Um, the link to StackOverflow does not seem to contain the same question. It does contain a stern warning not to use the $coef component of lm.ridge...
Is it perhaps the case that x1 and x2 have already been scaled to have standard deviation 1? In that case, x1*x2 won't be. Also notice that SPSS tends to use "Beta" for standardized regression coefficients, and (AFAIR) "b" for the regular ones. -pd > On 4 May 2017, at 16:28 , Nick Brown <nick.br...@free.fr> wrote: > > Hallo, > > I hope I am posting to the right place. I was advised to try this list by Ben > Bolker (https://twitter.com/bolkerb/status/859909918446497795). I also posted > this question to StackOverflow > (http://stackoverflow.com/questions/43771269/lm-gives-different-results-from-lm-ridgelambda-0). > I am a relative newcomer to R, but I wrote my first program in 1975 and have > been paid to program in about 15 different languages, so I have some general > background knowledge. > > > I have a regression from which I extract the coefficients like this: > lm(y ~ x1 * x2, data=ds)$coef > That gives: x1=0.40, x2=0.37, x1*x2=0.09 > > > > When I do the same regression in SPSS, I get: > beta(x1)=0.40, beta(x2)=0.37, beta(x1*x2)=0.14. > So the main effects are in agreement, but there is quite a difference in the > coefficient for the interaction. > > > X1 and X2 are correlated about .75 (yes, yes, I know - this model wasn't my > idea, but it got published), so there is quite possibly something going on > with collinearity. So I thought I'd try lm.ridge() to see if I can get an > idea of where the problems are occurring. > > > The starting point is to run lm.ridge() with lambda=0 (i.e., no ridge > penalty) and check we get the same results as with lm(): > lm.ridge(y ~ x1 * x2, lambda=0, data=ds)$coef > x1=0.40, x2=0.37, x1*x2=0.14 > So lm.ridge() agrees with SPSS, but not with lm(). (Of course, lambda=0 is > the default, so it can be omitted; I can alternate between including or > deleting ".ridge" in the function call, and watch the coefficient for the > interaction change.) > > > > What seems slightly strange to me here is that I assumed that lm.ridge() just > piggybacks on lm() anyway, so in the specific case where lambda=0 and there > is no "ridging" to do, I'd expect exactly the same results. > > > Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex > will not be easy to make, but I can share the data via Dropbox or something > if that would help. > > > > I appreciate that when there is strong collinearity then all bets are off in > terms of what the betas mean, but I would really expect lm() and lm.ridge() > to give the same results. (I would be happy to ignore SPSS, but for the > moment it's part of the majority!) > > > > Thanks for reading, > Nick > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-devel@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-devel -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Office: A 4.23 Email: pd....@cbs.dk Priv: pda...@gmail.com ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
