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Dear Professor R users,
My name is  Shan. I am currently a second year master student in statistics in the University of Saskatchewan. I use the gamma shared frailty model to modeling the recurrent data. Following is example of my code and results.

> fit<-coxph(Surv(start,stop,status)~ age+pre.counts+p1+frailty(id),data)
> summary(fit)
 n= 30744, number of events= 10460

                           coef            se(coef)          se2      Chisq       DF    p
age                     0.0110        0.00126      0.000799    75.5       1      0e+00
pre.counts           0.0367         0.00513     0.004348    51.0        1      9e-13
p1                       1.9276        0.06110      0.054815   995.5       1     0e+00
frailty(id)                                                               26967.2 11378 0e+00

                 exp(coef)  exp(-coef) lower .95 upper .95
age                1.01      0.989      1.01      1.01
pre.counts      1.04      0.964      1.03      1.05
p1                 6.87      0.145       6.10      7.75

Iterations: 8 outer, 56 Newton-Raphson
     Variance of random effect= 3.47   I-likelihood = -90236
Degrees of freedom for terms=     0.4     0.7     0.8 11378.3
Concordance= 0.935  (se = 0.003 )
Rsquare= 0.624   (max possible= 0.998 )
Likelihood ratio test= 30051  on 11380 df,   p=0
Wald test            = 1138  on 11380 df,   p=1
> a<-fit$loglik
> a
[1] -96715.59 -81690.25

My questions are:1. n= 30744 is the number observations  used in the fit. Why in the output shows the number of events= 10460 , what's the interpretation of this value?
                          2. The label Chisq in the output is Wald test or likelihood ratio test?
                          3.  The DF for the variables are all 1 except the frailty.  On the bottom of the results there is Degrees of freedom for terms=0.4     0.7     0.8 11378.3 , My question is what's the difference between this and the DF on top of the output? What does the degrees of freedom for terms mean?
                          4 .This algorithm is based on the penalized partial likelihood, so the results based on the penalized likelihood or unpenalized likelihood? If based on penalized likelihood where is the penalty terms?
                          5. The likelihood ratio test p=0, while the Wald test is 0=1. Why two test gives the totally different results? which should I choose?
                          6. I want to test whether the variable pre.counts significant or not. You mentioned in your reply to someone asked the similar question, you suggest to fit a model with fixed variance of frailty as following,

> fit<-coxph(Surv(start,stop,status)~ age+p1+frailty(id,theta=3.47),data)
    Variance of random effect= 3.47   I-likelihood = -90261
Degrees of freedom for terms=     0.4     0.8 11516.6
Concordance= 0.936  (se = 0.003 )
Rsquare= 0.627   (max possible= 0.998 )
Likelihood ratio test= 30301  on 11518 df,   p=0
Wald test            = 1077  on 11518 df,   p=1

                             you suggested to the test for additional pre.counts is (30301-30051)=250,on(11518-11380)=138 df.  My question is what is theory behind your method or why you choose to fixed the theta and than compared the two likelihood ratio test value? Why not just compared the loglikelihood  for model with pre.counts and the log likelihood value for the model without pre.counts?

Sorry about my long questions, but in order to interpret my results clearly i really need your help.


Thank you so much in advance.

Regards,
Shan





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