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Take a distribution function $F$. If $Q_F : [0, 1] \rightarrow \mathbb{R}$ is defined by $Q(p) = \inf\{x | p \leq F(x)\}$, is it true that

$$E_F[X] = \int_0^1 Q_F d\mu$$

where $\mu$ is the Lebesgue measure. I've seen this result a thousand times for ``well-behaved'' distribution functions, but is it true in general? My conjecture is yes, via some carefully-defined-but-ex-post-obvious pushforward-measure argument/change of variables. Can anyone help fill in the blanks?

My conjecture: We can simultaneously consider F as both a measure (as in a Lebesgue-Stieltjes integral) and a measurable mapping from supp(F) to [0,1]. I think that then the pushforward measure of F (using function F) is then equivalent to the Lebesgue measure on [0,1], and the standard change-of-variable formula applies, so

$$ \int_0^1 Q_F \ d(F_{*}F) = \int_{supp(F)} Q_F \circ F \ dF$$

But $F_{*}F = \mu$ (the standard Lebesgue measure) and $Q_F \circ F$ is simply the identity function, so we would have:

$$ \int_0^1 Q_F \ d\mu = \int_{supp(F)} \mu \ dF =: E_F[X].$$

But I am relatively untrained in all of this and have just tried to learn by Googling, so don't have much confidence that any of the above is rigorously correct.

housed_off_space
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    It can be proven that $Q_F$ regarded as a random variable on $((0,1),Leb)$ has cumulative distribution $F$, so the answer is yes. – No-one Apr 27 '22 at 20:27
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    If $X$ has distribution $F$, then for any uniform 0-1 distributed random variable $\theta$, $Q_f\circ\theta\sim X$. So $E[g(X)]=\int^1_0g(Q_F(\theta)),d\theta$ for any nice function $g$. – Mittens Apr 27 '22 at 23:40
  • This is interesting. A follow-up occurs to me: could one define $E[X]$ this way to begin with? In particular, one can always express the expectation as a (possibly improper) Riemann integral this way, right? $E[X] = \int_{0}^1 Q_F(p) dp$. Using the $\int X dP$ or $\int \mu dF$ definition, I don't know a clean Riemann analog for discontinuous $F$. – housed_off_space Apr 28 '22 at 03:00
  • @OliverDíaz nice meaning "measurable" here? – housed_off_space May 06 '22 at 18:33
  • @housed_off_space: measurable and bounded would work. It is enough to work with this functions (one can also start with measurable and nonnegative) The extensions to other integrable functions can also been done by monotone convergence or dominated convergence arguments. – Mittens May 06 '22 at 19:54

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