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.
Best Answer
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 $o(1)$