1 Nov 2005 04:58
Re: [R] nls() fit to Kahnemann/ Tversky function
Gabor Grothendieck <ggrothendieck <at> gmail.com>
2005-11-01 03:58:33 GMT
2005-11-01 03:58:33 GMT
Note that a simple logistic with a saturation level of 1 seems to do quite well. Below we have removed the last point in order to avoid the singularity: x <- p.kum[-10] y <- felt.prob.kum[-10] plot(log(y/(1-y)) ~ x) abline(lm(log(y/(1-y)) ~ x), col = "red") On 10/31/05, Mark Hempelmann <neo27 <at> t-online.de> wrote: > Dear WizaRds, > > I would like to fit a curve to ten points with nls() for one > unknown parameter gamma in the Kahnemann/ Tversky function, but somehow > it won't work and I am unable to locate my mistake. > > p.kum <- seq(0.1,1, by=0.1) > felt.prob.kum <- c(0.16, 0.23, 0.36, 0.49, 0.61, 0.71, 0.85, 0.89, 0.95, > 1) ## how to find a function that fits these points nicely? > plot(p.kum, felt.prob.kum) ## looks a little like an "S" > > gamma <- rep(0.5, 10) > nls.dataframe <- data.frame(p.kum,felt.prob.kum, gamma) > > nls.kurve <- nls( formula = felt.prob.kum ~ > p.kum^gamma/(p.kum^gamma+(1-p.kum)^gamma)^(1/gamma), data=nls.dataframe, > start=c(gamma=gamma), algorithm="plinear" ) > > summary(nls.kurve) >(Continue reading)
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