0

let $P$ be a transition matrix of a Markov chain with state space E, that is finite.

Does from $P^n$ irreducible for all $n\in\mathbb{N}$ follow that $P$ is irreducible and aperiodic?

The first thing is clear. If $P^n$ is irreducible for all $n\in\mathbb{N}$, then especially for $n=1$, so $P$ is irreducible.

But the aperiodicity is not clear to me. Do not see that and have no idea how to show it.

2 Answers2

0

A matrix $P$ is called primitive, if there is a natural number $m$, such that $P^m$ has only positive entries.

If a matrix is primitive, it is also irreducible and aperiodic.

So you only have to show, that a number $m$ with a positive matrix $P^m$ exists to complete the proof.

Peter
  • 84,454
  • Ups, to be honest I do not understand what do you want to tell me. –  Nov 04 '14 at 20:04
  • In case of markov chains this should be the case, if no probabilities $0$ appear. – Peter Nov 04 '14 at 20:04
  • Definition : A matrix $P$ is PRIMITIVE, if there is a natural number $m$, such that $P^m$ has only positive entries. Theorem : A matrix is PRIMITIVE if and only if it is irreducible and aperiodic. – Peter Nov 04 '14 at 20:06
  • 1
    I know the argument that shows that an irreducible and aperiodic matrix is primitive. But why is a primitive matrix irreducible and aperiodic? Is it because if $P^n>0$ then $P^k>0$ for every $k>n$ and so $p_{ii}^{(n)}>0, p_{ii}^{(n+1)}>0, p_{ii}^{(n+2)}>0$ etc so that $\left{n,n+1,n+2,...\right}\subseteq \left{n>0: p_{ii}^{(n)}>0\right}$ which means that the gcd only can be 1? –  Nov 04 '14 at 20:57
  • The period is equal to the number of eigenvalues with absolute value equal to the spectral radius. If the matrix is primitive, the perron-frobenius-theorem holds implying that there is only one eigenvalue with absolute value equal to the spectral radius, so the period must be $1$. – Peter Nov 04 '14 at 21:10
  • my argument is right, too? –  Nov 04 '14 at 21:20
  • I am not sure, whether the matrix must remain positive, when it is positive at some time. – Peter Nov 04 '14 at 21:24
  • But if the limit of the markov chain is a positive matrix, the argument clearly holds for a sufficient large $n$. – Peter Nov 04 '14 at 21:25
  • Why does it hold for large n? –  Nov 04 '14 at 21:37
  • Back to the original question: Which m does fullfill $ P^m> 0$? I did not find it. –  Nov 04 '14 at 22:49
0

Please look my answer to this question. In a nutshell, the following theorem should help you.

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 $A$.

Community
  • 1
Wormer
  • 266