{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

m507b-hw4soln-s08

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

This preview shows pages 1–2. Sign up to view the full content.

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 h ( X n ) 0 on D c , a.s.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}