# $\mathbb E[S_n'^4]\le C\cdot n^2$

Let $$X_1,...,X_n$$ be iid random variables on $$(\Omega,\mathcal A,\mathbb P),\ \ \mathbb E[X_1]=\mu\in\mathbb R,\ \ \mathbb E[X_1]^4<\infty,\ \ X_i':=X_i-\mu,\ \ S_n'=X_1'+...+X_n'.$$

Prove that $$\mathbb E[S_n'^4]\le C\cdot n^2$$ for $$C>0$$ constant.

I have tried everything I could think of already, but it didn't lead anywhere. My best shot was

$$\mathbb E[|S_n'|^4]=\int_0^\infty4t^3\ \mathbb P(|S_n|\ge t)\ \text{d}t \le 4\int_0^{\infty} t^3\cdot t^{-2} Var(S_n)\ \text{d}t= 4\int_0^{\infty} t\cdot Var(S_n)\ \text{d}t\\=4\int_0^{\infty} t n Var(X_1)\ \text{d}t$$ with Tchebychef. Can anyone help me with that?

Expand $$\mathbb{E} [(S_n')^4] = \mathbb{E} [(X_1' + \ldots + X_n')^4]$$. Notice that terms $$\mathbb{E}[X_i' (X_j')^3]$$ with $$i \neq j$$ will vanish using independence and $$\mathbb{E}[X'_i] = 0$$. For terms $$\mathbb{E} [(X_i')^2 (X_j')^2]$$ with $$i \neq j$$, notice that they are also independent and that by Cauchy-Schwarz $$\mathbb{E} [(X_i')^4] \geq \mathbb{E}[(X_i')^2]^2$$. So, in the end all non-vanishing terms can be bounded by the same constant $$C = \mathbb{E} [(X_i')^4]$$ and there are at $$n$$ terms $$\mathbb{E} [(X_i')^4]$$ and $$\binom{n}{2} \cdot \binom{4}{2}$$ terms $$\mathbb{E} [(X_i')^2 (X_j')^2]$$.