A  small note on using the left eigenvector of the one step transition matrix for a regular chain to determine the limiting distribution

In trying to solve problem 8.1, lecture notes, Math 504, I found that I need to determine ' w', which is the limiting distribution of the given markov chain. One way to determine the limiting distribution from P is by finding the normalized left eigenvector associated with eigenvalue 1 of the P matrix. This below shows how to do this for the P matrix given in the problem 8.1 and verify that it is the same as a row in the (P^N) matrix where N is a large value

First Setup the P matrix as shown in problem 8.1

pMat={{p,q,0},{p,0,q},{0,p,q}};
pMat//MatrixForm

Find the eigenvalues and associated with the left eigenvectors. First transpose the matrix

(t=Transpose[pMat])//MatrixForm

Now find the eigenvalues and vectors of the transposed matrix

({values,vec}=Eigensystem[t]);

The eigenvalues are

values//MatrixForm

And the corresponding eigenvectors

vec//MatrixForm

Now pick the eigenvector associated with eigenvalue 1. We see from the above that it is the 3 rd eigenvector (since p+q=1)

v=vec[[3]]

Normalize it. This is π (the limiting distribution)

w=v/Length[v]

Verify by comparing to row entry of (P^∞) Try some p and q values, say p=.5 and q=.5

w/.{p->.5,q->.5}

MatrixPower[pMat/.{p->.5,q->.5},100]

References

1. Lecture notes, Mathematics 504. Professor B. Gearhart
2. Wikepedia http://en.wikipedia.org/wiki/Markov_chain