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Unformatted text preview: For y 6= x, we borrow an idea from Theorem 1.5.
Let K = min{k : pk (x, y ) > 0}. Since pK (x, y ) > 0 there must be a sequence
y1 , . . . yK 1 so that
n p(x, y1 )p(y1 , y2 ) · · · p(yK 1 , y) >0 Since K is minimal all the yi 6= y , so Px (XK = y, Tx > K ) > 0 and hence
µx (y ) > 0.
Our next step is to prove
Theorem 1.21. Suppose p is irreducible and recurrent. Let Nn (y ) be the
number of visits to y at times n. As n ! 1
Nn (y )
1
!
n
Ey T y Why is this true? Suppose ﬁrst that we start at y . The times between returns,
t1 , t2 , . . . are independent and identically distributed so the strong law of large
numbers for nonnegative random variables implies that the time of the k th
return to y , R(k ) = min{n 1 : Nn (y ) = k }, has
R(k )
! Ey T y 1
k (1.15) If we do not start at y then t1 < 1 and t2 , t3 , . . . are independent and identically
distributed and we again have (1.15). Writing ak ⇠ bk when ak /bk ! 1 we have
R(k ) ⇠ kEy Ty . Taking k = n/Ey Ty we see...
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 Spring '10
 DURRETT
 The Land

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