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Unformatted text preview: = ⇢xy /(1 ⇢yy ) Proof. Accept for the moment the fact that for any nonnegative integer valued
random variable X , the expected value of X can be computed by
EX = 1
X P (X k) (1.6) k=1 We will prove this after we complete the proof of Lemma 1.11. Now the probability of returning at least k times, {N (y ) k }, is the same as the event that
k
the k th return occurs, i.e., {Ty < 1}, so using (1.5) we have
Ex N (y ) = 1
X P (N (y ) k ) = ⇢xy k=1 since P1 n=0 ✓n = 1/(1 1
X ⇢k 1 =
yy k=1 ⇢xy
1 ⇢yy ✓) whenever ✓ < 1. Proof of (1.6). Let 1{X k} denote the random variable that is 1 if X
0 otherwise. It is easy to see that
X= 1
X k=1 1{X Taking expected values and noticing E 1{X
EX = 1
X k and k} . k} P (X = P (X k ) gives k) k=1 Our next step is to compute the expected number of returns to y in a
di↵erent way.
P1
Lemma 1.12. Ex N (y ) = n=1 pn (x, y ). Proof. Let 1{Xn =y} denote the random variable that is 1 if Xn = y , 0 otherwise.
Clearly
1
X
N (y ) =
1{Xn =y} .
n=1 Taking expected values now gives Ex N (y ) = 1
X Px...
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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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