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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/
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