# Quantile function confusion about $\min\{x \text{ | }F(x) \geq p\}$

I am currently reviewing the quantile function for discrete random variables but I am a bit confused.

We use the following definition of the quantile function: $$\tilde{x}_p = \min\{x \in \mathbb{R}\mid F(x) \geq p\}$$ where $$F(x)=\operatorname{Pr}(X \leq x)$$

Why is it required to have $$F(x) \geq p$$ if $$F(x)$$ only gives the probability of $$\{X \leq x\}$$?
The $$\text{greater than}$$ relation confuses me.

Because the Distribution function gives me the probability that my discrete random variable $$X$$ has a value less or equal to some value $$x$$.

So I would assume that the quantile function has to give a $$x$$ for a given percentage $$p$$ which is less or equal than $$p$$.
But the definition says that the $$x$$ has to be the smallest value so that $$F(x)$$ is greater than or equals $$p$$?

Does it have something to do with the random variable being $$discrete$$?
If yes could someone elaborate that?

I read different textbooks but even those have different definitions of the quantile.

Consider the random variable $$X$$, which we define to be constant $$0$$. It has the distribution $$F(x) = \begin{cases} 0 & \text{if } x\leq 0 \\ 1 & \text{else} \end{cases}$$

Define the quantile function $$F^{-1}(\alpha) = \inf \{ x\in\mathbb{R} \,: F(x) \geq \alpha \}$$

Consider the median of $$X$$. It is supposed to be $$F^{-1}(1/2)$$. We know by intuition that it should be $$0$$, since that is the only value that $$X$$ will ever take.

Now compare these sets and their infimums \begin{align} S_0 &= \{x\,: F(x) = 1/2 \} = \emptyset &&\inf S_0 = -\infty \\ S_1 &= \{x\,: F(x)\geq 1/2\} = [0,\infty) && \inf S_1 = 0 \end{align}

So if you only allow equality, and not the greater-than-or-equal relation, you get the wrong result.

This fact stems from the discontinouities in the distribution function $$F(x)$$ and the convention that it is right-continous. If the distribution function was left-continous, we could do some other variant of the quantile function as well.

• The $F(x)$ that you have defined in your first displayed equation is not a right-continuous function, and for your $F(x)$, it is not true that $S_1 = \{x\colon F(x) \geq \frac 12\}$ is $[0, \infty)$; it is $(0,\infty)$ since your $F(0)$ equals $0$. Mar 10 at 2:11
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