# Average norm of a N-dimensional vector given by a normal distribution

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?

This amounts to integration in spherical coordinates $(r=\|x\|)$: $$E(\|x\|) = \frac{1}{(\sqrt{2\pi} \sigma)^N } \frac{N\pi^{N/2}}{\Gamma\big(\frac{N}{2}+1\big)}\int_0^\infty e^{-r^2/(2\sigma^2)} r^{N-1} \,dr \tag1$$

This is not so bad: substitute $t=r^2/(2\sigma^2)$, so that $dt = r/\sigma^2$. The resulting integral gives Euler's gamma function $\Gamma$. I'll skip the boring cancellations and get to the result: $$E(\|x\|) = \frac{\sqrt{2}\, \Gamma\big(\frac{N+1}{2}\big)}{\Gamma\big(\frac{N }{2}\big)}\,\sigma$$ As stated in this paper, where you can also find the inequalities $$\frac{N}{\sqrt{N+1}}\le \sigma^{-1}E(\|x\|)\le \sqrt{N}$$

• Thanks. Times $\sigma$? I guess so, since for 1 dimension I got this – jmacedo Jun 9 '14 at 17:19
• @joxnas Right, I internally normalized $\sigma=1$ and forgot to include it in the answer. – user147263 Jun 9 '14 at 17:40
• @Yes: I am confused regarding (1). In the $N=2$ case the Jacobian is $r(N−1)=r$. Also, in the two dimensional case the corresponding integral would be $E[\|X\|]=...\int...r^2dr$ but you have \$...\int..rdr. Please help meunderstand. – zoli 21 hours ago – zoli May 7 '15 at 8:43

The above answer contains mistakes, as has been noted in the comments. I needed recently to derive this so the general result is: $$\mathbb{E}\left[||x||_2^n\right] = 2^\frac{n-2}{2}\sigma^n N \frac{\Gamma\left(\frac{N+n}{2}\right)}{\Gamma\left(\frac{N+2}{2}\right)}$$

When $$\sigma=1$$, this is the first moment of Chi-distribution. Furthermore, $$$$\mathbb{E}_{x\sim\mathcal{N}(0,\sigma^2I)}\{ \|x\|_2^k \} = \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|\sigma x\|_2^k \}= \sigma^k \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \},$$$$ where $$\mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \}$$ is the $$k$$th moment of Chi-distribution, which has value $$$$\mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|x\|_2^k \} = 2^{k/2} \frac{\Gamma((N+k)/2)}{\Gamma(N/2)}.$$$$

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.