# Upper bound of a ratio of integrals

I'm wondering how to upper bound the following ratio of integrals: $$\frac{\int_{\Delta_a}(\prod_{i=1}^n\lambda_i)^{p-1}\prod_{i in terms of $$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, 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.

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