A question about Markov Chain.

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Suppose $X_n$ has Markov property. Show that for any $n,r in Bbb N,i in S,A subset S^n, B subset S^r$



$P[(X_n+1, cdots , X_n+r) in B mid X_n=i,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n=i]$.



If further $C subset S$ then,



$P[(X_n+1, cdots , X_n+r) in B mid X_n in C,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n in C]$.





How do I proceed using Markov property? Please help me in this regard.



Thank you very much.







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  • Maybe I'm wrong, but it seems there is a counterexample to the second statement given here (see the answer with 103 upvotes).
    – Shalop
    Aug 7 at 21:09















up vote
1
down vote

favorite














Suppose $X_n$ has Markov property. Show that for any $n,r in Bbb N,i in S,A subset S^n, B subset S^r$



$P[(X_n+1, cdots , X_n+r) in B mid X_n=i,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n=i]$.



If further $C subset S$ then,



$P[(X_n+1, cdots , X_n+r) in B mid X_n in C,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n in C]$.





How do I proceed using Markov property? Please help me in this regard.



Thank you very much.







share|cite|improve this question





















  • Maybe I'm wrong, but it seems there is a counterexample to the second statement given here (see the answer with 103 upvotes).
    – Shalop
    Aug 7 at 21:09













up vote
1
down vote

favorite









up vote
1
down vote

favorite













Suppose $X_n$ has Markov property. Show that for any $n,r in Bbb N,i in S,A subset S^n, B subset S^r$



$P[(X_n+1, cdots , X_n+r) in B mid X_n=i,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n=i]$.



If further $C subset S$ then,



$P[(X_n+1, cdots , X_n+r) in B mid X_n in C,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n in C]$.





How do I proceed using Markov property? Please help me in this regard.



Thank you very much.







share|cite|improve this question















Suppose $X_n$ has Markov property. Show that for any $n,r in Bbb N,i in S,A subset S^n, B subset S^r$



$P[(X_n+1, cdots , X_n+r) in B mid X_n=i,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n=i]$.



If further $C subset S$ then,



$P[(X_n+1, cdots , X_n+r) in B mid X_n in C,(X_0,cdots,X_n-1) in A]=P[(X_n+1,cdots,X_n+r) in B mid X_n in C]$.





How do I proceed using Markov property? Please help me in this regard.



Thank you very much.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 7 at 17:51
























asked Aug 7 at 17:43









Debabrata Chattopadhyay.

14312




14312











  • Maybe I'm wrong, but it seems there is a counterexample to the second statement given here (see the answer with 103 upvotes).
    – Shalop
    Aug 7 at 21:09

















  • Maybe I'm wrong, but it seems there is a counterexample to the second statement given here (see the answer with 103 upvotes).
    – Shalop
    Aug 7 at 21:09
















Maybe I'm wrong, but it seems there is a counterexample to the second statement given here (see the answer with 103 upvotes).
– Shalop
Aug 7 at 21:09





Maybe I'm wrong, but it seems there is a counterexample to the second statement given here (see the answer with 103 upvotes).
– Shalop
Aug 7 at 21:09
















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