Return to Revisions

The $A_i$'s are iid zero-mean random vectors in $\mathbb{R}^{d \times d}$, which is a Hilbert space with respect to the Frobenius norm $\|\cdot\|:=\|\cdot\|_F$. So, by a vector version of Rosenthal's inequality (see e.g. Theorem 5.2), for some real universal constant $K$, \begin{equation} E\Big\|\sum_i A_i\Big\|^4\le K\Big[\sum_iE\|A_i\|^4+\Big(\sum_iE\|A_i\|^2\Big)^2\Big]. \end{equation}