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The probability of reaching the absorbing states from a particular transient state?


Plotting absorbing state probabilities from state 1Defining a function of a distributionNicely illustrating the evolution and end-state of a discrete-time Markov chainHow to obtain the number of Markov Chain transitions in a simulation?Arranging “ranked” nodes of a graph symmetricallyState “i” goes to state “j”: list accessible states in a Markov-chainPart specification error with InterpolatingFunction when generating a Markov Modulated Poisson ProcessUsing Mathematica to calculate expected time to absorptionHidden Markov Model: emissions probabilities dependent on observable parameterEstimate process parameters of geometric Brownian motion with a two-state Markov chainPlotting absorbing state probabilities from state 1













4












$begingroup$


Can I use the data available from MarkovProcessProperties to compute the probability of reaching each of the absorbing states from a particular transient state?



In an earlier post, kglr showed a solution involving the probabilities from State 1. Can that solution be amended easily to compute the probabilities from any of the transient states?










share|improve this question











$endgroup$












  • $begingroup$
    Do you mean something like this? StationaryDistribution[DiscreteMarkovProcess[{1,0,0},{{0,1/2,1/2},{0,1,0},{0,0,1}}]]
    $endgroup$
    – Sjoerd Smit
    3 hours ago










  • $begingroup$
    I am looking for a solution like the one shown by kglr in the link, but which is more dynamic because it offers the possibility of specifying the particular transient state to be examined.
    $endgroup$
    – user120911
    2 hours ago
















4












$begingroup$


Can I use the data available from MarkovProcessProperties to compute the probability of reaching each of the absorbing states from a particular transient state?



In an earlier post, kglr showed a solution involving the probabilities from State 1. Can that solution be amended easily to compute the probabilities from any of the transient states?










share|improve this question











$endgroup$












  • $begingroup$
    Do you mean something like this? StationaryDistribution[DiscreteMarkovProcess[{1,0,0},{{0,1/2,1/2},{0,1,0},{0,0,1}}]]
    $endgroup$
    – Sjoerd Smit
    3 hours ago










  • $begingroup$
    I am looking for a solution like the one shown by kglr in the link, but which is more dynamic because it offers the possibility of specifying the particular transient state to be examined.
    $endgroup$
    – user120911
    2 hours ago














4












4








4





$begingroup$


Can I use the data available from MarkovProcessProperties to compute the probability of reaching each of the absorbing states from a particular transient state?



In an earlier post, kglr showed a solution involving the probabilities from State 1. Can that solution be amended easily to compute the probabilities from any of the transient states?










share|improve this question











$endgroup$




Can I use the data available from MarkovProcessProperties to compute the probability of reaching each of the absorbing states from a particular transient state?



In an earlier post, kglr showed a solution involving the probabilities from State 1. Can that solution be amended easily to compute the probabilities from any of the transient states?







markov-chains markov-process






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago







user120911

















asked 3 hours ago









user120911user120911

67228




67228












  • $begingroup$
    Do you mean something like this? StationaryDistribution[DiscreteMarkovProcess[{1,0,0},{{0,1/2,1/2},{0,1,0},{0,0,1}}]]
    $endgroup$
    – Sjoerd Smit
    3 hours ago










  • $begingroup$
    I am looking for a solution like the one shown by kglr in the link, but which is more dynamic because it offers the possibility of specifying the particular transient state to be examined.
    $endgroup$
    – user120911
    2 hours ago


















  • $begingroup$
    Do you mean something like this? StationaryDistribution[DiscreteMarkovProcess[{1,0,0},{{0,1/2,1/2},{0,1,0},{0,0,1}}]]
    $endgroup$
    – Sjoerd Smit
    3 hours ago










  • $begingroup$
    I am looking for a solution like the one shown by kglr in the link, but which is more dynamic because it offers the possibility of specifying the particular transient state to be examined.
    $endgroup$
    – user120911
    2 hours ago
















$begingroup$
Do you mean something like this? StationaryDistribution[DiscreteMarkovProcess[{1,0,0},{{0,1/2,1/2},{0,1,0},{0,0,1}}]]
$endgroup$
– Sjoerd Smit
3 hours ago




