Homogeneous Markov ChainTransforming an inhomogeneous Markov chain to a homogeneous oneProve Markov Chain by definitionMarkov-chain propertiestime-homogeneous continuous time Markov chainProving identity of Markov chain.Markov Process, Markov ChainInequality with a homogeneous markov chain and hitting probabilitiesMarkov chain transition matrix and homogeneityHow to construct a Markov Chain of higher order given another Markov ChainA die is rolled repreatedly and … is it a Markov Chain?
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Homogeneous Markov Chain
Transforming an inhomogeneous Markov chain to a homogeneous oneProve Markov Chain by definitionMarkov-chain propertiestime-homogeneous continuous time Markov chainProving identity of Markov chain.Markov Process, Markov ChainInequality with a homogeneous markov chain and hitting probabilitiesMarkov chain transition matrix and homogeneityHow to construct a Markov Chain of higher order given another Markov ChainA die is rolled repreatedly and … is it a Markov Chain?
$begingroup$
If $X_n$ is a homogeneous Markov chain, is it true that $X_n^2$ ($n$ is of the power $2$ not $X_n$) is also a homogeneous Markov chain? And why?
stochastic-processes
edited Mar 28 at 21:14
Shashi
7,3621628
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If $X_n$ is a homogeneous Markov chain, is it true that $X_n^2$ ($n$ is of the power $2$ not $X_n$) is also a homogeneous Markov chain? And why?
stochastic-processes
edited Mar 28 at 21:14
Shashi
7,3621628
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– Alex Vong
Mar 28 at 20:47
$begingroup$
I know that its not true, but I dont know the counter example.
$endgroup$
– Jennyfer Lahoud
Mar 28 at 20:50
add a comment |
$begingroup$
If $X_n$ is a homogeneous Markov chain, is it true that $X_n^2$ ($n$ is of the power $2$ not $X_n$) is also a homogeneous Markov chain? And why?
stochastic-processes
edited Mar 28 at 21:14
Shashi
7,3621628
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If $X_n$ is a homogeneous Markov chain, is it true that $X_n^2$ ($n$ is of the power $2$ not $X_n$) is also a homogeneous Markov chain? And why?
stochastic-processes
stochastic-processes
edited Mar 28 at 21:14
Shashi
7,3621628
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 21:14
Shashi
7,3621628
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 21:14
Shashi
7,3621628
edited Mar 28 at 21:14
Shashi
7,3621628
edited Mar 28 at 21:14
Shashi
7,3621628
7,3621628
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
asked Mar 28 at 20:43
Jennyfer LahoudJennyfer Lahoud
74
74
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jennyfer Lahoud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What have you tried?
$endgroup$
– Alex Vong
Mar 28 at 20:47
$begingroup$
I know that its not true, but I dont know the counter example.
$endgroup$
– Jennyfer Lahoud
Mar 28 at 20:50
add a comment |
$begingroup$
What have you tried?
$endgroup$
– Alex Vong
Mar 28 at 20:47
$begingroup$
I know that its not true, but I dont know the counter example.
$endgroup$
– Jennyfer Lahoud
Mar 28 at 20:50
$begingroup$
What have you tried?
$endgroup$
– Alex Vong
Mar 28 at 20:47
$begingroup$
What have you tried?
$endgroup$
– Alex Vong
Mar 28 at 20:47
$begingroup$
I know that its not true, but I dont know the counter example.
$endgroup$
– Jennyfer Lahoud
Mar 28 at 20:50
$begingroup$
I know that its not true, but I dont know the counter example.
$endgroup$
– Jennyfer Lahoud
Mar 28 at 20:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a simple example. Consider the random walk on $mathbb Z$ that stays at its position, goes to right or left with probability $1/3$ each. We consider the a particular transition probability
beginalign
p_n(0,2):=mathbb P(X_n^2=2 mid X_(n-1)^2=0 )
endalign
which, if $X_n$ is a homoegeneous Markov chain, should not depend on $n$. But guess what? It depends on $n$, we have
beginalign
p_1(0,2)=mathbb P(X_1=2mid X_0=0)=0
endalign
while
beginalign
p_2(0,2)=mathbb P(X_4=2mid X_1=0)>0
endalign
Now it remains to think about why we have $p_2(0,2)>0$... Do you see why?
