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hw11soln

# hw11soln - ORIE 3500/5500 Fall'10 HW 11 Solution Assignment...

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ORIE 3500/5500, Fall ’10 HW 11 Solution Assignment 11 Solution Problem 1 (a) We have X 1 - X 2 N ( μ 1 - μ 2 , σ 2 1 + σ 2 2 - 2 ρσ 1 σ 2 ) = N (545 - 555 , 110 2 + 120 2 - 2 × 0 . 5 × 110 × 120) = N ( - 10 , 13300) . Therefore, P ( | X 1 - X 2 | > 5) = P ( X 1 - X 2 > 5) + P ( X 1 - X 2 < - 5) = P Z > 5 + 10 13300 + P Z < - 5 + 10 13300 = P ( Z > 0 . 13) + P ( Z < . 04) = 1 - . 5517 + . 5160 = 0 . 9643 . (b) Given X 1 = 550, the conditional distribution of X 2 is normal with mean E ( X 2 | X 1 = 550) = ρ σ 2 σ 1 (550 - μ 1 ) + μ 2 = (0 . 5) 120 110 (550 - 545) + 555 = 557 . 73 and variance Var( X 2 | X 1 = 550) = (1 - ρ 2 ) σ 2 2 = (1 - 0 . 5 2 )120 2 = 10800 . Therefore, P ( | X 1 - X 2 | > 5 | X 1 = 550) = P ( X 2 > 555 | X 1 = 550) + P ( X 2 < 545 | X 1 = 550) = P Z > 555 - 557 . 73 10800 + P Z < 545 - 557 . 73 10800 = P ( Z > - 0 . 03) + P ( Z < - . 12) = . 5120 + 1 - . 5478 = 0 . 9642 . Also, given X 2 = 550, the conditional distribution of X 1 is normal with mean E ( X 1 | X 2 = 550) = ρ σ 1 σ 2 (550 - μ 2 ) + μ 1 = (0 . 5) 110 120 (550 - 555) + 545 = 542 . 71 and variance Var( X 1 | X 2 = 550) = (1 - ρ 2 ) σ 2 1 = (1 - 0 . 5 2 )110 2 = 9075 . 1

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ORIE 3500/5500, Fall ’10 HW 11 Solution Therefore, P ( | X 1 - X 2 | > 5 | X 2 = 550) = P ( X 1 > 555 | X 2 = 550) + P ( X 1 < 545 | X 2 = 550) = P Z > 555 - 542 . 71 9075 + P Z < 545 - 542 . 71 9075 = P ( Z > . 13) + P ( Z < . 02) = 1 - . 5517 + . 5080 = 0 . 9563 . Problem 2 (a) Since X N (0 . 2 , 0 , 4 2 ), P ( X 0 . 4) = P X - 0 . 2 0 . 4 0 . 4 - 0 . 2 0 . 4 = P ( N (0 , 1) 0 . 5) = 0 . 69146 . Since Y N (0 . 2 , 4 2 ), P ( | Y | ≤ 0 . 4) = P - 0 . 4 - 0 . 2 4 Y - 0 . 2 4 0 . 4 - 0 . 2 4 = P ( - 0 . 15 N (0 , 1) 0 . 05) = 0 . 07956 .
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hw11soln - ORIE 3500/5500 Fall'10 HW 11 Solution Assignment...

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