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**Unformatted text preview: **violations to meyer@ncsu.edu and since expectation is linear ( E [αX (i) + X (h)] = αE [X (i)] + E [X (h)] ), the
expected fraction of times that the pea occupies position j before move k is
E X (0) + X (1) + · · · + X (k − 1)
E [X (0)] + E [X (1)] + · · · + E [X (k − 1)]
=
k
k
pj (0) + pj (1) + · · · + pj (k − 1)
p(0) + p(1) + · · · + p(k − 1)
=
=
k
k
j
p(0) + Ap(0) + · · · + Ak−1 p(0)
=
k =
j D
E I + A + · · · + Ak−1
k T
H → [Gp(0)]j . p(0) j In other words, as the game progresses indeﬁnitely, the components of the Ces`ro
a
limit p = Gp(0) provide the expected proportion of times that the pea is under
each shell, and this is exactly what we wanted to know. IG
R Computing the Limiting Vector. Of course, p can be determined by ﬁrst
computing G with a full-rank factorization of I − A as described in (7.10.41),
but there is some special structure in this problem that can be exploited to make
the task easier. Recall from (7.2.12) on p. 518 that if λ is a simple eigenvalue for
A, and if x and y∗ are respective right-hand and left-hand eigenvectors associated with λ, then xy∗ /y∗ x is the projector onto N (λI − A) along R (λI − A).
We can use this because, for the shell game, λ = 1 is a simple eigenvalue for
A. Furthermore, we get an associated left-hand eigenvector for free—namely,
eT = (1, 1, 1, 1) —because each column sum of A is one, so eT A = eT . Consequently, if x is any right-hand eigenvector of A associated with λ = 1, then
(by noting that eT p(0) = p1 (0) + p2 (0) + p3 (0) + p4 (0) = 1) the limiting vector
is given by
xeT p(0)
x
x
p = Gp(0) =
.
(7.10.44)
=T=
xi
eT x
ex Y
P O
C In other words, the limiting vector is obtained by normalizing any nonzero solution of (I − A)x = 0 to make the components sum to one. Not only does
(7.10.44) show how to compute the limiting proportions, it also shows that the
limiting proportions are independent of the initial values in p(0). For example, a
simple calculation reveals that x = (1, 2, 2, 1)T is one so...

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