2 weeks ago
I'm trying to check that the proportional hazards assumption is satisfied with all my variables in my Cox model.
I used 2 methods to do this, but they give different results.
Add time-dependent variable to the original model (in this case, a product of a variable of interest and logarithm of time variable was added for each covariate).
prog phreg data=data covsandwich;
class drinker meds;
model time*event(0)=drinker meds comorbs age avg_glucose avg_creatinine
t_drinker t_meds t_comorbs t_age t_avg_glucose t_avg_creatinine;
t_drinker=drinker*log(time); t_meds=meds*log(time); t_comorbs=comorbs*log(time); t_age=age*log(time); t_avg_glucose=avg_glucose*log(time); t_avg_creatinine=avg_creatinine*log(time);
This method showed that proportional hazards assumption was satisfied for all variables because none of these time-dependent variables were significant.
for continuous variables - plots of Shoenfeld residuals against the time variable
for categorical variables - plots of log-minus-log of the survival function against the time variable, stratified by the category levels.
proc lifetest data=data plots=(s, lls); time time*event(0); strata drinker; run;
This method showed that all the continous variables satisfied the proportional hazards assumption (graghs showed straight line), whilst it did not for one of the categorical variable (drinker) as the lines of the survival function crossed.
The first method and second method showed slightly different results (all covariates satisfy the PHA in the first, whilst it doesn't for the second). Could this be because the second analysis is unadjusted? I wonder whether it's fair to conclude that the PH assumption is satisfied as indicated by the first method, or whether it isn't because of the second method.
Many thanks for your help in advance.
2 weeks ago
Unfortunately, it's not an exact science to determine if the PH assumption is violated. You have to make the call, especially knowing the KM curves cross for gender. One thing to consider is the p-value itself, if it's 0.0546, which is greater than 0.05 but close maybe it's still ok. It's a judgement call....
Another option is the ASSESS statement. I would suggest searching on Lexjansen.com to see how to use the ASSESS statement and then using that as another check.
2 weeks ago
You have proportional hazards if the difference of Log(-Log(kaplanmeier)) between the two groups under consideration is constant.
Not crossing Kaplan Meier curves does not imply proportional hazard rates. This just imply that one group has a higher risk than the other.