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-   -   Inverse of KT Probability Weighting Function (http://mymathforum.com/calculus/341883-inverse-kt-probability-weighting-function.html)

Omnipotent September 17th, 2017 07:33 AM

Inverse of KT Probability Weighting Function
 
http://oi67.tinypic.com/28uo45i.jpg

I'm trying to find the inverse function of above. I can't seem to get it right, and websites such as wolfram and symbolab aren't working for me as well.

Would be great to get some pointers. Many thanks! (Relates to my thesis)

Country Boy September 17th, 2017 11:01 AM

That would involve solving a 'polynomial' of degree . (I put "polynomial" in quotes because is not necessarily an integer. There is no general formula for that.

romsek September 17th, 2017 12:15 PM

$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$

$\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}$

$\omega^\gamma = \dfrac{1}{1+\left(\frac{1-p}{p}\right)^\gamma}$

$\omega^{-\gamma} = 1+\left(\frac{1-p}{p}\right)^\gamma$

$\omega^{-\gamma}-1 = \left(\frac{1-p}{p}\right)^\gamma$

$\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} = \dfrac{1-p}{p} = \dfrac 1 p - 1$

$\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$

$p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$

Omnipotent September 18th, 2017 01:52 PM

Quote:

Originally Posted by romsek (Post 580552)
$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$

$\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}$

$\omega^\gamma = \dfrac{1}{1+\left(\frac{1-p}{p}\right)^\gamma}$

$\omega^{-\gamma} = 1+\left(\frac{1-p}{p}\right)^\gamma$

$\omega^{-\gamma}-1 = \left(\frac{1-p}{p}\right)^\gamma$

$\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} = \dfrac{1-p}{p} = \dfrac 1 p - 1$

$\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$

$p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$

There is a mistake at the beginning, the 1/y is supposed to be for the denominator.

But, thanks nonetheless, I think I have an idea now

Omnipotent September 18th, 2017 04:24 PM

Quote:

Originally Posted by Country Boy (Post 580547)
That would involve solving a 'polynomial' of degree . (I put "polynomial" in quotes because is not necessarily an integer. There is no general formula for that.

How would I go about finding the inverse if y (gamma) was .65?

Omnipotent September 18th, 2017 04:59 PM

Quote:

Originally Posted by romsek (Post 580552)
$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$

$\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}$

$\omega^\gamma = \dfrac{1}{1+\left(\frac{1-p}{p}\right)^\gamma}$

$\omega^{-\gamma} = 1+\left(\frac{1-p}{p}\right)^\gamma$

$\omega^{-\gamma}-1 = \left(\frac{1-p}{p}\right)^\gamma$

$\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} = \dfrac{1-p}{p} = \dfrac 1 p - 1$

$\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$

$p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$


Nevermind, I seem to have confused myself again. Your help would be greatly appreciated!

romsek September 18th, 2017 05:35 PM

Quote:

Originally Posted by Omnipotent (Post 580669)
Nevermind, I seem to have confused myself again. Your help would be greatly appreciated!

It appears the correct expression doesn't admit a closed form solution for $p$ in terms of $\omega$

You'll have to use numeric methods.

romsek September 18th, 2017 09:13 PM

Toying with this a bit in Mathematica produced

$p \approx 0.477328 -0.502426 \cos (3.33061 \omega+0.141105), ~\omega \in [0,1]$

as a pretty good fit to an inverse of the curve with $\gamma = 0.65$

Omnipotent September 19th, 2017 08:06 AM

Quote:

Originally Posted by romsek (Post 580697)
Toying with this a bit in Mathematica produced

$p \approx 0.477328 -0.502426 \cos (3.33061 \omega+0.141105), ~\omega \in [0,1]$

as a pretty good fit to an inverse of the curve with $\gamma = 0.65$

Thank you very much for your help!

I tried graphing that alongside the original equation, but the results were a bit weird.

I'm trying to go for something like this:
http://oi63.tinypic.com/2zgsydj.jpg
Where the green line is the inverse, and the red line is the original. Is this worth pursuing?

romsek September 19th, 2017 08:19 AM

1 Attachment(s)
Quote:

Originally Posted by romsek (Post 580697)
Toying with this a bit in Mathematica produced

$p \approx 0.477328 -0.502426 \cos (3.33061 \omega+0.141105), ~\omega \in [0,1]$

as a pretty good fit to an inverse of the curve with $\gamma = 0.65$

looks pretty good to me

http://mymathforum.com/attachment.ph...1&d=1505837930


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