# Given $S_n=X_1+X_2+\dots+X_n$ show that $\mathbb{E}[S_n^4]=\mathcal{O}(n^2)$

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.

• 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. – Clement C. Nov 18 '19 at 1:30
• BrianMoehring and Clement C. you are right, the mean has to be zero. I have edited it. – Babado Nov 18 '19 at 1:35

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.
• @ClementC. The question has been edited to assume that the $\mathbb E[X_1]=0$, now is it correct? – Math1000 Nov 18 '19 at 1:37
• @pre-kidney could you explain how did you get $O(n^2)$. It's because $\binom{n}{2} \in O(n^2)$? – Babado Nov 18 '19 at 1:52