What is the average absolute deviation of a multivariate normal distribution?
We know that the average absolute deviation (or MAD) of a monovariate normal distribution of mean $\mu$ and variance $\sigma^2$ is equal to $\sigma\sqrt{2/\pi}$. Is their any equivalent for the multivariate case?
I think I need it to compute $\sum_{i=1}^n \|\mathbf{x}_i\| = \sum_{i=1}^n \sqrt{{x_1}_i^2 + \ldots + {x_d}_i^2}$, where $\mathbf{x} = (x_1,\ldots,x_d)^T\sim\mathcal{N}_d(\boldsymbol{\mu},\boldsymbol{\Sigma})$.
For simplicity we can consider $\boldsymbol{\mu}=\mathbf{0}$, $\boldsymbol{\Sigma}=\sigma^2\mathbf{I}$, and $d=2$.