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I am struggling to find (or at least set some bounds on) $$\arg \max_x \log(1+x)(1-I_x(a,b-a)),$$ where $I_x(a,b-a)$ is the regularized incomplete beta function, i.e

$$I_x(a,b-a) = \frac {\int_0^x t^{a-1} (1-t)^{b-a-1}dt}{B(a, b-a)}$$

For $0<x<1$ with $a>1,b-a>1$.

I've noticed that this can be represented as the product of the following integrals:

$$\arg \max_x \int_0^x \frac{\mathrm{d}\tau}{1+\tau} \cdot \int_x^1 \frac{\tau^{a-1}(1-\tau)^{b-a-1}}{B(a,b-a)}\mathrm{d}\tau $$

Yet, it didn't help much to obtain the maximum nor the arg max. I've tried through the first-order derivative, tried to bound it with arithmetic mean-geometric mean, and even tried to apply Jenssen inequality, however, without any success.

I got a hunch that since the first term does not have an explicit maximum, the only term that impacts on the maximum is the second integral, which reaches a maximum at $x\to \frac{a-1}{b-2}$, but didn't succeed to prove this bound.

Maybe L'hoptial rule can help get some insights on this?

Any ideas, help, and clues are much appreciated.

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  • $\begingroup$ You can obtain a numerical solution very easily. The derivative of your cost function is quite daunting, I may say. But it can be seen, as @LadaDudnikova wrote, that $\frac{a-1}{b-2}$ is not a maximizer. $\endgroup$ – Pantelis Sopasakis Apr 27 '19 at 23:46
  • $\begingroup$ Thanks, @Pantelis Sopasakis, it is certainly not the maximizer, and finding the true maximizer would be awesome, however, it gives a lower bound (which is also fine for me). Though, I still don't know how to prove that this bound holds. $\endgroup$ – sefi Apr 28 '19 at 7:31
  • $\begingroup$ Looks like you have a typo in the denominator in the product: $B(a,b)$ instead of $B(a,b-a)$ $\endgroup$ – Lada Dudnikova Apr 28 '19 at 10:19
  • $\begingroup$ @sefi any feasible point gives you a lower bound $\endgroup$ – Pantelis Sopasakis Apr 28 '19 at 19:41
  • $\begingroup$ math.stackexchange.com/q/3213068/477927, I wrote a question about your question) $\endgroup$ – Lada Dudnikova May 4 '19 at 7:26
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Not an answer, I have graphed this, and it's really a tricky function. But your assumption

$x \to \frac{a-1}{b-2}$

is not true. For example, $a=0.7, b = 1$, $a=10, b=14$ Good Luck!

enter image description here

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  • $\begingroup$ Thank you @Lada Dudnikova, indeed this is not a maximizer, but only a bound. Can we prove it as a bound? I also noticed that this bound becomes tighter as $a$ and $b$ grow. (I forgot to mention $a,b>1$). $\endgroup$ – sefi Apr 28 '19 at 5:00
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graphsOne more thing: I tried using McLoren series for logarithm $(1+x)$ at 0 and 1. I really don't understand why both estimations go so terribly wrong. Note that I substitute $a, b-a$ for $a,b$

https://www.desmos.com/calculator/yvvenwyafg

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  • $\begingroup$ Thank you @Lada Dudnikova! I also went in a similar direction, taking $\log(1+x) \approx x$, which is encouraging, yet, it didn't simplify things much for me either. Interestingly, though, $x\to\frac{a}{b+1}$, does falls pretty close. $\endgroup$ – sefi May 3 '19 at 21:26
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I found a decent approximation to my question using right Riemann_sum.

Recalling that with Riemann sum approximation, the absolute approxination error is bounded by $$\left\vert \int_a^b f(\tau) \mathrm{d}\tau - A_{right}\right\vert \leq \frac{M_1\cdot (b-a)}{2n},$$ where $M_1$ is the maximum value of the absolute value of $f^{\prime}(\tau)$ on the interval. In our case both the derivative of $\log(1+x)$ and the derivative of $(1-I_x(a,b-a))$ are moderate which is good for our purposes.

Accordingly , we seek for $$\arg\max_x \log(1+x)(1-I_x(a,b-a))$$ or $$\arg \max_x \int_0^x \frac{\mathrm{d}\tau}{1+\tau} \cdot \int_x^1 \frac{\tau^{a-1}(1-\tau)^{b-a-1}}{B(a,b-a)}\mathrm{d}\tau.$$

With $n=1$, this is approximately

\begin{align*}\arg \max_x &\frac{1}{1+x}\cdot x \cdot\frac{x^{a-1}(1-x)^{b-a-1}}{B(a,b-a)}\cdot(1-x)&\\ = \arg\max_x&\frac{x^{a}(1-x)^{b-a}}{(1+x)B(a,b-a)}\end{align*}

This expression, however, is much easier to handle. Its derivative is

$$\frac{\partial}{\partial x } = \frac{x^{a-1} (1-x)^{-a+b-1} (-a + (-a+b+1)x +(b-1) x^2)}{B(a,b-a)(x+1)^2} $$ which has roots in zero, in one, and more interestingly, in $$x \to \frac{\pm\sqrt{a^2+2 a b-6 a+b^2+2 b+1}+a-b-1}{2 (b-1)}$$

Any further suggestions and comments are much appreciated!

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