# How to create a toy survival (time to event) data with right censoring

I wish to create a toy survival (time to event) data which is right censored and follows some distribution with proportional hazards and constant baseline hazard.

I created the data as follows, but I am unable to obtain estimated hazard ratios that are close to the true values after fitting a Cox proportional hazards model to the simulated data.

What did I do wrong?

R codes:

library(survival)

#set parameters
set.seed(1234)

n = 40000 #sample size

#functional relationship

lambda=0.000020 #constant baseline hazard 2 per 100000 per 1 unit time

b_haz <-function(t) #baseline hazard
{
lambda #constant hazard wrt time
}

x = cbind(hba1c=rnorm(n,2,.5)-2,age=rnorm(n,40,5)-40,duration=rnorm(n,10,2)-10)

B = c(1.1,1.2,1.3) # hazard ratios (model coefficients)

hist(x %*% B) #distribution of scores

haz <-function(t) #hazard function
{
b_haz(t) * exp(x %*% B)
}

c_hf <-function(t) #cumulative hazards function
{
exp(x %*% B) * lambda * t
}

S <- function(t) #survival function
{
exp(-c_hf(t))
}

S(.005)
S(1)
S(5)

#simulate censoring

time = rnorm(n,10,2)

S_prob = S(time)

#simulate events

event = ifelse(runif(1)>S_prob,1,0)

#model fit

km = survfit(Surv(time,event)~1,data=data.frame(x))

plot(km) #kaplan-meier plot

#Cox PH model

fit = coxph(Surv(time,event)~ hba1c+age+duration, data=data.frame(x))

summary(fit)

cox.zph(fit)


Results:

Call:
coxph(formula = Surv(time, event) ~ hba1c + age + duration, data = data.frame(x))

n= 40000, number of events= 3043

coef exp(coef) se(coef)     z Pr(>|z|)
hba1c    0.236479  1.266780 0.035612  6.64 3.13e-11 ***
age      0.351304  1.420919 0.003792 92.63  < 2e-16 ***
duration 0.356629  1.428506 0.008952 39.84  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

exp(coef) exp(-coef) lower .95 upper .95
hba1c        1.267     0.7894     1.181     1.358
age          1.421     0.7038     1.410     1.432
duration     1.429     0.7000     1.404     1.454

Concordance= 0.964  (se = 0.006 )
Rsquare= 0.239   (max possible= 0.767 )
Likelihood ratio test= 10926  on 3 df,   p=0
Wald test            = 10568  on 3 df,   p=0
Score (logrank) test = 11041  on 3 df,   p=0


but true values are set as

B = c(1.1,1.2,1.3) # hazard ratios (model coefficients)


It is not clear to me how you generate your event times (which, in your case, might be $<0$) and event indicators:

time = rnorm(n,10,2)
S_prob = S(time)
event = ifelse(runif(1)>S_prob,1,0)


So here is a generic method, followed by some R code.

Generating survival times to simulate Cox proportional hazards models

To generate event times from the proportional hazards model, we can use the inverse probability method (Bender et al., 2005): if $V$ is uniform on $(0, 1)$ and if $S(\cdot \,|\, \mathbf{x})$ is the conditional survival function derived from the proportional hazards model, i.e. $$S(t \,|\, \mathbf{x}) = \exp \left( -H_0(t) \exp(\mathbf{x}^\prime \mathbf{\beta}) \vphantom{\Big(} \right)$$ then it is a fact that the random variable $$T = S^{-1}(V \,|\, \mathbf{x}) = H_0^{-1} \left( - \frac{\log(V)}{\exp(\mathbf{x}^\prime \mathbf{\beta})} \right)$$ has survival function $S(\cdot \,|\, \mathbf{x})$. This result is known as the inverse probability integral transformation''. Therefore, to generate a survival time $T \sim S(\cdot \,|\, \mathbf{x})$ given the covariate vector, it suffices to draw $v$ from $V \sim \mathrm{U}(0, 1)$ and to make the inverse transformation $t = S^{-1}(v \,|\, \mathbf{x})$.

Example [Weibull baseline hazard]

Let $h_0(t) = \lambda \rho t^{\rho - 1}$ with shape $\rho > 0$ and scale $\lambda > 0$. Then $H_0(t) = \lambda t^\rho$ and $H^{-1}_0(t) = (\frac{t}{\lambda})^{\frac{1}{\rho}}$. Following the inverse probability method, a realisation of $T \sim S(\cdot \,|\, \mathbf{x})$ is obtained by computing $$t = \left( - \frac{\log(v)}{\lambda \exp(\mathbf{x}^\prime \mathbf{\beta})} \right)^{\frac{1}{\rho}}$$ with $v$ a uniform variate on $(0, 1)$. Using results on transformations of random variables, one may notice that $T$ has a conditional Weibull distribution (given $\mathbf{x}$) with shape $\rho$ and scale $\lambda \exp(\mathbf{x}^\prime \mathbf{\beta})$.

R code

The following R function generates a data set with a single binary covariate $x$ (e.g. a treatment indicator). The baseline hazard has a Weibull form. Censoring times are randomly drawn from an exponential distribution.

# baseline hazard: Weibull

# N = sample size
# lambda = scale parameter in h0()
# rho = shape parameter in h0()
# beta = fixed effect parameter
# rateC = rate parameter of the exponential distribution of C

simulWeib <- function(N, lambda, rho, beta, rateC)
{
# covariate --> N Bernoulli trials
x <- sample(x=c(0, 1), size=N, replace=TRUE, prob=c(0.5, 0.5))

# Weibull latent event times
v <- runif(n=N)
Tlat <- (- log(v) / (lambda * exp(x * beta)))^(1 / rho)

# censoring times
C <- rexp(n=N, rate=rateC)

# follow-up times and event indicators
time <- pmin(Tlat, C)
status <- as.numeric(Tlat <= C)

# data set
data.frame(id=1:N,
time=time,
status=status,
x=x)
}


Test

Here is some quick simulation with $\beta = -0.6$:

set.seed(1234)
betaHat <- rep(NA, 1e3)
for(k in 1:1e3)
{
dat <- simulWeib(N=100, lambda=0.01, rho=1, beta=-0.6, rateC=0.001)
fit <- coxph(Surv(time, status) ~ x, data=dat)
betaHat[k] <- fit$coef } > mean(betaHat) [1] -0.6085473  • Thank you for your excellent answer. I realized I had messed up the event times by getting events status after I randomized event times, which didn't make sense.. silly me! – stats_newb Jan 27 '15 at 9:12 • May I ask is there any specific reason why you draw censoring time from an exponential distribution? – pthao May 16 '16 at 5:06 • @pthao: there is no particular reason (this was just an illustration where I used the exponential distribution) – ocram May 16 '16 at 5:43 • Is there any guideline for choosing the distribution for censoring times? – pthao May 16 '16 at 6:51 • @ocram Interestingly, when I run flexsurvreg(Surv(time, status) ~ x, data=dat, dist = "weibull") on the same simulated data, the coefficient appears as 0.6212. Why is this? – neither-nor Jun 4 '19 at 23:50 For Weibull distribution, S(t) =$e^{-(\lambda * e^(x * \beta)*t)^\rho}$"$^{(1/rho)}\$" will be only for log(v)

so, I modified like this

Tlat <- (- log(v))^(1 / rho) / (lambda * exp(x * beta))


if rho = 1, result will be same.