[R] AFT model time-dependent with weibull distribution

[Göran Broström]

On Sat, Aug 20, 2011 at 7:33 PM, JPF <xpfenech at gmail.com> wrote:
>
> Göran Broström wrote:
>>
>>
>> Good. Do you still need answers to your other questions?
>>
>>
>
> Yes. Could answer the following two questions:
>
> 1- Can I use phreg function to estimate  a model with time-dependent
> covariates? In case of a positive answer, how?

Yes. You do it in the way you did it with your first example. No 'id'
argument is needed (or allowed) in the call to phreg. The reason is
that in PH regression, the hazard function at any time depends only on
the covariate value at that time point. In the AFT model, on the other
hand, the hazard function at time  t  depends on all covariate values
in the interval (0, t). Therefore aftreg needs the 'id' argument, but
phreg does not.

>
> 2- I could not find any example that clearly explains how to interpret
> aftreg output. Specially, refering to the diference between survreg and
> aftreg output (intercept and sign of the estimates).

I think you already answered that question. You can read more about it
in the vignette about parametric models in eha.

G.
>
> I include below an example of output of a regression with coxph, survreg,
> phreg and aftreg and a time-independent variable. I would appreciate if you
> could explain it or provide an external example that explains how it works.
>
> n=26
> events=25
> time at risk=45
>
> a/
>
> coxph(Surv(time,s) ~ Z1,  data=data.frame(data))
>
>     coef     exp(coef)  se(coef)    z     p
> Z1 0.0249      1.03      0.00907 2.75 0.006
>
>
> b/
>
> phreg(Surv(time,s) ~ Z1,  data=data.frame(data), dist="weibull")
>
> Covariate          W.mean      Coef Exp(Coef)  se(Coef)    Wald p
> Z1                      43.689     0.033     1.033     0.009     0.000
>
> log(scale)                          0.641     1.899     0.065     0.000
> log(shape)                         1.172     3.230     0.158     0.000
>
> Max. log. likelihood      -22.135
> LR test statistic         13.1
> Degrees of freedom        1
> Overall p-value           0.000302689
>
>
> c/
>
> aftreg(Surv(time,s) ~ Z1,  data=data.frame(data), dist="weibull")
>
> Covariate          W.mean      Coef Exp(Coef)  se(Coef)    Wald p
> mas                  43.689       0.010    1.010     0.002     0.000
>
> log(scale)                          1.147     3.149     0.141     0.000
> log(shape)                         1.172     3.230     0.158     0.000
>
> Max. log. likelihood      -22.135
> LR test statistic         13.1
> Degrees of freedom        1
> Overall p-value         0.000302692
>
>
> d/
>
> survreg(Surv(time,s) ~ Z1,  data=data.frame(data), dist="weibull")
>
>                      Value     Std. Error     z        p
> (Intercept)      1.1476    0.13498     8.50  1.87e-17
> mas               -0.0101   0.00232    -4.34  1.45e-05
> Log(scale)      -1.1724    0.15787    -7.43 1.11e-13
>
> Scale= 0.310
>
> Weibull distribution
> Loglik(model)= -22.1   Loglik(intercept only)= -28.7
>        Chisq= 13.05 on 1 degrees of freedom, p= 3e-04
> Number of Newton-Raphson Iterations: 5
>
>
> Thank you very much,
>
> J
>
>
>
>
>
>
>
> --
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>
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>



-- 
Göran Broström