Why is $X/\|X\|_2$ uniformly distributed on a unit sphere when X is n-dimensional standard gaussian vector?

Question

In the proving the above, I see that since $X$ is multivariate gaussian then for any orthogonal matrix $Q$ we have that $QX$ is standard multivariate gaussian. Then I somehow reasoned that $Y=X/\|X\|_2$ is also rotational invariant and has norm 1. This Y should be uniformly distributed on a sphere but I cannot write down exactly why this should be true. Can someone give me a rigorous mathematical argument to prove this claim? Thanks

Answer 1

I will fix the dimension $n$ and use $S:=\{x\in\Bbb R^n:\|x\|=1\}$ to denote the unit sphere. Let $\sigma$ denote surface measure on $S$, and define $\bar\sigma:=[\sigma(S)]^{-1}\sigma$, the "uniform distribution" on $S$.

Let $f:S\to\Bbb R$ be bounded and Borel measurable. Then $$ \eqalign{ \Bbb E[f(Y)] &=\int_{\Bbb R^n} f(x/\|x\|)(2\pi)^{-n/2}e^{-\|x\|^2/2}\,dx\cr &=(2\pi)^{-n/2}\sigma(S)\int_S\int_0^\infty f(u)e^{-r^2}r^{n-1}\,dr\,\bar\sigma(du),\cr &=\int_S f(u)\,\bar\sigma(du),\cr } $$ because $$ \int_0^\infty e^{-r^2}r^{n-1}\,dr =2^{n/2-1}\int_0^\infty e^{-t}t^{n/2-1}\,dt=2^{n/2-1}\Gamma(n/2) $$ while $\sigma(S)=2\pi^{n/2}/\Gamma(n/2)$.