Find the period of a state in a Markov chain

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Let $X_n:n=0,1,2,ldots$ be a Markov chain with transition probabilities as given below:




enter image description here




Determine the period of each state.




The answer is "The only state with period $> 1$ is $1$, which has period $3$. I don't understand why other states like $2$, $3$, $5$, $6$ are not with period $3$, they can also take $3$ steps back to themselves, can't them?




stochastic-processes markov-chains




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    You're right. All states except state $4$ takes $3n$ steps back to itself.
    – GNU Supporter
    Dec 13 '15 at 12:17














up vote
1
down vote

favorite













Let $X_n:n=0,1,2,ldots$ be a Markov chain with transition probabilities as given below:




enter image description here




Determine the period of each state.




The answer is "The only state with period $> 1$ is $1$, which has period $3$. I don't understand why other states like $2$, $3$, $5$, $6$ are not with period $3$, they can also take $3$ steps back to themselves, can't them?




stochastic-processes markov-chains




share|cite|improve this question


















  • 2




    You're right. All states except state $4$ takes $3n$ steps back to itself.
    – GNU Supporter
    Dec 13 '15 at 12:17












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Let $X_n:n=0,1,2,ldots$ be a Markov chain with transition probabilities as given below:




enter image description here




Determine the period of each state.




The answer is "The only state with period $> 1$ is $1$, which has period $3$. I don't understand why other states like $2$, $3$, $5$, $6$ are not with period $3$, they can also take $3$ steps back to themselves, can't them?




stochastic-processes markov-chains




share|cite|improve this question















Let $X_n:n=0,1,2,ldots$ be a Markov chain with transition probabilities as given below:




enter image description here




Determine the period of each state.




The answer is "The only state with period $> 1$ is $1$, which has period $3$. I don't understand why other states like $2$, $3$, $5$, $6$ are not with period $3$, they can also take $3$ steps back to themselves, can't them?





stochastic-processes markov-chains





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edited Dec 13 '15 at 12:11









Math1000

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asked Dec 13 '15 at 11:08









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  • 2




    You're right. All states except state $4$ takes $3n$ steps back to itself.
    – GNU Supporter
    Dec 13 '15 at 12:17












  • 2




    You're right. All states except state $4$ takes $3n$ steps back to itself.
    – GNU Supporter
    Dec 13 '15 at 12:17







2




2




You're right. All states except state $4$ takes $3n$ steps back to itself.
– GNU Supporter
Dec 13 '15 at 12:17




You're right. All states except state $4$ takes $3n$ steps back to itself.
– GNU Supporter
Dec 13 '15 at 12:17










1 Answer
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The period of a state $i$ is
$$d(i) = mathrmlcdn : P_ii^n > 0 . $$
If two states $i$ and $j$ communicate, that is, there exist $m,n>0$ such that $P_ij^n>0$ and $P_ji^m>0$, then $d(i)=d(j)$. By inspection, states $1, 2, 3, 5, 6$ all communicate. So you are correct.






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    Should lcd be gcd?
    – Mick A
    Dec 13 '15 at 12:39










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The period of a state $i$ is
$$d(i) = mathrmlcdn : P_ii^n > 0 . $$
If two states $i$ and $j$ communicate, that is, there exist $m,n>0$ such that $P_ij^n>0$ and $P_ji^m>0$, then $d(i)=d(j)$. By inspection, states $1, 2, 3, 5, 6$ all communicate. So you are correct.






share|cite|improve this answer
















  • 1




    Should lcd be gcd?
    – Mick A
    Dec 13 '15 at 12:39














up vote
0
down vote













The period of a state $i$ is
$$d(i) = mathrmlcdn : P_ii^n > 0 . $$
If two states $i$ and $j$ communicate, that is, there exist $m,n>0$ such that $P_ij^n>0$ and $P_ji^m>0$, then $d(i)=d(j)$. By inspection, states $1, 2, 3, 5, 6$ all communicate. So you are correct.






share|cite|improve this answer
















  • 1




    Should lcd be gcd?
    – Mick A
    Dec 13 '15 at 12:39












up vote
0
down vote










up vote
0
down vote









The period of a state $i$ is
$$d(i) = mathrmlcdn : P_ii^n > 0 . $$
If two states $i$ and $j$ communicate, that is, there exist $m,n>0$ such that $P_ij^n>0$ and $P_ji^m>0$, then $d(i)=d(j)$. By inspection, states $1, 2, 3, 5, 6$ all communicate. So you are correct.






share|cite|improve this answer












The period of a state $i$ is
$$d(i) = mathrmlcdn : P_ii^n > 0 . $$
If two states $i$ and $j$ communicate, that is, there exist $m,n>0$ such that $P_ij^n>0$ and $P_ji^m>0$, then $d(i)=d(j)$. By inspection, states $1, 2, 3, 5, 6$ all communicate. So you are correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '15 at 12:28









Math1000

18.5k31644




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  • 1




    Should lcd be gcd?
    – Mick A
    Dec 13 '15 at 12:39












  • 1




    Should lcd be gcd?
    – Mick A
    Dec 13 '15 at 12:39







1




1




Should lcd be gcd?
– Mick A
Dec 13 '15 at 12:39




Should lcd be gcd?
– Mick A
Dec 13 '15 at 12:39

















 

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