This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ISyE 2027, Spring 2010 HW 9 Solution 1. Let Y = 1 + X , or X = Y 1 . F Y ( y ) = Pr ( Y y ) = Pr ( X y 1 ) = F X ( y 1 ) . f Y ( y ) = dF X ( y 1 ) dy = dF X ( x ) dx vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x = y 1 dx dy = f X ( x ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x = y 1 dx dy . The pdf of Y , f Y ( y ), becomes f Y ( y ) = f X ( x ) dx dy = 2 e 2 x 1 , x , = 2 e 2( y 1) , y 1 , and zero otherwise. 2. Let Y = 1 + 2 Z , or Z = ( Y 1) 2 4 . Y is one of 1 , 3 , 1 + 2 2 , 1 + 2 3 , and 5, since Z is one of , 1 , 2 , 3 , and 4. The pmf of Y , p Y ( y ), becomes p Y ( y ) = Pr ( Y = y ) = Pr (1 + 2 Z = y ) = Pr ( Z = ( y 1) 2 4 ) = ( 4 ( y 1) 2 4 )( 1 4 ) ( y 1) 2 4 ( 3 4 ) 4 ( y 1) 2 4 , y = 1 , 3 , 1 + 2 2 , 1 + 2 3 , 5 , , otherwise. 3. Let Y = sin ( U ). The distribution function of Y , F Y ( y ), becomes F Y ( y ) = Pr ( Y y ) = Pr (sin( U ) y ) = Pr ( U a 1 ) + Pr ( U a 2 ) , where a 1 and a 2 are such that sin...
View Full
Document
 Fall '08
 Zahrn

Click to edit the document details