Question 11.3.9: Prove that, if a 3-by-3 transition matrix has the property t......
Prove that, if a 3-by-3 transition matrix has the property that its column sums are 1, then (1/3, 1/3, 1/3) is a fixed probability vector. State a similar result for n-by-n transition matrices. Interpret these results for ergodic chains.
Let
P = \left(\begin{matrix} P_{11} & P_{12} & P_{13}\\ P_{21} & P_{22} & P_{23} \\ P_{31} & P_{32} & P_{33}\end{matrix} \right) ,
with column sums equal to 1. Then
(1/3, 1/3, 1/3)P = (1/3 \sum\limits_{j=1}^{3}{p_{j1}}, 1/3\sum\limits_{j=1}^{3}{p_{j2}}, 1/3\sum\limits_{j=1}^{3}{p_{j3}} )
= (1/3, 1/3, 1/3) .
The same argument shows that if P is an n×n transition matrix with columns that add to 1 then
w = (1/n, · · · , 1/n)
is a fixed probability vector. For an ergodic chain this means the the average number of times in each state is 1/n.