# $E(\|x\|^p_2)$ of a $n$ dimensional vector given by a gaussian distribution

Suppose that vector $x \in \mathcal{R}^n$, and $x$ conforms to the standard gaussian distribution $\mathcal{N}(0, I_n)$. Can we calculate $E(\|x\|^2_2)$? Can we further extend to the case of $E(\|x\|^p_2)$, when $p\in \mathcal{N}, p\geq 1$?

Note that there is already a link for the $p=1$ case, i.e., Average norm of a N-dimensional vector given by a normal distribution

• Is this supposed to be the $p$-th power of the $2$-norm or just the $p$-norm? Of maybe the $p$-th power of the $p$-norm? Your formula and the title are conflicting on this. – Dominik Dec 20 '16 at 9:18
• Actually p-th power euclidean norm. – Zhuanghua Liu Dec 20 '16 at 9:39

If $X$ is a standard $n$-dimensional Gaussian random variable, then $\|X\|_2^2 \sim \chi_n^2$. The chi-squared distribution has density $$2^{-n/2} \Gamma(n/2)^{-1} x^{n/2 - 1} e^{-x/2}I\{x > 0\} \, dx.$$
With this we can find an explicit formula for the expectation of $\|X\|_2^p$ $$E[\|X\|_2^p] = E[(\|X\|_2^2)^{p/2}] = \int_0^\infty 2^{-n/2} \Gamma(n/2)^{-1} x^{(p + n)/2 - 1} e^{-x/2} \, dx = 2^{-n/2} \Gamma(n/2)^{-1} 2^{(p + n)/2} \Gamma((p + n)/2) = 2^{p/2} \frac{\Gamma\Big(\frac{p + n}{2}\Big)}{\Gamma\Big(\frac{n}{2}\Big)}.$$