Limit distribution of a reducible Markov chain












0












$begingroup$


I had a simple question yesterday when I was trying to solve an exercise on a reducible,aperiodic Markov Chain. The state spase S was $$S={1,...,7}$$ and we could partition it into two closed classes ${5,6,7} bigcup {3,4}$ but the class ${1,2}$ was open (thus transient). The exercise was giving the transition matrix and was asking to find the stationary distributions. At the end it was also asking to find the limit distribution, given that the INITIAL distribution was $$P[X_0=1]=P[X_0=2]=frac{1}{2}$$ I only know what to do in case the initial probability gives mass 1 to some state (I find the absorption probabilities in each closed class, etc..) but when I have initial distribution which gives mass to both of these transient states I do not know what to do. Any helpful ideas/theorems about this??
Thanks a lot !



$$ begin{pmatrix}
1/3 & 1/6 & 1/3 & 0 & 0 & 1/6 \
3/5 & 0 & 0 & 1/5 & 1/5 & 0 & 0 \
0 & 0 & 1/2 & 1/2 & 0 & 0 & 0 \
0 & 0 & 1/4 & 3/4 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1/4 & 3/4 \
0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 end{pmatrix} $$






markov-chains







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$endgroup$












  • $begingroup$
    Giving the transition matrix would make things easier (to answer) here. An answer of the kind "take 1/2 of the limit distribution for the case of giving full probability to the state 5 and also take 1/2 of the limit distribution for the case of giving full probability to the state 6 and add them" is not really beautiful.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:03










  • $begingroup$
    OK,let's suppose we have the following example in page 1 : win.tue.nl/~resing/SOR/handout.pdf and you want to answer question 2) .
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:20












  • $begingroup$
    It may be useful to use mathjax, and include the example in the original post, math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dan_fulea
    Jan 18 at 18:21










  • $begingroup$
    I will write it afterwards. Do you have any idea about the solution?
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:26










  • $begingroup$
    It should be a "linear play". More exactly, in some notation i am used to, the initial distribution is a row vector, shape $Ntimes 1$, where $N$ (six or seven in our case) is the number of states, and $A$, the transition matrix, is a an $Ntimes N$ matrix. So we need to compute something like $lim_n qA^n$ for $nto infty$. This expression is linear in $q$. So if we know the answer for the canonical base $e_1=[1,0,0,0,dots]$, $e_2=[0,1,0,0,dots]$, ... then we need to linearly combine them using the weights in $q$.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:33


















0












$begingroup$


I had a simple question yesterday when I was trying to solve an exercise on a reducible,aperiodic Markov Chain. The state spase S was $$S={1,...,7}$$ and we could partition it into two closed classes ${5,6,7} bigcup {3,4}$ but the class ${1,2}$ was open (thus transient). The exercise was giving the transition matrix and was asking to find the stationary distributions. At the end it was also asking to find the limit distribution, given that the INITIAL distribution was $$P[X_0=1]=P[X_0=2]=frac{1}{2}$$ I only know what to do in case the initial probability gives mass 1 to some state (I find the absorption probabilities in each closed class, etc..) but when I have initial distribution which gives mass to both of these transient states I do not know what to do. Any helpful ideas/theorems about this??
Thanks a lot !



$$ begin{pmatrix}
1/3 & 1/6 & 1/3 & 0 & 0 & 1/6 \
3/5 & 0 & 0 & 1/5 & 1/5 & 0 & 0 \
0 & 0 & 1/2 & 1/2 & 0 & 0 & 0 \
0 & 0 & 1/4 & 3/4 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1/4 & 3/4 \
0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 end{pmatrix} $$






markov-chains







share|cite|improve this question











$endgroup$












  • $begingroup$
    Giving the transition matrix would make things easier (to answer) here. An answer of the kind "take 1/2 of the limit distribution for the case of giving full probability to the state 5 and also take 1/2 of the limit distribution for the case of giving full probability to the state 6 and add them" is not really beautiful.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:03










  • $begingroup$
    OK,let's suppose we have the following example in page 1 : win.tue.nl/~resing/SOR/handout.pdf and you want to answer question 2) .
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:20












  • $begingroup$
    It may be useful to use mathjax, and include the example in the original post, math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dan_fulea
    Jan 18 at 18:21










  • $begingroup$
    I will write it afterwards. Do you have any idea about the solution?
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:26










  • $begingroup$
    It should be a "linear play". More exactly, in some notation i am used to, the initial distribution is a row vector, shape $Ntimes 1$, where $N$ (six or seven in our case) is the number of states, and $A$, the transition matrix, is a an $Ntimes N$ matrix. So we need to compute something like $lim_n qA^n$ for $nto infty$. This expression is linear in $q$. So if we know the answer for the canonical base $e_1=[1,0,0,0,dots]$, $e_2=[0,1,0,0,dots]$, ... then we need to linearly combine them using the weights in $q$.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:33
















