Likelihood test for goodness-of-fit for Accelerated Failure Time (AFT) models

Question

Background

From my understanding to the log likelihood test in survival analysis that this test calculated twice, first time its calculated for null model (no covariants) and second time its calculated to model with covariants. Then, as much we get small difference between these two values we can say that this model fit to our data well.

Applying this practically

Now let say I have two Accelerated Failure Time (AFT) AFT1 and AFT2 that fitted on the same data and same covariants but with different distributions.

Example In R with survival package one can do the following:

Srv <- Surv(start, end, event, type="interval")# recurrent data
AFT1 <- survreg(Sur~ X1+cluster(ID)+X2,X3, data=test, dist="w") # first AFT model with Weibull distribution
AFT1 <- survreg(Sur~ X1+cluster(ID)+X2,X3, data=test, dist="exponential") # 2nd model with exponential distribution

let us check the results from two above models.

First model

Call:
survreg(formula = Sur ~ X1 +cluster(ID)+X2,X3, data=test, dist="w")

Coefficients:
  (Intercept)          X1       X2             X3       
  7.606778e+00  7.419714e-05 -3.154395e-03  3.155968e-05   

 Scale= 0.00198871 

 Loglik(model)= -29235.1   Loglik(intercept only)= -31109.4
     Chisq= 3748.56 on 4 degrees of freedom, p= 0 
 n=43416 (30 observations deleted due to missingness)

Second model

 Call:
 survreg(formula = Sur ~ X1 +cluster(ID)+X2,X3, data=test, dist="exponential")

  Coefficients:
   (Intercept)          X1       X2             X3       
  7.606512e+00  7.923720e-05 -3.297417e-03  2.840966e-05   

 Scale= 0.001035689 

 Loglik(model)= -29144.8   Loglik(intercept only)= -31248
    Chisq= 4206.25 on 4 degrees of freedom, p= 0 
 n=43416 (30 observations deleted due to missingness)

we can then check the difference in loglik value for the two models; Loglik(model) and Loglik(intercept only) and we will get

• Loglik for AFT1 = -31109.4 -(-29235.1 ) = −1874.3

• Loglik for AFT2 = -31248 -(-29144.8 ) = −2103.2

Can we say that model AFT2 fit the our data better than AFT1 model?

Thanks in advance

Answer 1

The exponential AFT model is a special case of the Weibull regression, so you can create a likelihood ratio test to see if there is evidence against the simpler one (exponential). For this you can use the values of the log-likelihoods of the two models.

What I find strange in the outputs you presented above is that in the case of the exponential model the scale should be fixed at 1.

Answer 2

According to this in chapter six

When maximum likelihood estimation is used to generate the log-likelihoods, then the closer that the log-likelihood(LL) is to zero, the better is the model fit.

So we can say that AFT1 better than AFT2

Answer 3

Since you're doing this in R, you don't need to make inference by looking at model 1 vs null and model 2 vs null: you can compare them directly:

m1 <- survreg(formula = Sur ~ X1 +cluster(ID)+X2,X3, data=test, dist="w")
m2 <- survreg(formula = Sur ~ X1 +cluster(ID)+X2,X3, data=test, dist="exponential")

anova(m1, m2)

This will also give you a P value for the estimate that the fit is better for m2 vs m1.