Hazards in AFT with Weibull distribution

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 Does this mean that the application of AFT with Weibull is limited to cases where the proportional hazard assumption for CoxPH holds?

One Answer

As this source notes:

the Weibull is the only distribution 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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