basic Quantile proves

Let this be my definition of a quantile funktion. X is a real-valued random variable. And let F be it's distribution function. then \begin{align*} F^{-1}(a):=\inf\{x\in \mathbb{R}: F(x) \ge a\}. \end{align*} Now I want to solve, how can I prove this simple equation: \begin{align*} F^{-1}(a)= x_0 \Leftrightarrow P(X \le x_o ) = a, \text{ almost surely} \end{align*}

or what assumpations are needed in order to make it happen?

If $F^{-1}(a)=x_0$ and $F$ is continuous at $x_0$ then $F(x_0)=P(X \leq x_0)=a$. It can happen that $F$ is not continuous at $x_0$. In this case either $F(x_0)=a$ or else:

1. $\lim_{x \to x_0^-} F(x)<a$
2. $F(x_0)>a$
3. There is no $y$ such that $F(y)=a$.

The only way to have $F$ be discontinuous at $x_0$ is if $P(X=x_0)>0$.

• by continous, do you mean in the statistical sense? or the analytical`? – Aud Jul 22 '16 at 18:56
• @Audrey32 $F$ is just a function from $\mathbb{R}$ to $\mathbb{R}$, so yes, I just mean continuous in the usual analytical sense. – Ian Jul 22 '16 at 19:04
• Thanks alot. Can you give an example of a non-pathological Distribution function that is not continous? – Aud Jul 22 '16 at 19:56
• Take any discrete distribution; its CDF is not continuous at the points which have positive probability. – Ian Jul 22 '16 at 20:13
• but what about a (statistically) fully continous distribution . Is it possible that a function like this is not continous in the analytical way? – Aud Jul 22 '16 at 20:47

There is a problem in the discrete case. If two consecutive points $x_1<x_2$ are in the support of $F$ and $F(x_1)<a<F(x_2)$ then $F^{-1}(a)=x_2$ but $F(x_2)\neq a$. What the problem asks is a special case of the first line and holds only if $a$ is in the support of $F$.

• I'm pretty sure that when you say "the support of $F$", you actually mean the support of the PMF $f$. That is, I think you actually mean that if $x$ is in the support of the PMF $f$ and $F(x^-)<a<F(x)$ then $a$ will not be in the range of $F$, in which case (as I said in my answer) the inversion is not possible. – Ian Aug 1 '16 at 16:59
• The support of a probability measure $P$ is any set $S$ with $P(S)=1$. – theoGR Aug 1 '16 at 17:43
• OK, except the support of $F$ (not of the induced measure $P$) is way larger than you want; it never has any reasonable notion of "consecutive points". For example the support of the binomial CDF is $[0,\infty)$. As I said, I think I ultimately know what you meant, but what you actually said is unfortunately also defined and doesn't mean what you meant. So it comes across as confusing. – Ian Aug 1 '16 at 17:55
• You can make the support of $P$ as large as you want by simply adding sets of measure $0$. – theoGR Aug 1 '16 at 18:05
• You're not reading what I'm saying: what you wrote in your answer is "the support of $F$". $F$ is a real-valued function of a real variable, so its "support" is already defined without any reference to measure theory at all. So when you say "the support of $F$", it makes no sense to speak of "two consecutive points" in the support of $F$. (Also, I said the induced measure $P$, which is a measure on $\mathbb{R}$, not the underlying measure on $\Omega$.) – Ian Aug 1 '16 at 18:07