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Thread: x/(x-1) = a

  1. #1
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    x/(x-1) = a

    Dear folks--
    I'm sure this is a very simple problem, but I have been wrestling with it for two hours without making any progress at all. Help!

    I am trying to solve for the tail index of a Pareto distribution. Call it x. I have the mean divided by the lower bound. Call it c.
    I know that x/(x-1) = c. So how do I express x in terms of c?

    Just the answer would be great, but it would be even better if there was some technique I could learn to solve problems resembling this one.

    Thanks! --Beanxx
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    Re: x/(x-1) = a

    Quote Originally Posted by Beanxx View Post
    I am trying to solve for the tail index of a Pareto distribution. Call it x. I have the mean divided by the lower bound. Call it c.
    I know that x/(x-1) = c. So how do I express x in terms of c?
    $ \begin{align*}\dfrac{x}{x-1}&=c \\x&=(x-1)c\\x-cx&=xc-c\\(1-c)x&=-c\\x&=\dfrac{-c}{1-c}\\x&=\dfrac{c}{c-1} \end{align*}$

    You must know $c\ne 1~\&~x\ne 1$.
    Last edited by Plato; Feb 26th 2018 at 02:26 PM.
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    Re: x/(x-1) = a

    There's some sort of typo rearing it's ugly head here seeing as you wrote down the correct answer.

    Quote Originally Posted by Plato View Post
    $ \begin{align*}\dfrac{x}{x-1}&=c \\x&=(x-1)c\\x&=xc-c\end{align*}$
    Starting from x = xc - c:
    x - xc = -c

    x(1 - c) = -c

    $x = \frac{-c}{1 - c} = \frac{c}{c - 1}$

    So Plato's answer is correct in the end.

    -Dan
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    Re: x/(x-1) = a

    Thanks so much, Plato & topsquark! That's terrific -- just what I needed. --Beanxx
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