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If $P$ is the transition matrix belonging to a markov chain, then what does it mean that $P^n$ is irreducible for every $n\in\mathbb{N}$?

For $n=1$ it means that all states communicate with each other, i.e. for all states $i,j$ it is $$ \mathbb{P}(\exists m\in\mathbb{N}: X_m=j|X_0=i)>0. $$

But what does it mean for $n\geq 2$?

Edit

Does it maybe mean that for any states $i,j\in E$ it is

$$ \mathbb{P}(\exists m\in\mathbb{N}: X_{n\cdot m}=j|X_0=i)>0? $$

mathfemi
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  • From wikipedia: A matrix is irreducible if it is not similar via a permutation to a block upper triangular matrix (that has more than one block of positive size). – Git Gud Nov 04 '14 at 12:11
  • Sorry but that does not help me at all. – mathfemi Nov 04 '14 at 12:13
  • Means that P is irreducible and aperiodic. – Did Nov 04 '14 at 12:13
  • @Did I have to follow from this that $P$ is irreducible and aperiodic. So cannot use it. – mathfemi Nov 04 '14 at 12:14
  • Did not say to use it, just to prove it. By the way, your question is "what does it mean?" and my comment says what it means. – Did Nov 08 '14 at 10:48

1 Answers1

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It means that all states make up a single subclass. In other words, it means that matrix $P$ is primitive (as also noted in comments by @Did it's aperiodic which I believe is another term for that).

This can easily be seen from the following theorem:

Theorem. (Gantmacher F.R. The Theory of Matrices, 1960, Vol. 2, P. 81, Theorem 9.) If $A \geqslant 0$ is an irreducible matrix and some power $A^q$ of $A$ is reducible, then $A^q$ is completely reducible, i.e., $A^q$ can be represented by means of a permutation in the form $A^q = \operatorname{diag} \{ A_1, \dots, A_d \}$, where $A_1, \dots, A_d$ are irreducible matrices having one and the same maximal characteristic value. Here $d$ is the greatest common divisor of $q$ and $h$, where $h$ is the index of imprimitivity of $A$.

Since $d$ is $1$ for any $q$, $h = 1$. QED.

Wormer
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