wackerly statsitcial inference 5.91

wackerly statsitcial inference 5.91 - 2 ) dy 1 dy 2 Ans: f...

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

View Full Document Right Arrow Icon
Question: 5.91 statistical inference ( Wackerly) Let Y1 and Y2 have the joint probability density function given by: f(y1,y2) = {K(1-2y2), 0 ≤y ≤1, 0 ≤ 1 = { 0, elsewhere Show that Cov(Y1, Y2) = 0. Does this surprise you that Cov(Y1, Y2) is zero? Why? Solve: Cov(Y1, Y2) = E ( Y1, Y2) – E(Y1)E(y2) We need: -E ( Y1, Y2) = ∫ - ∫- y 1 .y 2 .f(y1, y2) dy 1 dy 2 -E(Y1)= ∫ - ∫- y 1 .f(y 1 , y2 ) dy 1 dy 2 -E(y2)= ∫ - ∫- y 2 .f(y 1 , y
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 ) dy 1 dy 2 Ans: f 1 (y 1 ) = 1 4y1y 2 dy 2 = 2y 1 fy 2 (y 2 ) = 1 4y1y 2 dy 1 =2y 2 E(y1) = 2/3 E(Y2) = 2/3 E(Y1, Y2) = 4/9 Therefore, Cov(Y1, Y2) = E ( Y1, Y2) E(Y1)E(y2) Cov(Y1, Y2) = 4/9 (2/3)(2/3) = Note: As the variables are independent the variables are uncorrelated. f(y 1 , y 2 ) = f(y 1 )f(y 2 ) . If independent 4y 1 y 2 = (2.y 1 ) (2.y 2 ) . . true....
View Full Document

This note was uploaded on 10/25/2010 for the course AMP 1650 taught by Professor Charleslawarence during the Fall '10 term at Brown.

Page1 / 2

wackerly statsitcial inference 5.91 - 2 ) dy 1 dy 2 Ans: f...

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