# Asymptotic radius of the smallest enclosing ball

1. Let $X_1,..., X_n$ be i.i.d. $d$-dimensional standard normal random variables, and let $R_n$ be the radius of the smallest ball containing $X_1,...,X_n$. What is known about the distribution of $R_n$ as $n \to \infty$?

2. Are there some examples for the asymptotics of $R_n$ when the independence assumption does not hold? I would appreciate any reference or suggestion here, also for a non-Gaussian case.

• I don't think the ball was supposed to be centred at the origin. – Anthony Quas Nov 9 '16 at 17:11
• @AnthonyQuas no, otherwise it would be the maximal distance and this I think is well known. – Sinusx Nov 9 '16 at 20:35
• In fact, though, I suspect this is the right answer. If you look at the smallest ball centred at the origin, then this grows slowly (at rate $\sqrt{\log n}$). Hence there are many points close to the maximal radius. These should fairly closely fill out a thin spherical shell of radius $R_n$. – Anthony Quas Nov 9 '16 at 22:26

This is to confirm and give more details to Anthony Quas' comment.

Let's start with a bit of wishful thinking. Denote by $P_n$ the convex hul of the points $X_1,\dotsc, X_n$.$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bE}{\mathbb{E}}$ Denote by $B_R(p)$ the ball or radius $R$ centered at $p\in\bR^d$.Suppose that for any $n\geq 3$ there exists radii $R_n^-<R_n^+$ such that.

1. $B_{R_n^-}(0)\subset P_n \subset B_{R_n^+}(0)$ a.s. for large $n$.
2. $R_n^-\sim R_n^+$ as $n\to \infty$, i.e.,

$$\lim_{n\to\infty} \frac{R_n^+}{R_n^-}=1.$$

Then, almost surely,

$$\frac{R_n}{R_n^-}\to 1.$$

Fortunately such radii can be found and they have the growth rate suggested by Anthony Quas. This is all contained in a preprint of Fresen & Vitale arXiv: 1402.2718

Let me give a few more details. Denote by $\mu$ The standard Gaussian measure on $\bR^d$. For each $\delta>0$ we denote by $F_\delta$ the intersection of all closed half-spaces $S\subset \bR^d$ such that $\mu(S)\geq 1-\delta$.

The floating body is then the convex set $F_{1/n}$. Due to the radial symmetry of $\mu$ you can see that $F_{1/n}$ is a ball of radius $r_n$ centered at the orgin,where $r_n$ is found from the equality

$$\Psi(r_n)=\frac{1}{n}, \;\;\Psi(x):=\frac{1}{\sqrt{2\pi}}\int_{x}^\infty e^{-t^2/2} dt=\frac{1}{n}.$$

From the known estimates

$$\Big(\frac{1}{x}-\frac{1}{x^3}\Big)\gamma(x) < \Psi(x) <\frac{1}{x}\gamma(x),\;\;\;\;\gamma(x)=\frac{1}{\sqrt{2\pi}} e^{-x^2/2}.$$

We deduce that

$$r_n=\sqrt{2\log n} +o(1), \;\;\mbox{as}\; n\to\infty.$$

In the above preprint construct (explicit) sequences of positive real numbers $\newcommand{\ve}{\varepsilon}$ $\ve_1(n),\ve_2(n)$ converging to $0$ with the following property: with probability $1$, there exists a natural number $N$ such that for all $n>N$

$$\big(1-\ve_1(n)\;\big) F_{1/n}\subset P_n\subset \big(1+\ve_2(n)\;\big) F_{1/n}.$$

Thus proves that

$$\frac{R_n}{\sqrt{2\log n}}\to 1\;\; \mbox{a.s.}.$$