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May 31'24

Exercise

[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]

\label{PROB-2}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].

\smallskip { \small 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]. }