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?



Expectation of the norm of a random vector
$\newcommand{\si}{\sigma}$
Let us prove a stronger estimate of $EY$, and let us do that under less restrictive conditions. Namely, let us prove that 
\begin{equation\*}
 EY-\sqrt n=O(1/\sqrt n) \tag{1}
\end{equation\*}
assuming only that $EX\_1^4<\infty$ (instead of the $X\_i$'s being sub-Gaussian). 


Substituting 
$$U:=\frac{Y^2}n=\frac1n\,\sum\_1^n X\_i^2$$ 
for $u$ in the inequalities 
$$\frac{1+u-(u-1)^2}2\le\sqrt u\le\frac{1+u}2$$
for $u\ge0$, taking the expectations, and using that $EU=1$ and $E(U-1)^2=\operatorname{Var}\,U=\sigma^2/n$, where $\si^2:=\operatorname{Var}(X\_1^2)<\infty$,  

we have 
$$1-\frac{\sigma^2}{2n}\le\frac{EY}{\sqrt n}\le1,$$
so that (1) indeed follows. 


