# 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?

• $U$ is uniformly distributed on the unit sphere; $R^2$ has a $\chi^2_n$ distribution, and they are independent. I'll have to think about the efficient way to prove those and explain them. I've noticed a pattern here: often several people write correct answers and I wait until I've figured out how to express it simply and so they beat me to it. Exercises of this sort get far more eleborate in statistics course than in probability courses, it seems --- simply because statisticians need this kind of stuff. Mar 5 '14 at 17:50
• See math.stackexchange.com/q/444700/321264 for distribution of $U$. Jun 6 '20 at 20:17

$$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]= \frac 1 {Z^n}\int f \left(\frac x {|x|}\right) g(|x|) \exp \left(-\frac 12 |x|^2\right) dx_1\cdots dx_n$$ 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.
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.