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Jul 26, 2023 at 11:49 comment added fedja In other words, it is $(2\pi)^{n/2}n^{-(n+1)/2}Q(c)^n$ where $Q(c)=\min_{t>0}[\Phi(ct)\frac 1t e^{t^2/2}]$ and $\Phi(z)=P(\xi>z)$ ($\xi$ is the standard 1D normal r.v.) is the Gaussian tail function. That is correct up to a constant factor depending on $c\in(0,1)$ but not on $n$.
Jul 26, 2023 at 9:38 comment added fedja Depends on the precision you want. For the crude asymptotic just think Gaussian.
S Jul 26, 2023 at 9:00 review First questions
Jul 26, 2023 at 9:43
S Jul 26, 2023 at 9:00 history asked user509227 CC BY-SA 4.0