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

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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....
View Full Document

This note was uploaded on 12/11/2011 for the course MATH 280a taught by Professor Driver,b during the Fall '08 term at UCSD.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online