# Inverse distribution function (quantile function)

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 $$a, causing $$F_X$$ to be constant). This problem can be solved by defining, for $$p \in [0,1]$$, the ''generalized inverse distribution function": $$$$F^{-1}(p) = \inf \{x \in \mathbb{R}: F(x) \geq p \}.$$$$

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.

• No, consider your own example. – Yves Daoust Nov 9 '18 at 13:29