{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 410Hw08ans - STAT 410 Homework#8(due Friday October 23 by...

This preview shows pages 1–4. Sign up to view the full content.

STAT 410 Homework #8 Fall 2009 (due Friday, October 23, by 3:00 p.m.) 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 . 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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
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) Find the pdf of Y. This is the pdf of the lognormal distribution .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 12

410Hw08ans - STAT 410 Homework#8(due Friday October 23 by...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online