12
$\begingroup$

I am trying to prove the following inequality concerning the Beta Function: $$ \alpha x^\alpha B(\alpha, x\alpha) \geq 1 \quad \forall 0 < \alpha \leq 1, \ x > 0, $$ where as usual $B(a,b) = \int_0^1 t^{a-1}(1-t)^{b-1}dt$.

In fact, I only need this inequality when $x$ is large enough, but it empirically seems to be true for all $x$.

The main reason why I'm confident that the result is true is that it is very easy to plot, and I've experimentally checked it for reasonable values of $x$ (say between 0 and $10^{10}$). For example, for $x=100$, the plot is:

Plot of the function to be proven greater than 1

Varying $x$, it seems that the inequality is rather sharp, namely I was not able to find a point where that product is larger than around $1.5$ (but I do not need any such reverse inequality).

I know very little about Beta functions, therefore I apologize in advance if such a result is already known in the literature. I've tried looking around, but I always ended on inequalities trying to link $B(a,b)$ with $\frac{1}{ab}$, which is quite different from what I am looking for, and also only holds true when both $a$ and $b$ are smaller than 1, which is not my setting.

I have tried the following to prove it, but without success: the inequality is well-known to be an equality when $\alpha = 1$, and the limit for $\alpha \to 0$ should be equal to 1, too. Therefore, it would be enough to prove that there exists at most one $0 < \alpha < 1$ where the derivative of the expression to be bounded vanishes. This derivative can be written explicitly in terms of the digamma function $\psi$ as: $$ x^\alpha B(\alpha, x\alpha) \Big(\alpha \psi(\alpha) - (x+1)\alpha\psi((x+1)\alpha) + x\alpha \psi(x\alpha) + 1 + \alpha \log x \Big). $$ Dividing by $x^\alpha B(\alpha, x\alpha) \alpha$, this becomes $$ -f(\alpha) + \frac{1}{\alpha} + \log x, $$ where $f(\alpha) = -\psi(\alpha) + (x+1)\psi((x+1)\alpha) - x \psi(x\alpha)$ is, as proven by Alzer and Berg, Theorem 4.1, a completely monotonic function. Unfortunately, the difference of two completely monotonic functions (such as $f(\alpha)$ and $\frac{1}{\alpha} + C$) can vanish in arbitrarily many points, therefore this does not allow to conclude.

Many thanks in advance for any hint on how to get such a bound!

[EDIT]: As pointed out in the comments, the link to the paper of Alzer and Berg pointed to the wrong version, I have corrected the link.

$\endgroup$
  • $\begingroup$ There is no Theorem 4.1 in the quoted paper. There is a Theorem 4 there, but it does not talk about the digamma function. Can you please clarify? $\endgroup$ – GH from MO Dec 29 '18 at 21:58
  • 1
    $\begingroup$ @GHfromMO Thanks for pointing out the wrong link, the one I inserted did not send to the most updated version. I have now corrected it! $\endgroup$ – Ester Mariucci Dec 29 '18 at 22:14
6
$\begingroup$

You can get away with the usual distribution function mumbo-jumbo. The general lemma is as follows:

Let $\mu,\nu$ be non-negative measures and $f,g$ be non-negative functions such that there exists $s_0>0$ with the property that $\mu\{f>s\}\ge \nu\{g>s\}$ for $s\le s_0$ and the reverse inequality holds for $s\ge s_0$. Suppose also that $\int f^q\,d\mu=\int g^q\,d\nu<+\infty$ for some $q>0$. Then, as long as the integrals in question are finite, we have $\int f^p\,d\mu\ge \int g^p\,d\nu$ for $0<p\le q$ and the reverse inequality holds for $p\ge q$.

The proof of the lemma is rather straightforward. Let $p\le q$ (that is the case you are really interested in) $$ \int f^p\,d\mu-\int g^p\,d\nu=p\int_0^\infty s^p[\mu\{f>s\}-\nu\{g>s\}]\frac{ds}s \\ =p\int_0^\infty [s^p-s_0^{p-q}s^q][\mu\{f>s\}-\nu\{g>s\}]\frac{ds}s\ge 0\,. $$

