Cross Validated Asked by user11130854 on December 17, 2020

I am trying to understand the relationship between AFT and Cox regression in terms of the Weibull distribution. Specifically, one can compute a factor for h(t) from the coefficients of an AFT using the shape parameter of the Weibull distribution, e.g. see https://rpubs.com/kaz_yos/aft. Does this mean that the application of AFT with Weibull is limited to cases where the proportional hazard assumption for CoxPH holds?

As this source notes:

the Weibull is the

onlydistribution that is closed under both the accelerated life and proportional hazards families.

So if you are working with a distribution in the Weibull family (that includes simple exponential models), then you can model with either a proportional hazard (PH) or an accelerated failure time (AFT) model, and you will still end up with a Weibull distribution. In that sense, whenever you use an AFT model with an underlying Weibull distribution, the PH assumption will hold--and vice versa. For distributions outside the Weibull family, at most of one of AFT or PH can hold.

Answered by EdM on December 17, 2020

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