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Unformatted text preview: r case is impossible,
since Pjj = pj and thus Pjj = p1 j → 0.
Ruling out these two trivial cases, we have p0 > 0 and p1 < 1 − p0 . In this case, state 0
is recurrent (i.e., it is a trapping state) and states 1, 2, . . . , are in a transient class. To see
this, note that P10 = p0 > 0, so F11 (∞) ≤ 1 − p0 < 1, which means by deﬁnition that state
1 is transient. All states i > 1 communicate with state 1, so by Theorem 5.1.1, all states
j ≥ 1 are transient. Thus one can argue that the process has ‘no place to go’ other than 0
The following ugly analysis makes this precise. Note from Lemma 5.1.1 part 4 that
Pjtj = ∞.
t→∞ n≤t Since this sum is nondecreasing in t, the limit must exist and the limit must be ﬁnite This
Pn = 0
n≥t j j
Now we can write P1j = �≤n f1j Pjn−� , from which it can be seen that limt→∞ n≥t P1j =
From this, we see that for every ﬁnite integer �,...
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This document was uploaded on 03/19/2014 for the course EECS 6.262 at MIT.
- Spring '11