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6 exercise(s) shown, 0 hidden
BBy Bot
Jun 01'24

Replicate the results of Figure and Table by running the corresponding experiments.

BBy Bot
Jun 01'24
[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]X[/math], [math]Y\sim\mathcal{N}(0,1,\mathbb{R}^d)[/math]. Show the following.

  • [math]\forall\:d\geqslant1\colon\E(\|X-Y\|-\sqrt{2d})\leqslant1/\sqrt{2d}[/math].
  • [math]\forall\:d\geqslant1\colon\V(\|X-Y\|)\leqslant 3[/math].

Hint: Check firstly [math]\V((X_i-Y_i)^2)=3[/math] by establishing that [math]X_i-Y_i\sim\mathcal{N}(0,2,\mathbb{R})[/math] and by using a suitable formula for computing the fourth moment. Conclude then that [math]\V(\|X-Y\|^2)\leqslant3d[/math]. Adapt finally the arguments we gave above for [math]\E(\|X\|-\sqrt{d})[/math] and [math]\V(\|X\|)[/math].

BBy Bot
Jun 01'24

Replicate the results of Table. Make additionally a plot of the distribution of mutual distances (this should give a picture similar to Figure).

BBy Bot
Jun 01'24

Replicate the results of Table. Let your code also compute the averages and variances of scalar products of the normalized vectors, i.e., [math]\langle{}x/\|x\|,y/\|y\|\rangle{}[/math].

BBy Bot
Jun 01'24

Compute in a simulation norm, distance and scalar product of points that are drawn from the hypercube [math]H_d[/math] (coordinate-wise) uniformly, i.e., [math]x=(x_1,\dots,x_d)[/math] is drawn such that [math]x_i\sim\mathcal{U}([-1,1])[/math] for [math]i=1,\dots,d[/math]. Make plots and tables similar to Figure and [[#TAB-1 |Table\,--\,]]. Compare the experimental data with our theoretical results above.

BBy Bot
Jun 01'24
[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 \sbox\myboxB{$\m@th\overline{\copy\myboxB}$}% Overlined phantom \setlength\mylenA{\the\wd\myboxA}% calc width diff \addtolength\mylenA{-\the\wd\myboxB}% \ifdim\wd\myboxB\lt\wd\myboxA% \rlap{\hskip 0.35\mylenA\usebox\myboxB}{\usebox\myboxA}% \else \hskip -0.5\mylenA\rlap{\usebox\myboxA}{\hskip 0.5\mylenA\usebox\myboxB}% \fi} \newcommand{\smallfrac}[2]{\scalebox{1.35}{\ensuremath{\frac{#1}{#2}}}} \newcommand{\medfrac}[2]{\scalebox{1.2}{\ensuremath{\frac{#1}{#2}}}} \newcommand{\textfrac}[2]{{\textstyle\ensuremath{\frac{#1}{#2}}}} \newcommand{\nsum}[1][1.4]{% only for \displaystyle \mathop{% \raisebox {-#1\depthofsumsign+1\depthofsumsign} \newcommand{\tr}{\operatorname{tr}} \newcommand{\e}{\operatorname{e}} \newcommand{\B}{\operatorname{B}} \newcommand{\Bbar}{\xoverline[0.75]{\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]{\ensuremath{\mathop{\textstyle\sum}_{#1}^{#2}}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\card}{\#} \newcommand{\Conv}{\mathop{\scalebox{1.1}{\raisebox{-0.08ex}{$\ast$}}}}% \usepackage{pgfplots} \newcommand{\filledsquare}{\begin{picture}(0,0)(0,0)\put(-4,1.4){$\scriptscriptstyle\text{\ding{110}}$}\end{picture}\hspace{2pt}} \newcommand{\mathds}{\mathbb}[/math]

If you enjoy horror movies, then watch 2002's Hypercube.