In a survival analysis, I would like to prove that a variable Y is more associated with an outcome than another variable Y'. My goal is only to prove better association, not to use any model for prediction.
For instance, let's take 2 models, adjusted for known confounding factors:
m1 = coxph(Surv(start, stop, event) ~ Y' + X1 + X2 + X3, data=db)
m2 = coxph(Surv(start, stop, event) ~ Y + X1 + X2 + X3, data=db)
The Surv() object is some clinical outcome (cancer, diabetes, etc.) and Y' is the update of Y, which is an assessment score. For instance, you could consider Y as the BMI and Y' as weight/height^3, or as Trefethen's "new BMI".
My hypothesis is that Y' "explains more" the outcome than Y, and indeed, HR are far broader for Y'.
But since m1 and m2 are not nested, I am not aware of any test to compare them directly.
I heard about Harell's C statistic, but according to some ressources, it "measures of the ordinal predictive power of a model", and "it should not be taken seriously if it is calculated in the dataset in which the model was fit".
Then, is it possible to test the comparison of Y and Y' ? Is there any measure or coefficient of the superiority of one over the other ?
EDIT:
On @EdM advices, I computed a model with both Y and Y' and did some LRT and extracted the AIC. Here are the results:
m12 = coxph(Surv(start, stop, event) ~ Y' + Y + X1 + X2 + X3, data=db)
extractAIC(m12)[2]
# [1] 57421.23
extractAIC(m1)[2]
# [1] 57461.42
anova(m12,m1)
# loglik Chisq Df P(>|Chi|)
#1 -28708
#2 -28727 36.952 1 1.842e-07 ***
extractAIC(m2)[2]
# [1] 57692.19
anova(m12,m2)
# loglik Chisq Df P(>|Chi|)
#1 -28708
#2 -28842 267.72 1 < 2.2e-16 ***
YandY'are highly correlated, can I really perform a LRT to see how much one add to the other ? – Dan Chaltiel Aug 24 '18 at 13:01Y'doesn't significantly add anything toY, and also vice versa. See this page for discussion and links on testing non-nested models. You could try comparing predictions between the two competing models with bootstrapping. If you are dealing with a composite variable like BMI, consider looking at each of its components; see this page for suggestions. – EdM Aug 25 '18 at 14:08YorY'remove a lot of information fromm12, but I don't think I can compare the two and I am not sure AIC can change this. Though it seems that theanovaremoving the updated scoreY'have a lower p-value, but does this mean I am removing more information ? (I am never sure what I am allowed to think based on a p-value). – Dan Chaltiel Aug 27 '18 at 12:17YandY'is important and still needs explanation. Compared to the first few versions, the post now does a better job describing your objective and explaining what you mean by "more associated." This reply doesn't deserve the downvote that has been applied to it. (I'm not intimating that you are the downvoter.) – whuber Aug 27 '18 at 13:27