Solved – the parameterization of exponential distribution for survival in Stata

exponential distributionparameterizationrstatasurvival

I'm new to data analysis so this is kind of a simple question.

I would like to understand why I cannot reproduce a survival curve generated by a fitted exponential model from Stata. I use the coefficients and make my function in R to plot but it looks nothing similar. I believe it's one of those daft problems where I am not interpreting something properly. I illustrate below.

First, some data in Stata:

use http://www.ats.ucla.edu/stat/data/uis.dta, clear
gen id = ID
drop ID
stset time, failure(censor)

Then we can fit a null exponential model

streg, dist(exponential) nohr

Which gives the following output:

Exponential regression -- log relative-hazard form 

No. of subjects =          628                     Number of obs   =       628
No. of failures =          508
Time at risk    =       147394
                                                   LR chi2(0)      =     -0.00
Log likelihood  =    -1043.531                     Prob > chi2     =         .

------------------------------------------------------------------------------
          _t |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
       _cons |  -5.670383   .0443678  -127.80   0.000    -5.757342   -5.583424
------------------------------------------------------------------------------

I take the survivor function from Stata's documentation:

$$
S(t)=\exp(-\lambda_{j} t_{j})
$$

So, in R, I plot this out with the following:

S <- function(x) exp(-5.670383*x) # 'x' acts like 't'
curve(S, 0, 1000)

This curve is not equivalent to Stata's, given by:

stcurve, surv

Where did I go wrong in my interpretation? Is my equation using the correct parameterization?

P.S. Why am I reproducing these curves? A little to do with overlaying curves but now that I have this problem, I have to know where I went wrong.

Best Answer

From the documentation you can see that $\lambda = \exp(xb)$, so

S <- function(x) exp(-5.670383*x)

should have been:

S <- function(x) exp(-exp(-5.670383)*x)

As an aside, I would stick to Stata for such graphs, mainly because switching programs within a project makes it harder to document that project. Here is how I would create such a graph in Stata:

twoway function survival = exp(-exp(_b[_cons])*x), ///
     range(0 <some reasonalble maximum>)           ///
     xtitle("time")

Moreover, you can replace

gen id = ID
drop ID

with

rename ID id

Related Solutions

Survival Analysis – Parametrization of ‘survreg’ in the ‘survival’ R Package

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))
# TRUE

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 
coef(lung_model_cox)
#             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.

Best Answer

Related Solutions

Survival Analysis – Parametrization of ‘survreg’ in the ‘survival’ R Package

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