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Aug 1, 2021 at 18:34 comment added jlewk Yes; to overcome dependence I had in mind to work instead with $b_0(1\pm\epsilon)$ in the event $|\sqrt{\chi^2_d/d}-1|\le \epsilon$ that would not affect the probability of the event by more than $2e^{-\epsilon^2 d/2}$.
Aug 1, 2021 at 18:28 answer added Yuval Peres timeline score: 1
Aug 1, 2021 at 17:28 comment added Yuval Peres @jlewk : Note that $Z$ and $\chi_d^2$ are dependent.
Aug 1, 2021 at 10:23 comment added jlewk WLOG assume $\|a\|=1$ and $b=b_0/\sqrt d$. Write $X=g/\|g\|$ with $g\sim N(0,I_d)$. Then your event has the form $\{Z>b_0 \sqrt{\chi^2_d/d} \}$ with $Z=g^Ta \sim N(0,1)$ and $\chi^2_d$ chi-square with $d$ degrees of freedom. Exponential concentration of the chi-square should give a solution.
Aug 1, 2021 at 9:50 history asked Probabilist CC BY-SA 4.0