[Rd] lm() gives different results to lm.ridge() and SPSS
peter dalgaard
pdalgd at gmail.com
Fri May 5 10:02:10 CEST 2017
I asked you before, but in case you missed it: Are you looking at the right place in SPSS output?
The UNstandardized coefficients should be comparable to R, i.e. the "B" column, not "Beta".
-pd
> On 5 May 2017, at 01:58 , Nick Brown <nick.brown at free.fr> wrote:
>
> Hi Simon,
>
> Yes, if I uses coefficients() I get the same results for lm() and lm.ridge(). So that's consistent, at least.
>
> Interestingly, the "wrong" number I get from lm.ridge()$coef agrees with the value from SPSS to 5dp, which is an interesting coincidence if these numbers have no particular external meaning in lm.ridge().
>
> Kind regards,
> Nick
>
> ----- Original Message -----
>
> From: "Simon Bonner" <sbonner6 at uwo.ca>
> To: "Nick Brown" <nick.brown at free.fr>, r-devel at r-project.org
> Sent: Thursday, 4 May, 2017 7:07:33 PM
> Subject: RE: [Rd] lm() gives different results to lm.ridge() and SPSS
>
> Hi Nick,
>
> I think that the problem here is your use of $coef to extract the coefficients of the ridge regression. The help for lm.ridge states that coef is a "matrix of coefficients, one row for each value of lambda. Note that these are not on the original scale and are for use by the coef method."
>
> I ran a small test with simulated data, code is copied below, and indeed the output from lm.ridge differs depending on whether the coefficients are accessed via $coef or via the coefficients() function. The latter does produce results that match the output from lm.
>
> I hope that helps.
>
> Cheers,
>
> Simon
>
> ## Load packages
> library(MASS)
>
> ## Set seed
> set.seed(8888)
>
> ## Set parameters
> n <- 100
> beta <- c(1,0,1)
> sigma <- .5
> rho <- .75
>
> ## Simulate correlated covariates
> Sigma <- matrix(c(1,rho,rho,1),ncol=2)
> X <- mvrnorm(n,c(0,0),Sigma=Sigma)
>
> ## Simulate data
> mu <- beta[1] + X %*% beta[-1]
> y <- rnorm(n,mu,sigma)
>
> ## Fit model with lm()
> fit1 <- lm(y ~ X)
>
> ## Fit model with lm.ridge()
> fit2 <- lm.ridge(y ~ X)
>
> ## Compare coefficients
> cbind(fit1$coefficients,c(NA,fit2$coef),coefficients(fit2))
>
> [,1] [,2] [,3]
> (Intercept) 0.99276001 NA 0.99276001
> X1 -0.03980772 -0.04282391 -0.03980772
> X2 1.11167179 1.06200476 1.11167179
>
> --
>
> Simon Bonner
> Assistant Professor of Environmetrics/ Director MMASc
> Department of Statistical and Actuarial Sciences/Department of Biology
> University of Western Ontario
>
> Office: Western Science Centre rm 276
>
> Email: sbonner6 at uwo.ca | Telephone: 519-661-2111 x88205 | Fax: 519-661-3813
> Twitter: @bonnerstatslab | Website: http://simon.bonners.ca/bonner-lab/wpblog/
>
>> -----Original Message-----
>> From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Nick
>> Brown
>> Sent: May 4, 2017 10:29 AM
>> To: r-devel at r-project.org
>> Subject: [Rd] lm() gives different results to lm.ridge() and SPSS
>>
>> Hallo,
>>
>> I hope I am posting to the right place. I was advised to try this list by Ben Bolker
>> (https://twitter.com/bolkerb/status/859909918446497795). I also posted this
>> question to StackOverflow
>> (http://stackoverflow.com/questions/43771269/lm-gives-different-results-
>> from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first
>> program in 1975 and have been paid to program in about 15 different
>> languages, so I have some general background knowledge.
>>
>>
>> I have a regression from which I extract the coefficients like this:
>> lm(y ~ x1 * x2, data=ds)$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]]
>>
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>
>
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>
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
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