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Unformatted text preview: equence converges
to a limit ⇢1 . Letting n ! 1 in ⇢n = (⇢n 1 ) implies that ⇢1 = (⇢1 ), i.e.,
⇢1 is a solution of (x) = x. To complete the proof now let ⇢ be the smallest
solution. Clearly ⇢0 = 0 ⇢. Using the fact that is increasing, it follows that
⇢1 = (⇢0 ) (⇢) = ⇢. Repeating the argument we have ⇢2 ⇢, ⇢3 ⇢ and
so on. Taking limits we have ⇢1 ⇢. However, ⇢ is the smallest solution, so
we must have ⇢1 = ⇢.
To see what this says, let us consider a concrete example.
Example 1.53. Binary branching. Suppose p2 = a, p0 = 1
other pk = 0. In this case (✓) = a✓2 + 1 a, so (x) = x means
0 = ax2 x+1 a = (x 1)(ax (1 a)) a, and the 58 CHAPTER 1. MARKOV CHAINS The roots are 1 and (1 a)/a. If a 1/2, then the smallest root is 1, while if
a > 1/2 the smallest root is (1 a)/a.
Noting that a 1/2 corresponds to mean µ 1 in binary branching motivates the following guess:
II. If µ > 1, then there is positive probability of avoiding extinction.
Proof. In view of Lemma 1.30, we only have to show there is a root < 1. We...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

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