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hw#5%20solution1

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

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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 ] = ( 0 , 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 ) = 0 . 2 , 1 < u < 2 0 . 4 , | u | ≤ 1 0 , else . Hence, E [ X ] = Z -∞ u · f X ( u ) d u = Z 1 - 1 0 . 4 · u d u + Z 2 1 0 . 2 · u d u = 0 + 0 . 2 · u 2 2 2 1 = 3 10 . 1

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IE 300 / GE 331: Homework #5 Solutions 2 Alternative Solution It is easy to see that f Y ( v ) = 0 . 2 , 1 < | v | < 2 0 . 3 , | v | ≤ 1 0 , 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 · 0 . 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 ] . (b) The region over which the joint pdf must be integrated in order to find P {X 2 + Y 2 > 2 α 2 } is shown in the left-hand figure below.
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