20061ee131A_1_131A6SOL

# 20061ee131A_1_131A6SOL - EE 131A 1(51 Homework#6 Solution...

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EE 131A Homework #6 Winter 2006 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

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20061ee131A_1_131A6SOL - EE 131A 1(51 Homework#6 Solution...

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