Timeline for On the concentration of Lipschitz functions near its expectation, where the vector has identical but not independent, components

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May 31, 2020 at 5:01	comment	added	Yuval Peres		For every $c$ the inequality fails if $n$ is large enough.
May 30, 2020 at 19:37	comment	added	Learning math		Thank you for your answer, but do you mind elaborating on it a bit more? For the sum function, the Lipschitz norm is $\sqrt{n}.$ I wonder if your answer is valid for each $n \ge 1,$ or when $n \to \infty ?$ (The limiting case is also quite important for me) I see that $P[ \|\sum_{i=1}^{n} X - \sum_{i=1}^{n} EX\| \ge t=n ] = P[nX \ge n] =P[X \ge 1] \nrightarrow 0,$ unlike the RHS term $2 exp(- \frac{cn^2}{n}) = 2 exp (- cn)\to 0$ as $n \to \infty.$ So this does prove that the inequality isn't true in the limiting case. Is that what you had in mind? Thanks again!
May 30, 2020 at 17:00	history	answered	Yuval Peres	CC BY-SA 4.0