# Expectation of the norm of a random vector

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?

$$\newcommand{\si}{\sigma}$$ 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.