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 . (I put "polynomial" in quotes because 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: 2,303 Thanks: 1221 
$\omega = \left(\dfrac{p^\gamma}{p^\gamma+(1p)^\gamma}\right)^\frac 1 \gamma$ $\omega^\gamma = \dfrac{p^\gamma}{p^\gamma+(1p)^\gamma}$ $\omega^\gamma = \dfrac{1}{1+\left(\frac{1p}{p}\right)^\gamma}$ $\omega^{\gamma} = 1+\left(\frac{1p}{p}\right)^\gamma$ $\omega^{\gamma}1 = \left(\frac{1p}{p}\right)^\gamma$ $\left(\omega^{\gamma}1 \right)^{\frac 1 \gamma} = \dfrac{1p}{p} = \dfrac 1 p  1$ $\left(\omega^{\gamma}1 \right)^{\frac 1 \gamma} +1 = \dfrac 1 p$ $p = \dfrac{1}{\left(\omega^{\gamma}1 \right)^{\frac 1 \gamma} +1 }$ 
September 18th, 2017, 02:52 PM  #4  
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4  Quote:
But, thanks nonetheless, I think I have an idea now  
September 18th, 2017, 05:24 PM  #5 
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4  
September 18th, 2017, 05:59 PM  #6  
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4  Quote:
Nevermind, I seem to have confused myself again. Your help would be greatly appreciated!  
September 18th, 2017, 06:35 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,303 Thanks: 1221  
September 18th, 2017, 10:13 PM  #8 
Senior Member Joined: Sep 2015 From: USA Posts: 2,303 Thanks: 1221 
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 $\gamma = 0.65$ Last edited by romsek; September 18th, 2017 at 11:09 PM. 
September 19th, 2017, 09:06 AM  #9  
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4  Quote:
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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,303 Thanks: 1221  Quote:
 

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function, inverse, probability, weighting 
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