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I am interested in the paper entitled "Generating survival times to simulate Cox proportional hazards models with time-varying covariates " (Stat Med. 2012 Dec 20;31(29):3946-58.),and want to simulate such kind of survival data. The R code is shown below. However, the beta(t) for time-varying covariate is not as defined in the beginning. I have struggled for a longtime but cannot figure out what's wrong with my code. Can anyone help me to figure it out?

N=1000 #number of subjects
k=0.3 #assuming the time varying covariates is proporitonal to time x2=k*t
lambda=0.01
betat=0.3 
beta=-0.6
rateC=0.01
#time fixed variable
x1 <- sample(x=c(0, 1), size=N, replace=TRUE, prob=c(0.5, 0.5))
  #Exponential latent event times
  u <- runif(n=N)
  Tlat <- log(1+betat*k*(-log(u))/(lambda*exp(beta*x1)))/betat*k
  Cen<- rexp(n=N, rate=rateC)#censoring times
  # follow-up times and event indicators
  time <- pmin(Tlat, Cen)
  status <- as.numeric(Tlat <= Cen)
  # data set
  df.tfixed<-data.frame(id=1:N,
         time=time,
         status=status,
         x1=x1)
##time dependent continuous variable
ntp<-sample(1:6,N,replace=T)#number of follow up time points
mat<-matrix(ncol=3,nrow=1)
i=0
for(n in ntp){
i=i+1
ft<-runif(n,min=0,max=df.tfixed$time[i])
ft<-sort(ft)
seq<-rep(ft,each=2)
seq<-c(0,seq,df.tfixed$time[i])
matid<-cbind(matrix(seq,ncol=2,nrow=n+1,byrow=T),i)
mat<-rbind(mat,matid)
}
df.td<-data.frame(mat[-1,])
colnames(df.td)<-c("start","stop","id")
df.td$x2<-k*df.td$start
#combine the two data frames 
df<-merge(df.td,df.tfixed,by="id")
df$status=0
df$status[cumsum(as.vector(table(df$id)))]<-1
fit <- coxph(Surv(start,stop, status) ~ x1+x2, data=df)
fit$coef
    x1         x2 
-0.6195906  1.3083348 

The output shows that the beta for x1 is consistent with what I have defined, but the coefficient for x2 is not what I have defined (betat=0.3).

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1 Answer 1

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I figure you are looking for the model in equation 4 in this article. Then this code will do

> library(survival)
> set.seed(747)
> 
> N <- 1e4 # number of subjects
> k <- 0.3 # assuming the time varying covariates is proporitonal to time x2=k*t
> lambda <- 0.01
> betat <- 0.3 
> beta <- -0.6
> rateC <- 0.01
> 
> #####
> # simulate 
> x1 <- as.numeric(runif(N) > .5) # covariate
> 
> # simulate outcome (NB: OP forgot parentheses)
> u <- runif(n = N)
> Tlat <- log(1 + betat*k*(-log(u))/(lambda*exp(beta*x1))) / (betat * k)
> 
> # censoring and status indicator
> Cen<- rexp(n = N, rate = rateC)
> 
> # make data.frame
> df <- data.frame(
+   time = pmin(Tlat, Cen), 
+   status = Tlat <= Cen, 
+   id = 1:N, 
+   x1 = x1, 
+   x2 = rep(k, N))
> 
> # fit model -- cannot use coxph for x2 due to the non-parametric intercept
> fit <- coxph(Surv(time, status) ~ x1, data = df)
> fit # coefficient for x1 is right
Call:
coxph(formula = Surv(time, status) ~ x1, data = df)

      coef exp(coef) se(coef)   z      p
x1 -0.6083    0.5442   0.0234 -26 <2e-16

Likelihood ratio test=675  on 1 df, p=0
n= 10000, number of events= 7891 
> 
> # # won't work as we try to estimate a term that is equal for all individuals 
> # # at each time t
> # fit <- coxph(Surv(time, status) ~ x1 + tt(x2), data = df, 
> #              tt = function(x, t, ...) x * t)
> 
> # The non-parametric cumulative hazard though match with what we expect
> base <- basehaz(fit, centered = FALSE)
> plot(base$time, base$hazard, type = "l")
> lines(
+   base$time, 
+   lambda / (betat * k) * (exp(betat * k * base$time) - 1), col = "red")

enter image description here

> #####
> # estimate parametric model
> neg_loglike_func <- function(b){
+   # compute intermediates
+   lp1 <- exp(b[1] + b[2] * df$x1)    # time-invariant terms
+   lp2 <- exp(b[3] * df$x2 * df$time) # time-varying terms
+   log_haz <- log(lp1) + log(lp2)                  # instant hazard
+   cumhaz <- lp1 * (lp2 - 1)/(b[3] * df$x2 + 1e-8) # cumhaz
+   
+   # compute log likelihood terms
+   ll_terms <- ifelse(df$status == 1,  log_haz - cumhaz, -cumhaz)
+   -sum(ll_terms) # neg log likelihood
+ }
> 
> # fit model
> b <- c(0.01, -0.02, 0.01) # starting valus
> fit <- optim(b, neg_loglike_func)
> 
> exp(fit$par[1]) # lambda estimate
[1] 0.01035017
> fit$par[2:3]    # beta and betat estimate
[1] -0.6099467  0.2966224

You may be able to avoid the optim function above by e.g., using one of the parametric survival models in the rstpm2 package or polspline.

