Stochastic

# E a double root at x 1 in general when 1 the graph of

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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.

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