0












0








0





$begingroup$


I had a simple question yesterday when I was trying to solve an exercise on a reducible,aperiodic Markov Chain. The state spase S was $$S={1,...,7}$$ and we could partition it into two closed classes ${5,6,7} bigcup {3,4}$ but the class ${1,2}$ was open (thus transient). The exercise was giving the transition matrix and was asking to find the stationary distributions. At the end it was also asking to find the limit distribution, given that the INITIAL distribution was $$P[X_0=1]=P[X_0=2]=frac{1}{2}$$ I only know what to do in case the initial probability gives mass 1 to some state (I find the absorption probabilities in each closed class, etc..) but when I have initial distribution which gives mass to both of these transient states I do not know what to do. Any helpful ideas/theorems about this??
Thanks a lot !



$$ begin{pmatrix}
1/3 & 1/6 & 1/3 & 0 & 0 & 1/6 \
3/5 & 0 & 0 & 1/5 & 1/5 & 0 & 0 \
0 & 0 & 1/2 & 1/2 & 0 & 0 & 0 \
0 & 0 & 1/4 & 3/4 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1/4 & 3/4 \
0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 end{pmatrix} $$






markov-chains







share|cite|improve this question











$endgroup$




I had a simple question yesterday when I was trying to solve an exercise on a reducible,aperiodic Markov Chain. The state spase S was $$S={1,...,7}$$ and we could partition it into two closed classes ${5,6,7} bigcup {3,4}$ but the class ${1,2}$ was open (thus transient). The exercise was giving the transition matrix and was asking to find the stationary distributions. At the end it was also asking to find the limit distribution, given that the INITIAL distribution was $$P[X_0=1]=P[X_0=2]=frac{1}{2}$$ I only know what to do in case the initial probability gives mass 1 to some state (I find the absorption probabilities in each closed class, etc..) but when I have initial distribution which gives mass to both of these transient states I do not know what to do. Any helpful ideas/theorems about this??
Thanks a lot !



$$ begin{pmatrix}
1/3 & 1/6 & 1/3 & 0 & 0 & 1/6 \
3/5 & 0 & 0 & 1/5 & 1/5 & 0 & 0 \
0 & 0 & 1/2 & 1/2 & 0 & 0 & 0 \
0 & 0 & 1/4 & 3/4 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1/4 & 3/4 \
0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 end{pmatrix} $$






markov-chains




markov-chains






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 18:42







Petros Karajan

















asked Jan 18 at 17:41









Petros KarajanPetros Karajan

214




214












  • $begingroup$
    Giving the transition matrix would make things easier (to answer) here. An answer of the kind "take 1/2 of the limit distribution for the case of giving full probability to the state 5 and also take 1/2 of the limit distribution for the case of giving full probability to the state 6 and add them" is not really beautiful.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:03










  • $begingroup$
    OK,let's suppose we have the following example in page 1 : win.tue.nl/~resing/SOR/handout.pdf and you want to answer question 2) .
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:20












  • $begingroup$
    It may be useful to use mathjax, and include the example in the original post, math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dan_fulea
    Jan 18 at 18:21










  • $begingroup$
    I will write it afterwards. Do you have any idea about the solution?
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:26










  • $begingroup$
    It should be a "linear play". More exactly, in some notation i am used to, the initial distribution is a row vector, shape $Ntimes 1$, where $N$ (six or seven in our case) is the number of states, and $A$, the transition matrix, is a an $Ntimes N$ matrix. So we need to compute something like $lim_n qA^n$ for $nto infty$. This expression is linear in $q$. So if we know the answer for the canonical base $e_1=[1,0,0,0,dots]$, $e_2=[0,1,0,0,dots]$, ... then we need to linearly combine them using the weights in $q$.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:33




















  • $begingroup$
    Giving the transition matrix would make things easier (to answer) here. An answer of the kind "take 1/2 of the limit distribution for the case of giving full probability to the state 5 and also take 1/2 of the limit distribution for the case of giving full probability to the state 6 and add them" is not really beautiful.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:03










  • $begingroup$
    OK,let's suppose we have the following example in page 1 : win.tue.nl/~resing/SOR/handout.pdf and you want to answer question 2) .
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:20












  • $begingroup$
    It may be useful to use mathjax, and include the example in the original post, math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dan_fulea
    Jan 18 at 18:21










  • $begingroup$
    I will write it afterwards. Do you have any idea about the solution?
    $endgroup$
    – Petros Karajan
    Jan 18 at 18:26










  • $begingroup$
    It should be a "linear play". More exactly, in some notation i am used to, the initial distribution is a row vector, shape $Ntimes 1$, where $N$ (six or seven in our case) is the number of states, and $A$, the transition matrix, is a an $Ntimes N$ matrix. So we need to compute something like $lim_n qA^n$ for $nto infty$. This expression is linear in $q$. So if we know the answer for the canonical base $e_1=[1,0,0,0,dots]$, $e_2=[0,1,0,0,dots]$, ... then we need to linearly combine them using the weights in $q$.
    $endgroup$
    – dan_fulea
    Jan 18 at 18:33


















