why is this Markov Chain aperiodic

Question

I have this Matrix:

$$P=\begin{pmatrix} 0 & 1 \\ 0.3 & 0.7 \end{pmatrix}$$

this markov chain is said to be aperiodic, I dont understand how it comes to it. Period $\delta$ is the gcd of the set of all diagonal elements, right? if $\delta>1$, $P$ is periodic, if $\delta=1$, then aperiodic.

but here it is not $\delta=1$, is it? or do i have to transit the matrix to some certain form?

Answer 1

Since my comment provided sufficient clarification:

When there's a stationary state, your system will evolve towards that state. In your case, the two left eigenvectors are $(−1,1)$ and $(3,10)$ with corresponding eigenvalues $−0.3$ and $1$. Every other state of the system can be decomposed into those two states. The first state exihibits oscillating behaviour, but it dies out as $0.3<1$. The other state is stationary. So whatever your initial state, you'll evolve towards that stationary state.

Answer 2

Call the states $a$ and $b$. The chain is irreducible since $P(a,b)=1\ne0$ and $P(b,a)=0.3\ne0$. One can go from $b$ to $b$ with positive probability in one step (probability $0.7$) hence the period of $b$ is $1$. The chain is irreducible hence the period of every state is $1$ (this can also be proved directly). That is, the chain is aperiodic.

Answer 3

Intutive answer;

the markov chain is aperiodic since it is reached from any node to itself in not periodic interval e.g 2, 4, 6,......

here the P11 makes it as a aperiodic since the chain re goes to node after 2,3,5 e.g no periodicity