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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).

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