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September 17th, 2017, 08:33 AM   #1
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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)
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September 17th, 2017, 12:01 PM   #2
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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.
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September 17th, 2017, 01:15 PM   #3
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$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$

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

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

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

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

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

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

$p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$
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September 18th, 2017, 02:52 PM   #4
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Quote:
Originally Posted by romsek View Post
$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$

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

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

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

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

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

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

$p = \dfrac{1}{\left(\omega^{-\gamma}-1 \right)^{\frac 1 \gamma} +1 }$
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
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September 18th, 2017, 05:24 PM   #5
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Quote:
Originally Posted by Country Boy View Post
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?
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September 18th, 2017, 05:59 PM   #6
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Quote:
Originally Posted by romsek View Post
$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1-p)^\gamma}\right)^\frac 1 \gamma$

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

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

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

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

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

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

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

Nevermind, I seem to have confused myself again. Your help would be greatly appreciated!
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September 18th, 2017, 06:35 PM   #7
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Quote:
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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.
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September 18th, 2017, 10:13 PM   #8
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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 $\gamma = 0.65$
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Last edited by romsek; September 18th, 2017 at 11:09 PM.
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September 19th, 2017, 09:06 AM   #9
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Quote:
Originally Posted by romsek View Post
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?
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September 19th, 2017, 09:19 AM   #10
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Quote:
Originally Posted by romsek View Post
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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