s2 - Stat 5101 (Geyer) 2002 Midterm 2 Problem 1 This is...

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: Stat 5101 (Geyer) 2002 Midterm 2 Problem 1 This is like Problem 3.6.4 in DeGroot and Schervish. Since g 1 ( x | y ) = f ( x,y ) f 2 ( y ) (1) First we need to find the marginal of y . f 2 ( y ) = Z 1 f ( x,y ) dx = 2 2- log 2 1 1 + y Z 1 ( x + y ) dx = 2 2- log 2 1 1 + y x 2 2 + xy 1 = 2 2- log 2 1 2 + y 1 + y Plugging back into (1) we get g 1 ( x | y ) = 2 2- log 2 x + y 1+ y 2 2- log 2 1 2 + y 1+ y = x + y 1 2 + y , < x < 1 Alternate Solution (Conditional Probability is Renormalization) The conditional density g 1 ( x | y ) is a probability density though of as a function of the variable(s) in front of the bar (here x ) and is proportional to the joint density, thus g 1 ( x | y ) = c ( x + y ) , < x < 1 . (2) where c is some constant that does not depend on what we are thinking of as the variable (the stuff in front of the bar x ), though it may (and generally does) depend on the stuff behind the bar, which is being treated as constant. All we need to do is choose the constant c so that (2) integrates to one, integrating with respect to the variable x (the stuff behind the bar y...
View Full Document

This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Fall '02 term at Minnesota.

Page1 / 4

s2 - Stat 5101 (Geyer) 2002 Midterm 2 Problem 1 This is...

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