07hw02sol

# 07hw02sol - EE544 Homework Assignment 2 Solution 1(Pursley...

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Unformatted text preview: EE544 Homework Assignment 2 Solution 1. (Pursley 2.3) Let X â‰¡ X 1- X 2 . Since X 1 and X 2 are jointly Gaussian random variables, X is also a Gaussian random variable with Î¼ X â‰¡ E { X } = E { X 1- X 2 } = E { X 1 } - E { X 2 } = Î¼ 1- Î¼ 2 and Ïƒ 2 X â‰¡ E { [( X 1- X 2 )- E ( X 1- X 2 )] 2 } = E { [( X 1- EX 1 )- ( X 2- EX 2 )] 2 } = E { ( X 1- EX 1 ) 2 } + E { ( X 2- EX 2 ) 2 } - 2 E { ( X 1- EX 1 )( X 2- EX 2 ) } = Ïƒ 2 1 + Ïƒ 2 2 The last equation results because X 1 and X 2 are independent and thus uncorrelated. Therefore, P { error } = P { X 1 > X 2 } = P { X > } = Z âˆž f X ( x ) dx = Z âˆž 1 q 2 Ï€Ïƒ 2 X e- ( x- Î¼ X ) 2 2 Ïƒ 2 X dx which can be written, after change of variable, as P { error } = Z âˆž- Î¼ X Ïƒ X 1 âˆš 2 Ï€ e- u 2 2 du = Q (- Î¼ X Ïƒ X ) or, P { error } = 1- Z- Î¼ X Ïƒ X-âˆž 1 âˆš 2 Ï€ e- u 2 2 du = 1- Î¦(- Î¼ X Ïƒ X ) 2. (Pursley 2.4) Y is a Gaussian random variable with Î¼ Y â‰¡ E { Y } = E { X 1 + X 2 + X 3 } = Î¼ 1 + Î¼ 2 + Î¼ 3 and Ïƒ 2 Y â‰¡ E {| Y- EY | 2 } = E { [ 3 X i =1 | X i- EX i | ] 2 } = Î› 1 , 1 + Î› 1 , 2 + Î› 1 , 3 + Î› 2 , 1 + Î› 2 , 2 + Î› 2 , 3 + Î› 3 , 1 + Î› 3 , 2 + Î› 3 , 3 = 3 X i,j =1 Î› i,j The distribution function is therefore F Y ( y ) = Z y-âˆž f Y ( u ) du = Z y-âˆž 1 q 2 Ï€Ïƒ 2 Y e- ( u- Î¼ Y ) 2 2 Ïƒ 2 Y du = Z y- Î¼ Y Ïƒ Y-âˆž 1 âˆš 2 Ï€ e- v 2 2 dv = Î¦( y- Î¼ Y Ïƒ Y ) 1 3. (Pursley 3.16) (a) Z ( t ) = cX (...
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07hw02sol - EE544 Homework Assignment 2 Solution 1(Pursley...

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