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Let's assume I have a zero-mean Gaussian random vector $x$ with co-variance matrix $\Sigma$. Is there a closed form equation for $\mathbb E[||x||]= \mathbb E[\sqrt{x^T x}]$?

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  • $\begingroup$ The usual expression is a finite number of terms involving the variances $(\sigma_i)$ and the co-variances $(c_{ij})$. $\sum_i \sigma_i^2+\sum_i\sum_{j\ne i}c_{ij}$. $\endgroup$ – herb steinberg Jun 21 '18 at 21:57
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    $\begingroup$ Assume without loss of generality you've diagonalised $\Sigma$, so its eigenvalues are $\Sigma_i:=\Sigma_{ii}$. The pdf of $x$ is $\prod_i\frac{1}{\sqrt{2\pi\Sigma_i}}\exp -\frac{x_i^2}{2\Sigma_i}$. You want the mean of $\sqrt{\sum_i x_i^2}$, which sadly is hard in general. When $\Sigma$ is a multiple of the identity matrix, however, it's easy, because the length is $\sqrt{\Sigma_1}$ times a $\chi_n$-distributed variable of mean $\sqrt{2}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}$ if $x$ is $n$-dimensional. $\endgroup$ – J.G. Jun 22 '18 at 12:11
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The usual expression is a finite number of terms involving the variances $(\sigma_i)$ and the co-variances $(c_{ij})$. $E||x||^2=\sum_i \sigma_i^2+\sum_i\sum_{j\ne i}c_{ij}$.

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  • $\begingroup$ This is square of the norm which is very easy. I am interested in the expected value of the norm. $\endgroup$ – motam79 Jun 21 '18 at 22:17
  • $\begingroup$ Could you spell out precisely what you mean by "expected value of the norm"? Is it the description given by J.G.? $\endgroup$ – herb steinberg Jun 23 '18 at 1:48
  • $\begingroup$ J.G.' s answer is correct. Looks like there is not closed form solution for the expected value of the norm $\endgroup$ – motam79 Jun 25 '18 at 15:01

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