3
$\begingroup$

I am aware of concentration inequalities for subgaussian matrices $A$ of the form $\mathcal{P}(\|Ax\|^2 \geq (1+\epsilon)\|x\|^2) \leq \exp(-nc(\epsilon))$. Do these inequalities hold even if $x$ is a random vector instead of a deterministic one?

1 Answers 1

3

I believe that the answer to your question is: Yes.

We have:

$\mathcal{P}(\|AX\|^2 \geq (1+\epsilon)\|X\|^2) = \int_X \mathcal{P}(\|AX\|^2 \geq (1+\epsilon)\|X\|^2 | X=x) f(x) dx$

where, $X$ is a random variable with pdf $f(x)$

But, we know that:

$\mathcal{P}(\|AX\|^2 \geq (1+\epsilon)\|X\|^2 | X=x) \leq \exp(-nc(\epsilon))$

Thus, we get:

$\mathcal{P}(\|AX\|^2 \geq (1+\epsilon)\|X\|^2) \le \int_X \exp(-nc(\epsilon)) f(x) dx = \exp(-nc(\epsilon))$

  • 0
    this makes sense ... thank you2011-11-29