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A truck transports goods among $10$ points located on a circular route. These goods are carried only from one point to the next with probability $p$, or to the preceding point with probability $q=1-p$.

  1. Write the transition probability matrix.
  2. Find the limiting stationary distribution.
  3. Write your conclusions about this question.
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  • $\begingroup$ Hi, welcome to MSE. It is important to show your attempt and ask specific questions. $\endgroup$ – Tab1e May 17 at 0:49
  • $\begingroup$ What are your thoughts on the problem? What have you tried? Have you been able to find the matrix in question? Do you understand all the terminology being used here? $\endgroup$ – Ben Grossmann May 17 at 0:53
  • $\begingroup$ For 2: the fact that the Markov chain is transitive (and irreducible) tells us that the stationary distribution must be uniform. $\endgroup$ – Ben Grossmann May 17 at 0:58
  • $\begingroup$ yes, i tried and i found the matrix. and i understand all the terminology being used here.but i was looking for a simple idea to find the limiting and i have a sense that the $pi=1/10 $\endgroup$ – Ahmad Zaben May 17 at 0:58
  • $\begingroup$ i think that the limiting stationary distribution will be 1/10 but i want to prove that. $\endgroup$ – Ahmad Zaben May 17 at 1:01
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The transition matrix is the circulant matrix $M = q \cdot P + p \cdot P^T$, where $P$ is the permutation matrix in the link. Computing the stationary distribution can be done by computing the solution to the system $(M - I)x = 0$.

However, rather than solving this system of equations, we can more easily prove that your guess of the stationary distribution $\pi = (1/10,\dots,1/10)$ is correct by verifying that $\pi M = M$. To see that this holds, note that $\pi = \frac 1{10} (1,\dots,1)$, and that the entries of $(1,\dots,1)M$ are the column-sums of $M$. The only non-zero entries of a given column of $M$ are $p$ and $q$, which means that every entry of $(1,\dots,1)M$ will be $p+q = 1$, which means that we have $$ (1,\dots,1)M = (1,\dots,1)M \implies \pi M = \pi. $$ So, $\pi$ is indeed the stationary distribution.

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