Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 157
0,0,0,q,p,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-1
0,0,0,0,q,p, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-2_
0,0,0,0,0,q, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-3
0,0,0,0,0,0, p,0,0,0,1,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 4-0
0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-1__
0,0,0,0,0,0, 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-2
0,0,0,0,0,0, 0,0,q,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-3
0,0,0,0,0,0, 0,0,0,q,0,0, 0,0,1,0,0,0, 0,0,0,0,0,0; # 0-4_
0,0,0,0,0,0, 0,0,0,0,0,p, 0,0,0,1,0,0, 0,0,0,0,0,0; # 4-1
0,0,0,0,0,0, 0,0,0,0,0,q, p,0,0,0,0,0, 0,0,0,0,0,0; # 3-2
0,0,0,0,0,0, 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0; # 2-3__
0,0,0,0,0,0, 0,0,0,0,0,0, 0,q,0,0,0,0, 1,0,0,0,0,0; # 1-4
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,p,0, 0,1,0,0,0,0; # 4-2
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,q,p, 0,0,0,0,0,0; # 3-3_
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,q, 0,0,0,1,0,0; # 2-4
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,p,0,1,0; # 4-3
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,q,0,0,1]; # 3-4
v7 = (A**7) * v;
w = v7(11)+v7(16)+v7(20)+v7(23)
endfunction
Using this script, we get that when the American League has a p = 0.55 probability of winning
each game then their probability of winning the first-to-win-four series is 0.60829. When their
probability of winning any one game is p = 0.6 then their probability of winning the series is
0.71021.
(b) This Octave session
octave:1> v0=[1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0];
octave:2> x=(.01:.01:.99)’;
octave:3> y=(.01:.01:.99)’;
octave:4> for i=.01:.01:.99
> y(100*i)=markov(i,v0);
> endfor
octave:5> z=[x, y];
octave:6> gplot z
yields this graph. By eye we judge that if p > 0.7 then the team is close to assurred of the series.
1
line 1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
7 (a) They must satisfy this condition because the total probability of a state transition (including
back to the same state) is 100%.
(b) See the answer to the third item.
(c) We will do the 2×2 case; bigger-sized cases are just notational problems. This product
has these two column sums
( ) ( )
a1,1 a 1,2 b1,1 b 1,2
=
a 2,1 a 2,2 b 2,1 b 2,2
( )
a1,1 b 1,1 + a 1,2 b 2,1 a 1,1 b 1,2 + a 1,2 b 2,2
a 2,1 b 1,1 + a 2,2 b 2,1 a 2,1 b 1,2 + a 2,2 b 2,2
(a 1,1 b 1,1 + a 1,2 b 2,1 ) + (a 2,1 b 1,1 + a 2,2 b 2,1 ) = (a 1,1 + a 2,1 ) · b 1,1 + (a 1,2 + a 2,2 ) · b 2,1 = 1 · b 1,1 + 1 · b 2,1 = 1
and
(a 1,1 b 1,2 + a 1,2 b 2,2 ) + (a 2,1 b 1,2 + a 2,2 b 2,2 ) = (a 1,1 + a 2,1 ) · b 1,2 + (a 1,2 + a 2,2 ) · b 2,2 = 1 · b 1,2 + 1 · b 2,2 = 1
as required.