On the distribution of a normalized Gaussian vector

Solution 1:

First, the distribution of $\tilde{x}$ is concentrated on $\mathbb{S}^{n-1}$. In particular, it doesn't have a density w.r.t. the Lebesgue measure on $\mathbb{R}^n$. So instead you may work with the polar representation. Let $r=\|x\|$ and let $\theta_j$ be the angle between $\tilde{x}$ and the $x_j$-th axis, i.e. $x_j=r\cos(\theta_j)$ (denote $\theta\equiv(\theta_1,\ldots,\theta_{n-1})$). Then the density of $(r,\theta)$ is $$ g(r,\theta)=\left(\prod_{i=1}^{n}2\pi\sigma_i^2\right)^{-1/2}\exp\left(-\sum_{i=1}^n\frac{(r\cos(\theta_i)-\mu_i)^2}{2\sigma_i^2}\right)r^{n-1}, $$ where $r\in [0,\infty)$, $\theta\in [0,\pi]^{n-2}\times [0,2\pi)$, and $\sum_{i=1}^n \cos^2(\theta_i)=1$. Consequently, the density of $\theta$ is given by $$ f(\theta)=\int_0^{\infty}g(r,\theta)dr. $$ In particular, when $\mu=0$ and $\Sigma=\sigma I_n$, $$ f(\theta)=\frac{1}{(2\pi\sigma^2)^{n/2}}\int_0^{\infty}\exp\left(-\frac{r^2}{2\sigma^2}\right)r^{n-1}dr=\frac{\Gamma(n/2)}{2\pi^{n/2}}, $$ which is $1$ over the surface area of the unit $(n-1)$-sphere.