**Unformatted text preview: **) ⎟ ⎜ 1
⎟⎜
⎜
2
⎟=⎜
or ⎜
⎟⎜
⎜
p2 (k −1)
⎜ p3 (k ) ⎟ ⎜ 0
p3 (k ) =
+ p4 (k −1)
⎠⎝
⎝
2
p3 (k −1)
p4 (k )
0
p4 (k ) =
2 p2 (k ) = p1 (k −1) + 1/2 0 0 1/2 1/2 0 0 1/2 0 ⎞
⎛ p1 (k −1) ⎞ ⎟
⎜
⎟
⎜
0 ⎟ p2 (k −1) ⎟
⎟
⎜
⎟
⎟
⎜
⎟.
⎟
⎜
⎟
⎟ p3 (k −1) ⎟
⎜
1⎠
⎝
⎠
0 p4 (k −1) The matrix equation on the right-hand side is a homogeneous diﬀerence equation
p(k ) = Ap(k − 1) whose solution, from (7.10.4), is p(k ) = Ak p(0), and thus
Problem 1 is solved. For example, if you know that the pea is initially under
shell #2, then p(0) = e2 , and after six moves the probability that the pea is
in the fourth position is p4 (6) = A6 e2 4 = 21/64. If you don’t know exactly
where the pea starts, but you assume that it is equally likely to start under any
one of the four shells, then p(0) = (1/4, 1/4, 1/4, 1/4)T , and the probabilities Copyright c 2000 SIAM Buy online from SIAM
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636
Chapter 7
Eigenvalues and Eigenvectors
http://www.amazon.com/exec/obidos/ASIN/0898714540
for occupying the four positions after six moves are given by p(6) = A6 p(0), or
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Please report violations to [email protected] ⎞⎛
p1 (6)
11/32
p2 (6) ⎟ ⎜ 0
⎜
⎝
⎠=⎝
p3 (6)
21/32
0
p4 (6) ⎞⎛
⎞
⎛⎞
0
21/64
0
1/4
43
1 ⎜ 85 ⎟
43/64
0
21/32 ⎟ ⎜ 1/4 ⎟
⎠⎝
⎠=
⎝ ⎠.
0
43/64
0
1/4
256 85
21/64
0
11/32
1/4
43 Solution to Problem 2: There is a straightforward solution when A is a convergent matrix because if Ak → G as k → ∞, then p(k ) → Gp(0) = p, and
the components in this limiting (or steady-state) vector p provide the answer.
Intuitively, if p(k ) → p, then after awhile p(k ) is practically constant, so the
probability that the pea occupies a particular position remains essentially the
same move after move. Consequently, the components in limk→∞ p(k ) reveal
the proportion of time spent in e...

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