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?
Do matrix concentration inequalities hold when operating on other random vectors
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probability-theory
1 Answers
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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:
$\mathcal{P}(\|AX\|^2 \geq (1+\epsilon)\|X\|^2) \le \int_X \exp(-nc(\epsilon)) f(x) dx = \exp(-nc(\epsilon))$
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0this makes sense ... thank you – 2011-11-29