Let va minn 0 xn 2 a be the time of the rst visit to a

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Unformatted text preview: we will let Wn = Zn /µ , and compute EWn . Noting that Zn is the sum of Zn 1 independent random variables with the same distribution so E (Zn |Zn 1 ) = µZn 1 , E ((Zn Zn 1 )2 |Zn 1 ) = 2 Zn 1 . It follows from Lemma 5.7 that 2 E (Zn |Zn 1 ) = 2 Zn 1 + (µZn 1 )2 Taking expected values in the last equation and dividing both sides by µ2n we conclude that 2 2 EWn = EWn 1 + 2 /µn+1 2 Iterating that we have EW1 = 1+ 2 EWn 2 2 /µ2 , EW2 = 1+ =1+ 2 n+1 X µ k 2 . k=2 Thus if µ > 1, 2 EWN CW = 1 + 2 1 X k=2 µ k /µ2 + 2 /µ3 , and hence 175 5.6. EXERCISES To complete the proof now, we note that if V 0 then 1 1 E (V 2 ; V > M ) EV 2 M M The argument for the last inequality in Theorem 5.17 shows that E (V ; V > M ) (5.16) 2 EW 2 lim EWn CW n!1 Thus using (5.16) |EWn E W | E |Wn W | E |Wn ^ M W ^ M | + 2CW /M As n ! 1 the first term tends to 0. If M is large the second one is < ✏. This shows that lim sup |EWn E W | ✏ n!1 Since ✏ is arbitrary we have 1 = EWn ! EW . 5.6 Exercises Throughout the exercises we will use our standard notion for hitting times. Ta = min{n 1 : Xn = a} and Va = min{n 0 : Xn = a}. 5.1. Brother–sister mating. Conside...
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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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