The probability of reaching the absorbing states from a particular transient state?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 Feb 26 at 13:35
user120911user120911
75128
$endgroup$
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 Feb 26 at 13:35
user120911user120911
75128
$endgroup$
$begingroup$
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
Feb 26 at 13:56
$begingroup$
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
Feb 26 at 14:20
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 Feb 26 at 13:35
user120911user120911
75128
$endgroup$
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 Feb 26 at 13:35
user120911user120911
75128
asked Feb 26 at 13:35
user120911user120911
75128
edited Feb 26 at 13:45
user120911
asked Feb 26 at 13:35
user120911user120911
75128
asked Feb 26 at 13:35
user120911user120911
75128
asked Feb 26 at 13:35
user120911user120911
75128
75128
$begingroup$
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
Feb 26 at 13:56
$begingroup$
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
Feb 26 at 14:20
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
Feb 26 at 13:56
$begingroup$
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
Feb 26 at 14:20
$begingroup$
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
Feb 26 at 13:56
$begingroup$
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
Feb 26 at 13:56
$begingroup$
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
Feb 26 at 14:20
$begingroup$
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
Feb 26 at 14:20
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]
$beginarrayccccc
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
endarray$
Update: "Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6."
The 6th row of ltm contains the desired probabilities:
ltm[[6, Flatten@ab]]
0., 0.5, 0.5, 0.
answered Feb 26 at 14:41
kglrkglr
189k10206424
$endgroup$
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
1
$begingroup$
@user120911, if, say,8is the transient state you are interested in you can useltm[[8, Flatten@ab]
$endgroup$
– kglr
Feb 27 at 0:59
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
$begingroup$
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]
$beginarrayccccc
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
endarray$
Update: "Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6."
The 6th row of ltm contains the desired probabilities:
ltm[[6, Flatten@ab]]
0., 0.5, 0.5, 0.
answered Feb 26 at 14:41
kglrkglr
189k10206424
$endgroup$
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
1
$begingroup$
@user120911, if, say,8is the transient state you are interested in you can useltm[[8, Flatten@ab]
$endgroup$
– kglr
Feb 27 at 0:59
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]
$beginarrayccccc
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
endarray$
Update: "Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6."
The 6th row of ltm contains the desired probabilities:
ltm[[6, Flatten@ab]]
0., 0.5, 0.5, 0.
answered Feb 26 at 14:41
kglrkglr
189k10206424
$endgroup$
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
1
$begingroup$
@user120911, if, say,8is the transient state you are interested in you can useltm[[8, Flatten@ab]
$endgroup$
– kglr
Feb 27 at 0:59
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]
$beginarrayccccc
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
endarray$
Update: "Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6."
The 6th row of ltm contains the desired probabilities:
ltm[[6, Flatten@ab]]
0., 0.5, 0.5, 0.
answered Feb 26 at 14:41
kglrkglr
189k10206424
$endgroup$
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]
$beginarrayccccc
& 4 & 7 & 9 & 10 \
3 & 0.5 & 0.5 & 0. & 0. \
6 & 0. & 0.5 & 0.5 & 0. \
2 & 0.25 & 0.5 & 0.25 & 0. \
8 & 0. & 0. & 0.5 & 0.5 \
5 & 0. & 0.25 & 0.5 & 0.25 \
1 & 0.125 & 0.375 & 0.375 & 0.125 \
endarray$
Update: "Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6."
The 6th row of ltm contains the desired probabilities:
ltm[[6, Flatten@ab]]
0., 0.5, 0.5, 0.
answered Feb 26 at 14:41
kglrkglr
189k10206424
edited Feb 27 at 1:16
answered Feb 26 at 14:41
kglrkglr
189k10206424
answered Feb 26 at 14:41
kglrkglr
189k10206424
answered Feb 26 at 14:41
kglrkglr
189k10206424
189k10206424
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
1
$begingroup$
@user120911, if, say,8is the transient state you are interested in you can useltm[[8, Flatten@ab]
$endgroup$
– kglr
Feb 27 at 0:59
add a comment |
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
1
$begingroup$
@user120911, if, say,8is the transient state you are interested in you can useltm[[8, Flatten@ab]
$endgroup$
– kglr
Feb 27 at 0:59
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
$begingroup$
Suppose I had a very large transition matrix, and I was interested in only one transient state, say 6. How would I go about entering just that number in your code (a newbie question, I know, but I am having a little difficulty seeing where the number 6 goes). But please don't remove your current solution, which is terrific.
$endgroup$
– user120911
Feb 27 at 0:53
1
1
$begingroup$
@user120911, if, say,
8 is the transient state you are interested in you can use ltm[[8, Flatten@ab]$endgroup$
– kglr
Feb 27 at 0:59
$begingroup$
@user120911, if, say,
8 is the transient state you are interested in you can use ltm[[8, Flatten@ab]$endgroup$
– kglr
Feb 27 at 0:59
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
Feb 26 at 13:56
$begingroup$
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
Feb 26 at 14:20