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Iosif Pinelis
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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), and (2b) holds in general.

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 $X_i$ 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), and (2b) holds in general.

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.

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Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

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 $X_i$ 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), and (2b) holds in general.