Is $\mathbb E\left[\frac{d}{||x||_1^2}\right]=O(1)$ for all $d\in\mathbb R^+$, where $x\in S^{d-1}$ is a random $d$-dimensional unit vector?

Question

Let $x\in S^{d-1}$ be chosen uniformly at random from the $d$-dimensional unit sphere.

I want to show that there exists a universal constant $c\in\mathbb R$ such that $\mathbb E\left[\frac{d}{||x||_1^2}\right]\le c$ for all dimensions $d$.

Any thoughts?

---

Using numerical simulations, it seems that $\mathbb E\left[\frac{d}{||x||_1^2}\right]< \pi/2$, but I'm not sure how to prove it.

Answer 1

We may sample $x$ as follows: choose i.i.d. standard Gaussian $\xi_1,\ldots,\xi_d$ and put $$x_i=\frac{\xi_i}{\sqrt{\sum_{j=1}^d \xi_j^2}},\quad i=1,2,\ldots,d.$$ Then $$\frac1{\|x\|_1^2}=\frac{\sum \xi_j^2}{(\sum |\xi_j|)^2}. $$ By law of large numbers, the numerator is usually of order $d$ and the denominator of order $d^2$. To specify, by Chernoff bounds the probability that either $\sum \xi_j^2>10 d$ or $\sum |\xi_j|<d/10$ is exponentially small (in $d$), and when this event happens, $\frac1{\|x\|_1^2}$ is still bounded from above by 1. Thus the result.