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Suppose there is a Markov chain with K states {1,2,...,K} arranged clockwise on a circle. $Z_0$ = 1 The probability of moving clockwise is $p$ and the probability of moving counterclockwise is $1-p$. Then what's the probability that the Markov chain complete a clockwise circle before a counterclockwise one? In addition, I'm also interested in the long run average number of loops completed, i.e., $\lim_{n\to\infty} L(n)/n = ?$, where $L(n)$ is the number of loops completed by time n.

I don't know how I should approach this question. Any help is appreciated!

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  • $\begingroup$ When you talk about "number of loops" do you mean to keep track of the loop direction, so $L(n)$ is a winding number? $\endgroup$ – kimchi lover Jun 8 at 22:00
  • $\begingroup$ No just the number of loops including clockwise and counterclockwise $\endgroup$ – Chloe Jun 9 at 3:09
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The first question is equivalent to a random walk on $\{-K,-(K-1),\dots,0,\dots,K-1,K\}$ starting at $0$ and taking a step to the right with probability $p$ or to the left with probability $1-p$, and asking the probability that the walk reaches $K$ before $-K$; presumably this is a standard problem that has been answered before.

For the second question, the expected number of (signed) clockwise steps per turn is $p-(1-p) = 1-2p$, and so the expected number of (signed) clockwise loops in $n$ steps is $n(1-2p)/K$. It becomes more difficult if you care about unsigned loops, "starting over" every time a loop is completed.

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  • $\begingroup$ $p-(1-p) = 2p-1$. $\endgroup$ – Fabio Somenzi Jun 8 at 23:46

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