answered Mar 28 at 21:13
ShashiShashi
7,3621628
$endgroup$
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a simple example. Consider the random walk on $mathbb Z$ that stays at its position, goes to right or left with probability $1/3$ each. We consider the a particular transition probability
beginalign
p_n(0,2):=mathbb P(X_n^2=2 mid X_(n-1)^2=0 )
endalign
which, if $X_n$ is a homoegeneous Markov chain, should not depend on $n$. But guess what? It depends on $n$, we have
beginalign
p_1(0,2)=mathbb P(X_1=2mid X_0=0)=0
endalign
while
beginalign
p_2(0,2)=mathbb P(X_4=2mid X_1=0)>0
endalign
Now it remains to think about why we have $p_2(0,2)>0$... Do you see why?
answered Mar 28 at 21:13
ShashiShashi
7,3621628
$endgroup$
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
add a comment |
$begingroup$
Here is a simple example. Consider the random walk on $mathbb Z$ that stays at its position, goes to right or left with probability $1/3$ each. We consider the a particular transition probability
beginalign
p_n(0,2):=mathbb P(X_n^2=2 mid X_(n-1)^2=0 )
endalign
which, if $X_n$ is a homoegeneous Markov chain, should not depend on $n$. But guess what? It depends on $n$, we have
beginalign
p_1(0,2)=mathbb P(X_1=2mid X_0=0)=0
endalign
while
beginalign
p_2(0,2)=mathbb P(X_4=2mid X_1=0)>0
endalign
Now it remains to think about why we have $p_2(0,2)>0$... Do you see why?
answered Mar 28 at 21:13
ShashiShashi
7,3621628
$endgroup$
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
add a comment |
$begingroup$
Here is a simple example. Consider the random walk on $mathbb Z$ that stays at its position, goes to right or left with probability $1/3$ each. We consider the a particular transition probability
beginalign
p_n(0,2):=mathbb P(X_n^2=2 mid X_(n-1)^2=0 )
endalign
which, if $X_n$ is a homoegeneous Markov chain, should not depend on $n$. But guess what? It depends on $n$, we have
beginalign
p_1(0,2)=mathbb P(X_1=2mid X_0=0)=0
endalign
while
beginalign
p_2(0,2)=mathbb P(X_4=2mid X_1=0)>0
endalign
Now it remains to think about why we have $p_2(0,2)>0$... Do you see why?
answered Mar 28 at 21:13
ShashiShashi
7,3621628
$endgroup$
Here is a simple example. Consider the random walk on $mathbb Z$ that stays at its position, goes to right or left with probability $1/3$ each. We consider the a particular transition probability
beginalign
p_n(0,2):=mathbb P(X_n^2=2 mid X_(n-1)^2=0 )
endalign
which, if $X_n$ is a homoegeneous Markov chain, should not depend on $n$. But guess what? It depends on $n$, we have
beginalign
p_1(0,2)=mathbb P(X_1=2mid X_0=0)=0
endalign
while
beginalign
p_2(0,2)=mathbb P(X_4=2mid X_1=0)>0
endalign
Now it remains to think about why we have $p_2(0,2)>0$... Do you see why?
answered Mar 28 at 21:13
ShashiShashi
7,3621628
edited Mar 28 at 21:18
answered Mar 28 at 21:13
ShashiShashi
7,3621628
answered Mar 28 at 21:13
ShashiShashi
7,3621628
answered Mar 28 at 21:13
ShashiShashi
7,3621628
7,3621628
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
add a comment |
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
$begingroup$
Alright. Thank you
$endgroup$
– Jennyfer Lahoud
Mar 28 at 21:23
add a comment |
Jennyfer Lahoud is a new contributor. Be nice, and check out our Code of Conduct.
Jennyfer Lahoud is a new contributor. Be nice, and check out our Code of Conduct.
Jennyfer Lahoud is a new contributor. Be nice, and check out our Code of Conduct.
Jennyfer Lahoud is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What have you tried?
$endgroup$
– Alex Vong
Mar 28 at 20:47
$begingroup$
I know that its not true, but I dont know the counter example.
$endgroup$
– Jennyfer Lahoud
Mar 28 at 20:50