How can I get the symbolic steady state vector of a Markov Chain?
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Hello, does anyone know how to obtain the symbolic steady state vector (i.e. the long-term probability of each state) of this Markov Chain example in MATLAB?
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1035775/image.png)
At the end of this demonstration, it does not show how can I further get a steady state vector?
It will be very appreciated if you can help me with this problem.
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John D'Errico
2022 年 8 月 7 日
編集済み: John D'Errico
2022 年 8 月 7 日
Easy, peasy. For example, given a simple Markov process, described by the 3x3 transition matrix T.
T = [.5 .2 .3;.1 .4 .5;.1 .1 .8]
There are no absorbing states. We can see this is indeed the transition matrix of a Markov chain. One good test is the rows all sum to 1, and none of the elements are greater than 1, or less than zero.
sum(T,2)
What are the steady-state probabilities?
[V,D] = eig(T')
Take eigenvector that corresponds to the unit eigenvalue. In this case, it is the first eigenvector.
P = V(:,1)';
Normalize so the elements sum to 1.
format long g
P = P/sum(P)
Those are the steady state probabilites for this system. We can see that this does not change P.
P*T
I won't do your homework for you, but you can easily enough see how to proceed from here.
4 件のコメント
Walter Roberson
2022 年 8 月 8 日
John, with symbolic coefficients, is it going to be possible to find the entry with eigenvalue 1?
Bruno Luong
2022 年 8 月 8 日
You don't need to compute eigen value, you can compute this, possibly easier in symbolic way:
ss = null(T.'-eye(size(T))).';
ss = ss/sum(ss)
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