I know that whether or not a MC is irreducible does not depend on the initial distribution (it depends only on the transitional matrix). But is the same (independence) true for periodicity?
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The period of a state $i$ in a Markov chain is $\gcd\{n > 0: P(X_n=i \mid X_0=i) > 0\}$, and a state is aperiodic if its period is $1$. You can see that the period depends only on the quantities $$P(X_n=i \mid X_0=i) = \sum_{x_{n-1},\ldots,x_1} P(X_n=i \mid X_{n-1}=x_{n-1}) \cdot P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2}) \cdots P(X_2=x_2 \mid X_1=x_1)\cdot P(X_1=x_1 \mid X_0=i)$$ so it does not depend on the initial distribution.
angryavian
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Shouldn't P(X_0 = i) be multiplied to the RHS? – ajfbiw.s Nov 27 '16 at 01:06
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Hi @angryavain : Can you address the comment above? Basically, shouldn't there be a X0 = i term? Thanks. – ajfbiw.s Nov 28 '16 at 01:36
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@ajfbiw.s The left-hand side is $P(X_n=i \mid X_0=i)$ so you don't need to multiply by $P(X_0=i)$. If it were $P(X_n=i, X_0=i)$, then yes. – angryavian Nov 28 '16 at 05:19
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Thanks! I thought there was a normalization sign before the summation... my bad. – ajfbiw.s Nov 28 '16 at 18:19