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 $\mathbb{E} [(X_i')^2 (X_j')^2]$.


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