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Related to my finance research, I am trying to find a non-trivial upper bound of \begin{equation*} \left( \frac{ \left(1 - r^{\frac{1}{K-1}}\right)K}{ \left(1 - r^{\frac{K}{K-1}} \right) } -1\right)\cdot \frac{r}{1-r}, \end{equation*} when we consider all $r \in (0,1)$ and $K$ being any integer weakly greater than $2$. (A numerical exercise suggests that it can be bounded from above by $0.5$, and the bound is achieved as $r$ goes to $1$.)

What I know is that it is non-negative and bounded from above by $1$, because it comes from the share of one trader's profit relative to the sum of profits of two or more traders, all of which are positive.

Thanks in advance for your help!

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  • $\begingroup$ What does "weakly greater than $2$" mean, please? $\endgroup$
    – saulspatz
    Jun 24, 2020 at 14:10
  • $\begingroup$ Thanks for your comment. I meant that K can be any integer that is equal to or greater than 2, when we consider an upper bound. $\endgroup$
    – tarou2
    Jun 24, 2020 at 14:16

2 Answers 2

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Have you try proving that for any $K>0$:

$$\lim_{r\to1 ^{-}} \left( \frac{ \left(1 - r^{\frac{1}{K-1}}\right)K}{ \left(1 - r^{\frac{K}{K-1}} \right) } -1\right)\cdot \frac{r}{1-r} =0.5?$$

That should give you the $1/2$ bound that you seem to be obtaining. Should be easy since for your parameter values this function is continous and monotonously increasing in $r$.

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  • $\begingroup$ I agree with your point. I'm trying to prove that it is increasing in $r$, but my attempt has not been successful. $\endgroup$
    – tarou2
    Jun 24, 2020 at 16:09
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$$\begin{equation*} y=\left( \frac{ \left(1 - r^{\frac{1}{k-1}}\right)k}{ \left(1 - r^{\frac{k}{k-1}} \right) } -1\right)\, \frac{r}{1-r} \end{equation*}$$ The Taylor expansion built around $r=1$ is $$y=\frac{1}{2}-\frac{2 k-1 }{12 (k-1)}(1-r)-\frac{2 k-1 }{24 (k-1)}(1-r)^2+O\left((1-r)^3\right)$$ and you have $$y <\frac{1}{2}-\frac{2 k-1 }{12 (k-1)}(1-r)$$

In fact, it seems (at least up to degree $5$) that the expansion is an upper bound.

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