In our lecture, the generalized inverse of a function $F$ is defined as \begin{equation} F^{-}(u) := \inf_x \{ F(x) \ge u \}. \end{equation}

Then we are introduced to the so-called "Flip Flop Formula", i.e. for a non-decreasing function $F$ and its generalized inverse $F^{-}$, $$x < F^{-}(u) \Leftrightarrow F(x) < u, \forall x, u.$$

Here is the proof: since $F$ is non-decreasing, \begin{equation}\begin{aligned} & x < F^{-}(u) = \inf_x \{ F(x) \ge u \} \\ \Leftrightarrow & x \notin \{ F(x) \ge u \} \\ \Leftrightarrow & F(x) < u. \end{aligned} \end{equation}

The proof is straightforward, but the formula itself is somewhat abstract. What's the intuition behind it? It would be great if someone could provide something like a graphical illustration.

  • $\begingroup$ I believe I have a counterexample to the formula: let $F(x) = \lceil x \rceil$. Then $F^-(u) = \operatorname{inf} \{x : \lceil x \rceil \geq u \}$, which equals $u - 1$ if $u$ is an integer and $\lfloor u \rfloor$ otherwise. Let $u$ be a non-integer and $x = \lfloor u \rfloor$: then $x < F^-(u)$ is false (the two are equal), but $F(x) < u$ is true (as $F(x) = x < u$. $\endgroup$ Sep 29, 2021 at 17:54
  • $\begingroup$ If $A$ is a real interval with right endpoint at $\infty$, then the inference from $x \notin A$ to $x < \inf A$ is not valid, as $A$ might be open at its left endpoint. Was the formula defined only for a specific class of functions (e.g. continuous functions, or functions with discrete domains) such that $\{x : F(x) \geq u\}$ is guaranteed to be a closed set? $\endgroup$ Sep 29, 2021 at 17:58
  • $\begingroup$ @ConnorHarris Actually, yes! The course I'm taking is named Distribution Theory, so it's likely that every $F$ is implicitly assumed to be a cumulative distribution function. $\endgroup$
    – nalzok
    Oct 3, 2021 at 2:39
  • $\begingroup$ OK, in that case, the formula is valid for continuous distributions, or for discrete distributions with the convention that an integral over a delta function $\int_{-\infty}^x \delta(t)\, dt$ is $1$ if $t \geq 0$ and $0$ if $t < 0$. $\endgroup$ Oct 3, 2021 at 15:01

1 Answer 1


I have found a counterexample to the formula as stated, but it is valid if the case $x = F^-(u)$ is excluded.

Graphically, $F^-(u_0)$ is the point $x_0$ chosen so that the graph of $F$ to the left of $x_0$ is strictly below $u_0$, and the graph to the right of $x_0$ is non-strictly above $u_0$. That is, for any point $x \neq x_0$, then either $x < x_0$ and $F(x) < u_0$, or $x > x_0$ and $F(x) \geq u_0$.

$F(x_0)$ itself can be either below or above $u_0$: consider the functions $F_1(x) = \lceil x \rceil$ and $F_2(x) = \lfloor x + 1 \rfloor$, which have the same inverse $F^-(u) = \lceil u - 1 \rceil$. Thus, if $u$ is a non-integer, then $x = F^-(u) = \lfloor u \rfloor$ and so $F_1(x) = x < u < F_2(x) = x + 1$.

(I believe the formula is valid as written if the graph of $F$ has no segments with an open left endpoint: that is, if the right-handed limit $\lim_{x' \to x^+} F(x') = x$ for all $x$. This guarantees that $\{x: F(x) \geq u\}$ is topologically closed. Note that $F_1$ above violates this criterion.)


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