Let $X$ be $N(0,I_n)$, a $n$-dimensional multivariate standard normal. Let $Y = X/||X||$. I know $Y$ is distributed uniformly on the surface of a $n$ dimensional "circle" but how do I prove this?
The answer given Here is fine but I am looking for a finding joint density/ using Jacobian transform type answer if possible.
One way, painful but mechanical: Represent each point $x\in\mathbb R^n$ in "polar coordinates" $(r,\sigma)$ by $x=r\sigma$, where $r\ge0$ and $\sigma\in\mathbb R^n$ with $\|\sigma\|=1$, that is, $\sigma\in S^{n-1}$, and apply the usual Jacobian change-of-variables formula to obtain the joint density for $(r,\sigma)$ when $x$ is a standard Gaussian vector. The key thing here is that the value of the Jacobian determinant depends only on $r$ and not on $\sigma$.
Another way is to argue from the formula for the characteristic function of the vector $X$: $\phi(u)=E[\exp(i\langle u,X\rangle)] = \prod_{k=1}^n E[\exp(i u_k X_k)]=\prod_{k=1}^n\exp(-u_k^2/2)$, to obtain (in the end) $\phi(u)=\exp(-\|u\|^2/2)$, and to exploit the spherical symmetry of the formula for $\phi$.
