Timeline for Dealing with non-proportional hazards in a Cox model with many variables and a large dataset
Current License: CC BY-SA 4.0
8 events
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| Aug 24 at 23:43 | comment | added | Thomas |
Also, interpretation of the plots is clearly explained in ?cox.zph (in the Details section; emphasis mine): "The plot gives an estimate of the time-dependent coefficient $\beta(t)$. If the proportional hazards assumption holds then the true $\beta(t)$ function would be a horizontal line. The table component provides the results of a formal score test for slope=0, a linear fit to the plot would approximate the test."
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| Aug 11 at 11:44 | comment | added | Thomas |
For the record, in response to my second comment: from the timedep vignette (section 4), "The cox.zph plot is excellent for diagnosis but does not, however, produce a formal fit of $\beta(t)$." Two methods to fit $\beta(t)$ are then explained: (i) survsplit() (section 4.1), for "a step function for $\beta(t)$, i.e., different coefficients over different time interval"; and (ii) the time-transform functionality of coxph (i.e. tt(...) terms) (section 4.2), "If $\beta(t)$ is assumed to have a simple functional form".
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| Aug 11 at 2:15 | comment | added | EdM |
@Thomas if y =0 and the curve is flat, then the HR is 1 (no association with outcome) and PH presumably holds. What’s called the “smoothed Schoenfeld residuals plot” isn’t quite that; it’s a plot of the estimate of the regression coefficient as a function of time. See this answer.
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| Aug 10 at 22:37 | comment | added | Thomas | Also, I thought that the smoothed curves on the plots of scaled Schoenfeld residuals had to have $y = 0$. I doubt it now, since the model's coefficients (HR) can be stable over time without being 0 (1). Can you please confirm that the curve doesn't have to be flat and at $y = 0$ for proportional hazards assumption to hold? Thanks! | |
| Aug 10 at 22:29 | comment | added | Thomas |
I hadn't really understood that the smoothed curves on the plots of scaled Schoenfeld residuals corresponded to the model coefficients! I've just realized that we can even use plot(zph, hr = TRUE) to directly show of the change in hazard ratio over time. (The plot does not change, only the axis label; see hr description in ?survival:::plot.cox.zph) So if I understand correctly, the plots provide the same information as if I were using survsplit() with very narrow cut points? Simple tests suggest it is the case. Thanks again!
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| Aug 10 at 22:09 | comment | added | Thomas | Thank you so much for your response, and for so many others on CV. Your educational responses provide extremely valuable support to researchers who have no formal training in statistics. You help us do better science! | |
| Aug 10 at 22:05 | vote | accept | Thomas | ||
| Aug 10 at 19:59 | history | answered | EdM | CC BY-SA 4.0 |