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Unformatted text preview: MATH 507b ASSIGNMENT 2 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 4: (7.1) Let N = min { n : X n > } , with N = if there is no such n . By Theorem 7.6, EX EX N Z { X N > } X N dP P ( X N > ) = P (sup n X n > ) , so P (sup n X n > ) EX / . (7.2) By Theorem 4.1 applied to T 1 n we have E ( X T 1 n ) = EX = 0, meaning E ( S T 1 n ( p q )( T 1 n )) 2 = 2 E ( T 1 n ) . Let Z n = S T 1 n ( p q )( T 1 n ). Since T 1 n = T 1 for large enough n , we have Z n S T 1 ( p q ) T 1 a.s., so by Monotone Convergence and Fatous Lemma, 2 ET 1 = lim n 2 E ( T 1 n ) = lim n EZ 2 n E (lim n Z 2 n ) = E ( S T 1 ( p q ) T 1 ) 2 = E (1 ( p q ) T 1 ) 2 . By Example 7.1d we have ET 1 < , so the preceding inequality shows that ET 2 1 < . Multiplying out Z 2 n we get 0 = E ( X T 1 n ) = E S 2 T 1 n 2 S T 1 n ( p q )( T 1 n ) + ( p q ) 2 ( T 1 n ) 2 2 ( T 1 n ) . (1) We have to show that Dominated Convergence can be applied to each of the four terms on the right side of (1). The third term is dominated by T 2 1 and the fourth by 2 T 1 , both of which are integrable. For the first term we have 1 S T 1 n inf m S m , and by Example 7.1c, inf m S m is squareintegrable, so S 2 T 1 n 1+(inf m S m ) 2 gives the domination. For the second term, 2 S T 1 n ( T 1 n ) S 2 T 1 n + ( T 1 n ) 2 , so this term is dominated as well. Thus by (1) and Dominated Convergence, 0 = E S 2 T 1 2 S T 1 ( p q ) T 1 + ( p q ) 2 T 2 1 2 T 1 = 1 2( p q ) ET 1 + ( p q ) 2 ET 2...
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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.
 Spring '08
 Alexander
 Math, Probability

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