In the book "Spectral Analysis of Large Dimensional Random Matrices" by Bai and Silverstein, there is the following lemma:

Lemma B.26 (pg. 530) Let $A=(a_{ij})$ be an $n\times n$ nonrandom matrix and $X = (x_1,\dots, x_n)$ be a random vector of independent entries. Assume that $\mathbb E x_i = 0, \mathbb E|x_i|^2 = 1,$ and $\mathbb E|x_j |^l\le \nu_l.$ Then, for any $p\ge 1$: $$\mathbb{E} | X^*AX - \text{tr} A |^p \le C_p \left( (\nu_4\text{tr}(AA^*))^{p/2} + \nu_{2p}\text{tr}(AA^*)^{p/2} \right).$$

Though not stated explicitly in the lemma, a look at the proof suggests that $C_p$ is given as at least exponential in $p$.

Question: Is there a more refined understanding of what $C_p$ is as a function of $p$? It seems to depend on several inequalities, including the Marcinkiewicz–Zygmund inequality, whose constants have been studied at least to some extent.


It seems that it should be at most exponential in p. You could start to get an estimate on its size by following the lines of appendix B in this paper of Erdos, Yau and Yin: http://arxiv.org/abs/1001.3453 (and references therein for the precise constants in the Marcinkiewicz–Zygmund and Burkholder inequalities). The proof is a little simpler than the Bai-Silverstein argument due to their assumption of sub-exponential tails, which implies an estimate on the growth of the moments $\nu_{2p}$.

Their objective is to obtain the Hanson-Wright inequality. If you are after sharp tail estimates for $X^*AX$, you might be interested in this work of Rudelson and Vershynin: http://arxiv.org/abs/1306.2872 where they obtain a Bernstein-type tail bound involving both the Frobenius and operator norms of $A$.


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