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 $P(X=x_0)>0$.

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

  • $\begingroup$ 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. $\endgroup$ – Ian Aug 1 '16 at 16:59
  • $\begingroup$ The support of a probability measure $P$ is any set $S$ with $P(S)=1$. $\endgroup$ – theoGR Aug 1 '16 at 17:43
  • $\begingroup$ 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. $\endgroup$ – Ian Aug 1 '16 at 17:55
  • $\begingroup$ You can make the support of $P$ as large as you want by simply adding sets of measure $0$. $\endgroup$ – theoGR Aug 1 '16 at 18:05
  • $\begingroup$ 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 $\Omega$.) $\endgroup$ – Ian Aug 1 '16 at 18:07

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.