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If I estimate a Cox Proportional Hazards model and my covariate of interest is dependent (continuous or categorical), does the proportional hazards assumption still matter? I recently went to a presentation where the speaker said that when using a time-dependent covariate, the importance of satisfying this assumption didn't matter but didn't really offer any justification for this, nor did he offer a reference.

Any comments?

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You are still assuming that the effect of the value at each covariates/factor at each timepoint is the same, you simply allow the covariate to vary its value over time (but the change in the log-hazard rate associated with a particular value is still exactly the same across all timepoints). Thus, it does not change the assumption. Or was the presenter perhaps talking about also putting the covariate by time (or log(time)) interaction in the model as a time-dependent covariate? If you do that (for all covariates), then you have a model that might possibly approximate (a linear interaction cannot fully capture the possibly more complex things that may be going on in any one dataset, but may be okay for approximately capturing it) a model that does not make such an assumption.

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I may be wrong but I believe that Björn's answer is not completely correct. The proportional hazards assumption means that the ratio of the hazard for a particular group of observations (determined by the values of the covariates) to the baseline hazard (when all covariates are zero) is constant over time. If there are time-varying covariates this is not true, and therefore the Cox model no longer assumes proportional hazards.

Here is a quote I have recently come across from David Collett's book, Modelling Survival Data in Medical Research (2nd ed., 2003, p. 253), that may be helpful:

It is important to note that in the model given in equation $h_i(t) = \exp \left\{ \sum_{j=1}^p \beta_j x_{ji}(t) \right\} h_o(t)$, the values of the variables $x_{ji}(t)$ depend on the time $t$, and so the relative hazard $h_i(t)/h_0(t)$ is also time-dependent. This means that the hazard of death at time $t$ is no longer proportional to the baseline hazard, and the model is no longer a proportional hazards model.

The accepted answer to this question on CV may also be relevant.

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    $\begingroup$ This might, however, be more of a terminological rather than a practical distinction. Even if the presence of time-dependent covariate values means that the proportional hazards (PH) assumption does not hold, approaches based on partial likelihood for analyzing Cox PH models still can be used reliably with time-dependent covariates, as references linked from the CV question you cite make clear. The underlying assumption with time-dependent covariate values is as Björn stated: "the change in the log-hazard rate associated with a particular value is still exactly the same across all timepoints." $\endgroup$ – EdM Jul 9 '18 at 15:20
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    $\begingroup$ Thank you for your comment. I think you raise a good point. Perhaps one practical aspect where this question could be important is in whether or not one would need to test the assumption in an applied setting. In a model with only fixed-time covariates I believe it is advisable to test the proportional hazards assumption, for instance by checking that the Schoenfeld residuals for the different variables are approximately constant over time. I think, however, that this would not make sense with time-varying covariates, though I may be wrong. $\endgroup$ – George Costanza Jul 9 '18 at 19:14
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    $\begingroup$ You can test the assumption that "the change in the log-hazard rate associated with a particular value [of a covariate] is still exactly the same across all timepoints," which for Cox models with time-dependent covariates (assuming that the current value of the covariate determines the instantaneous hazard versus baseline) is the analog of the strict PH assumption. For example, this document (linked from the CV page you cite) shows how to do so by testing the significance of adding a type of covariate*time interaction term to the model. $\endgroup$ – EdM Jul 9 '18 at 22:08

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