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Let $\Phi$ be the CDF of a standard Gaussian distribution, i.e. $$\Phi(x):=\int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-y^2/2}dy,\quad \forall~ x\in \mathbb R.$$ Denote by $\Phi^{-1}$ its inverse function. For every $p>0$, define $F_p: \mathbb R\times \mathbb R_+\to \mathbb R_+$ by

$$F_p(m,\sigma):=\left(\int_0^1 \big|(\sigma-1)\Phi^{-1}(t)+m\big|^p dt\right)^{1/p},\quad \forall~ m\in \mathbb R,~ \sigma\in \mathbb R_+.$$

Does there exist constants $c(p)\equiv c <C\equiv C(p)$ (only depending on $p$) such that

$$cF_2(m,\sigma) \le F_p(m,\sigma) \le CF_2(m,\sigma),\quad \forall~ m\in \mathbb R,~ \sigma\in \mathbb R_+?$$

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$\newcommand\si\sigma$Yes.

Indeed, $$F_p:=F_p(m,\si)=\|(\si-1)Z+mR\|_p,$$ where $Z$ and $R$ are independent random variables, $Z$ is standard normal, $R$ is Rademacher (with $P(R=\pm1)=1/2$), and $\|X\|_p:=(E|X|^p)^{1/p}$.

So, the desired result follows by the central limit theorem and the Haagerup inequalities, which imply that, for each real $p>0$, $$G_p:=\Big\|\sum_{i=1}^n a_iR_i\Big\|_p$$ differs from $G_2$ by at most a positive real factor depending only on $p$, where the $a_i$'s are any nonzero real numbers and the $R_i$'s are independent Rademacher r.v.'s.

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    $\begingroup$ Dear Iosif, I really like this fancy alternative expression of $F_p$. There is only one step that is not perfectly clear to me : The "equivalence" between $G_p$ and $G_2$ is identified by a constant only depending on $p$ (not on the deterministic coefficients $a_i$). Does this include the case $(\sigma-1)z + mR$? Here $z$ is an arbitrary realisation of $Z$. My understanding is that $(\sigma-1)z$ cannot written as the product of a real number and a Rademacher (unless $\sigma=1$) $\endgroup$
    – Fawen90
    Commented Jun 12 at 17:54
  • $\begingroup$ @Fawen90 : We do not have to deal here with individual values of $Z$. Rather, we use the fact that, by the central limit theorem, $Z$ is a limit in distribution as $n\to\infty$ of $Z_n:=\sum_{i=1}^n \frac1{\sqrt n}\,R_i$. (Of course, we also need to use here uniform integrability of the $|Z_n|^p$'s.) $\endgroup$
    – Iosif Pinelis
    Commented Jun 12 at 17:59
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    $\begingroup$ Thank you Iosif. Now I fully understand your reasoning. This proof is so tricky and so elegant. I show the result for $p=1$ by computing explicitly the integral, while for general $p$ it's harder. Really beautiful $\endgroup$
    – Fawen90
    Commented Jun 12 at 18:05
  • $\begingroup$ @Fawen90 : Thank you for your appreciation. Haagerup's result is very impressive. $\endgroup$
    – Iosif Pinelis
    Commented Jun 12 at 18:21

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