Concentration of norm

Question

Let $(X_1,...,X_n)\in \mathbb{R}^n$ be a random vector with independent sub-gaussian coordinates $X_i$ that satisfy $\mathbb{E}X_i^2=1$. Then

$$||||X||_2-\sqrt{n}||_{\psi_2}\leq CK^2$$,

Where $K=max||X_i||_{\psi_2}$.

In the book "High dimensional probability", it is claimed that we can assume $K\geq 1$ and $C$ is a universal constant.

But it is not clear why we can do so. Because the above expression is not homogeneous. I tried to change variables but if I change $X_i$ then the property of unit variance no longer holds.

Answer 1

I am unable to show that $K\geq 1$ but the following reasoning yields that $K \geq 2^{-3/2}$ is required because of the assumption of unit variance.

Let $X_i$ be the random variable which realises the maximal subgaussian norm. The equivalent definitions of subgaussian offered as Prop 2.5.2 in "High dimensional probability" yields that $1 = \mathbb{E}X_i^2 \leq K' \sqrt{2}$ for some $K'$ which differs from $K$ by at most a absolute constant factor. Using the proof offered in "High dimensional probability" we get a $K' \leq 2K$. Putting the two inequalities together we get the (partial) result.

Answer 2

The reason why we can assume $K\geq 1$ is because of the assumption $EX^2=1$. To see this we use the definition of the $\psi_2$ norm. Which is simply

$$ ||X||_{\psi_2}=\inf \left\{ t| E\exp\left( \frac{X^2}{t^2} \right) \leq 2\right\}$$

By Jensen's inequality we have

$$\exp\left(E\frac{X^2}{t^2}\right)\leq E\exp\left( \frac{X^2}{t^2} \right) $$

And by assumption $EX^2=1$ so that

$$\exp\left(E\frac{X^2}{t^2}\right)=\exp\left(\frac{1}{t^2}\right)$$

Thus, by inclusion we have

$$||X||_{\psi_2} \geq \inf \left\{ t| \exp\left( \frac{1}{t^2} \right) \leq 2\right\}$$

Then we just notice that $\exp\left( \frac{1}{t^2} \right) \leq 2$ iff $t\geq\frac{1}{\sqrt{\ln 2}}$ by rearrangement. And lastly $\frac{1}{\sqrt{\ln 2}} \geq 1$. Combining everything we conclude that if $EX^2=1$ then $||X||_{\psi_2}\geq1$. Thus, we see that $||X_i||_{\psi_2}\geq 1$ for each i and therefore $K=\max_i ||X_i||_{\psi_2}\geq 1$.