assignment9

# Assignment9

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Unformatted text preview: r case is impossible, m m since Pjj = pj and thus Pjj = p1 j → 0. 1 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 or ∞. The following ugly analysis makes this precise. Note from Lemma 5.1.1 part 4 that � lim Pjtj = ∞. � t→∞ n≤t Since this sum is nondecreasing in t, the limit must exist and the limit must be ﬁnite This means that � lim Pn = 0 n≥t j j t→∞ � � n � n Now we can write P1j = �≤n f1j Pjn−� , from which it can be seen that limt→∞ n≥t P1j = j 0. 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.

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