hw7-sol

hw7-sol - AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY...

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

View Full Document Right Arrow Icon
AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY THEORY Homework Set # 7 – Solution Notes (1). (15 points) If E (3 X )=1 and var (2 X )=5 ,Fnd(a) . E [(2 + X ) 2 ] and (b). (4 + 3 X ) . If E (3 X ) = 1 and (2 X ) = 5, then we know that E ( X / 3 and 5 = (2 X )=4 ( X )= 4[ E ( X 2 ) - [ E ( X )] 2 ]=4 [ E ( X 2 ) - (1 / 3) 2 ], so E ( X 2 )=(5 / 4) + (1 / 9) = 49 / 36. Thus, E [(2 + X ) 2 ]= E [4 + 4 X + X 2 ]=4+4 E ( X )+ E ( X 2 )=4+4 · (1 / 3) + 49 / 36 = 241 / 36 and (4 + 3 X )=3 2 ( X )=9 · (5 / 4) = 45 / 4 (2). (15 points) The random variables X and Y have a joint density function given by f ( x, y ± 2 e - 2 x /x if 0 x< , 0 y x 0 otherwise Compute cov ( X,Y ) . cov ( E ( XY ) - E ( X ) E ( Y ) = ² ³ 0 ³ x 0 xy · 2 e - 2 x /xdydx ´ - ² ³ 0 ³ x 0 x · 2 e - 2 x /xdydx ´ · ² ³ 0 ³ x 0 y · 2 e - 2 x /xdydx ´ (3). (30 points) [See examples 5a, 5b of Ross, Chapter 6.] Let X and Y be continuous random variables with joint probability density function given by f ( x, y ± 1 25 if 0 5 and - x y x 0 otherwise Compute the following quantities. (a). (6 points) The marginal density, f Y ( y ) ,o f Y .(B eexp l i c i tabou ta l lca se s . ) (b). (6 points) The conditional density, f X | Y ( x | 2) f X given Y =2 .Beexp l i c i tab ou l lc a s !
Background image of page 1

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

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

This note was uploaded on 06/06/2010 for the course EC 11 taught by Professor All during the Spring '10 term at UCLA.

Page1 / 3

hw7-sol - AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY...

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