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I have a CTMC with six states $\{0,1,\ldots,5\}$. It turns out that states 3 and 4 are equivalent and so are states 1, 2 and 5. I would love to clump equivalent states into one.

$$Q_1=\matrix {& 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & -\lambda & \lambda & 0 & 0 & 0 & 0\cr 1 & 0 & -(\mu+\lambda) & \lambda & 0 & \mu & 0\cr 2 & 0 & 0 & -2\mu & \mu & 0 & \mu \cr 3 & 0 & 0 & 0 & 0 & 0 & 0 \cr 4 & 0 & 0 & 0 & 0 & 0 & 0 \cr 5 & 0 & 0 & 0 & \mu & 0 & -\mu \cr }$$

In that case, is it alright to use the alternative reduced CTMC, with rate matrix $Q_2$ as follows?

$$Q_2=\matrix {& 0 & 1, 2, 5 & 3,4 \cr 0& -\lambda & \lambda & 0 \cr 1, 2, 5 & 0 & -\mu & \mu \cr 3,4 & 0 & 0 & 0 }$$

The intuition is that in each of the three states $\{1,2,5\}$ the out-rate to states $\{3,4\}$ is the same and is equal to $\mu$. However, I am not sure about one thing: in a CTMC, the time one stays in a state $i$ is given by $\exp(q_{ii})$. I could not find a way to find the new rate of staying in the same state.

In general, when is it legal to collapse the state space of a CTMC? Are there conditions imposed on the values in the rate matrix?

Bravo
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  • "It turns out that states 3 and 4 are equivalent and so are states 1, 2 and 5." Certainly not, at least in the usual sense. For example, the probability to visit 4 is 0 when starting from 3 and 1 when starting from 4. And so on. You might want to explain in which sense these states would be "equivalent". – Did Nov 24 '13 at 10:36
  • Hi @did, I meant the equivalence was exogenous, i.e., not pertaining to the existing rates or the CTMC. The states are similar in my model, so I simply want to clump these states together and form a new CTMC (if that could be done). – Bravo Nov 24 '13 at 10:38
  • I see. Please refer to http://math.stackexchange.com/a/27538/6179 or http://math.stackexchange.com/a/115976/6179. The relevant key-word is "lumpability". – Did Nov 24 '13 at 10:53

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Please refer to this or to this. The relevant key-word is lumpability.

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Did
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