TITLE: Average norm of a N-dimensional vector given by a normal distribution
QUESTION [19 upvotes]: I'm interested in knowing what is the expected value of the norm of a vector obtained from a gaussian distribution in function of the number of dimensions $N$ and $\sigma$, i.e:
$$E[\|x\|_2],\quad x\sim\mathcal{N}(0,\sigma I_N)$$
I tried to search for this but didn't find anything. Can I get some help from you?

REPLY [7 votes]: When $\sigma=1$, this is the first moment of Chi-distribution. Furthermore, \begin{equation}
\mathbb{E}_{x\sim\mathcal{N}(0,\sigma^2I)}\{ \|x\|_2^k \} = \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|\sigma x\|_2^k \}= \sigma^k \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \}, 
\end{equation}
where $\mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \}$ is the $k$th moment of Chi-distribution, which has value 
\begin{equation}
    \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|x\|_2^k \} = 2^{k/2} \frac{\Gamma((N+k)/2)}{\Gamma(N/2)}. 
\end{equation}
Two comments:
1. I think $x\sim\mathcal{N}(0,\sigma I_N)$ should be replaced by $x\sim\mathcal{N}(0,\sigma^2I_N)$ in the original post. 
2. This result is different from that of Botev's. 
Let me know if I made any mistakes.