TITLE: Distribution and moments of $\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$ when $X_i$'s are i.i.d $N(0,\sigma^2)$
QUESTION [8 upvotes]: Suppose $X_1,X_2,\ldots,X_n$ are independent $N(0,\sigma^2)$ random variables.
For $i,j\in \{1,2,\ldots,n\}$, consider $$U=\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$$
Provided $n>1$, we know that $U$ has a Beta distribution when $i=j$ :
$$U=\frac{X_i^2/\sigma^2}{\sum_{i=1}^n X_i^2/\sigma^2} \sim \text{Beta}\left(\frac12,\frac{n-1}2\right) \quad,\,i=1,2,\ldots,n$$

What can we say regarding the distribution of $U$ when $i\ne j$? What are the moments of $U$ in this case?

For $n=2$, if we transform $(X_1,X_2)$ to polar coordinates $(R,\Theta)$, then
$$U=\frac{X_1X_2}{X_1^2+X_2^2}=\frac{R^2\cos\Theta\sin\Theta}{R^2}=\frac{\sin(2\Theta)}{2}$$
Since $\Theta$ is uniformly distributed on $(0,2\pi)$, it seems $\sin(2\Theta)$ has an $\text{Arcsine}(-1,1)$ distribution with pdf
$$f(x)=\frac1{\pi \sqrt{1-x^2}}\mathbf 1_{(-1,1)}(x)$$
So that $U$ has pdf
$$f_U(u)=2 f(2u)=\frac2{\pi\sqrt{1-4u^2}}\mathbf1_{\left(-\frac12,\frac12\right)}(u)$$
If $\boldsymbol X=(X_1,X_2,\ldots,X_n)^T$, we can think of $U$ as the product of $i$th and $j$th components of the vector $\frac{\boldsymbol X}{\lVert \boldsymbol X \rVert}$. And we know that $\frac{\boldsymbol X}{\lVert \boldsymbol X \rVert}$ is uniformly distributed on the surface of a unit sphere. I am not sure if this helps in any way.

REPLY [3 votes]: Due to the symmetry  $\sigma_1=...=\sigma_n=\sigma$, it is enough to consider $i=1$ and $j=2$.
While my first thought was n-dimensional spherical coordinates also, you can immediately do the integrals under consideration with a trick. We start with
$$I(t)=\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty {\rm d}x_1 \cdots {\rm d}x_n \, \frac{e^{-{t(x_1^2+...+x_n^2)}}}{(2\pi \sigma^2)^{n/2}} \, \frac{x_1^2x_2^2}{(x_1^2+...+x_n^2)^2} $$
$$
I''(t)=\frac{1}{(2\pi \sigma^2)^{n/2}}\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty {\rm d}x_1 \cdots {\rm d}x_n \, e^{-t(x_1^2+...+x_n^2)} \, {x_1^2x_2^2} \\
= \frac{\left(\pi/t\right)^{n/2-1}}{(2\pi \sigma^2)^{n/2}} \cdot \frac{\pi}{4t^3} = \frac{t^{-n/2-2}}{4(2 \sigma^2)^{n/2}} \, .$$
Then by integrating we get $$I(t)= \frac{t^{-n/2}}{(2 \sigma^2)^{n/2}} \cdot \frac{1}{n(n+2)} + c_1t + c_2$$
and since $I(\infty)=0$ it follows $c_1=c_2=0$ and the result is then obtained by setting $t=\frac{1}{2\sigma^2}$.
Evidently this can be generalized to k-th momenta with $k$ being even, while for odd $k$ the integrand is odd and the integral thus zero.