hw#5%20solution1 - IE 300 / GE 331 Spring 2009 Homework #5...

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Unformatted text preview: IE 300 / GE 331 Spring 2009 Homework #5 Solutions Problem 1. (a) For 1 < | v | < 2, the cross-section 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 cross-section 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 ] ....
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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.

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hw#5%20solution1 - IE 300 / GE 331 Spring 2009 Homework #5...

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