My Math Forum Inverse of KT Probability Weighting Function

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 September 17th, 2017, 08:33 AM #1 Member   Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4 Inverse of KT Probability Weighting Function 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)
 September 17th, 2017, 12:01 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 That would involve solving a 'polynomial' of degree $\gamma$. (I put "polynomial" in quotes because $\gamma$ is not necessarily an integer. There is no general formula for that.
 September 17th, 2017, 01:15 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,323 Thanks: 1232 $\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$ $\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}$ $\omega^\gamma = \dfrac{1}{1+\left(\frac{1-p}{p}\right)^\gamma}$ $\omega^{-\gamma} = 1+\left(\frac{1-p}{p}\right)^\gamma$ $\omega^{-\gamma}-1 = \left(\frac{1-p}{p}\right)^\gamma$ $\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} = \dfrac{1-p}{p} = \dfrac 1 p - 1$ $\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$ $p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$ Thanks from Omnipotent
September 18th, 2017, 02:52 PM   #4
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Quote:
 Originally Posted by romsek $\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$ $\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}$ $\omega^\gamma = \dfrac{1}{1+\left(\frac{1-p}{p}\right)^\gamma}$ $\omega^{-\gamma} = 1+\left(\frac{1-p}{p}\right)^\gamma$ $\omega^{-\gamma}-1 = \left(\frac{1-p}{p}\right)^\gamma$ $\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} = \dfrac{1-p}{p} = \dfrac 1 p - 1$ $\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$ $p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$
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

September 18th, 2017, 05:24 PM   #5
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 Originally Posted by Country Boy That would involve solving a 'polynomial' of degree $\gamma$. (I put "polynomial" in quotes because $\gamma$ 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?

September 18th, 2017, 05:59 PM   #6
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Quote:
 Originally Posted by romsek $\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$ $\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}$ $\omega^\gamma = \dfrac{1}{1+\left(\frac{1-p}{p}\right)^\gamma}$ $\omega^{-\gamma} = 1+\left(\frac{1-p}{p}\right)^\gamma$ $\omega^{-\gamma}-1 = \left(\frac{1-p}{p}\right)^\gamma$ $\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} = \dfrac{1-p}{p} = \dfrac 1 p - 1$ $\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$ $p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$

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

September 18th, 2017, 06:35 PM   #7
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 Originally Posted by Omnipotent 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.

 September 18th, 2017, 10:13 PM #8 Senior Member     Joined: Sep 2015 From: USA Posts: 2,323 Thanks: 1232 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 $\gamma = 0.65$ Thanks from Omnipotent Last edited by romsek; September 18th, 2017 at 11:09 PM.
September 19th, 2017, 09:06 AM   #9
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Quote:
 Originally Posted by romsek 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:

Where the green line is the inverse, and the red line is the original. Is this worth pursuing?

September 19th, 2017, 09:19 AM   #10
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Quote:
 Originally Posted by romsek 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

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