[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].