Timeline for violation of cox proportional hazard assumption
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2023 at 19:35 | comment | added | Devi Sita | Let us continue this discussion in chat. | |
| Nov 27, 2023 at 19:05 | comment | added | EdM | @SAphi11 you don't ignore the violation of PH. You report it and illustrate it. The Cox model is still informative despite the violation of PH: this plot illustrates the model's estimate of how the Cox regression coefficient changes over time, as explained here. As AdamO notes in a comment on your question and in a link from this answer, the Cox coefficient and associated hazard ratio reported by the original model is a type of time-averaged value if you want a single value. | |
| Nov 27, 2023 at 19:03 | comment | added | Devi Sita | I really appreciate your time! | |
| Nov 27, 2023 at 18:54 | comment | added | EdM | @SAphi11 I added a few more thoughts. You have a real, but potentially interesting and important, change in hazard ratio over time. Apply your understanding of the subject matter to try to come up with an explanation. | |
| Nov 27, 2023 at 18:53 | history | edited | EdM | CC BY-SA 4.0 |
in response to newly provided graph
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| Nov 27, 2023 at 16:40 | comment | added | EdM |
@SAphi11 look for a reasonably flat smoothed curve that doesn't deviate much from the coxph()coefficient value. Set resid=FALSE in the call to plot() on the cox.zph() output to remove individual residuals. You can show plots of Schoenfeld residuals without violating confidentiality: when you use plot() on the cox.zph() output, include a value for the ylab argument that doesn't include the actual name of the variable. There is then no way for anyone who sees the plot to figure out what your actual data and variable names are.
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| Nov 16, 2023 at 8:34 | vote | accept | Devi Sita | ||
| Nov 15, 2023 at 17:49 | history | answered | EdM | CC BY-SA 4.0 |