Suppose $X$ is a random vector denoted as $(X_1,\cdots,X_n)$, where $X_1,\cdots,X_n$ are iid random variables with sub-Gaussian distributions. For all $i$, suppose $E[X_i^2]=1$ for simplicity and $\|X_i\|_{\psi_2}=K$ where$\|\cdot\|_{\psi_2}$ is the sub-Gaussian norm.

Let $Y=\|X\|$ be the 2-norm of $X$. A known fact is $Y-\sqrt n$ can be const bounded. On the other hand, one may ask when $n\rightarrow \infty$, will $Y-\sqrt n \rightarrow 0$?

For example, consider the standard normal distribution. Then $Y^2$ is $\chi^2(n)$ distribution, and $$E[Y]=\sqrt 2 \frac {\Gamma\left(\frac {n+1} 2\right)} {\Gamma\left(\frac {n} 2\right)}$$ so $Y-\sqrt n\rightarrow 0$ when $n\rightarrow \infty$. So for arbitary random variable which satisfy the above condition, does $Y-\sqrt n\rightarrow 0$ always hold?

Suppose $X$ is a random vector denoted as $(X_1,\cdots,X_n)$, where $X_1,\cdots,X_n$ are iid random variables with sub-Gaussian distributions. For all $i$, suppose $E[X_i^2]=1$ for simplicity and $\|X_i\|_{\psi_2}=K$ where$\|\cdot\|_{\psi_2}$ is the sub-Gaussian norm.

Let $Y=\|X\|$ be the 2-norm of $X$. A known fact is $E[Y]-\sqrt n$ can be const bounded. On the other hand, one may ask when $n\rightarrow \infty$, will $E[Y]-\sqrt n \rightarrow 0$?

For example, consider the standard normal distribution. Then $Y^2$ is $\chi^2(n)$ distribution, and $$E[Y]=\sqrt 2 \frac {\Gamma\left(\frac {n+1} 2\right)} {\Gamma\left(\frac {n} 2\right)}$$ so $E[Y]-\sqrt n\rightarrow 0$ when $n\rightarrow \infty$. So for arbitary random variable $X_i$ which satisfy the above condition, does $E[Y]-\sqrt n\rightarrow 0$ always hold?