I want to find discrete-time Markov process (its stochastic transition matrix) equivalent to a given contiouse time Markov process.
I googled it, what I found is that if we denote $q_{ij}$ as the transition rate in the Markov process to go from state $i$ to state $j$, then the transition probability to transit from $i$ to $j$ in the discrete version would be $p_{ij} = \frac{q_{ij}}{v_i}$, where $v_i = \sum_{k}{q_{i k}}$.
Okay, based on this, we will get a stochastic transition matrix in which the diagonal elements are zero. But the point is base don this transition matrix in each time step we will see the transition occurs, while in general transition matrix, the diagonal elements are not necessarily zero which means this is probable that transition does not occur is some time steps.
Even if we compare this embedded discrete time Markov with the continuous one, in contentious one we have holdings times which have an exponential distribution, resulting in that the total number of transition in a fix time duration is not equal. So I do not think embedded discrete-time Markov chain obtained from this recipe is equivalent to the original continuous time. Am I right?
Could you please let me know what is the standard way to go from a continuous time process to its equivalent discrete-time Markov chain?
Thank you so much in advance!