hw09soln

hw09soln - ECE 310 Spring 2005 HW 9 solutions Problem E8.4...

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ECE 310, Spring 2005, HW 9 solutions Problem E8.4 a) Y = X 2 X = Y d x d y = 1 2 y f Y ( y ) = f X ( y ) · 1 2 y = ατ α y - α - 1 2 U ( p ( y ) - τ ) · 1 2 y = α 2 ( τ 2 ) α 2 y - α 2 - 1 U ( y - τ 2 ) So, Y Par ( α 2 , τ 2 ) . β = α 2 , τ 0 = τ 2 b) Y = X n X = Y 1 /n d x d y = 1 ny n - 1 n f Y ( y ) = f X ( y 1 /n ) · 1 ny n - 1 n = ατ α y - α - 1 n U ( y 1 /n - τ ) · 1 ny n - 1 n = α n ( τ n ) α n y - α n - 1 U ( y - τ n ) Therefore, f Y ( y ) Par ( α/n, τ n ). Problem E8.8 a) p ( q 2 ) = P ( X > τ ) = Z τ 1 2 e - x dx = - 1 2 e - x fl fl fl fl τ = 1 2 e - τ p ( q 1 ) = P (0 < X 6 τ ) = Z τ 0 1 2 e - x dx = - 1 2 e - x fl fl fl fl τ 0 = 1 2 (1 - e - τ ) By symmetry of the pdf, we get: p ( - q 1 ) = p ( q 1 ) = 1 2 (1 - e - τ ) p ( - q 2 ) = p ( q 2 ) = 1 2 e - τ b) We need 1 2 e - τ = 1 4 τ = ln 2 Problem E8.9 a) P ( Y = 0) = P ( - 1 3 6 x 6 0) = 1 3 b) F Y ( y ) = P ( Y 6 y ) = 0 , if y < 0 1 3 , if y = 0 c) F Y ( y ) = P ( Y 6 y ) = P ( X 6 y ) = y + 1 3 , if 0 < y 6 4 9 1 , if y > 4 9 d) f Y ( y ) = 1 3 δ ( y ) + 1 2 y U ( y ) U ( 4 9 - y ) 1

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Problem E8.16 We know V N (0 , σ 2 ), where σ 2 = 4 kTW and P = V 2 , so : F ρ ( x ) = Pr (10 log 10 V 2 6 x ) = Pr ( | V | 6 (10 x 10 ) 1 2 ) = Φ( 10 x 20 σ ) - Φ( - 10 x 20
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• Spring '05
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