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05_fall_quiz_soln

# 05_fall_quiz_soln - For y ∈[0 1 f X | Y x | y = 8< 3 x...

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Solutions to Quiz ECE 534 Fall 2005 Problem 1 (6 points) Let X have the pdf f X ( x ) = sin( x ) 2 x [0 , π ] 0 else (a) Find the cumulative distribution function F X . In particular, what is F X (2 π )? (b) Compute E [sin( X )]. (a) The support of f X is the interval [0 , π ]. Thus, F X ( x ) = 8 < : 0 x < 0 R x 0 sin( u ) 2 du = - cos( u ) 2 | x 0 = 1 - cos( x ) 2 0 x < π 1 x π In particular, F X (2 π ) = 1. (b) E [sin( X )] = R π 0 sin( x ) f X ( x ) dx = R π 0 sin 2 ( x ) 2 dx = π 4 . Problem 2 (12 points) Suppose X and Y are jointly continuous random variables distributed over the unit square with the joint pdf given by f X,Y ( x, y ) = 3 x 2 2 + 2 xy x, y [0 , 1] 0 else (a) Calculate E [ X ]. (b) Are X and Y independent? Briefly justify your answer. (c) Calculate the pdf, f Y ( y ), of Y . Be sure to specify it for -∞ < y < . (d) Calculate the conditional density f X | Y ( x | y ). Be sure to indicate what values of y it is well-defined for, and for such y , specify it for -∞ < x < . (a) E [ X ] = R 1 0 R 1 0 x ( 3 x 2 2 + 2 xy ) dxdy = R 1 0 3 x 3 2 dx R 1 0 1 dy + R 1 0 2 x 2 dx R 1 0 ydy = 3 8 + 2 3 1 2 = 17 24 (b) No, because the joint density cannot be expressed as a function of x times a function of y . (c) f Y ( y ) =  R 1 0 ( 3 x 2 2 + 2 xy ) dx = 1 2 + y y [0 , 1] 0 else (d) f X | Y ( x | y ) is not defined unless y [0 ,
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Unformatted text preview: For y ∈ [0 , 1], f X | Y ( x | y ) = 8 < : 3 x 2 2 +2 xy 1 2 + y x ∈ [0 , 1] else Problem 3 (8 points) Suppose a N (0 , 9) random variable is passed through the quantizer function f shown. The output is Y = f ( X ). ! 1 ! 2 1 2 1 0.5 ! 0.5 ! 1 Y=f(x) x (a) Express the pmf of Y in terms of the Q function. (Check your answer. Make sure your answer is positive for the right values of y .) (b) Express the variance of Y in terms of the Q function. Give as simple an answer as possible. (a) p Y ( y ) = 8 > > < > > : 1-Q (-2 3 ) = Q ( 2 3 ) y = ± 1 Q ( 1 3 )-Q ( 2 3 ) y = ± . 5 1-2 Q ( 1 3 ) y = 0 else (b) Var( Y ) = E [ Y 2 ] = P [ Y ∈ {-1 , 1 } ] + 1 4 P [ Y ∈ {-. 5 , . 5 } ] = 2 Q ( 2 3 ) + 1 4 2( Q ( 1 3 )-Q ( 2 3 )) = 1 . 5 Q ( 2 3 ) + 0 . 5 Q ( 1 3 )....
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