[math] \newcommand{\smallfrac}[2]{\frac{#1}{#2}} \newcommand{\medfrac}[2]{\frac{#1}{#2}} \newcommand{\textfrac}[2]{\frac{#1}{#2}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\e}{\operatorname{e}} \newcommand{\B}{\operatorname{B}} \newcommand{\Bbar}{\overline{\operatorname{B}}} \newcommand{\pr}{\operatorname{pr}} \newcommand{\dd}{\operatorname{d}\hspace{-1pt}} \newcommand{\E}{\operatorname{E}} \newcommand{\V}{\operatorname{V}} \newcommand{\Cov}{\operatorname{Cov}} \newcommand{\Bigsum}[2]{\mathop{\textstyle\sum}_{#1}^{#2}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\card}{\#} \newcommand{\mathds}{\mathbb}[/math]

Let [math]d\geqslant3[/math] and [math]n[/math] be such that [math]2\ln(n)\leqslant d[/math] holds. Show that

[[math]] \P\bigl[\bigl|\bigl\langle{}\medfrac{x^{\scriptscriptstyle(j)}}{\|x^{\scriptscriptstyle(j)}\|},\medfrac{x^{\scriptscriptstyle(k)}}{\|x^{\scriptscriptstyle(k)}\|}\bigr\rangle{}\bigr|\leqslant\medfrac{\sqrt{6\ln n}}{\sqrt{d-1}} \text{ for all }j\not=k\bigr]\geqslant 1-\medfrac{1}{n} [[/math]]

holds, when [math]x^{\scriptscriptstyle(1)},\dots,x^{\scriptscriptstyle(n)}[/math] are drawn uniformly at random from the [math]d[/math]--dimensional unit ball.

Hint: Use that [math]1\leqslant d^2/(d-2\ln n)^2[/math] holds and apply then the Theorem of Total Probability.