5
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ Somewhat related, but I love the coffee cups as an indication of difficulty. $\endgroup$
    – Dfrtbx
    Jan 30, 2018 at 18:42
  • $\begingroup$ True. Apparently this is supposed to be a some what hard problem. @Dfrtbx $\endgroup$
    – Drew Brady
    Jan 30, 2018 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, 2018 at 18:48
  • $\begingroup$ I cannot seem to find equation 2.1.5. $\endgroup$
    – Dfrtbx
    Jan 30, 2018 at 18:52
  • 1
    $\begingroup$ Oh, that is equation 2.15, not equation 2.1.5, haha $\endgroup$
    – Dfrtbx
    Jan 30, 2018 at 18:58

1 Answer 1

2
$\begingroup$

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 $o(1)$

$\endgroup$
2
  • $\begingroup$ I agree with everything until the "hence" part. Can you spell out the remaining details? $\endgroup$
    – Drew Brady
    Jul 10, 2020 at 18:01
  • $\begingroup$ @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. $\endgroup$ Aug 18, 2020 at 21:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .