9.17 - a 6 = 0 ,Y = aX + b,so P ( Y ≤ y ) = P ( aX + b...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 543 Probability Homework Y u Hu Homework due 9/17 2.1: 1(1),2,4 2.1: 1(1). Solution: since X is a random variable, It means X is a measurable function, so aX (a is a constant)is also measurable function. Hence aX is a random variable. 2.1: 2. Solution: (1)When a = 0 ,Y = b, so the distribution function of Y isF ( Y )( When Y b,then f ( Y ) = 1; otherwise F ( Y ) = 0) (2)When
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a 6 = 0 ,Y = aX + b,so P ( Y ≤ y ) = P ( aX + b ≤ y ) = P ( X ≤ y-b a ) = F ( y-b a ) 2.1: 4. Solution: Let H = λF + (1-λ ) G , then when x → -∞ ,H ( x ) → when x → + ∞ ,H ( x ) → 1. And since F and G are nondecreasing and right-continuous, so H is also nondecreasing and right-continuous. 1...
View Full Document

This note was uploaded on 12/25/2011 for the course MATH 543 taught by Professor Grant during the Fall '08 term at BYU.

Ask a homework question - tutors are online