$psi$-irreducibility of m-skeletons.
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In Proposition 5.4.5 of Meyn and Tweedie's Markov Chains and Stochastic Stability, it is said that if a chain $Phi$ is $psi$-irreducible and aperiodic, then every $m$-skeleton of it is also $psi$-irreducible and aperiodic. The authors did not give an explicit proof of this.
It is easy to show aperiodicity using induction, but I had a hard time to figure out the proof for $psi$-irreducibility. Is induction and contradiction a right direction to go? In general, if there is no aperiodicity of the chain, is it also true that all the $m$-skeletons are $psi$-irreducible?
markov-chains
asked Mar 19 '14 at 5:11
942kid942kid
111
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add a comment |
$begingroup$
In Proposition 5.4.5 of Meyn and Tweedie's Markov Chains and Stochastic Stability, it is said that if a chain $Phi$ is $psi$-irreducible and aperiodic, then every $m$-skeleton of it is also $psi$-irreducible and aperiodic. The authors did not give an explicit proof of this.
It is easy to show aperiodicity using induction, but I had a hard time to figure out the proof for $psi$-irreducibility. Is induction and contradiction a right direction to go? In general, if there is no aperiodicity of the chain, is it also true that all the $m$-skeletons are $psi$-irreducible?
markov-chains
asked Mar 19 '14 at 5:11
942kid942kid
111
$endgroup$
add a comment |
$begingroup$
In Proposition 5.4.5 of Meyn and Tweedie's Markov Chains and Stochastic Stability, it is said that if a chain $Phi$ is $psi$-irreducible and aperiodic, then every $m$-skeleton of it is also $psi$-irreducible and aperiodic. The authors did not give an explicit proof of this.
It is easy to show aperiodicity using induction, but I had a hard time to figure out the proof for $psi$-irreducibility. Is induction and contradiction a right direction to go? In general, if there is no aperiodicity of the chain, is it also true that all the $m$-skeletons are $psi$-irreducible?
markov-chains
asked Mar 19 '14 at 5:11
942kid942kid
111
$endgroup$
In Proposition 5.4.5 of Meyn and Tweedie's Markov Chains and Stochastic Stability, it is said that if a chain $Phi$ is $psi$-irreducible and aperiodic, then every $m$-skeleton of it is also $psi$-irreducible and aperiodic. The authors did not give an explicit proof of this.
It is easy to show aperiodicity using induction, but I had a hard time to figure out the proof for $psi$-irreducibility. Is induction and contradiction a right direction to go? In general, if there is no aperiodicity of the chain, is it also true that all the $m$-skeletons are $psi$-irreducible?
markov-chains
markov-chains
asked Mar 19 '14 at 5:11
942kid942kid
111
asked Mar 19 '14 at 5:11
942kid942kid
111
asked Mar 19 '14 at 5:11
942kid942kid
111
asked Mar 19 '14 at 5:11
942kid942kid
111
asked Mar 19 '14 at 5:11
942kid942kid
111
111
add a comment |
add a comment |
1 Answer
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I think it can be proved as follows:
For any $C in B^{+}(X)$, w.o.l.g. (see Theorem 5.2.2), we can assume that $C$ is a $nu_M$ small set with $nu_M(C) > 0$ . Since, $Phi$ is aperiodic (and because this property does not depend on the small set as shown by the proof of Theorem 5.4.4), $g.c.d.(E_C)=1$. By Lemma D.7.4, there exists $k_0$ s.t. for all $k geq k_0$, $C$ is $nu_k$ small with $nu_k = delta_k nu_M$ and with $delta_k >0$.
For $x in X$, since $Phi$ is $psi$-irreducible, there is some $r$ s.t. $P^r(x, C) >0$. Hence, by Chapman-Kolmogorov, for any $k geq k_0$ we get:
$$P^{r+k}(x, C) geq int_C P^r(x, dy) P^k(y, C) geq P^r(x, C) delta_k nu_M(C) >0 quad .$$
Therefore, for any $m$, there exist $i$ s.t. $P^{icdot m}(x, C) >0$, which yields the result.
answered Dec 3 '18 at 11:05
jeanjean
1
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$begingroup$
I think it can be proved as follows:
For any $C in B^{+}(X)$, w.o.l.g. (see Theorem 5.2.2), we can assume that $C$ is a $nu_M$ small set with $nu_M(C) > 0$ . Since, $Phi$ is aperiodic (and because this property does not depend on the small set as shown by the proof of Theorem 5.4.4), $g.c.d.(E_C)=1$. By Lemma D.7.4, there exists $k_0$ s.t. for all $k geq k_0$, $C$ is $nu_k$ small with $nu_k = delta_k nu_M$ and with $delta_k >0$.
