The unique stationary distribution in Exercise 9 is v =
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The unique stationary distribution in Exercise 9 is v = (0, 1, 0, 0). This is an instance of the following general resul
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The unique stationary distribution in Exercise 9 is v = (0, 1, 0, 0). This is an instance of the following general result: Suppose that a Markov chain has exactly one absorbing state. Suppose further that, for each non-absorbing state k, there is n such that the probability is positive of moving from state k to the absorbing state in n steps. Then the unique stationary distribution has probability 1 in the absorbing state. Prove this result.
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