Question on approximations of full logistic regression model

Thank you for your reply, Prof. Harrell.

I agree with you. Dropping only one variable does not actually help a lot.

I have one more question.
During analysis of this model I found that the confidence
intervals (CIs) of some coefficients provided by bootstrapping (bootcov 
function in rms package) was narrower than CIs provided by usual 
variance-covariance matrix and CIs of other coefficients wider.  My data 
has no cluster structure. I am wondering which CIs are better.
I guess bootstrapping one, but is it right?

I would appreciate your help in advance.
--
KH



(11/05/16 12:25), Frank Harrell wrote:

I think you are doing this correctly except for one thing.  The
validation
and other inferential calculations should be done on the full model.  Use
the approximate model to get a simpler nomogram but not to get standard
errors.  With only dropping one variable you might consider just running
the
nomogram on the entire model.
Frank


KH wrote:

Hi,
I am trying to construct a logistic regression model from my data (104
patients and 25 events). I build a full model consisting of five
predictors with the use of penalization by rms package (lrm, pentrace
etc) because of events per variable issue. Then, I tried to approximate
the full model by step-down technique predicting L from all of the
componet variables using ordinary least squares (ols in rms package) as
the followings. I would like to know whether I am doing right or not.

library(rms)
plogit<- predict(full.model)
full.ols<- ols(plogit ~ stenosis+x1+x2+ClinicalScore+procedure,
sigma=1)
fastbw(full.ols, aics=1e10)

  Deleted       Chi-Sq d.f. P      Residual d.f. P      AIC    R2
  stenosis       1.41  1    0.2354   1.41   1    0.2354  -0.59 0.991
  x2            16.78  1    0.0000  18.19   2    0.0001  14.19 0.882
  procedure     26.12  1    0.0000  44.31   3    0.0000  38.31 0.711
  ClinicalScore 25.75  1    0.0000  70.06   4    0.0000  62.06 0.544
  x1            83.42  1    0.0000 153.49   5    0.0000 143.49 0.000

Then, fitted an approximation to the full model using most imprtant
variable (R^2 for predictions from the reduced model against the
original Y drops below 0.95), that is, dropping "stenosis".

full.ols.approx<- ols(plogit ~ x1+x2+ClinicalScore+procedure)
full.ols.approx$stats

           n  Model L.R.        d.f.          R2           g       Sigma
104.0000000 487.9006640   4.0000000   0.9908257   1.3341718   0.1192622

This approximate model had R^2 against the full model of 0.99.
Therefore, I updated the original full logistic model dropping
"stenosis" as predictor.

full.approx.lrm<- update(full.model, ~ . -stenosis)

validate(full.model, bw=F, B=1000)

           index.orig training    test optimism index.corrected    n
Dxy           0.6425   0.7017  0.6131   0.0887          0.5539 1000
R2            0.3270   0.3716  0.3335   0.0382          0.2888 1000
Intercept     0.0000   0.0000  0.0821  -0.0821          0.0821 1000
Slope         1.0000   1.0000  1.0548  -0.0548          1.0548 1000
Emax          0.0000   0.0000  0.0263   0.0263          0.0263 1000

validate(full.approx.lrm, bw=F, B=1000)

           index.orig training    test optimism index.corrected    n
Dxy           0.6446   0.6891  0.6265   0.0626          0.5820 1000
R2            0.3245   0.3592  0.3428   0.0164          0.3081 1000
Intercept     0.0000   0.0000  0.1281  -0.1281          0.1281 1000
Slope         1.0000   1.0000  1.1104  -0.1104          1.1104 1000
Emax          0.0000   0.0000  0.0444   0.0444          0.0444 1000

Validatin revealed this approximation was not bad.
Then, I made a nomogram.

full.approx.lrm.nom<- nomogram(full.approx.lrm,

fun.at=c(0.05,0.1,0.2,0.4,0.6,0.8,0.9,0.95), fun=plogis)

plot(full.approx.lrm.nom)

Another nomogram using ols model,

full.ols.approx.nom<- nomogram(full.ols.approx,

fun.at=c(0.05,0.1,0.2,0.4,0.6,0.8,0.9,0.95), fun=plogis)

plot(full.ols.approx.nom)

These two nomograms are very similar but a little bit different.

My questions are;

1. Am I doing right?

2. Which nomogram is correct

I would appreciate your help in advance.

--
KH

-----
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
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