# Why are IID normal random variables spherically symmetrical? [duplicate]

Given a finite sequence of $s+1$ IID normal random variables $X_1, \ldots, X_{s+1}$ They are spherically symmetrical.

This means that the radial projection of the point $(X_1, \ldots, X_{s+1})$ onto the unit sphere $S^s$ has uniform distribution on $S^s$.

My questions are: what is meant by radial projection and how can I prove that the resulting distribution will be uniform?

• The radial projection is explained at stats.stackexchange.com/a/7984/919. Uniformity follows from the facts that (a) linear transformations of a multivariate Normal are multivariate Normal and (b) variance is a quadratic form. – whuber Jun 27 '17 at 18:20
• @whuber thank you,in the link given $(X_1/\lambda, X_2/\lambda, X_3/\lambda)$ is the radial projection on the sphere correct? – Monolite Jun 28 '17 at 15:36
• Yes, that's the radial projection. In general, the radial projection in $\mathbb{R}^n$ is defined on all nonzero vectors $\mathbf{x}=(x_1, x_2, \ldots, x_n)$ and equals $(x_1, x_2, \ldots, x_n)/||\mathbf{x}||$. An easy calculation shows the image of this map consists of unit vectors--that is, the sphere $S^{n-1}$--and since it obviously fixes $S^{n-1}$ itself, the image is the entirety of $S^{n-1}$. The inverse image of any point $\omega\in S^{n-1}$ is the open ray from the origin through $\omega$. – whuber Jun 28 '17 at 15:41