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Unformatted text preview: x <1 (1 + x ) 2 / 21 x < 1(1x ) 2 / 2 0 x < 1 1 x 1 Then we get p X ( x  C ) = 1 + x1 x < 1x x < 1 otherwise. 2. E [ X  C ] = Z xp X ( x  C ) dx = 0 7. We know that Y 1 and Y 2 are jointly Gaussian. So we calculate the mean vector and the covariance matrix. EY 1 = 2 EX 1 + EX 2 + 3 = 3 and EY 2 =1. var( Y 1 ) = 4var( X 1 )+var( X 2 ) = 5 and var( Y 2 ) = var( X 1 )+var( X 2 ) = 2 cov( Y 1 ,Y 2 ) = E [( Y 13)( Y 2 + 1)] = E [(2 X 1 + X 2 )( X 1X 2 )] = 1 Then p Y 1 Y 2 ( y 1 ,y 2 ) = 1 2 9 exp {1 18 [2( y 13) 22( y 13)( y 2 +1)+5( y 2 +1) 2 ] } 7...
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This note was uploaded on 03/05/2012 for the course EE 7615 taught by Professor Naragipour during the Fall '11 term at LSU.
 Fall '11
 Naragipour

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