1. ## 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

2. ## Re: x/(x-1) = a

Originally Posted by Beanxx
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 $c\ne 1~\&~x\ne 1$.

3. ## 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.

Originally Posted by Plato
\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

4. ## Re: x/(x-1) = a

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