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:

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 $x \in \mathbb{R}^n.$

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?

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    $\begingroup$ What is your definition of a subgaussian random vector? $\endgroup$ Apr 15, 2020 at 2:47
  • $\begingroup$ @IosifPinelis Thanks for your comment! A random variable is subgaussian if it satisfies: $P[|X| > t] \le 2 e^{-ct^2}, t>0.$ A random vector $X \in \mathbb{R}^n$ is called sub-gaussian if the one-dimensional marginals $<X,x>$ are sub-gaussian random variables for all deterministic $x \in \mathbb{R}^n.$ $\endgroup$ Apr 15, 2020 at 8:33

1 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.

  • $\begingroup$ I appreciate your answer, it's helpful! I've a few following remarks/questions though. First, apologies, as I made a subtle mistake in the constant in the first and second concentration inequality (the ones that I already gathered and not askign to prove) I stated before: the constant I stated before wasn't absolute, but rather depending on the subgaussian vector itself; to clarify this, I modified it correctly now: $P[ |\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}| \ge t] \le 2e^{-ct^2/K^4},K=max_{1 \le i \le n}||X_i||_{\psi_2}.$ [contd.] $\endgroup$ Apr 15, 2020 at 20:44
  • $\begingroup$ With this modification where the constant $CK^2$ in the second assumed inequality has a part depending on $n$, namely, $K^2 \equiv K(n)^2= (max_{1 \le i \le n}||X_i||_{\psi_2})^2,$ and this can go to $\infty$ as $n \to \infty$ (correct me if I'm wrong!). So, in this case:we've $ || N - \sqrt{n}|| \le CK^2, || \frac{N}{\sqrt{n}} - 1 || \le C \frac{K(n)^2}{\sqrt{n}}$ may not go to $0,$, but it will, if $X_i$'s are iid, not assumed in the question. Correct me if I'm wrong! $\endgroup$ Apr 15, 2020 at 20:50
  • $\begingroup$ [contd.] Also, regarding the part where you wrote: "whence $|ER_n|\le\frac{E(N^2-n)^2}{n^{3/2}}=\frac{Var(N^2)}{n^{3/2}}\le\frac{s^2}{n^{1/2}}\to 0$," I'm not sure I see why: $s^2= max_{1 \le i \le n} Var[X_i^2]= \mathbb{E}{X_i}^4 - 1$ is finite for each $n$ for sure, but why does it stay bounded as $n \to \infty,$ i.e. why do we assume $lim sup s^2(n)$ is not $\infty ?$ In this case, $\frac{s^2}{n^{1/2}}$ may not go to $0,$ rigth? Also, I didn't understand the Central Limit Theorem argument: I see one can apply it on $N^2,$ but how does it helps us see $N - \sqrt{n} \to N(0,1/2)?$ $\endgroup$ Apr 15, 2020 at 21:08
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    $\begingroup$ @Learningmath : If you don't have a boundedness condition on the moments of the $X_i$'s of a high enough order (at least on an average), then of course the results won't hold. (My condition was $s^2<\infty$.) Moreover, all your questions will obviously lose any meaning if you don't specify the rate of growth of the moments of the $X_i$'s that you want to allow. Therefore, if you can specify such conditions, then I suggest you state them in another, separately posted question. $\endgroup$ Apr 16, 2020 at 0:29
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    $\begingroup$ @Learningmath : As for the central limit theorem (CLT) argument, you do first apply it to $N^2$, and then just use (1) and the fact that $R_n\to0$ in probability/distribution to get the CLT for $N$. (That $R_n\to0$ in probability and hence in distribution follows by Markov's inequality, because $ER_n\to0$.) $\endgroup$ Apr 16, 2020 at 0:34

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