Now we use it with $f(t)=t(1-t)^x$, $d\mu=\frac{dt}{t(1-t)}$ on $(0,1)$, $g(t)=t$, $d\nu=\frac{dt}{t}$ on $(0,\frac1x)$. Since the maximum of $t(1-t)^x$ is attained at $t=\frac{1}{x+1}$, we see that the function $s\mapsto \mu\{f>s\}$ drops to $0$ before the function $s\mapsto \nu\{g>s\}$. Also, the first function has larger in absolute value negative derivative than the second one for each value of $s$ where it is still positive. To see it, notice that the set where $f>s$ is an interval $(u,v)=(u(s),v(s))$ that shrinks as $s$ increases and the left end $u$ of this interval satisfies $$ du\left(\frac 1u-\frac x{1-u}\right)=\frac{ds}s\,, $$ so trivially $$ \frac{du}{u(1-u)}\ge \frac{du}u>\frac {ds}s $$ The right end moving to the left can only increase the decay speed. Finally, for $q=1$, the integrals are equal (which also shows that the graphs of the distribution functions must indeed intersect), so for $0<p\le 1$ (which plays the role of $\alpha$), we have the desired inequality.

$\endgroup$
  • $\begingroup$ That's a wonderful and clever proof, many thanks! I'm marking this as accepted (although @egs's one also proves the desired inequality) also because it came first. $\endgroup$ – Ester Mariucci Jan 1 at 19:59
5
$\begingroup$

One can also use Jensen's inequality. Let (for $\sigma>0$) $G_\sigma$ denote a random variable with $\Gamma(1,\sigma)$-distribution, i.e. having Lebesgue density $$f_\sigma(t)=\frac{t^{\sigma-1}}{\Gamma(\sigma)} e^{-t}\;1_{(0,\infty)}(t)\;,$$ then $\mathbb{E}(G_\sigma)=\sigma$. Since $\alpha\in (0,1)$ the functions $t\mapsto t^\alpha$ resp. $t\mapsto t^{1-\alpha}$ on $\mathbb{R}_+$ are concave. By Jensen's inequality $$\frac{\Gamma(\alpha+\alpha x)}{\Gamma(\alpha x)}=\mathbb{E}(G_{x\alpha}^\alpha)\leq \left(\mathbb{E}(G_{x\alpha})\right)^\alpha=(x\alpha)^{\alpha}$$

and $$\frac{1}{\Gamma(\alpha)}=\mathbb{E} G_\alpha^{1-\alpha}\leq\left(\mathbb{E}(G_{\alpha})\right)^{1-\alpha}=\frac{1}{\alpha^{\alpha-1}}$$ Using that gives $$B(\alpha,x \alpha)=\frac{\Gamma(\alpha)\,\Gamma(x\alpha)}{\Gamma(\alpha +x\alpha)}\geq \frac{\Gamma(\alpha)}{\alpha^\alpha x^\alpha}\geq \frac{\Gamma(\alpha)}{\alpha\,\Gamma(\alpha)\,x^\alpha}=\frac{1}{\alpha x^\alpha},$$ as desired.

$\endgroup$
  • 1
    $\begingroup$ That's a very quick way to prove it, thanks! This proof is of particular interest to me because I was also aiming to prove \Gamma(\alpha) \geq \alpha^{\alpha+1}, which indeed is essentially equivalent to the inequality on Beta by using Stirling on both \Gamma(x\alpha) and \Gamma((x+1)\alpha). $\endgroup$ – Ester Mariucci Jan 1 at 20:03
4
$\begingroup$

This is an attempt to strengthen your claim.

If $x$ is large then $B(x,y)\sim \Gamma(y)x^{-y}$ and hence $$B(\alpha x,\alpha)\sim \Gamma(\alpha)(\alpha x)^{-\alpha};$$ where $\Gamma(z)$ is the Euler Gamma function.

On the other hand, for small $\alpha$, we have the expansion $$\Gamma(1+\alpha)=1+\alpha\Gamma'(1)+\mathcal{O}(\alpha^2).$$ Since $\alpha\Gamma(\alpha)=\Gamma(1+\alpha)$, it follows that $$\Gamma(\alpha)\sim \frac1{\alpha}-\gamma+\mathcal{O}(\alpha)$$ where $\gamma$ is the Euler constant.

We may now combine the above two estimates to obtain $$\alpha x^{\alpha}B(\alpha x,\alpha)\sim \alpha x^{\alpha}\left(\frac1{\alpha}-\gamma\right)(\alpha x)^{-\alpha}=\left(\frac1{\alpha}-\gamma\right)\alpha^{1-\alpha}\geq1$$ provided $\alpha$ is small enough. For example, $0<\alpha<\frac12$ works.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.