What is the expected magnitude, i.e. euclidean distance from the origin, of a vector drawn from a p-dimensional spherical normal $mathcal{N}_p(mu,Sigma)$ with $mu=vec{0}$ and $Sigma=sigma^2 I$, where $I$ is the identity matrix?

In the univariate case this boils down to $E[|x|]$, where $x sim mathcal{N}(0,sigma^2)$. This is the mean $mu_Y$ of a folded normal distribution with mean $0$ and variance $sigma^2$, which can be calculated as:

$mu_Y = sigma sqrt{frac{2}{pi}} ,, expleft(frac{-mu^2}{2sigma^2}right) – mu , mbox{erf}left(frac{-mu}{sqrt{2} sigma}right) stackrel{mu=0}{=} sigma sqrt{frac{2}{pi}}$

Since the multivariate normal is spherical, I thought about simplifying the problem by switching to polar coordinates. Shouldn't be the distance from the origin in any direction be given by a folded normal distribution? Could I integrate over all distances, multiply with the (infinitesimal) probability to encounter a sample with that distance (e.g. CDF(radius)-CDF(radius-h), $h rightarrow 0$) and finally make the leap to more than one dimension by multiplying with the "number of points" on a hypersphere of dimension $p$? E.g. $2 pi r$ for a circle, $4 pi r^2$ for a sphere? I feel that this might be a simple question, but I'm not sure how to analytically express the probability for $h rightarrow 0$.

Simple experiments suggest that the expected distance follows the form $csqrt{sigma}$, but I'm stuck on how to make the leap to a multivariate distribution. By the way, a solution for $p le 3$ would be fine.

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#### Best Answer

The sum of squares of $p$ independent standard normal distributions is a chi-squared distribution with $p$ degrees of freedom. The magnitude is the square root of that random variable. It is sometimes referred to as the chi distribution. (See this Wikipedia article.) The common variance $sigma^2$ is a simple scale factor.

Incorporating some of the comments into this answer:

The mean of the chi-distribution with $p$ degrees of freedom is $$ mu=sqrt{2},,frac{Gamma((p+1)/2)}{Gamma(p/2)} $$

Special cases as noted:

For $p=1$, the folded normal distribution has mean $frac{sqrt{2}}{Gamma(1/2)}=sqrt{frac{2}{pi}}$.

For $p=2$, the distribution is also known as the *Rayleigh distribution* (with scale parameter 1), and its mean is $sqrt{2}frac{Gamma(3/2)}{Gamma(1)}=sqrt{2}frac{sqrt{pi}}{2} = sqrt{frac{pi}{2}}$.

For $p=3$, the distribution is known as the *Maxwell distribution* with parameter 1; its mean is $sqrt{frac{8}{pi}}$.

When the common variance $sigma^2$ is not 1, the means must be multiplied by $sigma$.

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