AMS 210 Homework Solutions for Section 4.4
10a: When -1 is subtracted from the main diagonal entries, elimination on A-I, where A is a tridiagonal
Markov chain matrix, always involves adding the current column to the next column. The last column is
0’ed out in this process because the column sums are all 0 in A-I. p6 is a free variable. For 10a, 2p6 = p5 =
p4 = p3 = p2 = 2p1. Setting the sum of the pi‘s = 1, yields p6 = .1.
11a) Repeating 10a with n states, yields a similar situation as in 10a) with pn a free variable, the middle
pi‘s all equal to 2pn and p1 = pn. Setting the sum of the pi‘s = 1, yields p6 = 1/(2n-2).
12a) det(A- λI) = λ2 – (5/4)λ + 1/4:
λ1=1, λ 1 = [2/3,1/3], λ2=1/4, u2 = [1,-1].
16. Letting the states be I, E, S, M in that order,
é 1/ 2 1/ 4 0
ê
1/ 2 1/ 2 1/ 4
a) the transition matrix is A = ê
ê 0 1/ 4 1/ 2
ê 0
0 1/ 4
ë
0
0
0
1
ù
ú
ú;
ú
ú
û
é 6 4 2 ù
ê
ú
b) the fundamental matrix is N = ê 8 8 4 ú
êë 4 4 4 úû
c) 18 rounds from I to M.
é 0
ê
.4
17. A = ê
ê .4
ê 0
ë
a) 10/7, b) 5.
.4 .4 2 ù
ú
0 .4 .2 ú
and the fundamental matrix is N =
.4 0 .2 ú
0 0 1 úû
é 15 / 7 10 / 7 10 / 7 ù
ê
ú
ê 10 / 7 15 / 7 10 / 7 ú
êë 10 / 7 10 / 7 15 / 7 úû