Distribution of $U=\frac{X}{\| X \|}$ and $R^2 = \| X \|^2$ where $X=(X_1, \dots , X_n)$, $X_1, \dots, X_n \sim$ N(0,1) i.i.d. Independence? I have the following problem:
Let $X=(X_1, \dots , X_n)$,  $X_1, \dots, X_n \sim N(0,1)$ i.i.d. What is the distribution of $U=\frac{X}{\| X \|}$ and $R^2 = \| X \|^2$. Are $U$ and $R^2$ independent?
As $R^2=X_1^2 + \dots + X_n^2$ I think $R^2$ have $\chi_n^2$ distribution.
Could anyone help me with this?
 A: $$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]=
\frac 1 {Z^n}\int f \left(\frac x {|x|}\right) g(|x|) \exp 
\left(-\frac 12  |x|^2\right) 
dx_1\cdots dx_n
$$
now note $x = r x'$, with $|x'| = 1$ and $r>0$, and $M(r,n)$ beeing the size of the sphere of $\mathbb R ^n$ of radius $r$ yields
$$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]=
\frac 1 {Z^n}\int_0^\infty dr M(r,n)  g(r)\exp
\left(-\frac 12r^2\right) \int_{S^n(0,1)} dx' f(x') 
$$
so $\frac X {|X|}$ and $|X|$ are independant.
A: You are right about $X_1^2+\cdots+X_n^2$, and there's really nothing more you need to say about it, unless the definition of the $\chi^2_n$ distribution that you're working with is something other than the distribution of the sum of squares of independent standard normally distributed random variables.
For the distribution of $X/\|X\|$, I think I might first find the conditional distribution of $X/\|X\|$ given $\|X\|$, then observe that that conditional distribution does not depend on $\|X\|$.  That would establish two things:


*

*That the marginal (i.e. "unconditional") distribution of $X/\|X\|$ is the same as that conditional distribution; and

*That $\|X\|$ and $X/\|X\|$ are independent.

