30 we only have to show there is a root 1 we begin by

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Unformatted text preview: ily line of each child must die out, an event of probability ⇢k , so we can reason that ⇢= 1 X k=0 p k ⇢k (1.33) 57 1.10. INFINITE STATE SPACES* 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 1.4: Generating function for Binomial(3,1/2). P1 If we let (✓) = k=0 pk ✓k be the generating function of the distribution pk , then the last equation can be written simply as ⇢ = (⇢). P1 The equation in (1.33) has a trivial root at ⇢ = 1 since (⇢) = k=0 pk ⇢k = 1. The next result identifies the root that we want: Lemma 1.30. The extinction probability ⇢ is the smallest solution of the equation (x) = x with 0 x 1. Proof. Extending the reasoning for (1.33) we see that in order for the process to hit 0 by time n, all of the processes started by first-generation individuals must hit 0 by time n 1, so P (Xn = 0) = 1 X pk P (Xn 1 = 0)k k=0 0, then ⇢n = (⇢n 1 ) for From this we see that if ⇢n = P (Xn = 0) for n n 1. Since 0 is an absorbing state, ⇢0 ⇢1 ⇢2 . . . and the s...
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