Simple property of spherically symmetric probability distributions Define a vector random variable $X \in \mathbb{R}^p$ as distributed according to a spherically symmetric distribution if $eX$ has the same probability distribution as $X$ for every orthogonal transform $e$. 
Show that, when a spherically symmetric distribution has a density, it is a function of $x^T x$ only.
I tried to look at $X$'s characteristic function but couldn't find any angle of attack. There must be some theory I am not aware of which should be used for this problem.
 A: We can write the density function as a function of "spherical" coordinates,
$$ f(x) = f(r,n), $$
where $r \geqslant 0$, $n$ is a unit vector, and $x=rn$, so $r^2=x^T x$. (Really, this is a generalisation of polar coordinates, so the vector $n$ is a function of coordinates as angles, and treated as independent of the length $\sqrt{x^T x}$). Then the action of an orthogonal matrix $e$ on $x$ is to send $rn \mapsto r(en) $, so
$$ e:(r,n) \mapsto (r,en).  $$
Hence the density function maps to $f(r,en)$. But $eX$ has the same probability distribution as $X$, so $f(r,en) = f(r,n)$ for any orthogonal matrix $e$. Since for any unit vectors $m,n$, there is an $e$ so $en=m$ (the orthogonal group acts transitively on the set of unit vectors), $f(r,n)$ does not depend on the value of $n$ (or rather, for any $n$, it has the same value), and hence $f(r,n)=g(r)$ is a function of $r=\sqrt{x^T x}$ only.
(Also note that the measure on $p$-dimensional space transforms as $dx = dy$ if $y=ex$, since (absolute value of) the Jacobian of this transformation is $1$, so this does not change under such transformations either. It is perhaps easier to consider $f(x) \, dx$ as one thing, but this depends on the theory of probability and integration used.)
