Expected value of minimal coordinate when uniformly sampling from unit sphere

Question

It is known that given $X=(X_1, X_2, \ldots, X_n)$ where $X_i$ $\sim N(0,1)$ and independent of one another, then $X^{'}=X/\sqrt{X_1^2+\cdots+X_n^2}$ is uniformly distributed over the surface of the unit sphere.

Denote $X^{'} = (X^{'}_1, X^{'}_2, \ldots, X^{'}_n)$. Also, denote by $U =min_{ 1\leq i\leq n}{|X^{'}_i|}$.

I am trying to:

1. find the expected value of $U$ (as a function of $n$).
2. show that it is concentrated around its mean.

My try:

I know that $Z = X_1^2+\cdots+X_n^2$ is distributed according to chi-squared with n degrees of freedom. Furthermore, I know that Z is concentrated around its mean. Since $E(X_i^2) = 1$ we have that $E(Z) = n$, so with high probability we divide each $X_i$ with something which is "close to" $\sqrt{n}$. I am not sure how to proceed since Z and $X_i$ are dependent, or if it is a good direction at all.

Any help would be much appreciated!

Answer 1

(Not a justified answer, but an informed guess at this point).

For the random variables $(X_1, \ldots, X_n)$ it is possible to show that:

$$\sqrt{\frac 2\pi} \; n \cdot \min\{|X_1|, \ldots, |X_n|\} \to Exp(1)$$

the latter symbol denoting the exponential law with parameter 1 (this fact only depends on the value of the density near $0$ of the underlying iid random variables). This part is easy to make rigorous.

Here I would bet that

$$\sqrt{\frac 2\pi}\; n^{3/2} \cdot \min\{|X'_1|, \ldots, |X'_n|\} \to Exp(1)$$

based on

• the concentration properties of $\sqrt{X_1^2+ \ldots + X_n^2}$ near $\sqrt{n}$
• the fact this random variable will be "almost" independent of the mininum value of the $|X_i|, i=1...n$.

This should be the hard part of the argument.