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Let $(\Omega, \mathcal A, P)$ be a probability space. Let $X:\Omega \rightarrow \mathbb R$ be an $L^1(\Omega, \mathcal A, P)$ random variable.

We define the distribution function of $X$ by

$$F(x) = P(X \leq x)$$

and the quantile function of $X$ by

$$Q(\alpha)= \inf \{x \in \mathbb R : F(x) \geq \alpha \}$$

Is it true that we always have $Q \in L^2(]0,1[)$ ? Meaning that

$$\int_0^1 Q(\alpha)^2 \, d\alpha < \infty$$

It is true for the "famous" distributions like the normal distribution, or the Laplace distribution. And it is clear that when $X$ is not $L^1$, it does not hold (Cauchy distribution).

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    $\begingroup$ Does not $F(x)=\frac{1}{x^2}$ on $\Omega=[1,\infty)$ give a counter-example? In this case $Q(\alpha)=F^{-1}(\alpha)=\frac{1}{\sqrt{\alpha}}$,which is not in $L^2(0,1)$. $\endgroup$ Commented Nov 1, 2019 at 1:54
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    $\begingroup$ @JosiahPark: Or rather, $F(x) = 1-\frac{1}{x^2}$ to make it a distribution function. $\endgroup$ Commented Nov 1, 2019 at 13:04
  • $\begingroup$ @NateEldredge Yes, thanks. This question feels like a more standard real analysis problem. $\endgroup$ Commented Nov 1, 2019 at 14:12

2 Answers 2

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Let $F^{-1}:=Q$. It is well known that $Y:=F^{-1}(U)=Q(U)$ equals $X$ in distribution, where $U$ is any random variable uniformly distributed on $(0,1)$. Therefore, for any real $p>0$, $$\int_0^1|Q(u)|^p\,du=E|Y|^p=E|X|^p.$$ So, $Q\in L^p((0,1))$ iff $X\in L^p(\Omega,\mathcal A,P)$, for each real $p>0$.

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Sinve you mentioned famous examples, here is a famous counterexample: power law distributions.

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