In the Land of Oz example (Example 11.1), change the transition matrix by making \(R\) an absorbing state. This gives
\[ P = \begin{array}{c|ccc} & R & N & S \\ \hline R & 1 & 0 & 0 \\ N & \frac{1}{2} & 0 & \frac{1}{2} \\ S & \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ \end{array} \]
Find the fundamental matrix \(N\), and also \(Nc\) and \(NR\). Interpret the results. \end{document}
I first extracted the submatrix \(Q\) from the given transition matrix, which includes the transition probabilities between the non-absorbing states N and S. This submatrix \(Q\) is:
\[ Q = \begin{bmatrix} 0 & 0.5 \\ 0.25 & 0.5 \end{bmatrix} \]
I then identified the identity matrix \(I\) corresponding to the non-absorbing states, which is a 2x2 identity matrix since we have two non-absorbing states:
\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
To find the fundamental matrix \(N\), I calculated the inverse of \(I - Q\):
\[ N = (I - Q)^{-1} \]
For the matrix \(R\), which represents the transition probabilities from non-absorbing to absorbing states, given that there is only one absorbing state R, it is a column matrix:
\[ R = \begin{bmatrix} 0 \\ 0.25 \end{bmatrix} \]
Multiplying the fundamental matrix \(N\) by \(R\), I obtained \(NR\), which gives the expected number of transitions from each non-absorbing state until being absorbed by state R.
The R code to find the fundamental matrix \(N\) and the matrix \(NR\) is:
# Define Q
Q <- matrix(c(0, 0.5, 0.25, 0.5), nrow=2, byrow=TRUE)
# Define the identity matrix I
I <- diag(2)
# Compute the fundamental matrix N
N <- solve(I - Q)
# Define R
R <- matrix(c(0, 0.25), ncol=1)
# Compute NR
NR <- N %*% RIn interpreting the results, the matrix \(N\) shows the expected number of visits to each non-absorbing state before reaching the absorbing state. The matrix \(NR\) indicates the expected number of steps to reach the absorbing state from each non-absorbing state.