This preview shows pages 1–3. 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: IE 300 / GE 331 Spring 2009 Homework #5 Solutions Problem 1. (a) For 1 <  v  < 2, the crosssection of the joint pdf surface along the line v is a rectangle extending from u = 1 to u = 1. Therefore the conditional pdf of X given Y is Uniform [ 1 , 1]. Similarly for  v  ≤ 1, the crosssection extends from u = 1 to u = 2, so the conditional pdf of X given Y is Uniform [ 1 , 2]. (b) Using the uniformity of the conditional pdfs, we have: E [ X  Y = v ] = ( , 1 <  v  < 2 1 2 ,  v  ≤ 1 . = rect( v/ 2) , var[ X  Y = v ] = ( 1 / 3 , 1 <  v  < 2 3 / 4 ,  v  ≤ 1 . (c) It is easy to see that f X ( u ) = . 2 , 1 < u < 2 . 4 ,  u  ≤ 1 , else . Hence, E [ X ] = Z ∞∞ u · f X ( u ) d u = Z 1 1 . 4 · u d u + Z 2 1 . 2 · u d u = 0 + 0 . 2 · u 2 2 2 1 = 3 10 . 1 IE 300 / GE 331: Homework #5 Solutions 2 Alternative Solution It is easy to see that f Y ( v ) = . 2 , 1 <  v  < 2 . 3 ,  v  ≤ 1 , else . Hence, E [ X ] = Z ∞∞ E [ X  Y = v ] · f Y ( v ) d v = Z ∞∞ rect( v/ 2) · f Y ( v ) d v = Z 1 1 1 2 · . 3 d v = 3 10 . Problem 2. (a) f X , Y ( u,v ) = f X ( u ) f Y ( v ) = 1 2 π exp[ u 2 + v 2 2 ] ....
View
Full
Document
This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

Click to edit the document details