Ravi Varadhan wrote:
But you are still left with the problem of choosing the regularization 
parameter, i.e. how much to shrink the coefficients?  In other words, there is 
no free ride.

Ravi.

Ravi,

Choosing a penalty factor is extremely easy compared to variable selection. There is a unique optimum and, depending on your model, only a single number to solve for. With stepwise variable selection variables move in and out of the model in an extremely non-monotonic way.

One good way to choose the penalty is to use the effective AIC. An example is at http://biostat.mc.vanderbilt.edu/rms . AIC is good to use in that context where the solution path is simple, unlike stepwise selection.

In the end, shrinkage beats standard variable selection easily with regard to predictive accuracy, discrimination, and adequate adjustment for confounding.

Frank


____________________________________________________________________

Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University

Ph. (410) 502-2619
email: rvarad...@jhmi.edu


----- Original Message -----
From: Frank E Harrell Jr <f.harr...@vanderbilt.edu>
Date: Saturday, July 4, 2009 9:26 am
Subject: Re: [R] is AIC always 100% in evaluating a model?
To: Tal Galili <tal.gal...@gmail.com>
Cc: r-help@r-project.org, Ben Bolker <bol...@ufl.edu>


Tal Galili wrote:
 > Hi Ben,
 > I just wished to give a small remark about your claim:
> "it's best not to consider hypothesis testing (statistical significance) and
 > AIC in the same analysis."
> > Since in the case of forward selection for orthogonal matrix's, it can be
 > shown that AIC is like using a P to enter rule of 0.16.  For further
 > reference see:page 3 of: "A SIMPLE FORWARD SELECTION PROCEDURE BASED
 > ONFALSE DISCOVERY RATE CONTROL" BY YOAV BENJAMINI AND YULIA GAVRILOV,
> > > > Cheers,
 > Tal Galili
Tal, That is not limited to orthogonal designs. When used for one variable at a time variable selection. AIC is just a restatement of the P-value, and as such, doesn't solve the severe problems with stepwise variable selection other than forcing us to use slightly more sensible alpha values. As an aside, some statisticians try to deal with multiplicity problems caused by stepwise variable selection by making alpha smaller than 0.05. This increases bias by giving variables whose effects are estimated with error a greater relative chance of being selected. alpha typically needs to be 0.5 or greater to avoid problems with stepwise
 variable selection.
AIC was designed to compare two pre-specified models. Variable selection does not compete well with shrinkage methods that
 simultaneously model all potential predictors.
Frank > > > > > > On Sat, Jul 4, 2009 at 1:46 AM, Ben Bolker <bol...@ufl.edu> wrote: > >>
 >>
 >> alexander russell-2 wrote:
 >>> Hello,
>>> I'd like to say that it's clear when an independent variable can be ruled >>> out generally speaking; on the other hand in R's AIC with bbmle, if one
 >>> finds a better AIC value for a model without the given independent
 >>> variable,
>>> versus the same model with, can we say that the independent variable is
 >>> not
 >>> likely to be significant(in the ordinary sense!)?
 >>>
>>> That is, having made a lot of models from a data set, then the best two
 >>> are
 >>> say 78.2 and 79.3 without and with (a second independent variable
>>> respectively) should we say it's better to judge the influence of the 2nd
 >>> IV
 >>> as insignificant?
 >>> regards,
 >>> -shfets
 >>> _____________________________________
 >>>
 >>>
 >> Without meaning to sound snarky, it's best not to consider hypothesis
 >> testing (statistical significance) and AIC in the same analysis.
 >> If you want to decide whether predictor variables have a significant
>> effect on a response, you should consider their effect in the full model, >> via Wald test, likelihood ratio test, etc.. If you want to find the model
 >> with the best expected predictive capability (i.e. lowest expected
 >> Kullback-Leibler distance), you should use AIC.
 >>
 >>  Burnham and Anderson, among others, say this repeatedly.
 >>
 >>  In general, for a one-parameter difference, hypothesis testing
 >> is "more conservative" than AIC (e.g., critical log-likelihood difference
 >> for a p-value of 0.05 under the LRT test is 1.92, while the log-likelihood
 >> difference required to say that a model is expected to have better
>> predictive capability/lower AIC is 1) -- but since they are designed to
 >> answer
 >> such different questions, it's not even a fair comparison.
 >>
 >>  Ben Bolker
 >>
 >> --
 >> View this message in context:
>> >> Sent from the R help mailing list archive at Nabble.com.
 >>
 >> ______________________________________________
 >> R-help@r-project.org mailing list
>> >> PLEASE do read the posting guide >> >> and provide commented, minimal, self-contained, reproducible code.
 >>
> > > -- Frank E Harrell Jr Professor and Chair School of Medicine
                       Department of Biostatistics   Vanderbilt University
______________________________________________
 R-help@r-project.org mailing list
PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code.



--
Frank E Harrell Jr   Professor and Chair           School of Medicine
                     Department of Biostatistics   Vanderbilt University

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