Let $X_1, X_2, \dots$ be i.i.d. with $E(X_1)=0$ and $E(X_1^4)<\infty.$ Let $S_n=X_1+\dots +X_n$. Show that $|S_n/\sqrt{n}|^\alpha$ are uniformly integrable for any $0<\alpha<4$.
I am trying to answer the above problem. I know that I want to show that: $$ \lim\limits_{M\rightarrow \infty}\left[\sup\limits_n E\left( \left|\frac{S_n}{\sqrt{n}}\right|^\alpha 1_{\{\left|\frac{S_n}{\sqrt{n}}\right|^\alpha >M\}} \right) \right]=0. $$
I also know that it would be enough to show that: $$ \sup\limits_n E\left( \left|\frac{S_n}{\sqrt{n}}\right|^\alpha \right)<\infty $$
But I don't really see what sort of manipulations or tricks to use to show either one of these statements.
I would appreciate some help.
Thanks!
In fact, it is sufficient to show that (what you wrote probably was a typo.) $$\sup_n E\left(\left(\frac{S_n}{\sqrt{n}}\right)^4\right) < \infty. \tag{1}$$ Since \begin{align*} S_n^4 = (X_1 + \ldots + X_n)^4 = \sum_{i = 1}^n X_i^4 + \sum_{i \neq j} X_i^2 X_j^2 + R \end{align*} where $R$ denotes other terms whose expectations are $0$. Notice that there are $3n(n - 1)$ terms with the form $X_i^2 X_j^2$, whence $$E(S_n^4) = nE(X_1^4) + 3n(n - 1)E(X_1^2X_2^2).$$ Consequently, $$ E\left(\left(\frac{S_n}{\sqrt{n}}\right)^4\right) = \frac{1}{n}E(X_1^4) + \frac{3(n - 1)}{n} E(X_1^2X_2^2) \leq E(X_1^4) + 3E(X_1^2X_2^2) < \infty,$$ from which $(1)$ holds, and thus the sequence is uniformly integrable.