exercise:38aca55fd9: Difference between revisions
From Stochiki
(Created page with "<div class="d-none"><math> \newcommand{\indexmark}[1]{#1\markboth{#1}{#1}} \newcommand{\red}[1]{\textcolor{red}{#1}} \newcommand{\NOTE}[1]{$^{\textcolor{red}\clubsuit}$\marginpar{\setstretch{0.5}$^{\scriptscriptstyle\textcolor{red}\clubsuit}$\textcolor{blue}{\bf\tiny #1}}} \newcommand\xoverline[2][0.75]{% \sbox{\myboxA}{$\m@th#2$}% \setbox\myboxB\null% Phantom box \ht\myboxB=\ht\myboxA% \dp\myboxB=\dp\myboxA% \wd\myboxB=#1\wd\myboxA% Scale phantom...") |
No edit summary |
||
Line 1: | Line 1: | ||
<div class="d-none"><math> | <div class="d-none"><math> | ||
\newcommand{\smallfrac}[2]{\frac{#1}{#2}} | |||
\newcommand{\medfrac}[2]{\frac{#1}{#2}} | |||
\newcommand{\textfrac}[2]{\frac{#1}{#2}} | |||
\newcommand{\smallfrac}[2] | |||
\newcommand{\medfrac}[2] | |||
\newcommand{\textfrac}[2] | |||
\newcommand{\tr}{\operatorname{tr}} | \newcommand{\tr}{\operatorname{tr}} | ||
\newcommand{\e}{\operatorname{e}} | \newcommand{\e}{\operatorname{e}} | ||
\newcommand{\B}{\operatorname{B}} | \newcommand{\B}{\operatorname{B}} | ||
\newcommand{\Bbar}{\ | \newcommand{\Bbar}{\overline{\operatorname{B}}} | ||
\newcommand{\pr}{\operatorname{pr}} | \newcommand{\pr}{\operatorname{pr}} | ||
\newcommand{\dd}{\operatorname{d}\hspace{-1pt}} | \newcommand{\dd}{\operatorname{d}\hspace{-1pt}} | ||
Line 33: | Line 12: | ||
\newcommand{\V}{\operatorname{V}} | \newcommand{\V}{\operatorname{V}} | ||
\newcommand{\Cov}{\operatorname{Cov}} | \newcommand{\Cov}{\operatorname{Cov}} | ||
\newcommand{\Bigsum}[2] | \newcommand{\Bigsum}[2]{\mathop{\textstyle\sum}_{#1}^{#2}} | ||
\newcommand{\ran}{\operatorname{ran}} | \newcommand{\ran}{\operatorname{ran}} | ||
\newcommand{\card}{\#} | \newcommand{\card}{\#} | ||
\newcommand{\ | \newcommand{\mathds}{\mathbb} | ||
\renewcommand{\P}{\operatorname{P}} | |||
\renewcommand{\L}{\operatorname{L}} | |||
</math></div> | |||
Use [[guide:7885448c04#NaiveTailBound |Problem]] to give an alternative proof of the Johnson-Lindenstrauss Lemma that does not rely on the Gaussian Annulus Theorem. | |||
Revision as of 02:43, 2 June 2024
[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}
\renewcommand{\P}{\operatorname{P}}
\renewcommand{\L}{\operatorname{L}}
[/math]
Use Problem to give an alternative proof of the Johnson-Lindenstrauss Lemma that does not rely on the Gaussian Annulus Theorem.