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I am trying to figure out a formula for the unique $\lambda>1$ such that $$ \mathbb{E}\bigg[\frac{X}{\lambda -X}\bigg]=1 $$ where $X$ is a discrete random variable taking values in $\{\frac{1}{n},...,\frac{n-1}{n},1\}$, distributed w.r.t. some distribution $\mu$.

We can rewrite the expression above which yields $$ \sum_{k=1}^n \frac{\mu(\frac{k}{n})\frac{k}{n}}{\lambda-\frac{k}{n}} = 1. $$

I know that there are no closed solutions for the roots of such a function, since they are based on solving for zeros of a high degree polynomial. Still, I think I miss some obvious results here how to analyse such a function.

I'd appreciate any kind of help. Thank's a lot!

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  • $\begingroup$ Instead of writing $\displaystyle \sum_{k=1}^n \frac{\mu(\frac{k}{n})\frac{k}{n}}{x-\frac{k}{n}} = 1,$ why not write $\displaystyle \sum_{k=1}^n \frac{\mu(\frac{k}{n})\frac{k}{n}}{\lambda-\frac{k}{n}} = 1$ so that the notation is the same as in the expected $\vphantom{\displaystyle\int}$value that you wrote at the outset? $\qquad$ $\endgroup$ Jun 8, 2019 at 15:53
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    $\begingroup$ Thanks for the correction, I edited the post. $\endgroup$
    – Gabriel
    Jun 8, 2019 at 18:46

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