2
$\begingroup$

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 $d=2$.

$\endgroup$
2
  • 1
    $\begingroup$ The special case you ask about is $\sigma\chi_2$-distributed. $\endgroup$
    – J.G.
    Nov 10, 2020 at 17:46
  • $\begingroup$ 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? $\endgroup$
    – T.L
    Nov 12, 2020 at 8:14

0

You must log in to answer this question.

Browse other questions tagged .