3
$\begingroup$
  1. Let $X=(X_{t},\,t \in T)$ be a non-homogeneous, continuous time Markov process with a finite state space $S=\{1,...,K\}$.

  2. Let $\alpha_{i,j}(t)$ be the hazard rates of some $\varGamma$-distributed random variables.

My question: How can I simulate random paths of the Markov process with a transition intensity matrix which is built with the above hazard rates $\alpha_{i,j}(t)$?

For example: $S=\{1,2,3,4\}$ with given $\alpha_{i,j}(t)$ with $i,j \in S$.

Any idea is greatly appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

Suppose you're in state $i$. For each $j$, let $X_j$ be an independent Exponential random variable with mean 1.

Now solve $\int_{0}^{T_j}\alpha_{i,j}(t)\ dt=X_j$ for each $i$. Whichever of the $T_j$'s is smallest, you jump from state $i$ to state $j$ at time $T_j$.

$\endgroup$
4
  • $\begingroup$ Thank you very much for your answer. Do you have a reference (a book or paper) for your answer. I will now apply this procedure in MATHEMATICA. Again, thank you! $\endgroup$ May 31, 2013 at 5:44
  • $\begingroup$ No sorry I don't. Something like this has been in the back of my mind for a while. You can check it though: compute the probability that there's a transition from $i$ to $j$ in the time interval $(t,t+dt)$ conditioned that there has been no transition up to time t. $\endgroup$ May 31, 2013 at 6:04
  • $\begingroup$ I tested it with constant transition intensities alpha and it works fine. For general alpha's as described above, it seems that it does not work correctly. Maybe I make a mistake. However, I will now apply your method to simulate a 3-state model (illness-death model). In this case, the analytical solution is known and will compare it with numerical results. $\endgroup$ May 31, 2013 at 8:19
  • $\begingroup$ It works pretty fine with the reduced model. I generalized the model and solved the corresponding Chapman-Kolmogorov equation system and everything works pretty fine. Thank you VERY much, with best wishes Wolfgang $\endgroup$ May 31, 2013 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.