5
$\begingroup$

Is this a known special function:

$\int\nolimits_0^1 a^p(1-a)^{1-p}\\,b^{1-p}\\,(1-b)^p dp\qquad ?$

I am really only interested in maximizing this over $(a,b)$ in $[0,1] \times [0,1]$, so a pointer to a nice numerical evaluation is appreciated as much or more so than an unstable exact formula.
Thanks for any help

  • 0
    What's the range of your $p$?2010-11-09

1 Answers 1

8

You can get a closed-form answer.

$\int_0^1 a^p (1-a)^{1-p} b^{1-p} (1-b)^p dp = b(1-a) \int_0^1 \left(\frac{a(1-b)}{b(1-a)}\right)^p dp = \left. \frac{b(1-a)}{\ln \frac{a(1-b)}{b(1-a)}} \left(\frac{a(1-b)}{b(1-a)}\right)^p \right|_0^1 $ $= \frac{a(1-b) - b(1-a)}{\ln a + \ln (1-b) - \ln b - \ln (1-a)} = \frac{a-b}{\ln a + \ln (1-b) - \ln b - \ln (1-a)}.$

This holds if $a \neq b$ and if neither of $a$ or $b$ is 0 or 1. If $a = b$, then instead we have $a(1-a) \int_0^1 dp = a - a^{2}.$ And, of course, if $a$ or $b$ is 0 or 1 then the value of the integral is 0.

So, as far as maximizing, you can use the usual approach of finding where both partial derivatives are 0. I haven't worked through the calculations, but I strongly suspect that because of the symmetry in $a$ and $b$ that the maximum value will occur at $a = b$.

  • 0
    You're right, I missed the $b(1-a)$ in front of the integral sign.2010-11-09