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The Marcinkiewicz–Zygmund inequality states that

$$ {\displaystyle A_{p}E\left(\left(\sum _{i=1}^{n}\left\vert X_{i}\right\vert ^{2}\right)_{}^{p/2}\right)\leq E\left(\left\vert \sum _{i=1}^{n}X_{i}\right\vert ^{p}\right)\leq B_{p}E\left(\left(\sum _{i=1}^{n}\left\vert X_{i}\right\vert ^{2}\right)_{}^{p/2}\right)}. $$ I wonder if there is any sort of result for the ratio of the two constant $A_p/B_p$, in particular for the case $p=1$.

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You are missing the crucial independence and zero-mean conditions, without which the Marcinkiewicz–Zygmund inequalities will not hold in general.

These inequalities are actually as follows: for each real $p\ge1$, some positive real $A_p$ and $B_p$ and all $n$-tuples $X:=(X_1,\ldots,X_n)$ of independent zero-mean random variables (r.v.'s) \begin{equation*} A_p E\|X\|_2^p\le E|S_X|^p\le B_p E\|X\|_2^p, \tag{1}\label{1} \end{equation*} where $S_X:=\sum_1^n X_i$ and $\|X\|_2:=(\sum_1^n X_i^2)^{1/2}$.

Consider first the case when the $X_i$'s are symmetric(ally) distributed. Then, conditioning on the $|X_i|$'s, we see that the best constants $A_p$ and $B_p$ in \eqref{1} are the same as the best constants $A_p$ and $B_p$ in the case when the $X_i$'s are real scalar multiples of Rademacher r.v.'s (see details on this at the end of this answer). In the latter, scalar-multiples-of-Rademacher case, the best constants $A_p$ and $B_p$ were obtained by Haagerup; let us denote these symmetric-case constants by $A_{p,S}$ and $B_{p,S}$, respectively.

Given now arbitrary independent zero-mean r.v.'s $X_1,\ldots,X_n$, we can symmetrize them by considering the $n$-tuple $X-X'$ of symmetric independent r.v.'s $X_i-X'_i$, where $X':=(X'_1,\ldots,X'_n)$ is an independent copy of $X=(X_1,\ldots,X_n)$. Then, by Jensen's inequality, \begin{equation*} E\|X\|_2^p\le E\|X-X'\|_2^p\le2^p E\|X\|_2^p \end{equation*} and \begin{equation*} E|S_X|^p\le E|S_X-S_{X'}|^p\le2^p E|S_X|^p. \end{equation*} So, \begin{equation*} E|S_X|^p\le E|S_X-S_{X'}|^p=E|S_{X-X'}|^p \\ \le B_{p,S}E\|X-X'\|_2^p \le2^p B_{p,S}E\|X\|_2^p \end{equation*} and \begin{equation*} 2^p E|S_X|^p\ge E|S_X-S_{X'}|^p =E|S_{X-X'}|^p \\ \ge A_{p,S}E\|X-X'\|_2^p \ge A_{p,S}E\|X\|_2^p. \end{equation*} So, \eqref{1} holds for each real $p\ge1$ and all $n$-tuples $X:=(X_1,\ldots,X_n)$ of independent zero-mean r.v.'s with \begin{equation*} A_p=A_p^*:=2^{-p}A_{p,S}\quad\text{and}\quad B_p=B_p^*:=2^p B_{p,S}. \end{equation*} In particular, letting here $p=1$, we get \begin{equation*} A_1^*:=2^{-3/2}\quad\text{and}\quad B_1^*:=2. \end{equation*} However, again by Jensen's inequality, for all $p\in[1,2]$ and, in particular, for $p=1$, \begin{equation*} E|S_X|^p\le(ES_X^2)^{p/2}=E\|X\|_2^p, \end{equation*} so that the second inequality in \eqref{1} actually holds with $B_p=1$.

Details on the conditioning: Here the conditioning on the $|X_i|$'s can be done simply as follows: Note that the joint distribution of the symmetric r.v.'s $X_1,\ldots,X_n$ is the same as the joint distribution of the r.v.'s $R_1|X_1|,\ldots,R_n|X_n|$, where $R_1,\ldots,R_n$ are independent Rademacher r.v.'s (with $P(R_i=\pm1)=1/2$) that are also independent of $X_1,\ldots,X_n$. So, for $Z:=(R_1|X_1|,\ldots,R_n|X_n|)$, first we average $|S_Z|^p$ and $\|Z\|_2^p$ with respect to the $R_i$'s with values of the $|X_i|$'s fixed and thus get \begin{equation*} A_{p,S} E\big(\|Z\|_2^p\big|\,|X|\big) \le \big(E|S_Z|^p\big|\,|X|\big) \le B_{p,S} E\big(\|Z\|_2^p\big|\,|X|\big), \tag{2}\label{2} \end{equation*} where $|X|:=(|X_1|,\ldots,|X_n|)$, and $A_{p,S}$ and $B_{p,S}$ are Haagerup's constant factors, as before.
Next, we average the terms in \eqref{2} with respect to the $|X_i|$'s, to remove the conditioning on $|X|$ and thus get \begin{equation*} A_{p,S} E\|Z\|_2^p \le E|S_Z|^p \le B_{p,S} E\|Z\|_2^p. \end{equation*} Finally, we recall that $Z$ has the same distribution as $X$.


As seen from the above, the constant factors in the "weighted Rademacher" case (and hence/equivalently, in the symmetric case) are better (by the factor $2^p$) than the constant factors in the general zero-mean case. Also, the "weighted Rademacher" case seems to be the minimal one. Moreover, it is easy to see that even in the "non-weighted Rademacher" case (with the scalars multiplying the $R_i$'s all equal $1$), Haagerup's constant factors $A_{p,S}$ and $B_{p,S}$ cannot be improved, if you want them to hold for all $n$.

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  • $\begingroup$ Why $E|S_X|^p \leq E|S_X-S_{X'}|^p$ and similar? $\endgroup$ Commented Nov 15, 2023 at 21:46
  • $\begingroup$ I see why: $|x+y|^p-|x|^p \geq p|x|^{p-2} xy$ which follows from the convexity of $|x|^p$. $\endgroup$ Commented Nov 15, 2023 at 22:04
  • $\begingroup$ However, could you explain in more detail how to reduce the symmetric case to the Rademacher functions? $\endgroup$ Commented Nov 15, 2023 at 22:42
  • $\begingroup$ @GiorgioMetafune : I have added details on the conditioning. Please let me know if more details are needed. $\endgroup$ Commented Nov 16, 2023 at 2:12
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    $\begingroup$ @LSpice : Thank you for your edits. They were probably removed accidentally, as I edited in an offline editor a version of the answer preceding your edits. $\endgroup$ Commented Nov 16, 2023 at 14:01

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