exercise:8dbb0c4f40: Difference between revisions

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Let <math>d\geqslant3</math> and <math>n</math> be such that <math>2\ln(n)\leqslant d</math> holds. Show that
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\label{PROB-NORMALIZED-UA-THM} Let <math>d\geqslant3</math> and <math>n</math> be such that <math>2\ln(n)\leqslant d</math> holds. Show that


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


\smallskip
{
\small
''Hint:'' Use that <math>1\leqslant d^2/(d-2\ln n)^2</math> holds and apply then the Theorem of Total Probability.
''Hint:'' Use that <math>1\leqslant d^2/(d-2\ln n)^2</math> holds and apply then the Theorem of Total Probability.
}

Revision as of 02:48, 1 June 2024

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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.