$begingroup$
Do you mean something like this? StationaryDistribution[DiscreteMarkovProcess[{1,0,0},{{0,1/2,1/2},{0,1,0},{0,0,1}}]]
$endgroup$
– Sjoerd Smit
3 hours ago












$begingroup$
I am looking for a solution like the one shown by kglr in the link, but which is more dynamic because it offers the possibility of specifying the particular transient state to be examined.
$endgroup$
– user120911
2 hours ago




$begingroup$
I am looking for a solution like the one shown by kglr in the link, but which is more dynamic because it offers the possibility of specifying the particular transient state to be examined.
$endgroup$
– user120911
2 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

proc = DiscreteMarkovProcess[1, {{0., 0.5, 0., 0., 0.5, 0., 0., 0., 0., 0.}, 
{0., 0., 0.5, 0., 0., 0.5, 0., 0., 0., 0.},
{0., 0., 0., 0.5, 0., 0., 0.5, 0., 0., 0.},
{0., 0., 0., 1., 0., 0., 0., 0., 0., 0.},
{0., 0., 0., 0., 0., 0.5, 0., 0.5, 0., 0.},
{0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.},
{0., 0., 0., 0., 0., 0., 1., 0., 0., 0.},
{0., 0., 0., 0., 0., 0., 0., 0., 0.5, 0.5},
{0., 0., 0., 0., 0., 0., 0., 0., 1., 0.},
{0., 0., 0., 0., 0., 0., 0., 0., 0., 1.}}];

Graph[proc]


enter image description here



{tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@ 
{ "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};

TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
TableHeadings -> {Flatten@tr, Flatten@ab}]



$begin{array}{ccccc}
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
end{array}$







share|improve this answer











$endgroup$













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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    proc = DiscreteMarkovProcess[1, {{0., 0.5, 0., 0., 0.5, 0., 0., 0., 0., 0.}, 
    {0., 0., 0.5, 0., 0., 0.5, 0., 0., 0., 0.},
    {0., 0., 0., 0.5, 0., 0., 0.5, 0., 0., 0.},
    {0., 0., 0., 1., 0., 0., 0., 0., 0., 0.},
    {0., 0., 0., 0., 0., 0.5, 0., 0.5, 0., 0.},
    {0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.},
    {0., 0., 0., 0., 0., 0., 1., 0., 0., 0.},
    {0., 0., 0., 0., 0., 0., 0., 0., 0.5, 0.5},
    {0., 0., 0., 0., 0., 0., 0., 0., 1., 0.},
    {0., 0., 0., 0., 0., 0., 0., 0., 0., 1.}}];

    Graph[proc]


    enter image description here



    {tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@ 
    { "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};

    TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
    TableHeadings -> {Flatten@tr, Flatten@ab}]



    $begin{array}{ccccc}
    & 4 & 7 & 9 & 10 \
    3 & 0.5 & 0.5 & 0. & 0. \
    6 & 0. & 0.5 & 0.5 & 0. \
    2 & 0.25 & 0.5 & 0.25 & 0. \
    8 & 0. & 0. & 0.5 & 0.5 \
    5 & 0. & 0.25 & 0.5 & 0.25 \
    1 & 0.125 & 0.375 & 0.375 & 0.125 \
    end{array}$







    share|improve this answer











    $endgroup$


















      5












      $begingroup$

      proc = DiscreteMarkovProcess[1, {{0., 0.5, 0., 0., 0.5, 0., 0., 0., 0., 0.}, 
      {0., 0., 0.5, 0., 0., 0.5, 0., 0., 0., 0.},
      {0., 0., 0., 0.5, 0., 0., 0.5, 0., 0., 0.},
      {0., 0., 0., 1., 0., 0., 0., 0., 0., 0.},
      {0., 0., 0., 0., 0., 0.5, 0., 0.5, 0., 0.},
      {0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.},
      {0., 0., 0., 0., 0., 0., 1., 0., 0., 0.},
      {0., 0., 0., 0., 0., 0., 0., 0., 0.5, 0.5},
      {0., 0., 0., 0., 0., 0., 0., 0., 1., 0.},
      {0., 0., 0., 0., 0., 0., 0., 0., 0., 1.}}];

