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For the continuous time markov chain, the rate matrix Q can be defined with diagonal elements $$Q_{ii}=-\sum_{i\neq j}Q_{ij}$$ My question is this. Since the stationary distribution is defined such that $Q\vec{\pi}=0$ and the matrix Q is defined as above, is there any situation in which $\pi$ is not just defined as $\vec 1$?

yankeefan11
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    "Since the stationary distribution is defined such that $Q\vec{\pi}=0$..." Nope. Please reread your notes. – Did May 06 '18 at 17:43
  • So that is what my notes say. The internet says pi Q = 0 defines pi... – yankeefan11 May 06 '18 at 18:05
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    Hmmm... would you disclose the source of your notes then? – Did May 06 '18 at 18:09
  • From a class. No textbooks. But since I have a rate matrix, not a transition probability (exp(Rt)), then the stationary distribution would satisfy piQ=0 and not pi Q = pi, right? – yankeefan11 May 06 '18 at 18:25
  • "But since I have a rate matrix, not a transition probability " - sounds like you need to study the Kolmogorov forward/backward equations. – Math1000 May 18 '18 at 20:50

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