0

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.

Kaish
  • 867
  • 2
  • 11
  • 19
  • Doesn't my answer to [your previous question](http://stats.stackexchange.com/questions/46389/solving-the-kolmorgorov-forward-equation-for-transition-probabilities) 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. – whuber Jan 08 '13 at 23:47
  • @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? – Kaish Jan 08 '13 at 23:49
  • 1
    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.) – whuber Jan 08 '13 at 23:53
  • @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? – Kaish Jan 09 '13 at 00:02

0 Answers0