# Powers of Frobenius norm of sum of random matrices

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 $$E[\|A_i\|^4_F]$$ ?

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 $$K$$, $$$$E\Big\|\sum_i A_i\Big\|^4\le K\Big[\sum_iE\|A_i\|^4+\Big(\sum_iE\|A_i\|^2\Big)^2\Big].$$$$ In the iid case, we have \begin{align} E\Big\|\sum_i A_i\Big\|^4&\le K[nE\|A_1\|^4+n^2(E\|A_1\|^2)^2] \\ &\le K(n+n^2)E\|A_1\|^4\le 2Kn^2E\|A_1\|^4. \end{align}
More elementarily, denoting by $$(xy)$$ the inner product $$\text{tr}(xy^T)$$ of vectors $$x,y$$ in $$\mathbb{R}^{d \times d}$$ and letting $$[n]:=\{1,\dots,n\}$$, we have $$$$E\Big\|\sum_i A_i\Big\|^4=\sum_{(i,j,k,l)\in[n]^4}E(A_iA_j)(A_kA_l).$$$$ The summand $$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$$. So, in the iid case, $$$$E\Big\|\sum_i A_i\Big\|^4=nE\|A_1\|^4+n(n-1)(E\|A_1\|^2)^2+2n(n-1)E(A_1A_2)^2,$$$$ 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}