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Iosif Pinelis
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The $A_i$'s are independent 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} In the iid case, we have \begin{align} E\Big\|\sum_i A_i\Big\|^4&\le K[nE\|A_1\|^4+n^2(E\|A_1\|^2)^2] \\ &\le K(n+n^2)E\|A_1\|^4\le 2Kn^2E\|A_1\|^4. \end{align}

Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229