For $x in X$, since $Phi$ is $psi$-irreducible, there is some $r$ s.t. $P^r(x, C) >0$. Hence, by Chapman-Kolmogorov, for any $k geq k_0$ we get:
$$P^{r+k}(x, C) geq int_C P^r(x, dy) P^k(y, C) geq P^r(x, C) delta_k nu_M(C) >0 quad .$$
Therefore, for any $m$, there exist $i$ s.t. $P^{icdot m}(x, C) >0$, which yields the result.
answered Dec 3 '18 at 11:05
jeanjean
1
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I think it can be proved as follows:
For any $C in B^{+}(X)$, w.o.l.g. (see Theorem 5.2.2), we can assume that $C$ is a $nu_M$ small set with $nu_M(C) > 0$ . Since, $Phi$ is aperiodic (and because this property does not depend on the small set as shown by the proof of Theorem 5.4.4), $g.c.d.(E_C)=1$. By Lemma D.7.4, there exists $k_0$ s.t. for all $k geq k_0$, $C$ is $nu_k$ small with $nu_k = delta_k nu_M$ and with $delta_k >0$.
For $x in X$, since $Phi$ is $psi$-irreducible, there is some $r$ s.t. $P^r(x, C) >0$. Hence, by Chapman-Kolmogorov, for any $k geq k_0$ we get:
$$P^{r+k}(x, C) geq int_C P^r(x, dy) P^k(y, C) geq P^r(x, C) delta_k nu_M(C) >0 quad .$$
Therefore, for any $m$, there exist $i$ s.t. $P^{icdot m}(x, C) >0$, which yields the result.
answered Dec 3 '18 at 11:05
jeanjean
1
$endgroup$
add a comment |
$begingroup$
I think it can be proved as follows:
For any $C in B^{+}(X)$, w.o.l.g. (see Theorem 5.2.2), we can assume that $C$ is a $nu_M$ small set with $nu_M(C) > 0$ . Since, $Phi$ is aperiodic (and because this property does not depend on the small set as shown by the proof of Theorem 5.4.4), $g.c.d.(E_C)=1$. By Lemma D.7.4, there exists $k_0$ s.t. for all $k geq k_0$, $C$ is $nu_k$ small with $nu_k = delta_k nu_M$ and with $delta_k >0$.
For $x in X$, since $Phi$ is $psi$-irreducible, there is some $r$ s.t. $P^r(x, C) >0$. Hence, by Chapman-Kolmogorov, for any $k geq k_0$ we get:
$$P^{r+k}(x, C) geq int_C P^r(x, dy) P^k(y, C) geq P^r(x, C) delta_k nu_M(C) >0 quad .$$
Therefore, for any $m$, there exist $i$ s.t. $P^{icdot m}(x, C) >0$, which yields the result.
answered Dec 3 '18 at 11:05
jeanjean
1
$endgroup$
I think it can be proved as follows:
For any $C in B^{+}(X)$, w.o.l.g. (see Theorem 5.2.2), we can assume that $C$ is a $nu_M$ small set with $nu_M(C) > 0$ . Since, $Phi$ is aperiodic (and because this property does not depend on the small set as shown by the proof of Theorem 5.4.4), $g.c.d.(E_C)=1$. By Lemma D.7.4, there exists $k_0$ s.t. for all $k geq k_0$, $C$ is $nu_k$ small with $nu_k = delta_k nu_M$ and with $delta_k >0$.
For $x in X$, since $Phi$ is $psi$-irreducible, there is some $r$ s.t. $P^r(x, C) >0$. Hence, by Chapman-Kolmogorov, for any $k geq k_0$ we get:
$$P^{r+k}(x, C) geq int_C P^r(x, dy) P^k(y, C) geq P^r(x, C) delta_k nu_M(C) >0 quad .$$
Therefore, for any $m$, there exist $i$ s.t. $P^{icdot m}(x, C) >0$, which yields the result.
answered Dec 3 '18 at 11:05
jeanjean
1
answered Dec 3 '18 at 11:05
jeanjean
1
answered Dec 3 '18 at 11:05
jeanjean
1
answered Dec 3 '18 at 11:05
jeanjean
1
1
add a comment |
add a comment |
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