Is there a closed-form for $\mathbb{E}\left[\frac{\left|X_1X_2\right|}{X_1^2+X_2^2+\cdots+X_d^2}\right]$, where all the $X_i$ are normal i.i.d.?

Question

I would like to compute $$\mathbb{E}\left[\frac{\left|X_1X_2\right|}{X_1^2+X_2^2+\cdots+X_d^2}\right]$$ where all the $X_i$ ae i.i.d. according to a standard normal Gaussian distribution $\mathcal{N}(0,1)$.

It's easy to get an upper-bound on this quantity via: $$\mathbb{E}\left[\frac{\left|X_1X_2\right|}{X_1^2+X_2^2+\cdots+X_d^2}\right]\leqslant\mathbb{E}\left[\frac{\left|X_1X_2\right|}{X_3^2+\cdots+X_d^2}\right]=\mathbb{E}\left[\left|X_1\right|\right]^2\mathbb{E}\left[\frac{1}{X_3^2+\cdots+X_d^2}\right]=\frac{2}{\pi(d-4)}$$

This upper-bound works well for large $d$, which is what I'm interested in anyway, but I was curious if it was possible to get a closed-form for this expression. Neither computing directly the integral nor somehow breaking down the expression into independent smaller ones seems possible. In fact, I'm not even sure this has a closed form to begin with. Am I missing something trivial?

Answer 1

Write $X = RY,$ where $R = \|X\|$ and $Y$ is uniform over the unit sphere. We'll use the standard fact that $R$ and $Y$ are independent.

First observe that expressing things in polar coordinate, the expression we're looking for is just $\mathbb{E}[|Y_1Y_2|]$. But since $R$ and $Y$ are independent, we have $$ 2/\pi = \mathbb{E}[|X_1X_2|] = \mathbb{E}[R^2|Y_1Y_2|] = \mathbb{E}[R^2]\mathbb{E}[|Y_1Y_2|],$$ where I've also used the independence of $X_1$ and $X_2$. Finally, $$\mathbb{E}[R^2] = \mathbb{E}[\sum X_i^2] = d.$$ So we get that $\mathbb{E}[|Y_1Y_2|] = 2/(\pi d)$.

Answer 2

$(U_1,U_2)=\left(\frac{X_1^2}{X_1^2+\cdots,X_n^2},\frac{X_2^2}{X_1^2+\cdots,X_n^2}\right)$ has the Dirichlet density $$u_1^{-1/2}u_2^{-1/2}(1-u_1-u_2)^{\frac{n}{2}-2}\frac{\Gamma(\frac{n}{2})}{\Gamma(1/2)\Gamma(1/2)\Gamma(\frac{n}{2}-1)}$$ and you are asking for $$E(\sqrt{U_1U_2})=\frac{\Gamma(\frac{n}{2})}{\Gamma(1/2)\Gamma(1/2)\Gamma(\frac{n}{2}-1)}\times \int_{u_1+u_2<1}(1-u_1-u_2)^{\frac{n}{2}-2}du_1du_2=\frac{\Gamma(\frac{n}{2})}{\Gamma(1/2)\Gamma(1/2)\Gamma(\frac{n}{2}-1)}\times \frac{\Gamma(1)\Gamma(1)\Gamma(\frac{n}{2}-1)}{\Gamma(\frac{n}{2}+1)} =\frac{2}{n\pi}.$$