MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


  • $X_i$ are independent random variables.
  • $|X_i| < 1$
  • $E[X_i] = 0$
  • $X = \sum_i^n X_i$
  • $var(X)=\sigma$


$$ E(X^p)^{1/p} = O(\sqrt{p}\sigma +p)$$ for all even p

Things I've tried:

First note, that all terms with more than p/2 terms are 0 (since $E[X_i] = 0$).

Furthermore, note that $\sum_i X^{4} \leq \sum_i X^2$

So this ends up being some way to count the various terms involving exactly $t$ variables. I don't know how to count this. What should I try?

share|cite|improve this question
Where does this question come from? – Igor Rivin Aug 22 '12 at 15:41
Look at Lemma 4, page 172, in Billingsley's book Convergence of probability measures. Lemma is for p=4, but it works for all even p. If I am not mistaken, this lemma gives the bound you are looking for. – user12345678 Aug 22 '12 at 16:08
@waler: Can you identify which section that's from? In my copy of the second edition of that book, there is no Lemma 4 on or near page 172. – Mark Meckes Aug 22 '12 at 18:04
@Mark Meckes: 1968 edition, chapter 4. dependent variables, section 20: mixing processes, paragraph on moment inequalities. – user12345678 Aug 22 '12 at 18:40

Take a loot at R. Ibragimov and Sh. Sharakhmetov, The Exact Constant in the Rosenthal Inequality for Random Variables with Mean Zero, Theory Probab. Appl., 46(1), 127–132. (6 pages) Read More:

Abstract Let $\xi_1, \ldots, \xi_n$ be independent random variables with ${\bf E}\xi_i=0,$ ${\bf E}|\xi_i|^t<\infty$, $t>2$, $i=1,\ldots, n,$ and let $S_n=\sum_{i=1}^n \xi_i.$ In the present paper we prove that the exact constant ${\overline C}(2m)$ in the Rosenthal inequality $$ {\bf E}|S_n|^t\le C(t) \max \Bigg(\sum_{i=1}^n{\bf E}|\xi_i|^t,\ \Bigg(\sum_{i=1}^n {\bf E}\xi_i^2\Bigg)^{t/2}\Bigg) $$ for $t=2m,$ $m\in {\bf N},$ is given by $$ \overline C(2m)=(2m)! \sum_{j=1}^{2m} \sum_{r=1}^j \sum \prod_{k=1}^r \frac {(m_k!)^{-j_k}} {j_k!}, $$ where the inner sum is taken over all natural $m_1 > m_2 > \cdots > m_r > 1$ and $j_1, \ldots, j_r$ satisfying the conditions $m_1j_1+\cdots+m_rj_r=2m$ and $j_1+\cdots+j_r=j$. Moreover $$ \overline C(2m)={\bf E}(\theta-1)^{2m}, $$ where $\theta $ is a Poisson random variable with parameter 1.

In your case, one gets $$ {\bf E}|S_n|^{2m} \le \overline C(2m) \max( n, (var S_n)^{m})= \overline C(2m) \max( n, n^{m}c^m). $$ where $c=var(\xi_i)$. Thus for large $m$, $$ \Bigl( {\bf E}|S_n|^{2m}\Bigr)^{\frac 1{2m}} \le (\overline C(2m))^{1/(2m)} \sqrt{n} c. $$ Finally, they cite earlier papers where it was shown that $$ \overline C(t)= O( t/\log t). $$ Ans since is $(t/\log t)^{1/t}<\sqrt{t}$ for $t>2$ we are done.

share|cite|improve this answer
That is a nice paper but is overkill for what the OP asks. For the $t/ \log t$ result, see the first paper by Schechtman, Zinn, and myself. It also contains an easy proof of the bound $t/ (\log t)^{1/2}$. – Bill Johnson Aug 22 '12 at 17:11
Johnson, W. B.; Schechtman, G.; Zinn, J. Best constants in moment inequalities for linear combinations of independent and exchangeable random variables. Ann. Probab. 13 (1985), no. 1, 234–253. – Bill Johnson Aug 22 '12 at 17:14
@Bill Johnson: absolutely, severe overkill. result must follow also from some easier inequalities as well. – user12345678 Aug 22 '12 at 18:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.