      Graph[proc]


      enter image description here



      {tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@ 
      { "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};

      TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
      TableHeadings -> {Flatten@tr, Flatten@ab}]



      $begin{array}{ccccc}
      & 4 & 7 & 9 & 10 \
      3 & 0.5 & 0.5 & 0. & 0. \
      6 & 0. & 0.5 & 0.5 & 0. \
      2 & 0.25 & 0.5 & 0.25 & 0. \
      8 & 0. & 0. & 0.5 & 0.5 \
      5 & 0. & 0.25 & 0.5 & 0.25 \
      1 & 0.125 & 0.375 & 0.375 & 0.125 \
      end{array}$







      share|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        proc = DiscreteMarkovProcess[1, {{0., 0.5, 0., 0., 0.5, 0., 0., 0., 0., 0.}, 
        {0., 0., 0.5, 0., 0., 0.5, 0., 0., 0., 0.},
        {0., 0., 0., 0.5, 0., 0., 0.5, 0., 0., 0.},
        {0., 0., 0., 1., 0., 0., 0., 0., 0., 0.},
        {0., 0., 0., 0., 0., 0.5, 0., 0.5, 0., 0.},
        {0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.},
        {0., 0., 0., 0., 0., 0., 1., 0., 0., 0.},
        {0., 0., 0., 0., 0., 0., 0., 0., 0.5, 0.5},
        {0., 0., 0., 0., 0., 0., 0., 0., 1., 0.},
        {0., 0., 0., 0., 0., 0., 0., 0., 0., 1.}}];

        Graph[proc]


        enter image description here



        {tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@ 
        { "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};

        TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
        TableHeadings -> {Flatten@tr, Flatten@ab}]



        $begin{array}{ccccc}
        & 4 & 7 & 9 & 10 \
        3 & 0.5 & 0.5 & 0. & 0. \
        6 & 0. & 0.5 & 0.5 & 0. \
        2 & 0.25 & 0.5 & 0.25 & 0. \
        8 & 0. & 0. & 0.5 & 0.5 \
        5 & 0. & 0.25 & 0.5 & 0.25 \
        1 & 0.125 & 0.375 & 0.375 & 0.125 \
        end{array}$







        share|improve this answer











        $endgroup$



        proc = DiscreteMarkovProcess[1, {{0., 0.5, 0., 0., 0.5, 0., 0., 0., 0., 0.}, 
        {0., 0., 0.5, 0., 0., 0.5, 0., 0., 0., 0.},
        {0., 0., 0., 0.5, 0., 0., 0.5, 0., 0., 0.},
        {0., 0., 0., 1., 0., 0., 0., 0., 0., 0.},
        {0., 0., 0., 0., 0., 0.5, 0., 0.5, 0., 0.},
        {0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.},
        {0., 0., 0., 0., 0., 0., 1., 0., 0., 0.},
        {0., 0., 0., 0., 0., 0., 0., 0., 0.5, 0.5},
        {0., 0., 0., 0., 0., 0., 0., 0., 1., 0.},
        {0., 0., 0., 0., 0., 0., 0., 0., 0., 1.}}];

        Graph[proc]


        enter image description here



        {tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@ 
        { "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};

        TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
        TableHeadings -> {Flatten@tr, Flatten@ab}]



        $begin{array}{ccccc}
        & 4 & 7 & 9 & 10 \
        3 & 0.5 & 0.5 & 0. & 0. \
        6 & 0. & 0.5 & 0.5 & 0. \
        2 & 0.25 & 0.5 & 0.25 & 0. \
        8 & 0. & 0. & 0.5 & 0.5 \
        5 & 0. & 0.25 & 0.5 & 0.25 \
        1 & 0.125 & 0.375 & 0.375 & 0.125 \
        end{array}$








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 hours ago

























        answered 2 hours ago









        kglrkglr

        186k10202421




        186k10202421






























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