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Let,

$0<q<1$

$b>0$

With equation,

$(1+xb)^q(1-x)^{1-q}=1$

$0<x<1$

Is there a closed-form solution for $x$, given that $q(1+b)>1$? Note that the latter condition guarantees exactly one solution, and without it there may be no solution.

Alternatively, can we at least prove that $x<2((q(1+b)-1)/b)$

This equation comes up in the analysis of the Kelly Criterion, where $x$ would be the threshold fraction above which betting is asymptotically nonprofitable.

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  • $\begingroup$ Questions that begin with "say we have" or "so we have" have a very short half life on MO. $\endgroup$
    – Franz Lemmermeyer
    Commented Apr 22, 2017 at 12:06
  • $\begingroup$ What exactly are you looking for? There is almost certainly not a closed form solution... $\endgroup$
    – Igor Rivin
    Commented Apr 22, 2017 at 15:24
  • $\begingroup$ Yes, i was looking for a closed-form solution, or some interesting approximation. $\endgroup$
    – Omri
    Commented Apr 22, 2017 at 16:06
  • $\begingroup$ Also asked at math.SE: math.stackexchange.com/questions/2244777/… $\endgroup$
    – Martin Sleziak
    Commented Jun 11, 2017 at 14:38

1 Answer 1

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For an "interesting approximation" I would suggest running Newton's method (by hand) from the initial point $x=q,$ for one or two steps. Otherwise, here is a plot of the $x$ as a function of $b, q.$ Somewhat surprisingly, it looks convex (in your region $q(1+b) > 1)$! This should be provable.

enter image description here

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