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Unformatted text preview: STATS 219  Stochastic Processes HW 9, Autumn 2009 1. Exercise 4.6.8. Suppose { Z n } is a branching process with P ( N = 1) < 1 and Z = 1. Show that P ( lim n Z n = ) = 1 p ex , first in case m 1, then in case P ( N = 0) = 0 and finally using the preceding exercise, for m > 1 and P ( N = 0) > 0. ANS: Since P ( N = 1) < 1 we have by Propositions 4.6.3 and 4.6.5 that p ex = 1 when m 1. That is, in this case w.p.1. Z n = 0 for all n large enough, yielding the stated claim. In contrast, if P ( N = 0) = 0 then Z n is nondecreasing, so p ex = 0. Further, in this case Z n is bounded only if N ( k ) 1 = 1 for all k large enough, which with P ( N = 1) < 1 occurs with zero probability, again resulting with the stated claim. Finally, for m > 1 and P ( N = 0) > 0 we have from Exercise 4.6.7 that p ex = (0 , 1) and further, 1 p ex = 1 P ( M = 1) = P ( M = 0) = P ( lim n Z n = 0) = P ( lim n Z n = ) , as claimed. 2. Exercise 4.6.9. Let { Z n } be a branching process with Z = 1. Compute p ex in each of the following situations and specify for which values of the various parameters the extinction is certain. (a) The offspring distribution satisfies, for some 0 < p < 1, P ( N = 0) = p, P ( N = 2) = 1 p. ANS: We have that m = E N = 2(1 p ) with m 1 if and only if p 1 / 2. Thus, p ex = 1 when p 1 / 2 by Proposition 4.6.3 (for p > 1 / 2) and Proposition 4.6.5 (for p = 1 / 2, applicable since here P ( N = 1) = 0 < 1). Finally, if p < 1 / 2 then m > 1 so { Z n } is supercritical with P ( N = 0) = p > 0. We have shown in Exercise 4.6.7 that in this case p ex is the unique solution in (0 , 1) of 0 = x ( x ) = x P ( N = 0) P ( N = 2) x 2 = x p (1 p ) x 2 (taking the function ( x ) per equation (4.6.2) that corresponds to our law of N ). As x p (1 p ) x 2 = (1 p )(1 x )( x p/ (1 p )) , we conclude that p ex = p/ (1 p ) < 1 when p < 1 / 2. 1 (b) The offspring distribution is (shifted) Geometric, i.e. for some 0 < p < 1, P ( N = k ) = p (1 p ) k , k = 0 , 1 , 2 , . . . ANS: We have now that m = E N = k =1 kp (1 p ) k = (1 p ) /p (where to get the last identity differentiate in p the identity k =0 (1 p ) k = 1 /p and multiply both sides by p (1 p )). As in part (a), if p 1 / 2 then m 1 and consequently p ex = 1 (for here too P ( N = 1) = p (1 p ) < 1). In contrast, p < 1 / 2 yields a supercritical branching process with P ( N = 0) = p > 0, so again from Exercise 4.6.7 we have that p ex is the unique solution in (0 , 1) of 0 = x ( x ) = x X k =0 P ( N = k ) x k = x p X k =0 (1 p ) k x k = x p 1 (1 p ) x . Thus, p ex is the unique root in (0 , 1) of the quadratic equation 0 = x (1 (1 p ) x ) p = x p (1 p ) x 2 , and as you have seen in part (a), it follows that p ex = p/ (1 p ). Thus, though the law of N in part (b) is different from its law in part (a), both result with same values of p ex (for all choices of p )....
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 Spring '09

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