I need to simulate data from a cox model with specified values of the parameters: $$\lambda_0(t)exp(X \beta)$$

For reasons I don't want to get into, the scale of my X's is very large. They range between 0 and 100. When I apply the algorithm of Bender et. all (2005) explained very well in this post on Cross Validated, I cannot get reasonable parameter estimates. I believe it is because the denominator of the expression,

$$T = S^{-1}(V \,|\, \mathbf{x}) = H_0^{-1} \left( - \frac{\log(V)}{\exp(\mathbf{x}^\prime \mathbf{\beta})} \right)$$

becomes very large and forces $- \frac{\log(V)}{\exp(\mathbf{x}^\prime \mathbf{\beta})}$ to 0.

Here's the example code, shamelessly ripped from user ocram:

# baseline hazard: Weibull
# N = sample size    
# lambda = scale parameter in h0()
# rho = shape parameter in h0()
# beta = fixed effect parameter

    simulWeib <- function(N, lambda, rho, beta){
  # covariate --> N Bernoulli trials
  x <- runif(N, 0, 1)# , size = 1, prob = .5)

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

  # data set

Along with the subsequent test to see how well beta is estimated:

coefs = rep(0, times = 1e3)
for(k in 1:1e3){
  dat <- simulWeib(N=100, lambda=0.01, rho=1, 
               beta = 2)
  fit <- coxph(Surv(time) ~ x, data=dat)
  coefs[k] <- fit$coef

I get an estimate of around 0.015. I can get very precise estimate when the scale of $X^{\prime} \beta$ is smaller.

So is there any way to simulate from a cox PH model with a large X, or equivalently large $\beta$?


1 Answer 1


You could do the internal calculations on the log scale, then exponentiate the result:

$$H^{-1}_0(exp( log(-log(v)) - x'\beta ))$$

That should stabilize it somewhat.

The other option is to only use very small values of $\beta$ with your $x$ variable. In realistic situations $\beta$ would tend to be small in absolute value if $x$ is on a broad scale like that. Is it realistic that 2 subjects could really have a hazard ratio of $e^{200}$?

  • $\begingroup$ Thank you for the sanity check. I mixed up the scale of my coefficients. You're right that there's no reason for a hazard ratio of that size. $\endgroup$
    – Eli
    Feb 26, 2018 at 23:39

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