# 410Hw07ans - STAT 410 Homework #7 (due Monday, July 20, by...

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STAT 410 Homework #7 Summer 2009 (due Monday, July 20, by 4:00 p.m.) Warm-up: 4.2.3 By Chebyshev’s Inequality, P ( | W n μ | ε ) 2 2 W ε σ n = 2 ε p n b 0 as n for all ε > 0. Therefore, μ W P n . 1. a) 4.1.4 b) 4.1.8 a) 4.1.4 Let X 1 , X 2 , X 3 , X 4 be four iid random variables having the same pdf f ( x ) = 2 x , 0 < x < 1, zero elsewhere. Find the mean and variance of the sum Y of these four random variables. E ( X ) = 1 0 2 x x x d = 1 0 2 2 x x d = 3 2 = μ . E ( X 2 ) = 1 0 2 2 x x x d = 1 0 3 2 x x d = 4 2 = 2 1 . Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 2 3 2 2 1 - = 18 1 = σ 2 . E ( X 1 + X 2 + … + X n ) = n μ E ( X 1 + X 2 + X 3 + X 4 ) = 3 8 . Var ( X 1 + X 2 + … + X n ) = n σ 2 Var ( X 1 + X 2 + X 3 + X 4 ) = 9 2 .

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b) 4.1.8 Determine the mean and variance of the mean X of a random sample of size 9 from a distribution having pdf f ( x ) = 4 x 3 , 0 < x < 1, zero elsewhere. E ( X ) = 1 0 3 4 x x x d = 1 0 4 4 x x d = 5 4 = 0.80 = μ . E ( X 2 ) = 1 0 3 2 4 x x x d = 1 0 5 4 x x d = 6 4 = 3 2 . Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 2 5 4 3 2 - = 75 2 = σ 2 . E ( X ) = μ = 0.80 . Var ( X ) = σ 2 / n = 675 2 .
2. a) 4.1.9 b) 4.1.13 a) 4.1.9 Let X and Y be random variables with μ 1 = 1, μ 2 = 4, σ 1 2 = 4, σ 2 2 = 6, ρ = 2 1 . Find the mean and variance of Z = 3 X – 2 Y. E ( 3 X – 2 Y ) = 3 E ( X ) – 2 E ( Y ) = 3 μ 1 – 2 μ 2 = 5 . Var ( 3 X – 2 Y ) = Cov ( 3 X – 2 Y , 3 X – 2 Y ) = 9 Var ( X ) – 12 Cov ( X , Y ) + 4 Var ( Y ) = 9 σ 1 2 – 12 ρ σ 1 σ 2 + 4 σ 2 2 = 60 – 12 6 30.606123 . b) 4.1.13 Determine the correlation coefficient of the random variables X and Y if Var ( X ) = 4, Var ( Y ) = 2, and Var ( X + 2 Y ) = 15. Var ( X ) = 4, Var ( Y ) = 2, Var ( X + 2 Y ) = 15. 15 = Var ( X + 2 Y ) = Var ( X ) + 4 Cov ( X , Y ) + 4 Var ( Y ) = 4 Cov ( X , Y ) + 12. Cov ( X , Y ) = 0.75. ρ = 2 4 75 . 0 0.265165 .

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3. 4.1.22 + (c) Let X be N ( μ , σ 2 ) and consider the transformation X = ln ( Y ) or, equivalently, Y = e X . (a) Find the mean and the variance of Y by first determining E ( e X ) and E [ ( e X ) 2 ] , by using the mgf of X. X is N ( μ , σ 2 ) M X ( t ) = E ( e X t ) = exp { μ t + σ 2 t 2 / 2 } X = ln ( Y ) Y = e X E ( Y ) = E ( e X ) = M X ( 1 ) = exp { μ + σ 2 / 2 } E ( Y 2 ) = E [ ( e X ) 2 ] = E ( e 2 X ) = M X ( 2 ) = exp { 2 μ + 2 σ 2 } Var ( Y ) = E ( Y 2 ) – [ E ( Y ) ] 2 = exp { 2 μ + 2 σ 2 } – exp { 2 μ + σ 2 } = e 2 μ + 2 σ 2 e 2 μ + σ 2 = e 2 μ ( e 2 σ 2 e σ 2 ) = e 2 μ + σ 2 ( e σ 2 1 ). (b)
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## This note was uploaded on 10/29/2009 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.

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410Hw07ans - STAT 410 Homework #7 (due Monday, July 20, by...

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