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sol.1 - x <-1 (1 + x ) 2 / 2-1 x < 1-(1-x ) 2 / 2 0 x...

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EE 7620 Solutions to Problem Set # 1 2.18. In general for Y = aX 3 + b , a > 0, the answer is given by: p Y ( y ) = 1 3 a [( y - b ) /a ] 2 / 3 p X ˆ y - b a ! 1 / 3 X is a Gaussian r.v. with zero mean and unit variance, i.e., p X ( x ) = 1 2 π e - x 2 / 2 . Hence p Y ( y ) = 1 3 a 2 π [( y - b ) /a ] 2 / 3 e - 1 2 ( y - b a ) 2 / 3 1
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2.20
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2.26
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5. 1. Z Z p XY ( x,y ) dxdy = 1 Then Z 1 y =0 Z 1 - y x =0 Cdxdy = 1 = C = 2 . 2. P (2 X Y ) = Z Z A p XY ( x,y ) dxdy = Z Z A 2 dxdy = 2area( A ) = 2(1 / 3) = 2 / 3 . x y 1 1 y=2x A 3. P ( X α ) = 0 for α < 0 If 0 α < 1, P ( X α ) = Z Z B 2 dxdy = 2 α - α 2 . For α 1, P ( X α ) = 1. Therefore, F X ( α ) = 0 α < 0 2 α - α 2 0 α < 1 1 α 1 . 5
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From F X ( α ) we can evaluate p X ( α ). p X ( α ) = ( 2 - 2 α 0 α < 1 0 otherwise 4. By symmetry we have F Y ( β ) = 0 β < 0 2 β - β 2 0 β < 1 1 β 1 . From F Y ( β ) we can evaluate p Y ( β ). p X ( β ) = ( 2 - 2 β 0 β < 1 0 otherwise We see that p XY ( x,y ) 6 = p X ( x ) p Y ( y ). Therefore, X and Y are dependent. 6. 1. F X ( x | C ) = P ( X x | 0 Y < 1) = P ( X x, 0 Y < 1) P (0 Y < 1) x y 2 -2 2 0 -1 1 A A Now P ( C ) = Z Z A p XY ( x,y ) dxdy = . 25area( A ) = . 25 . 6
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For x < - 1, P ( X x, 0 Y < 1) = 0. For - 1 x < 0, P ( X x, 0 Y < 1) = (1 + x ) 2 / 8 . For 0 x < 1, P ( X x, 0 Y < 1) = 1 - (1 - x ) 2 / 8 . And for x 1, P ( X x, 0 Y < 1) = P (0 Y < 1) = . 25. Thus F X ( x | C ) =
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Unformatted text preview: x <-1 (1 + x ) 2 / 2-1 x < 1-(1-x ) 2 / 2 0 x < 1 1 x 1 Then we get p X ( x | C ) = 1 + x-1 x < 1-x 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 1-3)( Y 2 + 1)] = E [(2 X 1 + X 2 )( X 1-X 2 )] = 1 Then p Y 1 Y 2 ( y 1 ,y 2 ) = 1 2 9 exp {-1 18 [2( y 1-3) 2-2( y 1-3)( 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.

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sol.1 - x <-1 (1 + x ) 2 / 2-1 x < 1-(1-x ) 2 / 2 0 x...

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