The probability of reaching the absorbing states from a particular transient state?Plotting absorbing state...
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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
$begingroup$
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?
markov-chains markov-process
asked 2 hours ago
user120911user120911
67228
$endgroup$
add a comment |
$begingroup$
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?
markov-chains markov-process
asked 2 hours ago
user120911user120911
67228
$endgroup$
$begingroup$
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
1 hour 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
1 hour ago
add a comment |
$begingroup$
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?
markov-chains markov-process
asked 2 hours ago
user120911user120911
67228
$endgroup$
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
markov-chains markov-process
asked 2 hours ago
user120911user120911
67228
asked 2 hours ago
user120911user120911
67228
edited 2 hours ago
user120911
asked 2 hours ago
user120911user120911
67228
asked 2 hours ago
user120911user120911
67228
asked 2 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
1 hour 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
1 hour ago
add a comment |
$begingroup$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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]
{tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@
{ "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};
TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
TableHeadings -> {Flatten@tr, Flatten@ab}]
$begin{array}{ccccc}
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
end{array}$
answered 1 hour ago
kglrkglr
186k10202421
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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]
{tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@
{ "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};
TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
TableHeadings -> {Flatten@tr, Flatten@ab}]
$begin{array}{ccccc}
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
end{array}$
answered 1 hour ago
kglrkglr
186k10202421
$endgroup$
add a comment |
$begingroup$
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]
{tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@
{ "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};
TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
TableHeadings -> {Flatten@tr, Flatten@ab}]
$begin{array}{ccccc}
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
end{array}$
answered 1 hour ago
kglrkglr
186k10202421
$endgroup$
add a comment |
$begingroup$
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]
{tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@
{ "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};
TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
TableHeadings -> {Flatten@tr, Flatten@ab}]
$begin{array}{ccccc}
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
end{array}$
answered 1 hour ago
kglrkglr
186k10202421
$endgroup$
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]
{tr, ab, ltm} = MarkovProcessProperties[proc, #] & /@
{ "TransientClasses", "AbsorbingClasses", "LimitTransitionMatrix"};
TeXForm @ TableForm[ltm[[Flatten@tr, Flatten@ab]],
TableHeadings -> {Flatten@tr, Flatten@ab}]
$begin{array}{ccccc}
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
end{array}$
answered 1 hour ago
kglrkglr
186k10202421
edited 48 mins ago
answered 1 hour ago
kglrkglr
186k10202421
answered 1 hour ago
kglrkglr
186k10202421
answered 1 hour ago
kglrkglr
186k10202421
186k10202421
add a comment |
add a comment |
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$begingroup$
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
1 hour 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
1 hour ago