I am fitting AFT models using the command survreg from the R package survival. I want to do some further plots of the hazard function but I do not understand what is the parametrization of the AFT model used in this package. The help of this command only indicates:


Fit a parametric survival regression model. These are location-scale models for an arbitrary transform of the time variable; the most common cases use a log transformation, leading to accelerated failure time models.

However, I would like to know if the parametrization in terms of the hazard function is




but this is not indicated in the manual



This might be helpful: https://cran.r-project.org/web/packages/SurvRegCensCov/vignettes/weibull.pdf

Quoting from the first page:

Weibull accelerated failure time regression can be performed in R using the survreg function. The results are not, however, presented in a form in which the Weibull distribution is usually given. Accelerated failure time models are usually given by $\log T=Y=\mu+\alpha^\top z +\sigma W$ ,where $z$ are set of covariates, and $W$ has the extreme value distribution. Given transformations $\gamma = 1/\sigma$, $\lambda =\exp(−\mu/\sigma)$, $\beta =−\alpha/\sigma$, we have a Weibull model with baseline hazard of $h(x|z) = (\gamma \lambda t^{\gamma−1}) exp(\beta^\top z)$.

So, in the AFT model as parametrized in the survreg function, larger values of $\alpha^\top z$ correspond to an increase in expected survival time (longer survival), whereas in the Cox model as parametrized in coxph, larger values of $\beta^\top z$ correspond to an increase in the hazard (shorter survival), and when the AFT error follows the Weibull distribution, they are related by $\beta^\top z = -(\alpha^\top z)/ \sigma$

To confirm directly that the AFT model in R uses $\alpha^\top z$, compare the linear predictors from a fitted AFT model using predict.survreg to the linear predictors calculated 'by hand':

library(survival); library(tidyverse);
# fit aft to lung data using defaults
lung_model_aft <- survreg(Surv(time, status) ~ age + sex + factor(ph.ecog), lung)

all(near(predict(lung_model_aft, type = 'lp'), model.matrix(lung_model_aft) %*% lung_model$coefficients))

We can also check for similarity to the fitted Cox model, but we should only expect them to be similar, not identical, since the Cox model estimates the baseline hazard nonparametrically but the AFT model estimates it parametrically:

- coef(lung_model_aft)[-1] / lung_model_aft$scale
#             age              sex factor(ph.ecog)1 
#     0.009938741     -0.541893431      0.397749179 
#factor(ph.ecog)2 factor(ph.ecog)3 
#     0.906020113      1.857645211 
#             age              sex factor(ph.ecog)1 
#      0.01079468      -0.54583052       0.41004801 
#factor(ph.ecog)2 factor(ph.ecog)3 
#      0.90330301       1.95454338 

Calculating the hazard The general expression for the hazard function at time $x$ is $h(x) = f_T(x)/\Pr(T > x)$, where $f_T(x)$ is the pdf of $T$ at $x$. When $T = \exp\{\mu + \alpha^\top z +\sigma W\}$, then T's distribution is determined by the distribution of $W$. And when $W$ is the extreme value distribution, then $T$ given $z$ is Weibull, and the hazard function is as given above.

The form of the hazard will be different when $W$ is differently distributed. For example, when $W$ is standard normal, then $T$ given $z$ is log-normal. So, $f_T(x) = \frac 1 {x\sigma\sqrt{2\pi}}\ \exp\left(-\frac{\left(\ln x-\mu -\alpha^\top z\right)^2}{2\sigma^2}\right)$ and $\Pr(T > x) = 1 - \Phi\left( \frac{\ln x - \mu - \alpha^\top z} \sigma \right)$, and $$h(x|z) = \dfrac{\frac 1 {x\sigma\sqrt{2\pi}}\ \exp\left(-\frac{\left(\ln x-\mu -\alpha^\top z\right)^2}{2\sigma^2}\right)}{1 - \Phi\left( \frac{\ln x - \mu - \alpha^\top z} \sigma \right)}$$

For other distributions of $W$, the form of the hazard will be different yet. As you may know, the choice of Weibull $T$ (equivalently, Extreme Value $W$) is the only choice that is both an AFT model as well as a proportional hazards model.

I'm not aware of functionality in R to automatically calculate and extract the hazard curves for all observations. So you would likely need to write up a function yourself.

  • $\begingroup$ So, just to confirm, if I want to plot the hazard associated to individual $i$, I need to plot the hazard $h_0(t\exp(-x^T\beta/\sigma))\exp(-x^T\beta/\sigma)$? Where $h_0$ is the baseline hazard (not necessarily Weibull). $\endgroup$ – DIVOC Jul 16 '20 at 17:21
  • $\begingroup$ I see now that my answer was incomplete. I have edited it to include how to calculate the hazard for a general error term $W$. With regard to your comment: generally the baseline hazard is called 'baseline' because it doesn't include any covariates, that is, it common to all observations. In contrast, your expression $h_0(t\exp(-x^\top \beta/\sigma))$ is an expression of covariates $x$, so I don't quite understand what you mean. Hopefully my edited answer clears up any confusion. $\endgroup$ – psboonstra Jul 16 '20 at 19:20
  • $\begingroup$ Thanks. Your answer is enough for me to proceed with the task I need to to. $\endgroup$ – DIVOC Jul 16 '20 at 19:33

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