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solution6 - EL630 Homework6 Solutions 1 a 29 29 29 X a j j...

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Unformatted text preview: EL630: Homework6 Solutions 1) a) ( 29 ( 29 ( 29 X a j j aX j Y j Y e E e e E e E ϖ ϖ μ μ ϖ ϖ ϖ = = = Φ + ) ( ) ( . Now, the second factor here is just the characteristic function of ) , ( 2 σ N with ϖ replaced by ϖ a . Thus, 2 / 2 / 2 2 2 2 2 2 ) ( ϖ σ μ ϖ ϖ σ ϖ μ ϖ a j a j Y e e e-- = = Φ . Note that this is the characteristic function of ) , ( 2 2 a N σ μ . So we conclude that Y is a normal r.v. with mean μ and variance 2 2 σ a . b) ( 29 ∫ ∫ ∞ ∞- -- ∞ ∞-- = = = Φ dx e dx e e e E x j x x j X j Z 2 2 2 2 2 2 2 1 2 / 2 1 2 1 ) ( ϖ σ σ ϖ ϖ π σ π σ ϖ Now we use the following integral formula. A dx e x A π = ∫ ∞ ∞-- 2 Which holds for any complex number A with positive real part. Note that if 2 2 1 σ = A then this formula just says that the integral of the normal density is 1. Using this formula we get ϖ σ ϖ σ π π σ ϖ 2 2 2 1 1 2 1 2 1 ) ( j j Z- =- = Φ ....
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solution6 - EL630 Homework6 Solutions 1 a 29 29 29 X a j j...

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