Let ($X_n$)$_{n\in\mathbb{N}}$ be a sequence of independent, identically distributed and $\mathbb{P}$-a.s bounded. Let furthermore be $\mathbb{E}$[X$_1$] = 0. We define $S_n$ = $n^{-1}$$\sum_{i=1}^n X_i$. Show that
$\mathbb{E}$[$S_n^{4}$] $\le$ $\frac{4}{n^2}$$\mathbb{E}$[$X_1^{4}$]
Tip: Use Hölder's Inequality
I tried to apply the tip on $S_n^{4}$ with $S_n^{2}$$S_n^{2}$ but I am not much more advanced. I will be grateful for any help or piece of advice.
Ok - This is not as smooth as what you want. I'm sure someone can help with a slicker solution.
Perfunctory use of Hölder's.
Assume $X_i$ and $X_j$ are independent. Then, $X^{2}_{i}$ and $X^{2}_{j}$ are independent as well. Now, $$ E(X^{2}_{i} X^{2}_{j}) \leq \sqrt{E(X^{4}_{i})} .\sqrt{E(X^{4}_{j})} = E(X^{4}_{1}) $$ taking $p=q=2$.
Now, expanding $(\sum_{i=1}^{n} X_i)^4$ and observing that only multinomial coefficients with expectation non-zero are when all terms have even powers. I.e. $E (X_{i}^{4})$ is non-zero, and $E (X_{i}^{2}X_{j}^{2})$ is non-zero. Other terms fall off as they at least have one single power term.
Note: possible multinomial expansions are {4, (2,2), (3,1), (2, 1, 1) and (1,1,1,1)}.
Thus, $$E(\sum_{i=1}^{n} X_i)^4 = \sum_{i=1}^{n}E(X_i^4) + \sum_{i\neq j} 6.E(X_i^2.X_j^2) \leq \sum_{i=1}^{n}E(X_i^4) + \sum_{i\neq j} 6.E(X_i^4).$$
Note now that $X_i$ are i.i.d. and by noting that the number of terms of $i \neq j$ is $ n.(n-1)/2,$
$$E(\sum_{i=1}^{n} X_i)^4 \leq {n}E(X_1^4) + 3n(n-1).E(X_i^4).$$ Divide by $n^4$ to get
$$E(\frac{1}{n}\sum_{i=1}^{n} X_i)^4 \leq 3n^{-2}.E(X_i^4) .$$