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EE562a_HW_1_Solutions

# EE562a_HW_1_Solutions - EE562A Homework 1 Solutions...

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EE562A Homework 1 Solutions Prepared by Hooman Shirani-Mehr Problem 1: a) From probability theorem: P ( X < 1) = Z 1 -∞ f ( x ) dx = Z 1 0 2 e - 2 x dx = 1 - e - 2 b) E [ Y ] = E [ X 2 + 1] = Z -∞ f ( x ) dx = Z 0 ( x 2 + 1)2 e - 2 x dx = 3 2 The integral can be evaluated by partial integration. Problem 2: a) M X ( s ) = Z X f ( x ) e sx dx = Z 0 λe - λx e sx dx = Z 0 λe ( s - λ ) xdx = λ λ - s , s < λ b) Using the properties of moment generating function: m Y = M Y (0) = 2 m 2 Y + σ 2 Y = M Y (0) = 8 σ 2 Y = 4 Problem 3: Following the Jacobian method in Discussion 1, given U, V, we have ( X 1 , Y 1 ) = ( V U, p U (1 - V 2 )) and ( X 2 , Y 2 ) = ( V U, - p U (1 - V 2 )) , and it turns out: | ˜ J 1 | = fl fl fl det ∂X 1 ∂U ∂X 1 ∂V ∂Y 1 ∂U ∂Y 1 ∂V fl fl fl = 1 1 - V 2 = | ˜ J 2 | And we know, f XY ( X, Y ) = 1 2 πσ 2 exp h - X 2 + Y 2 2 σ 2 i Then, f UV ( U, V ) = 2 X i =1 f XY ( X i , Y i ) | ˜ J i | = 1 1 - V 2 1 2 πσ 2 exp h - U 2 σ 2 i = f U ( U ) .f V ( V )

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where f U ( U ) = 1 2 σ 2 exp h - U 2 σ 2 i , U 0 f V ( V ) = 1 π 1 - V 2 - 1 < V < 1 Since f UV ( U, V ) = f U ( U ) .f V ( V ) , U and V are independent.
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