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Unformatted text preview: n ] E [ Y m ] . Furthermore, y m is an odd funtion and f ( y ) is a even function, so E [ Y m ] = 0 . As a consequence, E [ X n Y m ] = 0 , and this statement is true. (c) E [ X Y ] = 0 . Sol: Since X and Y are independent, E [ X Y ] = E [ X ] = 0 . This statement is true. (d) E [exp( X + Y )] = 1 =e where e is the exponential. Sol: E [exp( X + Y )] = E [exp( X )] E [exp( Y )] = 8 < : Z 1 2 exp( x )exp( x 2 2 ) 9 = ; 2 = 8 < : e Z 1 2 exp( ( x 1) 2 2 ) 9 = ; 2 = e , so this statement is false. (e) The probability density function of the random variable X + Y is an even function. Sol: From Theorem 6.10, the random variable W = X + Y is a Gaussian random variable with zero mean and variance 2 , so the PDF of W is an even function. This statement is true. 1 Solution: . = Similarly,...
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This note was uploaded on 02/21/2012 for the course EE 101 taught by Professor 張捷力 during the Spring '07 term at National Taiwan University.
- Spring '07