Approximating Moment of Sum of RVs

Question

Given

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

Prove:

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

Answer 1

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: http://epubs.siam.org/doi/abs/10.1137/S0040585X97978762

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.