hw09soln - ECE 310, Spring 2005, HW 9 solutions Problem...

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ECE 310, Spring 2005, HW 9 solutions Problem E8.4 a) Y = X 2 X = Y dx dy = 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 2 ) . β = α 2 , τ 0 = τ 2 b) Y = X n X = Y 1 /n dx dy = 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 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 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
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This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell.

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hw09soln - ECE 310, Spring 2005, HW 9 solutions Problem...

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