Let $X_n,n\in\mathbb{N}$ be i.i.d. zero-mean random variables in some separable Hilbert space with $E\|X_n\|^8<\infty$ and $Y_n=\frac{1}{n}\sum_{i=1}^nX_n$. I need to find bounds on $E\|Y_n\|^4$. The CLT tells me that $Y_n = O_p(n^{-1/2})$. Can I say something about $E\|Y_n\|^4$ ?
As I understand, $\|\sqrt{n}Y_n\|^4\xrightarrow{d}\|G\|^4$, where $G$ is some Gaussian process, so that $\|Y_n\|^4 = O_p(n^{-2})$, but I don't see how do deal with expected value of something bounded in probability.
Update: I did the following computations $$\begin{aligned} E\|Y_n\|^4 & = \frac{1}{n^4}E\left(\sum_{i=1}^n\sum_{j=1}^n\langle X_i,X_j\rangle\right)^2 \\ & = \frac{1}{n^4}\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{l=1}^nE[\langle X_i,X_j\rangle\langle X_k,X_l\rangle] \end{aligned}$$ but then it is a dissaster, to consider all cases for indices... For example, if $i\ne j\ne k\ne l$, $E[\langle X_i,X_j\rangle\langle X_k,X_l\rangle] = E[\langle X_i,X_j\rangle]E[\langle X_k,X_l\rangle]=0$ due to independence. Is there an elegant way to tackle it?
We have
$$\mathbb{E}(\|Y_n\|^4) = \frac{1}{n^4} \sum_{j,k,l,m=1}^n \mathbb{E}(\langle X_j, X_k \rangle \langle X_l, X_m \rangle).$$
It follows from the independence of the random variables and the fact that $X_n$ has mean zero (i.e. $\mathbb{E}(\langle X_n,x \rangle)=0$ for all $x \in H$) that
$$\mathbb{E}(\langle X_j, X_k \rangle \langle X_l, X_m \rangle) \leq \begin{cases} \mathbb{E}(\|X_1\|^4) & j=k=l=m \\ \mathbb{E}(\|X_1\|^2)^2 & (j=k \neq l=m) \, \text{or} \, (j=l \neq k=m) \, \text{or} \, (j=m \neq k=l) \\ 0 & \text{otherwise}. \end{cases}$$
Consequently,
$$\mathbb{E}(\|Y_n\|^4) \leq \frac{1}{n^4} \sum_{j=1}^n \mathbb{E}(\|X_1\|^4) + \frac{3 \mathbb{E}(\|X_1\|^2)^2}{n^4} \sum_{j=1}^n (n-1) = O(n^{-2}).$$