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Math Help - 2nd Markov chain question

  1. #1
    Junior Member
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    2nd Markov chain question

    Consider the markov chain with transition matrix

    P=
    [0 1]
    [1 0]

    Show that P^n does not tend to a limit, but that

    A_n = (I + P + P^2 +...+ P^n)/(n+1)

    does tend to a limit.

    thank you!
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  2. #2
    MHF Contributor matheagle's Avatar
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    P^{2n+1} is the identity matrix.

    While P^{2n} is the orginal matrix.

    So, there is no limit.

    You're bouncing back and forth between these two states with probability one.

    I+P is the matrix of all ones, and so is P^{2}+P^3 and so on .....

    Thus you will have a limit of .5 in all four positions of A_n
    Last edited by matheagle; August 23rd 2009 at 11:21 PM.
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  3. #3
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    You can show that A_n has a limit by showing that all the entries of A_n have a limit.

    The elements of the diagonal of A_n make this sequence:

    (1/1, 1/2, 2/3, 2/4, 3/5, 3/6, ...)

    As you can see, for n odd, they are always 0.5, for n even they are (1+ n/2) / (n+1).

    lim (1 + n/2) / (n+1) = lim 1/(n+1) + lim n / (2*(n+1) = 0 + lim n/2n = 0.5

    The elements outside of the diagonal are also 0.5 for n odd (above 0) and for n even they are (n/2) / (n+1), whose limit is 0.5 as shown above.

    Therefore lim A_n =
    [0.5 0.5]
    [0.5 0.5]
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