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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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