Concentration of the Norm for Sub-gaussians

Question

I am reading Theorem 3.1.1 in HDP book by Vershynin. The theorem states that

$ \text{Let } X=\left(X_1,\ldots,X_n \right) \text{be a random vector with independent, sub-gaussian coordinates } X_i \text{ that satisfy } \mathbb{E}X_i^2=1. \text{Then}$ $$ \| \| X\|_2-\sqrt{n}\|\|_{\psi_2} \leq CK^2$$ $ \text{where } K=\max_i{\|X_i\|_{\psi_2}} \text{ and } C \text{ is an absolute constant.}$

The $\psi_2$ norm is the Orlicz norm with Orlicz function $\psi(x)=e^{x^2}-1. $

I found a place that I don't understand in the proof.

The whole proof only showed that $ \| X \|_2 -\sqrt{n} $ is a sub-gaussian random variable. And in the last sentence, the author just said it is equivalent to the conclusion of the theorem.

I would like to ask about the equivalence in the last sentence.

I've tried to look at the centering property of sub-gaussian, but it seems that $\sqrt n \neq \mathbb{E}\|X\|_2 $. Any hint or idea is appreciated.

Answer 1

There is a bit of "circular feeling" reasoning you have to do which isn't (for me at least) immediately obvious. In short, there are two things at play:

1. First, from the proof we have the concentration inequality $$\mathbb{P}\left\{ \big| ||X||_2 - \sqrt{n} \big|\geq t\right\} \leq 2 \exp\left(-\frac{ct^2}{K^4}\right) \\ = 2 \exp\left(-\frac{ct^2}{(K^2)^2}\right)$$ which holds for all $t \geq 0$. As you mention, this implies the absolute value term is sub-Gaussian with paramter $K^2$. From Proposition 2.5.2, we know that there is an equivalent (up to a constant factor) $K_1=c_1K^2$ such that $\mathbb{E}\exp\left(X^2/K_1^2\right) \leq 2$.
2. From the definition of the Orlicz norm, $$\big|\big| ||X||_2 - \sqrt{n}\big|\big|_{\psi_2} $$ which specifies the norm as the infimum or minimal positive $t$ with $\mathbb{E}\exp\left(X^2/t^2\right) \leq 2$. From this, we conclude that the norm must not be more than $K_1$. We related $K_1$ to $K^2$ above and the result follows.