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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 ﬁrstgeneration 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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 Spring '10
 DURRETT
 The Land

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