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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?

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1 Answer 1

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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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this makes sense ... thank you – NSR Nov 29 '11 at 6:04

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