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coxph reference hazard rate

2 messages · Georges Dupret, David Winsemius

Original
Georges Dupret
# 
Hi,

In the following results I interpret exp(coef) as the factor that multiplies
the base hazard rate if the corresponding variable is TRUE. For example,
when the bucket is ks008 and fidelity <= 3, then the rate, compared to the
base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case
does the base hazard rate correspond to? I would expect the reference to be
the first factor value, i.e. bucket jpc001 with fidelity <= 3, but its
exp(coef) is not one. I verified the contrasts, and the row corresponding to
the first factor value is zero everywhere; moreover, I didn't change the
default, so a bad setting of the contrasts doesn't seem to be the issue.

Best,

ge

Call:
coxph(formula = Surv(time, event = (censored == 
    FALSE)) ~ bucket:factor(fidelity > 3), data = week.15)


                                         coef exp(coef) se(coef)      z  p
bucketjpc001:factor(fidelity > 3)FALSE -1.606     0.201  0.00624 -257.5  0
bucketks006:factor(fidelity > 3)FALSE  -1.613     0.199  0.00627 -257.5  0
bucketks007:factor(fidelity > 3)FALSE  -1.620     0.198  0.00626 -258.8  0
bucketks008:factor(fidelity > 3)FALSE  -1.611     0.200  0.00625 -257.6  0
bucketks009:factor(fidelity > 3)FALSE  -1.620     0.198  0.00626 -258.9  0
bucketks010:factor(fidelity > 3)FALSE  -1.619     0.198  0.00626 -258.6  0
bucketjpc001:factor(fidelity > 3)TRUE  -0.156     0.856  0.00596  -26.2  0
bucketks006:factor(fidelity > 3)TRUE   -0.171     0.842  0.00600  -28.6  0
bucketks007:factor(fidelity > 3)TRUE   -0.168     0.845  0.00602  -28.0  0
bucketks008:factor(fidelity > 3)TRUE   -0.167     0.846  0.00600  -27.8  0
bucketks009:factor(fidelity > 3)TRUE   -0.170     0.844  0.00599  -28.4  0
bucketks010:factor(fidelity > 3)TRUE       NA        NA  0.00000     NA NA

Likelihood ratio test=294562  on 11 df, p=0  n= 1173838, number of events=
629383 

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David Winsemius
# 
As has been answered several times on rhelp ... the baseline hazard is  
for a case with the "mean value". It's not a meaningful case with all  
factor variables. There can be no cases where fidelity>3 has a  
fractional value.  You should be using predict() and survfit() to  
display estimates for particular meaningful cases. You might want to  
use a different rhs formula as well because the one offered omits the  
main effects. Choosing ~bucket*factor(fidelity > 3) would include  
those terms.