Suppose that $X_1,\ldots,X_n$ are independent random variables with $\operatorname E X_k=0$ and $\operatorname E |X_k|^p<\infty$ with $1<p<2$ for each $1\le k\le n$. I am interested in the inequalities that establish a lower bound for the $p$-th absolute moment of $S_n=\sum_{k=1}^nX_k$ in terms of the $p$-th absolute moments of $X_1,\ldots,X_n$.

I was able to find an upper bound for $E|S_n|^p$. von Bahr and Esseen (1965) among other results established that $$ \operatorname E|S_n|^p\le2\sum_{k=1}^n\operatorname E|X_k|^p. $$ But I can't seem to find an inequality that establishes a lower bound for $\operatorname E|S_n|^p$. My questions are as follow:

Are there any inequalities that establish a lower bound for $\operatorname E|S_n|^p$ in terms of the $p$-th absolute moments of $X_1,\ldots,X_n$? Is it true that $\operatorname E|S_n|^p\ge C\sum_{k=1}^n\operatorname E|X_k|^p$ with some positive constant $C$?

Any help is much appreciated!

  • $\begingroup$ Do you wish to assume than the $X_i$ are not only independent, but also identically distributed? Because that is currently missing. $\endgroup$
    – wolfies
    Feb 4, 2015 at 11:27
  • $\begingroup$ @wolfies I'm interested in the case when $X_i$ are not necessarily identically distributed. $\endgroup$
    – Cm7F7Bb
    Feb 4, 2015 at 11:32
  • $\begingroup$ I would suggest you change the title from 'moment' to 'fractional moment'. The term $r^{th}$ moment is conventionally taken to refer to integer values of $r$. $\endgroup$
    – wolfies
    Feb 4, 2015 at 11:43
  • $\begingroup$ Gaussians show that you do not have better than an $\ell_2$ lower bound. Mean zero independent RVs are 3-unconditional and $L_p$ has cotype 2 when $p<2$, so you indeed do have an $\ell_2$ lower estimate. See, e.g., the book by Albiac and Kalton. $\endgroup$ Feb 4, 2015 at 13:47
  • $\begingroup$ @BillJohnson Thank you very much for the comment and the reference. So it is not possible to get a better lower bound than the lower bound in the Marcinkiewicz–Zygmund inequality, right? $\endgroup$
    – Cm7F7Bb
    Feb 5, 2015 at 10:24

1 Answer 1


If the $X_i$ are i.i.d. Gaussian with variance $1$, then you have $$ c_p := \mathbb{E} |X_k|^p = \frac{2^{p/2} \Gamma(\frac{p+1}{2})}{\sqrt{\pi}}.$$ The variable $S_n$ is also Gaussian with variance $n$, therefore you have $$\mathbb{E} |S_n|^p = c_p n^{p/2}.$$

Hence, $\frac{\sum_{k=1}^n \mathbb{E} |X_k|^p}{\mathbb{E} |S_n|^p} = n^{1-p/2} \rightarrow \infty$ for $1<p<2$. At least, it means that you cannot hope for a constant $C$ as you expected.

  • $\begingroup$ Is it true that $\mathbb{E} |S_n|^p \geq c n^{p/2-1} \sum \mathbb{E} |X_i|^p$? $\endgroup$
    – Omer
    Feb 4, 2015 at 13:59
  • $\begingroup$ @Omer If the $X_i$ are i.i.d with finite variance (let say $1$), the answer is yes. Indeed, $S_n / \sqrt{n} \rightarrow \mathcal{N}(0,1)$ with the central limit theorem. Hence we have $\mathbb{E} [\lvert S_n/\sqrt{n} \rvert^p] \rightarrow \mathbb{E}[\lvert \mathcal{N}(0,1) \rvert^p]$ and from that $$\mathbb{E}[|S_n|^p] \sim Cn^{p/2}.$$ $\endgroup$
    – Goulifet
    Feb 4, 2015 at 14:45

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