Dgdrxrt

What is a romance in Latin?

One verb to replace 'be a member of' a club

How to prevent "they're falling in love" trope

What exploit Are these user agents trying to use?

In 'Revenger,' what does 'cove' come from?

Is it inappropriate for a student to attend their mentor's dissertation defense?

How do I gain back my faith in my PhD degree?

A category-like structure without composition?

Expand and Contract

Is "remove commented out code" correct English?

If human space travel is limited by the G force vulnerability, is there a way to counter G forces?

Avoiding direct proof while writing proof by induction

Why didn't Boeing produce its own regional jet?

Examples of smooth manifolds admitting inbetween one and a continuum of complex structures

How dangerous is XSS?

Should I tell management that I intend to leave due to bad software development practices?

Solving a recurrence relation (poker chips)

Size of subfigure fitting its content (tikzpicture)

What do you call someone who asks many questions?

Cursor Replacement for Newbies

Forgetting the musical notes while performing in concert

Reverse dictionary where values are lists

How do conventional missiles fly?

How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?

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?

1

$begingroup$

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

share|cite|improve this question

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 | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

1 Answer
1

active

oldest

votes

2

$begingroup$

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?

share|cite|improve this answer

edited Mar 28 at 21:18

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

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Jennyfer Lahoud is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166401%2fhomogeneous-markov-chain%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

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?

share|cite|improve this answer

edited Mar 28 at 21:18

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 | 

2

$begingroup$

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?

share|cite|improve this answer

edited Mar 28 at 21:18

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?

share|cite|improve this answer

edited Mar 28 at 21:18

answered Mar 28 at 21:13

ShashiShashi

7,3621628

$endgroup$

share|cite|improve this answer

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