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hw5sol - Math 280A Fall 2011 Homework 5 Solutions 1 Problem...

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Math 280A, Fall 2011 Homework 5 Solutions 1. Problem 16 (page 116). If X and Y are independent random variables, then X and 1 B ( Y ) are also independent random variables, for each Borel set B . Also X is integrable by hypothesis, and so is 1 B ( Y ) (being bounded). The product X · 1 B ( Y ) is therefore integrable. Finally, E [ X 1 { Y B } ] = E [ X · 1 B ( Y )] = E [ X ] · E [1 B ( Y )] = E [ X ] · P [ Y B ] . 2. Problem 21 (page 117). I shall let X n denote the n th binary digit of X , so that X 1 , X 2 , . . . is an iid sequence of Bernoulli random variables with P [ X n = 1] = P [ X n = 0] = 1 / 2 for all n . (a) As noted in class, for each k = 0 , 1 , 2 , . . . , { l n = k } = { X n = 0 , X n +1 = 0 , . . . , X n + k - 1 = 0 , X n + k = 1 } so that P [ l n = k ] = 2 - k - 1 , k = 0 , 1 , 2 , . . . by independence. (b) Let us fix k 0 and define A n := { l n = k } for n = 1 , 2 , . . . . Observe that A 1 depends only on X 1 , X 2 , . . . , X k +1 ; A k +2 depends only on X k +2 , X k +3 , . . . , X 2 k +2 ; etc. The point is that the sequence of events A ( n - 1)( k +1)+1 , n = 1 , 2 , 3 , . . . , is an iid sequence with P [ A ( n - 1)( k +1)+1 ] = 1 / 2 k +1 for all n . Since n 2 - k - 1 = + , the second Borel-Cantelli lemma tells us that P [ A ( n - 1)( k +1)+1 i.o.( n )] = 1 .
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