MHB Markov Chain - Is state 2 periodic?

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The Markov chain defined by the given transition matrix is irreducible, as all states communicate with each other. State 2 is non-periodic because the probability of returning to it after any number of steps is greater than zero. Consequently, all states in the chain are also non-periodic. Additionally, since all states are aperiodic, the chain is considered ergodic. The discussion confirms the properties of the Markov chain regarding its irreducibility and ergodicity.
mathmari
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Hey! :o

Given the Markov chain $\{X_n, n \geq 1\}$ and the following probability transition matrix:
$\begin{pmatrix}
0 & 1/3 & 2/3\\
1/4 & 3/4 & 0\\
2/5 & 0 & 3/5
\end{pmatrix}$

All states communicate, so the chain is irreducible, isn't?

Could you tell me if the state $2$ is periodic?
 
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mathmari said:
Hey! :o

Given the Markov chain $\{X_n, n \geq 1\}$ and the following probability transition matrix:
$\begin{pmatrix}
0 & 1/3 & 2/3\\
1/4 & 3/4 & 0\\
2/5 & 0 & 3/5
\end{pmatrix}$

All states communicate, so the chain is irreducible, isn't?

Could you tell me if the state $2$ is periodic?

Yes, the chain is irreducible!... the fact that $P_{2,2} \ne 0$ makes possible the return in the state 2 after any number of steps so that the state 2 is non periodic. In fact none of the states of the TM is periodic... Kind regards $\chi$ $\sigma$
 
chisigma said:
Yes, the chain is irreducible!... the fact that $P_{2,2} \ne 0$ makes possible the return in the state 2 after any number of steps so that the state 2 is non periodic. In fact none of the states of the TM is periodic... Kind regards $\chi$ $\sigma$

Ok! And is the chain ergodic?
 
mathmari said:
Ok! And is the chain ergodic?

The MC has all aperiodic states so that it is ergodic...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
The MC has all aperiodic states so that it is ergodic...

Kind regards

$\chi$ $\sigma$

Ok! Thank you for your answer! :o
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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