# What is the average absolute deviation of a multivariate normal distribution?

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 $$d=2$$.

• The special case you ask about is $\sigma\chi_2$-distributed.
– J.G.
Nov 10, 2020 at 17:46
• Si I am looking for the mean of samples drawn from a Rayleigh distribution? We have $\frac{1}{n}\sum_i \sqrt{x_i^2 + y_i^2} = \sigma \sqrt{\pi/2}$ with $x$ and $y$ two variables following a centered normal distribution of variance $\sigma^2$ right?
– T.L
Nov 12, 2020 at 8:14