# 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 Jun 9, 2014 at 17:19
• @joxnas Right, I internally normalized $\sigma=1$ and forgot to include it in the answer.
– user147263
Jun 9, 2014 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, 2015 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: $$\mathbb{E}\left[||x||_2^n\right] = 2^\frac{n-2}{2}\sigma^n N \frac{\Gamma\left(\frac{N+n}{2}\right)}{\Gamma\left(\frac{N+2}{2}\right)}$$

• I don't have necessary knowledge to assert which of these solutions is the correct one. Perhaps someone with some reputation can confirm this is correct and the above solution is wrong? Jan 6, 2020 at 17:21
• @jmacedo Both mine and Liyuan's are correct, in fact there are equivalent as I mentioned in a comment below his answer, feel free to choose any of the two answers. Jan 6, 2020 at 21:06

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.

• The two formulas are equivalent -> G(N/2)=G((N+2)/2)*2/N, but I did not see the simplification when I derived this. Nov 29, 2019 at 12:25