# [R] Getting AIC from lrm in Design package

Kyle Werner kylewerner10 at gmail.com
Sun Oct 25 17:55:14 CET 2009

```David,

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
<dwinsemius at 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
>
>

```