I'm considering a ratio of incomplete Selberg integral: $$f_n(a,b)=\frac{\int_{\Delta_a}\prod_{i=1}^nx_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-x_i)^{-1/2}\prod_{i<j}|x_i-x_j|}{\int_{\Delta_b}\prod_{i=1}^nx_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-x_i)^{-1/2}\prod_{i<j}|x_i-x_j|}$$ where $$\Delta_a=\{(x_1,...,x_n):0<x_1<a,0<x_2<a,...,0<x_n<a\}$$ and $1>a>b>0$, $\alpha>\frac{n+1}{2}$.

My question is, how can we upper bound $f_n(a,b)$? Is there a bound like some powers of $Ca/b$ for some constant $C>0$?

  • $\begingroup$ I have changed $\Delta_a$ to be a hypercube instead of the previous hyper rectangle. $\endgroup$ – neverevernever May 5 '19 at 15:54
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    $\begingroup$ if $C$ may depend on $n$, then of course we have $C_n(a/b)^M$ upper bound, where $a=n(\alpha-\frac{n+1}2)+{n\choose 2}$. $\endgroup$ – Fedor Petrov May 5 '19 at 21:31
  • $\begingroup$ Sorry I do not quite get it. $\endgroup$ – neverevernever May 7 '19 at 0:34
  • $\begingroup$ 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: $\endgroup$ – ofer zeitouni May 13 '19 at 10:56
  • $\begingroup$ $n^{-2} \log f_n(a,b)\to -\inf_{\mu: \mu([0,a])=1} I(\mu)+\inf_{\mu:\mu([0,b])=1} I(\mu)$. Just for completeness, the function $I$ is the following: $$I(\mu)=-\int\int \log|x-y|\mu(dx)\mu(dy)-\frac{1}{2}\int \log(x) \mu(dx)$$ $\endgroup$ – ofer zeitouni May 13 '19 at 11:00

Denote the numerator by $I(a)$. The change of variables $x_i=ay_i$ gives $$ I(a)=a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\times \\ \times\int_{[0,1]^n}\prod_{i=1}^ny_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-ay_i)^{-1/2}\prod_{i<j}|y_i-y_j|. $$ The latter integral may be estimated using the two-sided estimates $1\leqslant (1-ay_i)^{-1/2}\leqslant (1-y_i)^{-1/2}$. So, we get $$c_1(n) a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\leqslant I(a)\leqslant c_2(n) a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2},$$ where $c_1(n)$ and $c_2(n)$ are corresponding integrals (actually $c_2(n)=I(1)$). It implies $$ \frac{I(a)}{I(b)}\leqslant \frac{c_2(n)}{c_1(n)}\cdot (a/b)^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}. $$

You may get some explicit bounds for $c_2(n)/c_1(n)$ , for example as follows. We have $$ c_2(n)=I(1)=\int_{[0,1]^n}\prod_{i=1}^ny_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-y_i)^{-1/2}\prod_{i<j}|y_i-y_j| dy_1\ldots dy_n, $$ and $c_1(n)$ the same integral without $\prod_{i=1}^n(1-y_i)^{-1/2}$. In the integral for $c_2(n)$, change the variables: take $y_i=1-t_i^2$. You get $$ c_2(n)=2^n\int_{[0,1]^n}\prod_{i=1}^n(1-t_i^2)^{\alpha-\frac{n+1}{2}}\prod_{i<j}|t_i^2-t_j^2| dt_1\ldots dt_n. $$ Use the bounds $1-t_i^2=(1-t_i)(1+t_i)\leqslant 2(1-t_i)$, $|t_i^2-t_j^2|=|t_i-t_j|(t_i+t_j)\leqslant 2|t_i-t_j|$. We get $$ c_2(n)\leqslant 2^{n(\alpha-\frac{n+1}{2})+\frac{n(n+1)}2} \int_{[0,1]^n}\prod_{i=1}^n(1-t_i)^{\alpha-\frac{n+1}{2}}\prod_{i<j}|t_i-t_j| dt_1\ldots dt_n=\\ =2^{n(\alpha-\frac{n+1}{2})+\frac{n(n+1)}2} c_1(n), $$ the last equality follows from the change of variables $1-t_i=z_i$.

Is it ok for you or you need the estimates which are explicit w.r.t. $n$?

  • $\begingroup$ Yeah... I think I need to bound the ratio $c_1(n)/c_2(n)$... How can we do that? $\endgroup$ – neverevernever May 8 '19 at 3:31
  • $\begingroup$ That is very helpful! Thank you! $\endgroup$ – neverevernever May 15 '19 at 3:16

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