Unformatted text preview: f (Xm ) !
f (x)⇡ (x)
n m=1
x The key idea here is that by breaking the path at the return times to x we get a
seqeunce of random variables to which we can apply the law of large numbers.
Sketch of proof. Suppose that the chain starts at x. Let T0 = 0 and Tk =
min{n > Tk 1 : Xn = x} be the time of the k th return to x. By the strong
Markov property, the random variables
Yk = Tk
X m=Tk f (Xm ) 1 +1 are independent and identically distributed. By the cycle trick in the proof of
Theorem 1.20
X
EYk =
µx (y )f (y )
x Using the law of large numbers for i.i.d. variables TL
L
X
1X
1X
f (Xm ) =
Yk !
µx (y )f (y )
L m=1
L
x
k=1 Taking L = Nn (x) = max{k : Tk n} and ignoring the contribution from the
last incomplete cycle (Nn (x), n]
Nn (x)
n
X
1X
Nn (x)
1
f (Xm ) ⇡
·
Yk
n m=1
n
Nn (x)
k=1 Using Theorem 1.21 and the law of large numbers the above
! X
1X
µx (y )f (y ) =
⇡ (y )f (y )
Ex T x y
y 44 1.8 CHAPTER 1. MARKOV CHAINS Exit Distributions To motivate developments, we begin with an example.
Example 1.39. Two year college. At a local two year...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

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