In the Wikipedia page, you have the following definition for period:

$k = \gcd\{ n > 0: \Pr(X_n = i \mid X_0 = i) > 0\}$

And, for a Markov Chain to be aperiodic, $k = 1$.

It is stated that this happens only and if and only if there exists $n$ such that for all $n' ≥ n$:

$\Pr(X_{n'} = i \mid X_0 = i) > 0$

Is there proof of this last statement anywhere? I can't find it and I don't feel this is intuitive or obvious nor I can prove it.

For example, if $n > 0: \Pr(X_n = i \mid X_0 = i) > 0$ was true for two primes, like 3 and 5, k would already be 1, would this be considered aperiodic too?


1 Answer 1


For those that come looking for this, I found the answer. Both definitions are equivalent, as once you have two co-primes, you will be able to find all natural numbers after a certain N.

This is the coin problem and the solution is the Frobenius number:


Here is also a proof I found:



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