# maximum of certain Gaussian processes

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).

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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

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

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You can accept your own answer. – Nate Eldredge Jul 22 '14 at 20:20