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20071ee131A_1_131a07_7sol

20071ee131A_1_131a07_7sol - EE 131A 1(51 Homework#7...

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EE 131A Homework #7 Winter 2007 Solution K. Yao 1. (51) P [ Y = 3 . 5 d ] = P [ Y = - 3 . 5 d ] = Z - 3 d -∞ 1 2 ae ax dx = 1 2 e - 3 ad P [ Y = 2 . 5 d ] = P [ Y = - 2 . 5 d ] = Z - 2 d - 3 d 1 2 ae ax dx = 1 2 { e - 2 ad - e - 3 ad } P [ Y = 1 . 5 d ] = P [ Y = - 1 . 5 d ] = Z - d - 2 d 1 2 ae ax dx = 1 2 { e - ad - e - 2 ad } P [ Y = 0 . 5 d ] = P [ Y = - 0 . 5 d ] = Z 0 - d 1 2 ae ax dx = 1 2 { 1 - e - ad } P [ | X | > 4 d ] = 2 Z 4 d 1 2 ae - ax dx = e - 4 ad . 2. (54) a. F Y ( y ) = P [ | X | ≤ y ] = P [ - y X y ] = ( 0 y < 0 F X ( y ) - F X ( - y - ) y 0 . Assuming X is a continuous random variable, f Y ( y ) = F 0 Y ( y ) = f X ( y ) + f X ( - y ) . b. P [ y < Y y + dy ] = P [ y < X y + dy ] + P [ - y - dy < X ≤ - y ] f Y ( y ) dy = f X ( y ) dy + f X ( - y ) | dy | f Y ( y ) = f X ( y ) + f X ( - y ) . c. If f X ( x ) is an even function of x, then f X ( x ) = f X ( - x ) and thus f Y ( y ) = 2 f X ( y ) 3. (57) a. F Y ( y ) = 0 y < - a F X ( y ) - a y < a 1 y a From the sketch of F Y ( y ) we see that f Y ( y ) = F 0 Y ( y ) = F X ( - a ) δ ( y + a ) + f X ( y ) + [1 - F X ( a - )] δ ( y - a ) for | y | ≤ a and f Y ( y ) = 0 elsewhere. b. If f X ( x ) = β 2 e - β | x | then F X ( x ) = ( 1 2 e βx x < 0 1 - 1 2 e - βx x 0 , so, F Y ( y ) = 0 y < - a 1 2 e βy - a y < 0 1 - 1 2 e - βy 0 y < a 1 y a, and f Y ( y ) = 1 2 e - βa δ ( y + a )+ β 2 e - β | y | + 1 2 e - βa δ ( y - a
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