$begingroup$
Giving the transition matrix would make things easier (to answer) here. An answer of the kind "take 1/2 of the limit distribution for the case of giving full probability to the state 5 and also take 1/2 of the limit distribution for the case of giving full probability to the state 6 and add them" is not really beautiful.
$endgroup$
– dan_fulea
Jan 18 at 18:03




$begingroup$
Giving the transition matrix would make things easier (to answer) here. An answer of the kind "take 1/2 of the limit distribution for the case of giving full probability to the state 5 and also take 1/2 of the limit distribution for the case of giving full probability to the state 6 and add them" is not really beautiful.
$endgroup$
– dan_fulea
Jan 18 at 18:03












$begingroup$
OK,let's suppose we have the following example in page 1 : win.tue.nl/~resing/SOR/handout.pdf and you want to answer question 2) .
$endgroup$
– Petros Karajan
Jan 18 at 18:20






$begingroup$
OK,let's suppose we have the following example in page 1 : win.tue.nl/~resing/SOR/handout.pdf and you want to answer question 2) .
$endgroup$
– Petros Karajan
Jan 18 at 18:20














$begingroup$
It may be useful to use mathjax, and include the example in the original post, math.meta.stackexchange.com/questions/5020/…
$endgroup$
– dan_fulea
Jan 18 at 18:21




$begingroup$
It may be useful to use mathjax, and include the example in the original post, math.meta.stackexchange.com/questions/5020/…
$endgroup$
– dan_fulea
Jan 18 at 18:21












$begingroup$
I will write it afterwards. Do you have any idea about the solution?
$endgroup$
– Petros Karajan
Jan 18 at 18:26




$begingroup$
I will write it afterwards. Do you have any idea about the solution?
$endgroup$
– Petros Karajan
Jan 18 at 18:26












$begingroup$
It should be a "linear play". More exactly, in some notation i am used to, the initial distribution is a row vector, shape $Ntimes 1$, where $N$ (six or seven in our case) is the number of states, and $A$, the transition matrix, is a an $Ntimes N$ matrix. So we need to compute something like $lim_n qA^n$ for $nto infty$. This expression is linear in $q$. So if we know the answer for the canonical base $e_1=[1,0,0,0,dots]$, $e_2=[0,1,0,0,dots]$, ... then we need to linearly combine them using the weights in $q$.
$endgroup$
– dan_fulea
Jan 18 at 18:33






$begingroup$
It should be a "linear play". More exactly, in some notation i am used to, the initial distribution is a row vector, shape $Ntimes 1$, where $N$ (six or seven in our case) is the number of states, and $A$, the transition matrix, is a an $Ntimes N$ matrix. So we need to compute something like $lim_n qA^n$ for $nto infty$. This expression is linear in $q$. So if we know the answer for the canonical base $e_1=[1,0,0,0,dots]$, $e_2=[0,1,0,0,dots]$, ... then we need to linearly combine them using the weights in $q$.
$endgroup$
– dan_fulea
Jan 18 at 18:33












1 Answer
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$begingroup$

As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} $$
For the case $ P[X_0=2]=1 $ , since $q_{2,E_1}=frac{10}{17}$ and $ q_{2,E_2}=frac{7}{17}$, we get the limit vector : $$begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$ Hence, the final answer to our question is : $$ frac{1}{2}begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} + frac{1}{2} begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$






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    $begingroup$

    As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} $$
    For the case $ P[X_0=2]=1 $ , since $q_{2,E_1}=frac{10}{17}$ and $ q_{2,E_2}=frac{7}{17}$, we get the limit vector : $$begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$ Hence, the final answer to our question is : $$ frac{1}{2}begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} + frac{1}{2} begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$






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      0












      $begingroup$

      As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} $$
      For the case $ P[X_0=2]=1 $ , since $q_{2,E_1}=frac{10}{17}$ and $ q_{2,E_2}=frac{7}{17}$, we get the limit vector : $$begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$ Hence, the final answer to our question is : $$ frac{1}{2}begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} + frac{1}{2} begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} $$
        For the case $ P[X_0=2]=1 $ , since $q_{2,E_1}=frac{10}{17}$ and $ q_{2,E_2}=frac{7}{17}$, we get the limit vector : $$begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$ Hence, the final answer to our question is : $$ frac{1}{2}begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} + frac{1}{2} begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$






        share|cite|improve this answer









        $endgroup$



        As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} $$
        For the case $ P[X_0=2]=1 $ , since $q_{2,E_1}=frac{10}{17}$ and $ q_{2,E_2}=frac{7}{17}$, we get the limit vector : $$begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$ Hence, the final answer to our question is : $$ frac{1}{2}begin{pmatrix} 0, & 0, & frac{11}{51}, & frac{22}{51}, & frac{18}{221}, & frac{24}{221}, & frac{36}{221} end{pmatrix} + frac{1}{2} begin{pmatrix} 0, & 0, & frac{10}{51}, & frac{20}{51}, & frac{21}{221}, & frac{28}{221}, & frac{42}{221} end{pmatrix} $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 19:02









        Petros KarajanPetros Karajan

        214




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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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