This post gives a nice and quick proof that $\mathbb{E}[\| X\|_2] = (1 - o_n(1)) \sqrt{n}$ when $X$ is a multivariate isotropic Gaussian. I was made aware of this proof by my adviser, and its based on Chapter 3.1 of Vershynin's excellent book. $ \newcommand{\abs}[1]{| #1 |} \newcommand{\bigabs}[1]{\left| #1 \right|} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\mathbb{E}} \renewcommand{\Pr}{\mathbb{P}} \newcommand{\calE}{\mathcal{E}} \newcommand{\calF}{\mathcal{F}} \newcommand{\calD}{\mathcal{D}} \newcommand{\calN}{\mathcal{N}} \newcommand{\calL}{\mathcal{L}} \newcommand{\calM}{\mathcal{M}} \newcommand{\calN}{\mathcal{N}} \newcommand{\bbP}{\mathbb{P}} \newcommand{\bbQ}{\mathbb{Q}} \newcommand{\ip}[2]{\langle #1, #2 \rangle} \newcommand{\bigip}[2]{\left\langle #1, #2 \right\rangle} \newcommand{\T}{\mathsf{T}} \newcommand{\Tr}{\mathrm{Tr}} \newcommand{\ind}{\mathbf{1}} \newcommand{\calL}{\mathcal{L}} \newcommand{\norm}[1]{\lVert #1 \rVert} $

The proof is short. Let $X \sim \calN(0, I_n)$. We will assume for simplicity that $n \geq 5$. We first derive: $$ \begin{align*} \E[ (\norm{X}_2 - \E[\norm{X}_2])^2 ] &= \int_0^\infty \Pr( (\norm{X}_2 - \E[\norm{X}_2])^2 \geq t ) \; dt \\ &= \int_0^\infty \Pr( \abs{ \norm{X}_2 - \E[\norm{X}_2]} \geq \sqrt{t} ) \; dt \\ &\stackrel{(a)}{\leq} 2 \int_0^\infty e^{-t/2} \; dt \\ &= 4 \:. \end{align*} $$ In step (a), we used the fact that the function $f(x) := \norm{x}_2$ is a 1-Lipschitz function and hence the random variable $\norm{X}_2$ is a sub-Gaussian random variable with variance 1; this is a well known result. On the other hand, the left hand side is: $$ \E[ (\norm{X}_2 - \E[\norm{X}_2])^2 ] = n - (\E[\norm{X}_2])^2 \:. $$ Rearranging, we have that: $$ \begin{align*} \E[\norm{X}_2] &\geq \sqrt{n-4} \\ &= \sqrt{n} + \sqrt{n-4} - \sqrt{n} \\ &\geq \sqrt{n} + \sqrt{n} - \frac{2}{\sqrt{n-4}} - \sqrt{n} \\ &= \sqrt{n} - \frac{2}{\sqrt{n-4}} \\ &= \left(1 - \frac{2}{\sqrt{n(n-4)}}\right) \sqrt{n} \\ &= (1-o_n(1)) \sqrt{n} \:. \end{align*} $$ The inequality above uses the fact that $\sqrt{x}$ is a concave function. On the other hand by Jensen's inequality we have that $\E[\norm{X}_2] \leq \sqrt{n}$. The claim that $\E[\norm{X}_2] = (1-o_n(1)) \sqrt{n}$ now follows.