David, Thank you for your reply. I am not using glm, but instead lrm. I am consulting the documentation to try to parse out what the output "Model L.R." actually means: http://lib.stat.cmu.edu/S/Harrell/help/Design/html/lrm.fit.html ("model likelihood ratio chi-square")
>From my read of the documentation, it appears that the "Model L.R." output by lrm is not the deviance, but 2 * ln( [model deviance] / [null deviance] ) This is why I believe I am correct to be talking about likelihood ratios instead of deviance, but I am unsure of this - which really relates back to the core of my question. Whether or not I have the sign wrong in the AIC formulation is dependent upon whether or not the - is incorporated into the calculation of the "Model L.R.", above, which is one part of my original question. The AIC formula is, generally, AIC = -2*ln(likelihood ratio) + 2k, with the best model (assuming the same observations) having the lowest AIC. I hope that my understanding of this fundamental formula is correct, but please let me know if not. Thanks. On Sun, Oct 25, 2009 at 10:51 AM, David Winsemius <dwinsem...@comcast.net> wrote: > > On Oct 25, 2009, at 9:24 AM, Kyle Werner wrote: > >> I am trying to obtain the AICc after performing logistic regression >> using the Design package. For simplicity, I'll talk about the AIC. I >> tried building a model with lrm, and then calculating the AIC as >> follows: >> >> likelihood.ratio <- >> unname(lrm(succeeded~var1+var2,data=scenario,x=T,y=T)$stats["Model >> L.R."]) #Model likelihood ratio??? >> model.params <- 2 #Num params in my model >> AIC = -2*log(likelihood.ratio) + 2 * model.params > > You might want to check your terminology. A single model has a deviance. You > construct a likelihood ratio as twice the logged ratio between two > likelihoods (deviances) (which then turns into a difference on the log > scale). And don't you have the sign wrong on that expression for AIC? I > thought you penalized (i.e. subtracted) for added degrees of freedom? (There > is an implicit base model, so if you define AIC as a difference between L1 + > 2p1 and L2+2p2 you would get (L1-L2) + (0 -2p2) = (LR - 2p2). See p 202 of > Harrell's "Regression Modeling Strategies".) > >> >> However, this is almost certainly the wrong interpretation. When I >> replace var1 and var2 by runif(N,0,1) (that is, random variables), I >> obtain better (lower) AICs than when I use real var1 and var2 that are >> known to be connected with the outcome variable. Indeed, when I use >> GLM instead of LRM, the real model has a lower AIC than that with the >> predictors replaced by random values. Therefore, it appears that lrm's >> "Model L.R." is not actually the model likelihood ratio, but instead >> something else. >> >> Going to the Design documentation, lrm.fit states that "Model L.R." is >> the model likelihood-ratio chi-square. Does this mean that it is >> returning 2*log(likelihood)? If so, AIC becomes >> AIC = -[Model L.R.] + 2*model.params >> >> Can anyone confirm that this final formula for obtaining the AIC from >> lrm is correct? > > > I would not confirm it. What sources are you consulting? > > -- > David Winsemius, MD > Heritage Laboratories > West Hartford, CT > > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
