The Stationary VectorDate: 11/13/2000 at 00:59:36 From: Jessica Searing Subject: Markov Processes What does a stationary vector tell you and how do you find it? Date: 11/13/2000 at 11:20:00 From: Doctor Anthony Subject: Re: Markov Processes The stationary vector is a position from which no further change occurs. It will correspond to the eigenvector with eigenvalue equal to 1. Below is an example where weather conditions from one day to the following day are considered. The transition matrix is FROM -------- SUN CLOUD RAIN SUN [ 0 1/4 1/4 ] TO CLOUD [ 1/2 1/2 1/4 ] -- RAIN [ 1/2 1/4 1/2 ] If we take higher and higher powers of the matrix, it will eventually settle down so that whatever the starting conditions are on day 1, the effect will not influence the nth day. In this situation the columns of the matrix are all the same, so that any starting vector gives the same resulting vector. This vector is in fact the eigenvector corresponding to an eigenvalue of 1. (Note that EVERY transition matrix where the columns each sum to 1 will ALWAYS have an eigenvalue of 1.) To find the eigenvector [a,b,c] corresponding to the eigenvalue 1 we have to solve: (0-1)a + (1/4)b + (1/4)c = 0 (1/2)a + (1/2 - 1)b + (1/4)c = 0 -a + b/4 + c/4 = 0 a/2 - b/2 + c/4 = 0 a -b c ---------- = ---------- = ---------- |1/4 1/4| |-1 1/4| |-1 1/4| |-1/2 1/4| |1/2 1/4| |1/2 -1/2| a -b c ---- = ---- = ---- 3/16 -3/8 3/8 a/3 = b/6 = c/6 a/1 = b/2 = c/2 [a] [1/5] [b] = [2/5] [c] [2/5] Therefore 1/5 of the days are sunny and each of the other two types occurs on 2/5 of the days. We can check that [1/5, 2/5, 2/5] is a stable vector by multiplying this vector by the original matrix SUN [ 0 1/4 1/4 ][1/5] [1/5] CLOUD [ 1/2 1/2 1/4 ][2/5] = [2/5] RAIN [ 1/2 1/4 1/2 ][2/5] [2/5] and see that the following day is unchanged. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/