4
$\begingroup$

I'm interested in concentration of the following random matrix sum in spectral norm

$\frac{1}{m}\sum_{k=1}^m b_k^2\mathbf{a}_k\mathbf{a}_k^*$

Here $\mathbf{a}_k\in\mathbb{R}^n$ are i.i.d. standard normal Gaussian vectors distributed as $\mathcal{N}(0,\mathbf{I}_n)$ and $b_k$ are standard normal random variables distributed as $\mathcal{N}(0,1)$ and independent of the $\mathbf{a}_k$'s. I'm interested in showing

I'm interested in showing

$\|\frac{1}{m}\sum_{k=1}^m b_k^2\mathbf{a}_k\mathbf{a}_k^*-\mathbf{I}\|\le \delta$

holds with high probability as long as $m\ge c(\delta) n$ with $c(\delta)$ a constant depending only on $\delta$. Here for a matrix $\mathbf{X}$ the spectral norm is given by $\|\mathbf{X}\|$. I can get a looser result as long as $m\ge c \frac{n\log n}{\delta^2}$. However, I would like to get rid of the extra log factor. I should also say that for the lower tail I can also get rid of the log factor i.e. prove

$\frac{1}{m}\sum_{k=1}^m b_k^2\mathbf{a}_k\mathbf{a}_k^*\succeq(1-\delta)\mathbf{I}$

holds with high probability as long as $m\ge c \frac{n}{\delta^2}$. So really need only the upper tail. Any proof that getting rid of the extra log for the upper tail is not possible would also be welcome.

Improve this question
$\endgroup$
1
  • $\begingroup$ May you kindly comment how to avoid the log factor for the lower bound? $\endgroup$
    – John
    Commented May 12, 2023 at 12:43

1 Answer 1

Reset to default
2
$\begingroup$

Removing the log factor for the upper bound is not possible actually.

$\frac{1}{m}\sum_{k=1}^m b_k^2(\mathbf{a}_k^*u)^2\ge \underset{k}{\max}b_k^2\|\mathbf{a}_k\|_{\ell_2}^2\gtrsim \frac{n(\log n)}{m}$

Improve this answer
$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .