Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 155
> endfor
octave:4> z=[x, y];
octave:5> gplot z
yields this plot. There is no threshold value — no probability above which the curve rises sharply.
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
line 1
0
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
5 (a) From these equations
0.90 · p T (n) + 0.01 · p C (n) = p T (n + 1)
0.10 · p T (n) + 0.99 · p C (n) = p C (n + 1)
we get this matrix.
( ) ( )
0.90 0.01 pT (n)
0.10 0.99 p C (n)
=
( )
pT (n + 1)
p C (n + 1)
(b) This is the result from Octave.
n = 0 1 2 3 4 5
0.30000
0.70000
0.27700
0.72300
0.25653
0.74347
0.23831
0.76169
0.22210
0.77790
0.20767
0.79233
6 7 8 9 10
0.19482
0.80518
0.18339
0.81661
0.17322
0.82678
0.16417
0.83583
0.15611
0.84389
(c) This is the s T = 0.2 result.
n = 0 1 2 3 4 5
0.20000
0.80000
0.18800
0.81200
0.17732
0.82268
0.16781
0.83219
0.15936
0.84064
0.15183
0.84817
6 7 8 9 10
0.14513
0.85487
0.13916
0.86084
0.13385
0.86615
0.12913
0.87087
0.12493
0.87507
(d) Although the probability vectors start 0.1 apart, they end only 0.032 apart. So they are alike.
6 These are the p = .55 vectors,