1. Introduction to Frailty Models
STA635 Project by Benjamin Hall

2. Cox proportional hazards model
Model: hyi (t) = h0(t)e1X 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% .5 2 3
20% 10 8

4. Simulation Example, continued
Let’s simulate this example and apply the Cox model:
Have R simulate 100 people according to the previous table’s 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
.38
[2,]
.07
[3,]
.23
[4,]
.15
[5,]
.11
[6,]
.89
[7,] 2
.22
[8,]
.04
...

6. Simulation Example, continued
Now we run coxph on our data:
> myfit1
Call:
coxph(formula = Surv(time, status) ~ treat)
coef exp(coef) se(coef) z p
treat
Likelihood ratio test=1.37 on 1 df, p= 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1X 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
Let’s apply the frailty model to our simulated data:
> myfit2
Call:
coxph(formula = Surv(time, status) ~ treat + frailty(id))
coef se(coef) se2 Chisq DF p
treat e-01
frailty(id) e-05
Iterations: 5 outer, 17 Newton-Raphson
Variance of random effect= I-likelihood =
Degrees of freedom for terms=
Likelihood ratio test=117 on 44.0 df, p=1.37e-08 n= 445
Notice that the LRT now has a highly significant p-value.

10. Frailty Model in R
Now let’s try implementing the frailty model to a real data set, the kidnet data set.
Here are the results for the regular Cox Model:
> kfit1
Call:
coxph(formula = Surv(time, status) ~ age + sex, data = kidney)
coef exp(coef) se(coef) z p
age
sex
Likelihood ratio test=7.12 on 2 df, p= n= 76
Here the LRT is significant with a p-value of 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:
> kfit2
Call:
coxph(formula = Surv(time, status) ~ age + sex + frailty(id), data = kidney)
coef se(coef) se2 Chisq DF p
age
sex
frailty(id)
Iterations: 7 outer, 49 Newton-Raphson
Variance of random effect= I-likelihood =
Degrees of freedom for terms=
Likelihood ratio test=46.8 on 14.1 df, p=2.31e-05 n= 76
Now the LRT is even more significant.

12. Resources Therneau and Grambsch, Modeling Survival Data, Chapter 9
Wienke, Andreas, “Frailty Models”,
Govindarajulu, “Frailty Models and Other Survival Models”,

13. R Code
library(survival) #GEN_TIME gen_time <- function(group, treat) { if (group == 1) { return (round(rexp(1, 1-(.5*treat)),2))} if (group == 2) { return (round(rexp(1, 2-treat),2))} if (group == 3) { return (round(rexp(1, 10-2*treat),2))}} # PERSON DATA person_data <- function() { treat <- rbinom(1,1,.5) x <- runif(1) t1 <- matrix(NA, nrow=1, ncol=25) if (x < .4) { group <- 1 } if (x > .4 & x < .8) { group <- 2} if (x > .8) { group <- 3} elapse <- 0 count <- 1 while (elapse < 1) { t1[(count+3)] <- gen_time(group, treat) elapse <- elapse + t1[(count+3)] count <- count + 1} count <- count - 1 t1[1] <- group t1[2] <- treat t1[3] <- count for (i in (count+4):25) { t1[i] <- 0 } if (count == 1) { t1[count+3] <- 1 } if (count > 1) { t1[count+3] <- 1-t1[count+2] } return (t1)}
m1 <- matrix (NA, nrow=100, ncol=25)
for (i in 1:100) {
m1[i,] <- person_data() }
samp_size <- sum(m1[,3])
samp <- matrix(NA, nrow = samp_size, ncol= 5)
colnames(samp) <- c("id", "group", "treat", "time", "status")
count2 <- 1
for (j in 1:m1[i,3]) {
samp[count2, 1] <- i
samp[count2, 2] <- m1[i,1]
samp[count2, 3] <- m1[i,2]
samp[count2, 4] <- m1[i,j+3]
samp[count2, 5] <- 1
if(j==m1[i,3]) { samp[count2, 5] <- 0 }
count2 <- count2 + 1}}
myfit1 <- coxph(Surv(samp[,4], samp[,5]) ~ samp[,3])
myfit2 <- coxph(Surv(samp[,4], samp[,5]) ~ samp[,3] + frailty(samp[,1]))
myfit1
myfit2