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If $Q$ is a generator matrix of a continuous time Markov chain (CTMC), and I need to use this matrix to solve the Kolmogorov forward equation, I would need to start by integrating it. But I haven't got a clue how to do it. Can someone show me please?

I know something like, let's assume $i$ represents the current state of a CTMC. Then, the forward equation basically tells us that we can work out $X(i + 1)$ by doing

$$X(i + 1) = X(i) \cdot (Id_2 + Q)$$

To look at the difference in time, we can subtract $X(i)$ from both sides and get

$$X(i + 1) - X(i) = X(i) Q$$

Thinking of this in terms of functions instead of matrices, we can say that this can be written as

$$ P'(t) = X(i) Q dt$$

But I don't get how you get to this bit and how you can integrate from here.

I would really appreciate any help. Thank you.

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  • $\begingroup$ Doesn't my answer to your previous question also address this one? Although that answer asks the same question for a specific generator $\mathbb{Q}$, it explains the procedure for obtaining $\mathbb{P}$ and even for checking the answer. $\endgroup$ – whuber Jan 8 '13 at 23:47
  • $\begingroup$ @whuber Yeah that's what I was working through when I came up with a load of questions and I didn't know if you would be able to check them to answer them. I need to go to sleep right now, but if I write them up in either the edit on here or on the other question, could you answer them please? $\endgroup$ – Kaish Jan 8 '13 at 23:49
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    $\begingroup$ I suspect your questions might be purely mathematical ones, so give some consideration as to the correct place to post them: here or on the math site. (The main computational issue concerns diagonalizing $\mathbb{Q}$, which is a purely linear algebraic problem.) $\endgroup$ – whuber Jan 8 '13 at 23:53
  • $\begingroup$ @whuber Ok, thank you for your help. Do you know roughly what time you will be on tomorrow? Because you seem to explain the difficulties I have with this module well so I'd like to ask you some (not all :) ) questions if thats ok? $\endgroup$ – Kaish Jan 9 '13 at 0:02

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