Asymptotically tight concentration of norms of subgaussian random vectors with independent coordinates, as the dimension $n \to \infty?$

Question

Let $X=(X_1 \dots X_n)\in \mathbb{R}^n,$ be a subgaussian random vector so that $X_i$'s are independent, $\mathbb{E}X_i = 0, \mathbb{E}X_i^2=1.$ Before we pose our question, let's state the following:

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Definition notation and background: A random variable $V \in \mathbb{R}$ is called subgaussian, if $P[|V|\ge t]\le 2 e^{-c(V)t^2}, c(V)>0$ depends on $V$ only. One can show that: $\mathbb{E}\left[exp({\frac{V^2}{t^2}})\right ]\le 2$ for some $t>0,$ which makes us define the Orlicz norm: $||V||_{\psi_2}:= inf_{t > 0}\mathbb{E}\left[exp({\frac{V^2}{t^2}})\right ]\le 2.$ Note that: then we have: $P[|V|\ge t]\le 2 e^{-ct^2/||V||_{\psi_2}^2}, c>0$ is an absolute constant. A random vector $X\in \mathbb{R}^n$ is called subgaussian if $<X,x>_{\mathbb{R}^n}$ is subgaussian random variable for every constant $x \in \mathbb{R}^n.$

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With the above, let's state the question, $c, C$ below are absolute:

We know the following about the concentration of $ ||X||_2,(\mathbb{E}X_i=0, \mathbb{E}X_i^2=1$)

$$P[ |\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}| \ge t] \le 2e^{-ct^2/K^4}, K=max_{1 \le i \le n}||X_i||_{\psi_2}.$$

Or equivalently:

$$||\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}||_{\psi_2}\le CK^2 > 0, K=max_{1 \le i \le n}||X_i||_{\psi_2}$$

where ${\psi_2}$ denotes the subgaussian norm. But we note that the right side of the inequality is dimension-free, i.e. there's no function of $n$ on the right that goes to zero as $n \to \infty.$ So my question is: do we necessarily have tighter concentration of the norm around $\sqrt{n}$ as $n \to \infty?$ That is: do we have:

$$(1a) \hspace{1mm} lim_{n \to \infty} |\mathbb{E}||X||_2 - \sqrt{n}|=0? $$ $$(1b) \hspace{1mm} lim_{n \to \infty} | \frac{\mathbb{E}||X||_2}{\sqrt{n}} - 1|=0? $$ $$ (2a) \hspace{1mm} lim_{n \to \infty}||\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}||_{\psi_2}=0?$$ $$ (2b) \hspace{1mm} lim_{n \to \infty}||\hspace{1mm}{\frac{||X||_2}{\sqrt{n}} - 1 } \hspace{1mm}||_{\psi_2}=0?$$ Next, if we don't assume that the co-ordinates are independent, are (1a,b) and (2a,b) still true?

Answer 1

Consider first the case when the $X_i$'s are independent. In view of your definition of a sub-gaussian random vector, we appear to have the condition $$s^2:=\sup_i Var(X_i^2)<\infty.$$

Let $N:=\|X\|_2$. We have this key identity: $$N-\sqrt n=\frac{N^2-n}{2\sqrt n}-R_n,\tag{1}$$ where $$R_n:=\frac{(N^2-n)^2}{2\sqrt n(N+\sqrt n)^2}.$$ Moreover, $$0\le R_n\le\frac{(N^2-n)^2}{n^{3/2}},$$ whence $$|ER_n|\le\frac{E(N^2-n)^2}{n^{3/2}}=\frac{Var(N^2)}{n^{3/2}}\le\frac{s^2}{n^{1/2}}\to0$$ (as $n\to\infty$). So, by (1), $$EN-\sqrt n=-ER_n\to0,$$ so that your condition (1a) holds, which also obviously implies (1b).

Your condition (2b) immediately follows from your second displayed inequality, $\|N-\sqrt n\|_{\psi_2}\le C$, which implies $\|\frac N{\sqrt n}-1\|_{\psi_2}\le C/n$.

Your condition (2a) does not hold even when the $X_i$'s are iid standard normal -- because then, by (1) and the central limit theorem (say), the distribution of $N-\sqrt n$ converges to $N(0,1/2)$.

Thus, in the "independent" case, your conditions (1a), (1b), and (2b) hold, whereas (2a) does not hold in general.

Consider now the "dependent" case, when the $X_i$'s are not necessarily independent. Let e.g. $X_i=X_1$ for all $i$, where $X_1$ is any zero-mean unit-variance random variable such that $a:=E|X_1|$ is strictly less than $1$. Then $N=\sqrt n\,|X_1|$. So, $\frac{EN}{\sqrt n}=a-1\not\to0$, so that (1b) fails to hold, and hence (1a) fails to hold. Also, here $\|\frac N{\sqrt n}-1\|_{\psi_2}=\||X_1|-1\|_{\psi_2}\not\to0$, so that (2b) fails to hold, and hence (2a) fails to hold.

Thus, in the "dependent" case, none of your conditions (1a), (1b), (2a), (2b) holds in general.