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.


1 Answer 1


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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