Chapter 11.4 Exercise 1

Another way to think about the steady state vector \(w\) of \(P\) is to solve for the eigenvector associated with eigenvalue 1 of \(P^T\)

(P = matrix(c( 0.5 , 0.5, 0.25, 0.75), nrow=2, ncol=2, byrow=TRUE))

##      [,1] [,2]
## [1,] 0.50 0.50
## [2,] 0.25 0.75

(EIG = eigen(t(P)))

## eigen() decomposition
## $values
## [1] 1.00 0.25
## 
## $vectors
##            [,1]       [,2]
## [1,] -0.4472136 -0.7071068
## [2,] -0.8944272  0.7071068

(ee1 = EIG$vectors[,1])

## [1] -0.4472136 -0.8944272

The eigenvector \(ee1\) represents the eigenvector associated with \(\lambda =1\). A rescaled version of \(ee1\) will represent the vector \(w\) which is the fixed probability vector.

(w = ee1/sum(ee1))

## [1] 0.3333333 0.6666667

(w %*% P)

##           [,1]      [,2]
## [1,] 0.3333333 0.6666667

Note that the following condition holds:

\[w P = w\]

Thus, the fundamental limit theorem states that \(P^n\) will converge to \(W\) where each row is the same as \(w\).

\[W = \left[\begin{array}{rr} 1/3 & 2/3 \\ 1/3 & 2/3 \\ \end{array} \right] \] Finally, we know \[ W y = \left[\begin{array}{rr} 1/3 & 2/3 \\ 1/3 & 2/3 \\ \end{array} \right] \left[ \begin{array}{r} 1 \\ 0 \\ \end{array} \right] = \left[ \begin{array}{r} 1/3 \\ 1/3 \\ \end{array} \right] \]

This also proves the result.