Timeline for Ratio of Selberg integral
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2019 at 3:16 | vote | accept | neverevernever | ||
| May 13, 2019 at 11:07 | comment | added | ofer zeitouni | Note that unsurprisingly, the leading order asymptotics does not depend on $\alpha$. | |
| May 13, 2019 at 11:00 | comment | added | ofer zeitouni | $n^{-2} \log f_n(a,b)\to -\inf_{\mu: \mu([0,a])=1} I(\mu)+\inf_{\mu:\mu([0,b])=1} I(\mu)$. Just for completeness, the function $I$ is the following: $$I(\mu)=-\int\int \log|x-y|\mu(dx)\mu(dy)-\frac{1}{2}\int \log(x) \mu(dx)$$ | |
| May 13, 2019 at 10:56 | comment | added | ofer zeitouni | I can't write a detailed answer at this time, so here is instead a sketch. If you normalize by $\Delta_1$ both integrals, then what you are asking is the ratio of probabilities that all eigenvalues are smaller than $a$ and the same with $b$. Since you have a large deviation principle at scale $n^2$ for the empirical measure of eigenvalues, with rate function $I(\mu)$, with a function $I$ that is explicit (involving the non-commutative entropy of $\mu$ - see section 2.6 in Anderson-Guionnet-Zeitouni's book on RMT), you can read of the answer: | |
| May 7, 2019 at 5:58 | answer | added | Fedor Petrov | timeline score: 3 | |
| May 7, 2019 at 0:34 | comment | added | neverevernever | Sorry I do not quite get it. | |
| May 5, 2019 at 21:31 | comment | added | Fedor Petrov | if $C$ may depend on $n$, then of course we have $C_n(a/b)^M$ upper bound, where $a=n(\alpha-\frac{n+1}2)+{n\choose 2}$. | |
| May 5, 2019 at 15:54 | comment | added | neverevernever | I have changed $\Delta_a$ to be a hypercube instead of the previous hyper rectangle. | |
| May 5, 2019 at 15:53 | history | edited | neverevernever | CC BY-SA 4.0 |
edited body
|
| May 4, 2019 at 15:43 | history | asked | neverevernever | CC BY-SA 4.0 |