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I need to construct a stationary distribution for $\{X_n, n\in \mathcal{N}\}$ with state space $E=\{0,1,2,3,4,5\}$ and transition matrix: \begin{bmatrix} 0.8 & 0.2 & 0&0&0&0\\ 0&0.1&0.4&0.5&0&0\\ 0&0.3&0.3&0.4&0&0\\ 0&0.6&0.3&0.1&0&0\\ 0&0&0&0&0.7&0.3\\ 0&0&0&0&0.5&0.5 \end{bmatrix}

In the previous steps of this problem I am asked to find the stationary distribution of all positive recurrent classes. For the positive recurrent class $\{1,2,3\}$ I was able to find the stationary distribution to be $1/3$ for each state, and for $\{4,5\}$ to be $5/8$ and $3/8$, respectively. However, I need to put these distributions together to find the stationary distribution for $X_n$ and I am not sure how to do so. This is just a practice problem as I study for my exam. Thank you for any help

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States $\{0,1,2,3\}$ are isolated from $\{4,5\}$, so the stationary distribution isn't unique: any weighted combination of the stationary distributions for the subsystems will be stationary. The probabilities will evolve to a combination of the two, with weights equal to the total probability each subsystem started with. For instance, if the initial probabilities are $(1/3,1/3,0,0,0,1/3),$ the limiting probabilities are $(0,2/9,2/9,2/9,5/24,3/24).$

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  • $\begingroup$ Thank you and you were right about the limiting probabilities of $\{4,5\}$. I corrected my mistake. $\endgroup$
    – EM823823
    Dec 24, 2020 at 0:12
  • $\begingroup$ @EM823823 Yeah, I saw you had corrected it. One more thing: the 3rd row of your matrix doesn't add up to one... should be (.3,.3,.4) rather than (.3,.4,.4) (Or I assume that's what it should be, based on the stationary distribution you found.) $\endgroup$
    – spaceisdarkgreen
    Dec 24, 2020 at 0:25
  • $\begingroup$ It should be (0.3 0.3 0.4), indeed. I made a typo when I typed the matrix. I am sorry about that! Thanks for pointing that out to me. $\endgroup$
    – EM823823
    Dec 24, 2020 at 1:17

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