I'm wondering how to upper bound the following ratio of integrals: $$\frac{\int_{\Delta_a}(\prod_{i=1}^n\lambda_i)^{p-1}\prod_{i<j}|\lambda_i-\lambda_j|}{\int_{\Delta_b}(\prod_{i=1}^n\lambda_i)^{p-1}\prod_{i<j}|\lambda_i-\lambda_j|}$$ in terms of $a,b,p,n$, where $$\Delta_a=\{(\lambda_1,....,\lambda_n):\sum_{i=1}^n\lambda_i\leq a,\lambda_1\geq0,...,\lambda_n\geq0\}$$ and $1\geq a>b>0, p>0$. For example, when $n=1$, the ratio can be calculated exactly since $\prod_{i<j}|\lambda_i-\lambda_j|=1$, which yields a bound of $(a/b)^p$. However, for $n\geq 2$, I have no idea how to do. I conjecture something like $(a/b)^{np}$ up to some multiplicative constant but do not know how to prove.

I encountered this problem when trying to bound the ratio of small ball probability of the trace of matrix valued beta distribution which is needed in a Bayesian analysis.


1 Answer 1


Let \begin{equation} J_a:=\int_{\Delta_a} \Big(\prod_{i=1}^nx_i\Big)^{p-1}\prod_{i<j}|x_i-x_j|\,\prod_{i=1}^n dx_i. \end{equation} By the change of variables $x_i=ay_i$, we see that \begin{equation} J_a=a^{np+n(n-1)/2}J_1. \end{equation} Hence, your ratio of integrals is \begin{equation} J_a/J_b=(a/b)^{np+n(n-1)/2}. \end{equation} In particular, for $n=1$ we get your result, $(a/b)^p$.

  • $\begingroup$ Thank you! That is very helpful! $\endgroup$ Apr 4, 2019 at 15:01

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