hw10soln

# hw10soln - ECE 310 Spring 2005 HW 10 solutions Problem E9.1...

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ECE 310, Spring 2005, HW 10 solutions Problem E9.1 E X = Z 0 x f X ( x ) dx = Z 1 0 x · 2 x dx = 2 x 3 3 fl fl fl fl 1 0 = 2 3 E ( X 2 ) = Z 1 0 x 2 · 2 x dx = 2 x 4 4 fl fl fl fl 1 0 = 1 2 Therefore, VAR( X ) = E ( X 2 ) - ( E X ) 2 = 1 2 - 4 9 = 1 18 Problem E9.2 E X = Z 0 xf X ( x ) dx = Z 1 2 x 2 dx = - 2 x fl fl fl fl 1 = 2 E ( X 2 ) = Z -∞ x 2 f X ( x ) dx = Z 1 2 x dx = 2 ln x fl fl fl fl 1 = Therefore, VAR( X ) = E ( X 2 ) - ( E X ) 2 = Problem E9.3 E X = i =2 X i =0 i p i = 0 · 0 . 2 + 1 · 0 . 5 + 2 · 0 . 3 = 1 . 1 VAR( X ) = E ( X - E X ) 2 = 2 X i =0 p i · ( i - E X ) 2 = . 2 · ( - 1 . 1) 2 + . 5 · ( - 0 . 1) 2 + . 3 · (0 . 9) 2 = . 49 Problem E9.13 Using properties ( E 4) and ( E 5): E ( 1 2 X + Y ) = 1 2 E X + E Y = 1 2 Z 1 0 x dx + 2 = 2 . 25 Problem E9.15 p X ( x ) = 1 X y =0 p X,Y ( x, y ) = p X,Y ( x, y = 0) + p X,Y ( x, y = 1) E ( XY ) = 1 X x =0 1 X y =0 xyp X,Y ( x, y ) = 1 · 0 . 4 = 0 . 4 Problem E9.21 COV( aX + b,cY + d ) = E ( ( aX + b )( cY + d ) ) - E ( aX + b ) E ( cY + d ) = E ( acXY + adX + bcY ) + bd - ( acE ( X ) E ( Y ) + adE ( X ) + bcE ( Y ) + bd ) = acE ( XY ) - acE ( X ) E ( Y ) = ac COV( X, Y ) 1

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Problem E9.22 a) E ( P ) = E ( V I ) = E ( V 2 /R ) = E ( V 2 ) R = VAR( V ) - =0 z }| { ( E ( V )) 2 R = 4 k TW b) E ( V 1 + V 2 ) 2 = E ( V 1 + V 2 + 2 V 1 V 2 ) = E ( V 2 1 ) + E ( V 2 2 ) + 2 E ( V 1 V 2 ) =0 = E ( V 2 1 ) + E ( V 2 2 ) + 2(COV( V 1 , V 2 ) | {z } =0 - E ( V 1 )
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• Spring '05
• HAAS
• Trigraph, dx

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