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In a Markov chain (you can add additional conditions here, such as discrete-time, homogeneous, finite-state, .... But the less additional condition, the better ), what sufficient and/or necessary condition can make every initial distribution have a limit distribution?

Note that here the limit distributions for different initial distributions may be different. Added: What I was thinking when posting the question is to include the case when there does not exist the limiting distribution same for all initial distributions, but there exists a limit distribution for every initial distribution.

Thanks and regards!

My question comes from my comment to Michael Hardy's reply.

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Tim
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  • There must not be eigenvalues $\lambda$ with $|\lambda|\ge 1$ and $\lambda\ne1$. – Hagen von Eitzen Dec 10 '12 at 16:35
  • @HagenvonEitzen: Thanks! do you mean all eigenvalues of transition matrix have absolute value strictly less than 1? Why is that? – Tim Dec 10 '12 at 16:37
  • @HagenvonEitzen I think that the eigenvalues of a stochastic matrix can not exceed one by the Perron-Frobenius theorem http://en.wikipedia.org/wiki/Stochastic_matrix – Learner Dec 10 '12 at 16:44
  • @Learner OK, so we can simply say: There must not be eigenvalues $\lambda$ with $|\lambda|=1$ and $\lambda\ne 1$. – Hagen von Eitzen Dec 10 '12 at 16:48
  • @Tim Yes, e.g. Michael Hardy's example is a case where an eigenvalue $-1$ occurs. – Hagen von Eitzen Dec 10 '12 at 16:50
  • @HagenvonEitzen: can you explain why? – Tim Dec 10 '12 at 18:12
  • @Tim Consider the Jordan normal form. Block belonging to $\lambda=1$ correspond to limit distributions, parts belonging to blocks with $|\lambda|<1$ will simply §fade away" in the long rund, but vectors corrsponidng to other blocks with $|\lambda|=1$ keep oscillating. – Hagen von Eitzen Dec 10 '12 at 20:07

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I think that for a finite-state discrete-time Markov chain, sufficient conditions are that the chain be irreducible aperiodic and positive recurrent. In that case, it will be ergodic and will possess a unique limiting distribution.

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  • +1 Thanks! I did know that. What I was thinking when posting the question is to include the case when there does not exist the limiting distribution same for all initial distributions, but there exist a limit distribution for every initial distribution. – Tim Dec 10 '12 at 16:39