Before edits

I cannot see how what you do is related to the cited article. Further, I am struggling to see how x2 enters into the model. Maybe post equations for the model you are trying to simulate from?

What may help you is that all of your observation end up dying at the end it seems which seems odd given that you have # censoring times in the Cen object. I show this below

library(survival)
set.seed(747)

N=1000 # number of subjects
k=0.3 # assuming the time varying covariates is proporitonal to time x2=k*t
lambda=0.01
betat=0.3 
beta=-0.6
rateC=0.01


#####
# Time fixed variable
x1 <- sample(x=c(0, 1), size=N, replace=TRUE, prob=c(0.5, 0.5))

#Exponential latent event times
u <- runif(n=N)
Tlat <- log(1+betat*k*(-log(u)) / (lambda*exp(beta*x1))) / betat*k
Cen<- rexp(n=N, rate=rateC) #censoring times

# follow-up times and event indicators
time <- pmin(Tlat, Cen)
status <- as.numeric(Tlat <= Cen)
# data set
df.tfixed<-data.frame(id=1:N, time=time, status=status, x1=x1)


#####
# time dependent continuous variable
ntp<-sample(1:6,N,replace=T) #number of follow up time points
mat<-matrix(ncol=3,nrow=1)
i=0
for(n in ntp){
  i=i+1
  ft<-runif(n,min=0,max=df.tfixed$time[i])
  ft<-sort(ft)
  seq<-rep(ft,each=2)
  seq<-c(0,seq,df.tfixed$time[i])
  matid<-cbind(matrix(seq,ncol=2,nrow=n+1,byrow=T),i)
  mat<-rbind(mat,matid)
}
df.td<-data.frame(mat[-1,])
colnames(df.td)<-c("start","stop","id")
df.td$x2<-k*df.td$start


#combine the two data frames 
df<-merge(df.td,df.tfixed,by="id")
df$status=0
df$status[cumsum(as.vector(table(df$id)))]<-1

fit <- coxph(Surv(start,stop, status) ~ x1+x2, data=df)
fit$coef
>         x1         x2 
> -0.6040089  1.8471387 

# All observations dies exaclty once -- no censoring
all(tapply(df$status == 1, df$id, sum) == 1)
> [1] TRUE

It comes down to this line df$status[cumsum(as.vector(table(df$id)))]<-1.

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  • 1
    $\begingroup$ Sorry I was looking at the wrong article. $\endgroup$ Commented Nov 12, 2017 at 16:21
  • $\begingroup$ thank you so much for your code @Benjamin Christoffersen. However, I still have a question that is can I display x2 at different follow-up time points in a counting process form. In the present format, X2 has its value at baseline and all these values are the same. This cannot be the case in real practice. Thus, what I am trying to to is to simulate x2 with random values at baseline and also in randomly selected follow up time points. $\endgroup$
    – Z. Zhang
    Commented Nov 12, 2017 at 22:09
  • $\begingroup$ The article you link does not provide a setup for where the covariate $x_2$ is time-varying unless it it is binary variable (c.f., section 3.3). While it is not hard to implement then this is not what your original question seems to be about. Particularly you write: ... "Generating survival times to simulate Cox proportional hazards models with time-varying covariates " (Stat Med. 2012 Dec 20;31(29):3946-58.),and want to simulate such kind of survival data. $\endgroup$ Commented Nov 12, 2017 at 22:34
  • $\begingroup$ May I have your email address? I may have more questions that need your help. $\endgroup$
    – Z. Zhang
    Commented Nov 13, 2017 at 6:13
  • $\begingroup$ I am trying to code the same problem as @Z.Zhang for simulating failure times from the proportional hazards model $\lambda(t|X) = \lambda_0(t) e^{X\beta_1 + Z(t)\beta_t}$ where $Z(t) = kt$. The paper indicates that you can sample from this model where the time-varying covariate is continuous. In your answer @Benjamin Christoffersen, you included the $Z(t)$ covariate in the coxph function as x2 but the estimated parameter for x2 was very biased. Is there a way to include this time-varying covariate and obtain the correct estimate using the coxph function? $\endgroup$
    – J McVittie
    Commented Oct 8, 2023 at 4:44

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