Markov chain problems, absorbing states and eigenvalues Problem set B Answers

\(T=\begin{pmatrix}0.7 & 0.2 & 0 \\ 0 & 0.5 & 0.4 \\ 0.3 & 0.3 & 0.6\end{pmatrix}\)

Three vertices; loop at K 0.24; K to L 0.76; L to K 0.32; loop at L 0.28; L to H 0.4; H to L 0.52; loop at H 0.48.

\(T=\begin{pmatrix}0.92 & 0.6 \\ 0.08 & 0.4\end{pmatrix}\)

1. (a) 0.34
2. (b) 0.326

1. (a) 0.8x + 0.45y = x; 0.2x + 0.55y = y; x+y=1
2. (b) x = 9/13, y = 4/13

1. (a) \(T=\begin{pmatrix}0.6 & 0.2 & 0.1 \\ 0.3 & 0.5 & 0.2 \\ 0.1 & 0.3 & 0.7\end{pmatrix}\)
2. (b) (323, 390, 487)
3. (c) (318, 388, 494)

0.155

0.368

1. (a) \(T=\begin{pmatrix}1 & 0.45 & 0 & 0 & 0 \\ 0 & 0 & 0.45 & 0 & 0 \\ 0 & 0.55 & 0 & 0.45 & 0 \\ 0 & 0 & 0.55 & 0 & 0 \\ 0 & 0 & 0 & 0.55 & 1\end{pmatrix}\)
2. (b) 0.329

1. (a) Eigenvalues 1 and 0.75; eigenvectors \(\begin{pmatrix}1 \\ 4\end{pmatrix}\) and \(\begin{pmatrix}1 \\ -1\end{pmatrix}\) (or multiples).
2. (b) \(P=\begin{pmatrix}1 & 1 \\ 4 & -1\end{pmatrix},\ D=\begin{pmatrix}1 & 0 \\ 0 & 0.75\end{pmatrix}\)

1. (a) \(f_n=0.375+0.525(0.6)^{n}\)
2. (b) 0.375

1. (a) 1, 0.5, 0.35
2. (b) A multiple of \(\begin{pmatrix} 8 \\ 7 \\ 11 \end{pmatrix}\)
3. (c) \(\begin{pmatrix} \frac{4}{13} \\ \frac{7}{26} \\ \frac{11}{26} \end{pmatrix}\)