Simulating time to event data with a lagged treatment effect

General
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1

Hi all,

My post is more of a preamble to before I need to use Stan. I’m currently trying to simulate some time to event data from a Weibull distribution. The trick is that I want to simulate the data when the treatment has some proposed effect on the hazard after a lag time L. I could really use a second pair of eyes to tell me if this makes any sense at all.

Some brief theory when survival time has a Weibull distribution

Suppose that under the placebo condition, survival time follows a Weibull distribution T_0 \sim \textrm{Weibull}(k, \, \lambda) where k>0 and \lambda > 0 are the shape and scale parameters respectively. Let t be the time from randomisation.

Let h_0(t) be the hazard function under the placebo condition. Then for t \geq 0,

\begin{align*} h_0(t) = \frac{k}{\lambda} \left(\frac{t}{\lambda} \right)^{k-1} \end{align*}

Let S_0(t) be the survival function under the placebo condition. Then for t \geq 0,

\begin{align*} S_0(t) = \exp\left[- \int_{0}^{t} h_0 (u) du \right] = \exp\left[{-\left(\frac{t}{\lambda} \right)^{k}}\right] \end{align*}

Let H_0(t) be the cumulative hazard function under the placebo condition. Then for t \geq 0,

\begin{align*} H_0(t) = \int_{0}^{t} h_0 (u) du = -\ln{[S_0(t)]} \end{align*}

Delayed treatment effect

Suppose that the treatment exerts its effect by scaling the hazard function by \theta < 1 (on the hazard ratio scale) after a lag period L from randomisation.

h_1(t) = \begin{cases} h_0(t), & 0 \leq t < L \\ \theta \, h_0(t), & L \leq t \end{cases}

The survival function under treatment S_1(t) is then:

\begin{align*} S_1(t) &= \exp \left[-\int_{0}^{t} h_1(u) du \right] \\ &= \begin{cases} S_0(t), & 0 \leq t < L \\ \exp{\left[-\left(H_0(L) + \int_{L}^{t} h_1(u)du \right) \right]}, & L \leq t \end{cases} \\ &= \begin{cases} S_0(t), & 0 \leq t < L \\ \exp{\left[-\left(H_0(L) + \theta H_0(t) - \theta H_0(L) \right) \right]}, & L \leq t \end{cases} \\ &= \begin{cases} S_0(t), & 0 \leq t < L \\ \exp{\left[-\left( \theta H_0(t) + (1-\theta)H_0(L) \right) \right]}, & L \leq t \end{cases} \\ &= \begin{cases} S_0(t), & 0 \leq t < L \\ \exp{\left[\theta\ln S_0(t) + (1-\theta)\ln S_0(L) \right]}, & L \leq t \end{cases} \\ &= \begin{cases} S_0(t), & 0 \leq t < L \\ \left[S_0(t)\right]^{\theta} \, \left[S_0(L)\right]^{1-\theta}, & L \leq t \end{cases} \end{align*}

Proposed idea

With the closed form equation to S_1(t), I can generate a random survival time by generating a random uniform random variable u from a \textrm{Unif}(0, 1). Solve for the particular value of t^{*} that satisfies:

u = 1 - S_1(t^{*})

i.e. the method of inverse transform sampling.

# u is a random variate from the uniform U(0,1)
# shape is the weibull shape parameter
# scale is the weibull scale parameter
# lag is the lag period after which the hazard will be scaled. Defaults to 0
# hr is the scaling factor on the hazard function. Defaults to 1
rweibull.delayed <- function(u, shape, scale, lag=0, hr=1) {
  
  # u could be a vector
  
  # First check if F^-1 (u) = t < lag
  # If it is, then return this t
  t = qweibull(u, scale=scale, shape=shape, lower.tail = T)
  
  # If more than lag, return a t from the part of the survival function that has been modified by the scaled hazard
  t = ifelse(t > lag, 
             (uniroot(function(x) 1 - (pweibull(x, shape=shape, scale=scale, lower.tail = F)^hr) * (pweibull(lag, shape=shape, scale=scale, lower.tail = F)^(1-hr)) - u,
                    c(0, 1e9), tol = 0.00001))$root,
             t)
  
  return(t)
  
}
1 Like
2

Sorry this didn’t get answered sooner.

This is unlikely to be actually true, so it’s more of a supposition that a Weibull distribution will be a reasonable approximation, which is something that should be tested, not assumed.

Are you sure there’s really a discrete change point like this? If so, the best thing to do is model it. Rather than using a fixed hazard function, let the hazard function change.

This looks like it still has integrals inside of it. Assuming you can solve them, then what you’re proposing makes sense.

3

You could try a shifted lognormal as that contains the lag you want to model explicitly