Introduction%20to%20Frailty%20Models

1
Introduction to Frailty Models

• STA635 Project by Benjamin Hall

2
Cox proportional hazards model

• Model hyi (t) h0(t)e?1X1... ?kXk is the
  hazard function of the ith individual
• Assumption The hazard function for each
  individual is proportional to the basine hazard,
  h0(t). This assumption implies that the hazard
  function is fully determined by the covariate
  vector.
• Problem There may be unobserved covariates that
  cause this assumption to be violated.

3
Simluation Ex Unobserved covariate

• Consider a situation with the following
  population
• Obviously Drug A is effective for the entire
  population.
• But what happens in the Cox Model if the group is
  unobservable?

Group Proportion of Population Hazard Rate with placebo Hazard Rate with Drug A
1 40 1 .5
2 40 2 1
3 20 10 8
4
Simulation Example, continued

• Lets simulate this example and apply the Cox
  model
• Have R simulate 100 people according to the
  previous tables probabilities and randomly
  assign them to treatment or placebo
• For each person, have R simulate of incidents
  within period of length 1. At the end of period
  of length 1 right-censoring occurs.

5
Simulation Example, continued

• Here is some of the data generated by R (see
  final slide for code)

id group treat time status
1, 1 3 0 .38 1
2, 1 3 0 .07 1
3, 1 3 0 .23 1
4, 1 3 0 .15 1
5, 1 3 0 .11 1
6, 1 3 0 .89 0
7, 2 3 1 .22 1
8, 2 3 1 .04 1
... ... ... ... ... ...
6
Simulation Example, continued

• Now we run coxph on our data
• gt myfit1
• Call
• coxph(formula Surv(time, status) treat)
• coef exp(coef) se(coef)
  z p
• treat -0.128 0.88 0.11
  -1.17 0.24
• Likelihood ratio test1.37 on 1 df, p0.242 n
  445
• Notice that the LRT has a p-value of .242 which
  is not significant. But we know that treatment is
  effective for everyone. What is happening?

7
Simulation Example, continued

• The problem is that we have heterogeneity in the
  data due to the unobservable groups.
• Since we cannot include group in our model, the
  assumption of proportional hazards is violated.
• What can we do to solve this problem? Use a
  frailty model.

8
Frailty Model

• Frailty models can help explain the unaccounted
  for heterogeneity.
• Frailty Model hyi (t) z ? h0(t)e?1X1... ?kXk
  is the hazard function of the ith individual
• The distribution of z is specified to be, say,
  Gamma. (Note z must be non-negative since the
  hazard is non-negative.)
• In this situation, the shared frailty model is
  appropriate, that is multiple observations of the
  same individual always has the same value of z.

9
Frailty Model in R

• Lets apply the frailty model to our simulated
  data
• gt myfit2
• Call
• coxph(formula Surv(time, status) treat
  frailty(id))
• coef se(coef)
  se2 Chisq DF p
• treat -0.147 0.160
  0.111 0.85 1.0 3.6e-01
• frailty(id)
  93.89 43.5 1.4e-05
• Iterations 5 outer, 17 Newton-Raphson
• Variance of random effect 0.294
  I-likelihood -1887.9
• Degrees of freedom for terms 0.5 43.5
• Likelihood ratio test117 on 44.0 df, p1.37e-08
  n 445
• Notice that the LRT now has a highly significant
  p-value.

10
Frailty Model in R

• Now lets try implementing the frailty model to a
  real data set, the kidnet data set.
• Here are the results for the regular Cox Model
• gt kfit1
• Call
• coxph(formula Surv(time, status) age sex,
  data kidney)
• coef exp(coef) se(coef)
  z p
• age 0.00203 1.002 0.00925 0.220
  0.8300
• sex -0.82931 0.436 0.29895 -2.774
  0.0055
• Likelihood ratio test7.12 on 2 df, p0.0285 n
  76
• Here the LRT is significant with a p-value of
  .0285 even without considering frailty.

11
Frailty Model in R

• However, a frailty model seems applicable in this
  situation since their are multiple oberservations
  (i.e. 2 kidneys) per person. Below considers
  frailty
• gt kfit2
• Call
• coxph(formula Surv(time, status) age sex
  frailty(id), data kidney)
• coef se(coef) se2
  Chisq DF p
• age 0.00525 0.0119 0.0088 0.2
  1 0.66000
• sex -1.58749 0.4606 0.3520 11.9
  1 0.00057
• frailty(id)
  23.1 13 0.04000
• Iterations 7 outer, 49 Newton-Raphson
• Variance of random effect 0.412
  I-likelihood -181.6
• Degrees of freedom for terms 0.5 0.6 13.0
• Likelihood ratio test46.8 on 14.1 df,
  p2.31e-05 n 76
• Now the LRT is even more significant.

12
Resources

• Therneau and Grambsch, Modeling Survival Data,
  Chapter 9
• Wienke, Andreas, Frailty Models,
  http//www.demogr.mpg.de/papers/working/wp-2003-03
  2.pdf
• Govindarajulu, Frailty Models and Other Survival
  Models, www.ms.uky.edu/statinfo/nonparconf/govin
  darajulu.ppt

13
R Code

• library(survival)
• GEN_TIME
• gen_time lt- function(group, treat)
• if (group 1)
• return (round(rexp(1, 1-(.5treat)),2))
• if (group 2)
• return (round(rexp(1, 2-treat),2))
• if (group 3)
• return (round(rexp(1, 10-2treat),2))
• PERSON DATA
• person_data lt- function()
• treat lt- rbinom(1,1,.5)
• x lt- runif(1)
• t1 lt- matrix(NA, nrow1, ncol25)
• if (x lt .4) group lt- 1
• if (x gt .4 x lt .8) group lt- 2
• if (x gt .8) group lt- 3
• elapse lt- 0

m1 lt- matrix (NA, nrow100, ncol25) for (i in
1100) m1i, lt- person_data() samp_size lt-
sum(m1,3) samp lt- matrix(NA, nrow samp_size,
ncol 5) colnames(samp) lt- c("id", "group",
"treat", "time", "status") count2 lt- 1 for (i in
1100) for (j in 1m1i,3) sampcount2, 1
lt- i sampcount2, 2 lt- m1i,1 sampcount2, 3
lt- m1i,2 sampcount2, 4 lt- m1i,j3 sampcount
2, 5 lt- 1 if(jm1i,3) sampcount2, 5 lt- 0
count2 lt- count2 1 myfit1 lt-
coxph(Surv(samp,4, samp,5) samp,3) myfit2
lt- coxph(Surv(samp,4, samp,5) samp,3
frailty(samp,1)) myfit1 myfit2