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.

  • $\begingroup$ Somewhat related, but I love the coffee cups as an indication of difficulty. $\endgroup$ – Dfrtbx Jan 30 '18 at 18:42
  • $\begingroup$ True. Apparently this is supposed to be a some what hard problem. @Dfrtbx $\endgroup$ – Drew Brady Jan 30 '18 at 18:45
  • $\begingroup$ 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. $\endgroup$ – Dfrtbx Jan 30 '18 at 18:48
  • $\begingroup$ I cannot seem to find equation 2.1.5. $\endgroup$ – Dfrtbx Jan 30 '18 at 18:52
  • 1
    $\begingroup$ Oh, that is equation 2.15, not equation 2.1.5, haha $\endgroup$ – Dfrtbx Jan 30 '18 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.