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By Klartag B.

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By the conclusion of that proposition, √ √ 2 Prob{||X| − n | ≥ ε n} ≤ Ce−cε n (0 ≤ ε ≤ 1). (26) Let Z be a gaussian random vector in Rn , independent of X, with E Z = 0 and Cov(Z ) = n −α0 Id. 1(i), we know that √ Prob{|Z | ≥ 1} ≤ Prob{|Z | ≥ 20n · n 1−α0 } ≤ e−n . Consequently, the event −1 ≤ |X + Z | − |X| ≤ 1 holds with probability greater than 1 − e−n . By applying (26) we obtain that for 0 ≤ ε ≤ 1, √ √ (27) Prob{||X + Z | − n | ≥ ε n} √ 1 √ 2 n ≤ C e−c ε n ≤ e−n + Prob ||X| − n | ≥ ε − √ n (in obtaining √ the last inequality √ in (27), one needs to consider separately the cases ε < 2/ n and ε ≥ 2/ n).

2 seem surprisingly good: Marginals of almost-proportional dimension are allegedly close to gaussian. 2 hides, first, in the requirement that ε > C/ log n, and second, in the use of the rather weak T -distance. Note added in proofs: We were recently able to improve some of the quantitative estimates that were described in this article. The logarithmic bounds may be replaced with a power-law dependence on the dimension. MG/0611577. References 1. : The central limit problem for convex bodies. Trans.

Ii) Y has a spherically-symmetric distribution. √ √ 2 (iii) Prob{| |Y | − n | ≥ ε n} ≤ Ce−cε n for any 0 ≤ ε ≤ 1. Proof: Recall that √ Vol( n Dn ) ≤ Cˆ n (25) for some universal constant Cˆ > 1. 1 and Cˆ is the constant from (25). Throughout this proof, α0 , C0 , C1 and Cˆ will stand for the universal constants just mentioned. We assume that inequality (24) – the main assumption of this proposition – holds, with the constant C1 as was just defined. 1, based on (24), since C0 n ≤ C1 n log n. By the conclusion of that proposition, √ √ 2 Prob{||X| − n | ≥ ε n} ≤ Ce−cε n (0 ≤ ε ≤ 1).