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Thread: Markov Chain

  1. #1
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    Markov Chain

    A question which I shouldn't really be finding to be as difficult as I am.


    I suspect my answer to part i to be correct. Part ii confuses me as the chain we've been given isn't Ergodic however I'm guessing the question is just a general one and not aimed at the chain? Still, I'm not quite sure what to say. And for part iii, I don't know what method to use.

    Thanks.
    Last edited by ZTM1989; July 28th 2009 at 04:45 PM.
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  2. #2
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    You can't get to state 4 from any state (let alone from every other state). So the chain is obviously not ergodic. I don't know what they're talking about.

    For part (iii) I would set up the transition matrix and then set up the steady-state equations.

    Usually setting up the transition matrix is the hardest part. In this case it is pretty simple because the transition probabilites are given to you.


    \left(\begin{array}{ccccc}0&0.5&0.5&0&0\\0.7&0&0.3  &0&0\\1&0&0&0&0\\0.6&0&0&0&0.4\\0&0.4&0.6&0&0\end{  array}\right)


    Let  \pi_{i} be the long-run proportion of time that X spends in state i

     \pi_{1}=0.7\pi_{2}+\pi_{3}+0.6\pi_{4}
     \pi_{2}=0.5\pi_{1}+0.4\pi_{5}
     \pi_{3}=0.5\pi_{1}+0.3\pi_{2}+0.6\pi_{5}
     \pi_{4}=0
     \pi_{5}=0.4\pi_{4}

    \pi_{1}+\pi_{2}+\pi_{3}+\pi_{4}+\pi_{5}=1

    now solve for  \pi_{1} (which shouldn't be too bad because \pi_{4}=\pi_{5}=0 )
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  3. #3
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    Thanks a lot, you've been of great help

    PS. Would obviously still encourage some more opinions on part ii.
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  4. #4
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    Sorry made an error on the image, the communicating classes are {1,2,3},{4},{5}, right?

    Image Fixed.
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by ZTM1989 View Post
    Sorry made an error on the image, the communicating classes are {1,2,3},{4},{5}, right?

    Image Fixed.
    I always get confused by classes, but it seems right to me.

    You can go from state 1 to state 2 to state 3 and then back to state 1. So {1,2,3} is a communicating class. And it's recurrent because the process will keep re-entering each state over and over again.

    State 4 doesn't communicate with any other state. So it's own class. And it's transient because there is a postive probability that the process. starting in state 4, will never enter state 4 again. (The probability is, of course, 1).

    Same argument for state 5.
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