For $i= 1, \ldots, n$, let $A\_i \in \mathbb{R}^{d \times d}$ be random i.i.d. matrices with $E [A\_i] =0$.
Can we relate (upper bound) $E[\|\sum\_{i=1}^n A\_i \|\_F^4]$ to $E[\|A\_i\|^4\_F]$ ?



Powers of Frobenius norm of sum of random matrices
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](https://projecteuclid.org/download/pdf_1/euclid.aop/1176988477)), 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} 




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More elementarily, denoting by $(xy)$ the inner product $\text{tr}(xy^T)$ of vectors $x,y$ in $\mathbb{R}^{d \times d}$ and letting $[n]:=\{1,\dots,n\}$, we have 
\begin{equation}
 E\Big\|\sum\_i A\_i\Big\|^4=\sum\_{(i,j,k,l)\in[n]^4}E(A\_iA\_j)(A\_kA\_l). 
\end{equation}
The summand $E(A\_iA\_j)(A\_kA\_l)$ is nonzero only if (i) $i=j=k=l$ or (ii) $i=j\ne k=l$ or (iii) $i=k\ne j=l$ or (iv) $i=l\ne j=k$. So, in the iid case, 
\begin{equation}
 E\Big\|\sum\_i A\_i\Big\|^4=nE\|A\_1\|^4+n(n-1)(E\|A\_1\|^2)^2+2n(n-1)E(A\_1A\_2)^2, 
\end{equation}
so that, by the Schwarz inequality, 
\begin{align}
 nE\|A\_1\|^4+n(n-1)(E\|A\_1\|^2)^2&\le E\Big\|\sum\_i A\_i\Big\|^4 \\ 
 &\le nE\|A\_1\|^4+3n(n-1)(E\|A\_1\|^2)^2 \\
 &\le n(3n-2)E\|A\_1\|^4. 
\end{align}


