Mixing
Solving Markov chain in MATLAB
$begingroup$
I am sitting in a class for Markov chains/processes, having no real background in probability but high school knowledge. Now I gotta go through it and I found it better understandable than I had expected.
But now we got a MATLAB snippet from our teacher, and while applying that code:
P =[...]
v = null(transpose(P-eye(length(P))));
pi = v.*(ones(length(P),1)*(1./sum(v)))
I found two questions:
- For what is the nullspace needed?
- What is the mathematical background for the second line (" pi = ...")
Having a transition matrix P and a stationary distribution vector $vecpi$ I can set up the system as $pi P = pi $ and rewrite is as $pi(P-I)=0$ if I do want to solve it in MATLAB.
Thank you for any help!
markov-chains matlab
asked Jan 28 '17 at 10:31
MaethorMaethor
63
$endgroup$
add a comment |
$begingroup$
I am sitting in a class for Markov chains/processes, having no real background in probability but high school knowledge. Now I gotta go through it and I found it better understandable than I had expected.
But now we got a MATLAB snippet from our teacher, and while applying that code:
P =[...]
v = null(transpose(P-eye(length(P))));
pi = v.*(ones(length(P),1)*(1./sum(v)))
I found two questions:
- For what is the nullspace needed?
- What is the mathematical background for the second line (" pi = ...")
Having a transition matrix P and a stationary distribution vector $vecpi$ I can set up the system as $pi P = pi $ and rewrite is as $pi(P-I)=0$ if I do want to solve it in MATLAB.
Thank you for any help!
markov-chains matlab
asked Jan 28 '17 at 10:31
MaethorMaethor
63
$endgroup$
add a comment |
$begingroup$
I am sitting in a class for Markov chains/processes, having no real background in probability but high school knowledge. Now I gotta go through it and I found it better understandable than I had expected.
But now we got a MATLAB snippet from our teacher, and while applying that code:
P =[...]
v = null(transpose(P-eye(length(P))));
pi = v.*(ones(length(P),1)*(1./sum(v)))
I found two questions:
- For what is the nullspace needed?
- What is the mathematical background for the second line (" pi = ...")
Having a transition matrix P and a stationary distribution vector $vecpi$ I can set up the system as $pi P = pi $ and rewrite is as $pi(P-I)=0$ if I do want to solve it in MATLAB.
Thank you for any help!
markov-chains matlab
asked Jan 28 '17 at 10:31
MaethorMaethor
63
$endgroup$
I am sitting in a class for Markov chains/processes, having no real background in probability but high school knowledge. Now I gotta go through it and I found it better understandable than I had expected.
But now we got a MATLAB snippet from our teacher, and while applying that code:
P =[...]
v = null(transpose(P-eye(length(P))));
pi = v.*(ones(length(P),1)*(1./sum(v)))
I found two questions:
- For what is the nullspace needed?
- What is the mathematical background for the second line (" pi = ...")
Having a transition matrix P and a stationary distribution vector $vecpi$ I can set up the system as $pi P = pi $ and rewrite is as $pi(P-I)=0$ if I do want to solve it in MATLAB.
Thank you for any help!
markov-chains matlab
markov-chains matlab
asked Jan 28 '17 at 10:31
MaethorMaethor
63
asked Jan 28 '17 at 10:31
MaethorMaethor
63
asked Jan 28 '17 at 10:31
MaethorMaethor
63
asked Jan 28 '17 at 10:31
MaethorMaethor
63
asked Jan 28 '17 at 10:31
MaethorMaethor
63
63
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For your first question. If I add a 1:
v = null(transpose(P-1*eye(length(P))));
you see that it is the nullspace $Ker(P-lambda I)$ associated to eigenvalue $lambda=1$, which is reduced to a single vector as you know for Markov matrices.
The second question: Here is the purpose of this operation: "null" outputs a normalized vector, with the meaning of quadratic norm ("2-norm") $sum x_i^2$ equal to 1. But for a probability vector, it is the 1-norm $sum |x_i|=sum x_i$ that must be equal to 1.
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
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
$begingroup$
For your first question. If I add a 1:
v = null(transpose(P-1*eye(length(P))));
you see that it is the nullspace $Ker(P-lambda I)$ associated to eigenvalue $lambda=1$, which is reduced to a single vector as you know for Markov matrices.
The second question: Here is the purpose of this operation: "null" outputs a normalized vector, with the meaning of quadratic norm ("2-norm") $sum x_i^2$ equal to 1. But for a probability vector, it is the 1-norm $sum |x_i|=sum x_i$ that must be equal to 1.
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
add a comment |
$begingroup$
For your first question. If I add a 1:
v = null(transpose(P-1*eye(length(P))));
you see that it is the nullspace $Ker(P-lambda I)$ associated to eigenvalue $lambda=1$, which is reduced to a single vector as you know for Markov matrices.
The second question: Here is the purpose of this operation: "null" outputs a normalized vector, with the meaning of quadratic norm ("2-norm") $sum x_i^2$ equal to 1. But for a probability vector, it is the 1-norm $sum |x_i|=sum x_i$ that must be equal to 1.
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
add a comment |
$begingroup$
For your first question. If I add a 1:
v = null(transpose(P-1*eye(length(P))));
you see that it is the nullspace $Ker(P-lambda I)$ associated to eigenvalue $lambda=1$, which is reduced to a single vector as you know for Markov matrices.
The second question: Here is the purpose of this operation: "null" outputs a normalized vector, with the meaning of quadratic norm ("2-norm") $sum x_i^2$ equal to 1. But for a probability vector, it is the 1-norm $sum |x_i|=sum x_i$ that must be equal to 1.
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
$endgroup$
For your first question. If I add a 1:
v = null(transpose(P-1*eye(length(P))));
you see that it is the nullspace $Ker(P-lambda I)$ associated to eigenvalue $lambda=1$, which is reduced to a single vector as you know for Markov matrices.
The second question: Here is the purpose of this operation: "null" outputs a normalized vector, with the meaning of quadratic norm ("2-norm") $sum x_i^2$ equal to 1. But for a probability vector, it is the 1-norm $sum |x_i|=sum x_i$ that must be equal to 1.
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
edited Jan 28 '17 at 11:09
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
answered Jan 28 '17 at 10:46
Jean MarieJean Marie
31k42255
31k42255
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
add a comment |
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
$begingroup$
Okay I see your points, I will try to apply this to the example I have and see if I can understand it better with that knowledge and by adding some more steps inbetween
$endgroup$
– Maethor
Jan 28 '17 at 11:06
add a comment |
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