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# zz-24 - Utah State University ECE 6010 Stochastic Processes...

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Utah State University ECE 6010 Stochastic Processes Homework # 3 Solutions 1. Show that cov( aX + b, cY + d ) = ac cov( X, Y ). cov( aX + b, cY + d ) = E [( aX + b - E [ aX + b ]) ( cY + d - E [ cY + d ])] E [ aX + b ] = x + b & E [ cY + d ] = y + d Therefore, cov( aX + b, cY + d ) = E [ a ( X - μ x ) · c ( Y - μ y )] = ac E [( X - μ x )( Y - μ y )] = ac cov( X, Y ) 2. Suppose X ∼ N (0 , σ 2 ). Use the ch.f. of X to find an expression for E [ X n ], n Z + . φ X ( u ) = e jμu - 1 2 u 2 σ 2 = e 1 2 u 2 σ 2 E [ X n ] = i - n d n du n φ X ( u ) u =0 = i - n d n du n e 1 2 u 2 σ 2 u =0 = d n du n i - n 1 - σ 2 u 2 2 1 1! + σ 4 u 4 2 2 2! - σ 6 u 6 2 3 3! + · · · u =0 = 0 n odd σ n n ! 2 n/ 2 ( n/ 2)! n even = 0 n odd 1 · 3 · 5 . . . ( n - 1) σ n n even 3. Suppose X and Y are the indicator functions of events A and B , respectively. Find ρ ( X, Y ), and show that X and Y are independent if and only if ρ ( X, Y ) = 0. X = 1 x A 0 x / A Y = 1 y B 0 y / B ρ ( X, Y ) = cov( X, Y ) p var( X )var( Y ) = E [ X, Y ] - E [ X ] E [ Y ] p var( X )var( Y ) = P ( x A and y B ) - P ( x A ) P ( y B p var( X )var( Y ) So from the equation above, ρ = 0 P ( x A and y B ) = P ( x A ) P ( y B ) X, Y are independent X, Y independent P ( x A and y B ) = P ( x A ) P ( y B ) ρ = 0 Therefore, ρ ( X, Y ) = 0 X and Y are independent. 1

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4. Suppose φ ( u ) is a ch.f. Show that | φ ( u ) | 2 is also a ch.f. φ ( u ) = E [ e jux ] = Z -∞ e jux f X ( x ) dx | φ ( u ) | 2 = φ ( u ) · φ * ( u ) = E [ e jux ] E [ e - jux ] = Z -∞ e jux f X ( x ) dx Z -∞ e - jux f X ( x ) dx Let Y = - X
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