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Let $\{X_n\}_{n\geq 1}$ be a sequence of i.i.d. random variables with mean zero and finite fourth moment. Let $S_n=X_1+X_2+\dots+X_n$.

Show that $$\mathbb{E}[S_{_n}^4]=\mathcal{O}(n^2)$$

I feel like this is more of a calculus problem but I don't know how to show this (simple) fact.

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    $\begingroup$ This is not true as stated. Are they zero-mean? If not... take the trivial $X_i = 1$ a.s.,so that $S_n = n$ a.s., as a counterexample. $\endgroup$ – Clement C. Nov 18 '19 at 1:30
  • $\begingroup$ BrianMoehring and Clement C. you are right, the mean has to be zero. I have edited it. $\endgroup$ – Babado Nov 18 '19 at 1:35
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Rather than being a "calculus problem", it actually ends up being a bit of combinatorics.

When you apply the multinomial theorem to expand $S_n^4$, the only monomials that will result in a non-zero expectation must have each distinct variable appearing with even degree. This leaves either $X_i^4$ or $X_i^2X_j^2$. There are $O(n)$ of the first kind and $O(n^2)$ of the second kind.

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    $\begingroup$ That's a good answer to a different question. As stated, there is no zero-mean assumption, and the thing to prove is false. $\endgroup$ – Clement C. Nov 18 '19 at 1:30
  • $\begingroup$ @ClementC. The question has been edited to assume that the $\mathbb E[X_1]=0$, now is it correct? $\endgroup$ – Math1000 Nov 18 '19 at 1:37
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    $\begingroup$ @ClementC. it was clear that this was the question the OP meant to ask, and I figured it was a matter of time before the question was fixed rather than make a big deal about it... $\endgroup$ – pre-kidney Nov 18 '19 at 1:38
  • $\begingroup$ @pre-kidney could you explain how did you get $O(n^2)$. It's because $\binom{n}{2} \in O(n^2)$? $\endgroup$ – Babado Nov 18 '19 at 1:52
  • $\begingroup$ @Babado Yes, precisely. $\endgroup$ – pre-kidney Nov 18 '19 at 2:16

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