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Let $\{X_n:n=0,1,2,\ldots\}$ be a Markov chain with transition probabilities as given below:

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Determine the period of each state.

The answer is "The only state with period $> 1$ is $1$, which has period $3$. I don't understand why other states like $2$, $3$, $5$, $6$ are not with period $3$, they can also take $3$ steps back to themselves, can't them?

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The period of a state $i$ is $$d(i) = \mathrm{lcd}\{n : P_{ii}^n > 0 \}. $$ If two states $i$ and $j$ communicate, that is, there exist $m,n>0$ such that $P_{ij}^n>0$ and $P_{ji}^m>0$, then $d(i)=d(j)$. By inspection, states $1, 2, 3, 5, 6$ all communicate. So you are correct.

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    $\begingroup$ Should lcd be gcd? $\endgroup$ – Mick A Dec 13 '15 at 12:39

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