MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathbf{a}_k\in\mathbb{C}^n$ for $k=1,2,\ldots,m$ be i.i.d. standard complex normal random vectors with distribution $c\mathcal{N}(0,\mathbf{I})$. I am interested in a tight upper bound on the following quantities with high probabilities (say with probability at least $1-\frac{1}{n}$ or something similar): \begin{align} \underset{\mathbf{x}\in \mathbb{C}^ns.t.\|\mathbf{x}\|_{\ell_2}=1}{\text{max}}\frac{1}{m}\sum_{k=1}^m |\mathbf{a}_k^*\mathbf{x}|^4\le ? \end{align} \begin{align} \underset{\mathbf{x},\mathbf{y}\in \mathbb{C}^ns.t.\|\mathbf{x}\|_{\ell_2}=\|\mathbf{y}\|_{\ell_2}=1} {\text{max}}\frac{1}{m}\sum_{k=1}^m |\mathbf{a}_k^*\mathbf{x}|^2Re(\mathbf{y}^*\mathbf{a}_k\mathbf{a}_k^*\mathbf{x})\le? \end{align}

Of course I know how to do this for fixed quantities of $\mathbf{x}$ and $\mathbf{y}$. However, the corresponding probabilities do not allow for a covering argument. I am hoping that the upper bound for all $\mathbf{x},\mathbf{y}$ is comparable to the result for fixed $\mathbf{x},\mathbf{y}$ up to constant/log factors. If anybody knows of a counter argument which shows that this is not the case that would also be very helpful. Assume that $m \ge c n$ for a sufficiently large numerical constant $c$. I would also be ok with an argument which assumes $m \ge c n (\log n)^\alpha$ for some small $\alpha$ like $\alpha=1,2,3,4$ (the smaller of course the better).

share|cite|improve this question
Whether your hope is true or not, certainly depends on the relation between $m$ and $n$. Let's take two extremes: 1) $m=1$. Then the discrepancy factor is like $n^2$. 2) $m=\infty$. Then the law of large numbers tells you that your quantity is essentially the same for all $x$. – fedja May 22 '14 at 17:57
Thanks! I forgot to mention the relationship between m and n. I have added this in the above – mohi May 22 '14 at 18:05
up vote 1 down vote accepted

I actually found a simple counter example. Setting $x=\frac{\mathbf{a}_1}{\|\mathbf{a}_1\|_{\ell_2}}$ already rules out my claim.

share|cite|improve this answer
You can accept your own answer. – Nate Eldredge Jul 22 '14 at 20:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.