Why is this Markov chain's stationary distribution not (1/2, 0, 0, 1/2)?

Question

I have the Markov chain

 1,  0,  0,  0
1/2, 0, 1/2, 0
 0, 1/2, 0, 1/2
 0,  0,  0,  1

I understand how to build the system with which I am supposed to find that stationary distribution. But my coursebook says that the answer is in fact (p, 0, 0, 1-p), without p being mentioned before and, intuitively, I believe that the answer should be (0.5, 0, 0, 0.5). Where are they getting this answer from?

Answer 1

Let $x = [p;\ 0;\ 0;\ 1 - p]$ be a row vector. It is not hard to verify that for any $0 \leq p \leq 1$, $$ x = xP $$ where $P$ is the transition probability matrix.

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Note that your Markov chain is not recurrent (or persistent). There is no way out from the first state and the fourth state. Therefore, the chain does not possess a unique steady distribution.