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20082ee131A_1_HW6SOL

20082ee131A_1_HW6SOL - EE 131A Homework#6 Solution Spring...

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EE 131A Homework #6 Solution Spring 2008 K. Yao 1. a. Start by using integration by parts to perform 0 f S ( s ) ds = 1 , where one can take u = s and dv = s exp( - s 2 / (2 σ 2 )) ds. Then a = 2 σ 3 . b. We have s 2 = 2 x/m or s = 2 x/m. Thus, f X ( x ) = f S ( s ) | dx/ds | s = 2 x/m = 2 x/πm exp( - x/ ( 2 )) σ 3 , 0 < x < . 2. a. P [ X > 4] = 1 - F X (4) = 1 - Φ( 4 - 5 4 ) = 1 - Φ( - 1 4 ) = Φ( 1 4 ) = 0 . 598 P [ X 7] = 1 - F X (7) = 1 - Φ( 7 - 5 4 ) = 1 - Φ( 1 2 ) = 0 . 308 P [6 . 72 < X < 10 . 16] = Φ( 10 . 16 - 5 4 )Φ( 6 . 72 - 5 4 ) = Φ(1 . 29) - Φ(0 . 43) = 0 . 235 P [2 < X < 7] = Φ( 7 - 5 4 )Φ( 2 - 5 4 ) = Φ( 1 2 ) - Φ( - 3 4 ) = 0 . 465 P [6 X 8] = Φ( 8 - 5 4 )Φ( 6 - 5 4 ) = Φ( 3 4 ) - Φ( 1 4 ) = 0 . 175 b. P [ X < a ] = 0 . 8869 Φ( a - 5 4 ) = 0 . 8869 = 1 - Q ( x ) Q ( x ) = 0 . 1131 x = 1 . 21 = a - 5 4 a = 9 . 84 c. P [ X > b ] = 1 - Φ( b - 5 4 ) = 0 . 11131 Q ( x ) = 0 . 11131 x = 1 . 22 = b - 5 4 b = 9 . 88 d. P [13 < x c ] = 0 . 0123 Φ( c - 5 4 ) - Φ( 13 - 5 4 ) = Φ( c - 5 4 ) - Φ(2) = 0 . 0123 Φ( c - 5 4 ) = 0 . 0123 + 0 . 9772 = 0 . 9895 Q ( c - 5 4 ) = 1 - Φ( c - 5 4 ) = 0 . 0105 c - 5 4 = 2 . 3 c = 14 . 2 3. a. F Y ( X + N y | X = +1) = F N ( y - 1) F Y ( X + N y | X = - 1) = F N ( y + 1) f Y ( y | X = +1) = f N ( y - 1) = 1 σ 2 π e - ( y - 1) 2 2 σ 2 f Y ( y | X = - 1) = f N ( y + 1) = 1 σ 2 π e - ( y +1) 2 2 σ 2 b. f Y ( y | X = - 1) P [ X = - 1] > f Y ( y | X = +1) P [ X = +1] decide ”0” 1 σ 2 π e - ( y +1) 2 2 σ 2 (3 p 1 ) > 1
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