1
$\begingroup$

Can i find some constants $\mathbf a$ and $\mathbf b$ in $\mathbb{R}_+^d$ such that for all $\mathbf{x} \in \mathbb{R}_{+}^{d}$, the inverse beta function :

$$IB(\mathbf x) = \frac{\Gamma\left(\lvert \mathbf x\rvert\right)}{\prod\limits_{i=1}^d \Gamma(x_i)}$$

satisfies $$\mathbf a^{\mathbf x} \le IB(\mathbf x) \le \mathbf b^{\mathbf x}\text{ ?}$$

Or is there some other known bounds on this function that have another shape ? this is the shape i think of, but maybe another shape can be usefull too.

If it is usefull, you can take the restiction that all $x_i$ except the first are integers.

[Please re-tag the question if you think these are not the right tags]

Improve this question
$\endgroup$
6
  • $\begingroup$ Lettiing $x_1\downarrow0$, we see that any lower bound of the form $\mathbf a^{\mathbf x} $ with $\mathbf a\in(0,\infty)^n$ will not hold. Also, if $\mathbf b$ is not allowed to depend on $d$, then any upper bound of the form $\mathbf b^{\mathbf x}$ will not hold either. $\endgroup$
    – Iosif Pinelis
    Commented Nov 28, 2020 at 22:53
  • $\begingroup$ $\mathbf a,\mathbf b$ are allowed to depend on $d$. I don get why you first point holds ? $\endgroup$
    – lrnv
    Commented Nov 29, 2020 at 17:25
  • $\begingroup$ If $x_1\downarrow 0$, then $\Gamma(x_1)=\Gamma(x_1+1)/x_1\sim1//x_1\to\infty$, whence $IB(x)\to0$. Also, if the coordinates of $\mathbf b$ are allowed to depend on $d$, just choose $\mathbf b=(d,\dots,d)$. Also, how about using Stirling bounds on the gamma function? $\endgroup$
    – Iosif Pinelis
    Commented Nov 29, 2020 at 21:34
  • $\begingroup$ I did try to bound the IB function wiht the stirling gbounds, but this does not help me in the surrounding problem. Bounding hte IB function might not be exactly what i need. Thanks anyway ! $\endgroup$
    – lrnv
    Commented Dec 1, 2020 at 12:11
  • $\begingroup$ I am afraid that your question cannot be satisfactorily answered unless you specify the bounds you need. $\endgroup$
    – Iosif Pinelis
    Commented Dec 1, 2020 at 17:34

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .