exercise:093e852bc9: Difference between revisions
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Let <math>X</math>, <math>Y\sim\mathcal{N}(0,1,\mathbb{R}^d)</math>. Show the following. | |||
<ul style="list-style-type:lower-roman"><li> <math>\forall\:d\geqslant1\colon\E(\|X-Y\|-\sqrt{2d})\leqslant1/\sqrt{2d}</math>. | <ul style="list-style-type:lower-roman"><li> <math>\forall\:d\geqslant1\colon\E(\|X-Y\|-\sqrt{2d})\leqslant1/\sqrt{2d}</math>. | ||
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''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>. | ''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>. | ||
Latest revision as of 01:41, 1 June 2024
[math]
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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].