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Unformatted text preview: hile the period of y is d < c. Let k
be such that pk (x, y ) > 0 and let m be such that pm (y, x) > 0. Since
pk+m (x, x) pk (x, y )pm (y, x) > 0 we have k + m 2 Ix . Since x has period c, k + m must be a multiple of c. Now
let ` be any integer with p` (y, y ) > 0. Since
pk+`+m (x, x) pk (x, y )p` (y, y )pm (y, x) > 0 k + ` + m 2 Ix , and k + ` + m must be a multiple of c. Since k + m is itself a
multiple of c, this means that ` is a multiple of c. Since ` 2 Iy was arbitrary, we
have shown that c is a divisor of every element of Iy , but d < c is the greatest
common divisor, so we have a contradiction.
Lemma 1.18 easily settles the question for the inventory chain (Example
1.6)
012345
0 0 0 .1 .2 .4 .3
1 0 0 .1 .2 .4 .3
2 .3 .4 .3 0 0 0
3 .1 .2 .4 .3 0 0
4 0 .1 .2 .4 .3 0
5 0 0 .1 .2 .4 .3
Since p(x, x) > 0 for x = 2, 3, 4, 5, Lemma 1.17 implies that these states are
aperiodic. Since this chain is irreducible it follows from Lemma 1.18 that 0 and
1 are aperiodic.
Con...
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 Spring '10
 DURRETT
 The Land

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