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Say we have,

$0<q<1$

$b>0$

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

$0<x<1$

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

Omri
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    Also asked at MO: https://mathoverflow.net/questions/267875/kellys-criterion-loss-threshold I think this answer gives a very reasonable advice about cross-posting. Of course, there are also several other posts about [meta-tag:cross-posting] on meta. – Martin Sleziak Jun 11 '17 at 14:38

1 Answers1

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We see $x=1$ is not a solution. Now write $y=\frac{1}{1-x}$, and see that

\begin{align} (1+xb)^q(1-x)^{1-q}&=1\\ (1+xb)^q\left(\frac{1}{1-x}\right)^q\left(\frac{1}{1-x}\right)^{-1}&=1\\ \left(\frac{1+xb}{1-x}\right)^q&=\frac{1}{1-x}\\ \left(\frac{1+b+-b(1-x)}{1-x}\right)^q&=\frac{1}{1-x}\\ \left(-b+\frac{1+b}{1-x}\right)^q&=\frac{1}{1-x}\\ \left(-b+(1+b)y\right)^q&=y\\ \end{align}

so if we were to write $q$ as a fraction $\frac nm$ with $n,m$ positive integers, then we have the equation

$$\left(-b+(1+b)y\right)^n=y^m$$

which is simply a polynomial of degree $\max(n,m)=m$ (the maximum is $m$ since $0<q<1$), so if either one of $m$ is larger than $4$ then it's unlikely that a closed form for the roots exist (and so the same for $x$). Although there may be no closed form, there are still $m$ solutions (including complex ones and the trivial $y=1$).

  • Note that I corrected to: $0<x<1$, and given that $q(1+b)>1$. The latter condition guarantees exactly one solution, and without it there may be no solution. – Omri Apr 21 '17 at 10:48