{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

View Full Document Right Arrow Icon
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 0 f ( x, y ) dx = 2 2 - log 2 · 1 1 + y Z 1 0 ( x + y ) dx = 2 2 - log 2 · 1 1 + y x 2 2 + xy 1 0 = 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 , 0 < 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 ) , 0 < 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 being constant).
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}