m507b-hw4soln-s08

m507b-hw4soln-s08 - MATH 507b ASSIGNMENT 4 SOLUTIONS SPRING...

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MATH 507b ASSIGNMENT 4 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (2.1) Let A σ ( X 0 ,..,X n ) and B σ ( X n ,X n +1 ,... ). Using Theorem 1.2 Chapter 4 (p. 224) and the Markov property, since 1 A ∈ F n we get P μ ( A B | X n ) = E μ ( E μ (1 A 1 B | F n ) | X n ) = E μ (1 A E μ (1 B | F n ) | X n ) = E μ (1 A E μ (1 B | X n ) | X n ) = E μ (1 B | X n ) E μ (1 A | X n ) . Here the last equality uses the fact that E μ (1 B | X n ) σ ( X n ), so it can be pulled out of the conditional expectation. (2.2)(i) Let a be absorbing, let δ > 0, let D = { X n = a for some n 1 } , let A = { x : P x ( D ) δ ), and h ( x ) = P x ( D ). We want to apply Theorem 2.3 with B m = { a } for all m and A n = A for all n . In fact we have P ( m = n +1 { X m = a } | X n ) = P X n ( D ) δ on { X n A } , so that theorem says P ( { X n A i.o. }\{ X n = a i.o. } ) = 0. Since a is absorbing, we have { X n = a i.o. } = D , so this is equivalent to P ( { X n A i.o. } ∩ D c ) = 0. In other words, on D c , except for a null set we have X n A only finitely often, or equivalently, h ( X n ) < δ for all sufficiently large n . Since δ is arbitrary, this shows
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This note was uploaded on 06/23/2008 for the course MATH 507A taught by Professor Alexander during the Spring '08 term at USC.

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m507b-hw4soln-s08 - MATH 507b ASSIGNMENT 4 SOLUTIONS SPRING...

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