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So I do think my cox regression residuals are linear. Please correct me if they're not. My reasoning behind this is that the end of the martingale red line "drops off" simply because of few data points and thus, more weight towards the right side.

Would you agree? Also, if you wouldn't consider this normal, how would I go about identifying the non-linear variables?

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It's not clear how the plot of martingale residuals (top) was obtained, and the label of the x-axis is missing. If these are overall residuals, then heed the warning of Therneau and Grambsch (Section 4.2.2)

... martingale residuals can not play all the roles that linear model residuals do; in particular, the overall distribution of the residuals does not aid in the global assessment of fit.

There are ways to use martingale residuals to evaluate the functional form of a continuous predictor, but that doesn't seem to be what you've done. See this page.

The deviance residuals in the bottom plot are transformations of the martingale residuals. From Section 4.3 of Therneau and Grambsch:

The deviance residual was designed to improve on the martingale residual for revealing individual outliers, particularly in plotting applications. In practice it has not been as useful as anticipated.

So neither of your plots seems to get directly to evaluating the linearity of the model.

The way to "go about identifying the non-linear variables" is one at a time. For each continuous predictor you plot the martingale residuals against the corresponding values of that predictor. Then a "flat" smoothed curve indicates a reasonable fit of that predictor. This only makes sense for continuous predictors. Categorical predictors are necessarily "linear," although they might violate the proportional hazards assumption.

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  • $\begingroup$ The x axis are just fitted values $\endgroup$
    – Antonio
    Dec 17, 2022 at 15:53
  • $\begingroup$ @Antonio then the answer is correct. Plots of overall martingale residuals against fitted (linear-predictor) values doesn't evaluate model fit adequately, and their transformation to deviance residuals doesn't add much except for a potentially better way to identify outliers. $\endgroup$
    – EdM
    Dec 17, 2022 at 15:59
  • $\begingroup$ @Antonio I added a paragraph at the end to explain one approach. Also see this page and its links. $\endgroup$
    – EdM
    Dec 17, 2022 at 16:19

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