Find the period of a state in a Markov chain
$begingroup$
Let ${X_n:n=0,1,2,ldots}$ be a Markov chain with transition probabilities as given below:
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
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
$endgroup$
add a comment |
$begingroup$
Let ${X_n:n=0,1,2,ldots}$ be a Markov chain with transition probabilities as given below:
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
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
$endgroup$
3
$begingroup$
You're right. All states except state $4$ takes $3n$ steps back to itself.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 13 '15 at 12:17
add a comment |
$begingroup$
Let ${X_n:n=0,1,2,ldots}$ be a Markov chain with transition probabilities as given below:
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
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
$endgroup$
Let ${X_n:n=0,1,2,ldots}$ be a Markov chain with transition probabilities as given below:
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
stochastic-processes markov-chains
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
edited Dec 13 '15 at 12:11
Math1000
19.4k31746
19.4k31746
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
asked Dec 13 '15 at 11:08
whoisitwhoisit
396112
396112
3
$begingroup$
You're right. All states except state $4$ takes $3n$ steps back to itself.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 13 '15 at 12:17
add a comment |
3
$begingroup$
You're right. All states except state $4$ takes $3n$ steps back to itself.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 13 '15 at 12:17
3
3
$begingroup$
You're right. All states except state $4$ takes $3n$ steps back to itself.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 13 '15 at 12:17
$begingroup$
You're right. All states except state $4$ takes $3n$ steps back to itself.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 13 '15 at 12:17
add a comment |
1 Answer
1
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$begingroup$
The period of a state $i$ is
$$d(i) = mathrm{lcd}{n : P_{ii}^n > 0 }. $$
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.
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
$endgroup$
1
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
add a comment |
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1 Answer
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$begingroup$
The period of a state $i$ is
$$d(i) = mathrm{lcd}{n : P_{ii}^n > 0 }. $$
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.
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
$endgroup$
1
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
add a comment |
$begingroup$
The period of a state $i$ is
$$d(i) = mathrm{lcd}{n : P_{ii}^n > 0 }. $$
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.
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
$endgroup$
1
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
add a comment |
$begingroup$
The period of a state $i$ is
$$d(i) = mathrm{lcd}{n : P_{ii}^n > 0 }. $$
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.
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
$endgroup$
The period of a state $i$ is
$$d(i) = mathrm{lcd}{n : P_{ii}^n > 0 }. $$
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.
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
answered Dec 13 '15 at 12:28
Math1000Math1000
19.4k31746
19.4k31746
1
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
add a comment |
1
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
1
1
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
$begingroup$
Should lcd be gcd?
$endgroup$
– Mick A
Dec 13 '15 at 12:39
add a comment |
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3
$begingroup$
You're right. All states except state $4$ takes $3n$ steps back to itself.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 13 '15 at 12:17