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Unformatted text preview: begin by discarding a trivial case. If p0 = 0, then (0) = 0, 0 is the smallest root, and there is no probability of dying out. If p0 > 0, then (0) = p0 > 0. Di↵erentiating the definition of , we have 0 (x) = 1 X k=1 pk · kxk 1 so 0 (1) = 1 X kpk = µ k=1 If µ > 1 then the slope of at x = 1 is larger than 1, so if ✏ is small, then (1 ✏) < 1 ✏. Combining this with (0) > 0 we see there must be a solution of (x) = x between 0 and 1 ✏. See the figure in the proof of (7.6). Turning to the borderline case: III. If µ = 1 and we exclude the trivial case p1 = 1, then extinction occurs with probability 1. Proof. By Lemma 1.30 we only have to show that there is no root < 1. To do this we note that if p1 < 1, then for y < 1 0 (y ) = 1 X k=1 pk · kxk R1 so if x < 1 then (x) = (1) for all x < 1. x 0 1 < 1 X pk k = 1 k=1 (y ) dy > 1 (1 x) = x. Thus (x) > x Note that in binary branching with a = 1/2, (x) = (1 + x2 )/2, so if we try to solve (x) = x we get 0=1 2...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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