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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.

Drew Brady
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  • 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
  • @Dfrtbx It is on the top of page 28. It says that for subgaussian random variable $U$, $|U|{L_p} \leq C|U|{\psi_2}\sqrt{p}$. I've also updated the text of my question to clarify the notation. – Drew Brady Jan 30 '18 at 18:56
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    Oh, that is equation 2.15, not equation 2.1.5, haha – Dfrtbx Jan 30 '18 at 18:58
  • Yes my bad, @Dfrtbx, my sincerest apologies! I'll update the text of the problem accordingly. – Drew Brady Jan 30 '18 at 19:01
  • Unfortunately I seem to be a little out of my depth, but I'm glad that I could help you clean up the question a tad. If this remains unanswered I will revisit it later. – Dfrtbx Jan 30 '18 at 19:20

1 Answers1

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Using $e^x \geq 1+x$ \begin{align*} \mathbb{E}exp\left (\frac{(\|X\|_2-\sqrt{n})^2}{(\mathbb{E}\|X\|_2-\sqrt{n})^2}\right ) & \geq 1+\mathbb{E}\frac{(\|X\|_2-\sqrt{n})^2}{(\mathbb{E}\|X\|_2-\sqrt{n})^2} \\ & = 1+\frac{\mathbb{E}\|X\|_2^2-2\sqrt{n}\mathbb{E}\|X\|_2+n}{(\mathbb{E}\|X\|_2)^2-2\sqrt{n}\mathbb{E}\|X\|_2+n} \\ &\geq 2 \end{align*} hence $$ |\mathbb{E}\|X\|_2-\sqrt{n}| \leq \|\|X\|_2-\sqrt{n}\|_{\psi_2}\leq CK^2 $$

other solutions

replace $CK^2$ with $o(1)$

mse
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  • I agree with everything until the "hence" part. Can you spell out the remaining details? – Drew Brady Jul 10 '20 at 18:01
  • @DrewBrady Recall the definition of $ | X | _{\psi_2} $ that $\mathbb{E} \exp (X^2/| X | _{\psi_2} ^2) \leq 2$. It is then less than or equal to the LHS of the second line. So we got something like $\mathbb{E} \exp (X^2/a^2) \leq \mathbb{E} \exp (X^2/b^2)$. Hence, $b^2 \leq a^2$. The remaining part is taking square root on both sides. – mentalabuse Aug 18 '20 at 21:51