# Concentration of the norm (sub-gaussianity)

I'm trying to solve the following problem (exercise 3.1.4 of these notes)

Suppose $X = (X_1, \dots, X_n) \in \mathbf{R}^n$ is a random vector with independent, sub-gaussian coordinates $X_i$, each of which satisfy $\mathbf{E} X_i^2 = 1$. Show that: $$\sqrt{n} - CK^2 \leq \mathbf{E}\|X\|_2 \leq \sqrt{n} + CK^2.$$ Can $CK^2$ be replaced by $o(1)$, a quantity that vanishes as $n \to \infty$?

Notation: $\|\cdot\|_{\psi_2}$ refers to the sub-gaussian norm.

What I've tried:

The first statement is equivalent to showing that $|\mathbf{E} \|X\|_2 - \sqrt{n}| \leq CK^2$. From Theorem 3.1.1 of the notes above, I know that $\|\|X\|_2 - \sqrt{n}\|_{\psi_2} \leq CK^2$. Thus, it would suffice to establish that $$|\mathbf{E} \|X\|_2 - \sqrt{n}| \leq \|\|X\|_2 - \sqrt{n}\|_{\psi_2}$$ By Jensen's inequality, $$|\mathbf{E} \|X\|_2 - \sqrt{n}| \leq \mathbf{E} |\|X\|_2 - \sqrt{n}| = \|\|X\|_2 - \sqrt{n}\|_{L_1}.$$ But by equation 2.15 (of the same notes): $$|\mathbf{E} \|X\|_2 - \sqrt{n}| \leq \|\|X\|_2 - \sqrt{n}\|_{L_1} \leq C' \|\|X\|_2 - \sqrt{n}\|_{\psi_2} \leq C' \cdot CK^2.$$

Question: I'm not sure if this the tightest way to solve the first part of the problem. As you can see, I have to incur another absolute constant. Also, any help with the statement regarding whether $CK^2$ can be $o(1)$ would be appreciated. I have no idea.

• Somewhat related, but I love the coffee cups as an indication of difficulty. – Dfrtbx Jan 30 '18 at 18:42
• True. Apparently this is supposed to be a some what hard problem. @Dfrtbx – Drew Brady Jan 30 '18 at 18:45
• If anyone else is confused like I was, $\|\cdot\|_{\psi_2}$ refers to the sub-gaussian norm and is defined in def. 2.5.6. – Dfrtbx Jan 30 '18 at 18:48
• I cannot seem to find equation 2.1.5. – Dfrtbx Jan 30 '18 at 18:52
• Oh, that is equation 2.15, not equation 2.1.5, haha – Dfrtbx Jan 30 '18 at 18:58