From Wiki: If the CDF ''F'' is strictly increasing and continuous then $ F^{-1}( p ), p \in [0,1], $ is the unique real number $ x $ such that $ F(x) = p $. In such a case, this defines the "inverse distribution function" or [[quantile function]].

Some distributions do not have a unique inverse (for example in the case where $f_X(x)=0$ for all $a<x<b$, causing $F_X$ to be constant). This problem can be solved by defining, for $ p \in [0,1] $, the ''generalized inverse distribution function": \begin{equation} F^{-1}(p) = \inf \{x \in \mathbb{R}: F(x) \geq p \}. \end{equation}

If $S \subset \mathbb{R}$ is a bounded set, such that $x$ takes values only in $S$, i.e. $x \in S$, then for $0< p< 1$, is the above definition identical to:

$$ F^{-1}(p) = \sup \{x \in S: F(x) \leq p \}? $$

I think it's not, due to the following reason:

Suppose $a\in S$, and we define: $$ F(x) = \begin{cases} 0.5 \; \text{if}\; x < a\\ 1 \; \text{if}\; x \geq a\\ \end{cases} $$ In the above example, for $p = 0.5$, the first definition gives $F^{-1}(0.5) = a$, while the second definition seems to be invalid, as supremum may not exist.

  • 1
    $\begingroup$ No, consider your own example. $\endgroup$ – Yves Daoust Nov 9 '18 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.