How to prove that $P(Z_1 = j| Z_0 = i) = P(Z_n+1 = j| Z_n = i)?$
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Currently I am reading this Markov Chains' notes.
At page $112,$ section $4.2$ Transition matrix, I fail to understand how author obtains the following.
As seen above, the random evolution of a Markov chain $(Z_n)_nin mathbbN$ is determined
by the data of
$$P_i,j := P(Z_1 = j | Z_0 = i), i, j in mathbbS,$$
which coincides with the probability $P(Z_n+1 = j | Z_n = i)$ which is independent of $n in mathbbN.$
I do not understand how author obtains the following equality
$$P(Z_1 = j| Z_0 = i) = P(Z_n+1 = j| Z_n = i).$$
I think Markov property might be applied here, but I do not know how.
Any hint is appreciated.
probability markov-chains markov-process
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up vote
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down vote
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Currently I am reading this Markov Chains' notes.
At page $112,$ section $4.2$ Transition matrix, I fail to understand how author obtains the following.
As seen above, the random evolution of a Markov chain $(Z_n)_nin mathbbN$ is determined
by the data of
$$P_i,j := P(Z_1 = j | Z_0 = i), i, j in mathbbS,$$
which coincides with the probability $P(Z_n+1 = j | Z_n = i)$ which is independent of $n in mathbbN.$
I do not understand how author obtains the following equality
$$P(Z_1 = j| Z_0 = i) = P(Z_n+1 = j| Z_n = i).$$
I think Markov property might be applied here, but I do not know how.
Any hint is appreciated.
probability markov-chains markov-process
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Currently I am reading this Markov Chains' notes.
At page $112,$ section $4.2$ Transition matrix, I fail to understand how author obtains the following.
As seen above, the random evolution of a Markov chain $(Z_n)_nin mathbbN$ is determined
by the data of
$$P_i,j := P(Z_1 = j | Z_0 = i), i, j in mathbbS,$$
which coincides with the probability $P(Z_n+1 = j | Z_n = i)$ which is independent of $n in mathbbN.$
I do not understand how author obtains the following equality
$$P(Z_1 = j| Z_0 = i) = P(Z_n+1 = j| Z_n = i).$$
I think Markov property might be applied here, but I do not know how.
Any hint is appreciated.
probability markov-chains markov-process
Currently I am reading this Markov Chains' notes.
At page $112,$ section $4.2$ Transition matrix, I fail to understand how author obtains the following.
As seen above, the random evolution of a Markov chain $(Z_n)_nin mathbbN$ is determined
by the data of
$$P_i,j := P(Z_1 = j | Z_0 = i), i, j in mathbbS,$$
which coincides with the probability $P(Z_n+1 = j | Z_n = i)$ which is independent of $n in mathbbN.$
I do not understand how author obtains the following equality
$$P(Z_1 = j| Z_0 = i) = P(Z_n+1 = j| Z_n = i).$$
I think Markov property might be applied here, but I do not know how.
Any hint is appreciated.
probability markov-chains markov-process
probability markov-chains markov-process
asked Sep 5 at 7:46
Idonknow
3,115643109
3,115643109
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1 Answer
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You cannot prove this. Many books consider what are called homogeneous Markov chain. For these what you are trying to prove is part of the definition. There are non-homogeneous MC's so Markov property does not imply this property.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You cannot prove this. Many books consider what are called homogeneous Markov chain. For these what you are trying to prove is part of the definition. There are non-homogeneous MC's so Markov property does not imply this property.
add a comment |Â
up vote
1
down vote
You cannot prove this. Many books consider what are called homogeneous Markov chain. For these what you are trying to prove is part of the definition. There are non-homogeneous MC's so Markov property does not imply this property.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You cannot prove this. Many books consider what are called homogeneous Markov chain. For these what you are trying to prove is part of the definition. There are non-homogeneous MC's so Markov property does not imply this property.
You cannot prove this. Many books consider what are called homogeneous Markov chain. For these what you are trying to prove is part of the definition. There are non-homogeneous MC's so Markov property does not imply this property.
answered Sep 5 at 8:05
Kavi Rama Murthy
26.3k31437
26.3k31437
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