[Rd] lm() gives different results to lm.ridge() and SPSS

Nick Brown nick.brown at free.fr
Fri May 5 01:58:32 CEST 2017

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, 

----- 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. 



## Load packages 

## Set seed 

## 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 

[,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 
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