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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).
 Spring '10
 DURRETT
 The Land

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