$ $Q$ is discontiunous at $p_0$ with $Q(p_0)=x_1$ and $lim_{x \rightarrow x_1+}F(x)= x_0$.

Thanks!

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1 Answer 1

up vote 1 down vote accepted

I think your last shoud be $Q(p_0)=x_1$ and $lim_{p\rightarrow p_0+} Q(p) =x_0$ with the rest OK

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