On Oct 25, 2009, at 12:55 PM, Kyle Werner wrote:
David,
Thank you for your reply. I am not using glm, but instead lrm.
Does not matter. "lrm" is giving you the same output as would glm with
a logistic link.
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] )
Believe it to be 2 * ln( [null deviance] / [model deviance] ), which
is why you and I differ as to sign. On page 183 Harrell formulates the
L.R stat as:
-2log(L at H0/ Lat MLEs)
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.
I have offered my opinion as to your misunderstanding, and have
suggested specific references to what Harrell has written about the
subject.
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
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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